He tratado de evaluar $$\int\sqrt{\tan(x)} dx$$ pero, según wolfram, debo de haber cometido un error en alguna parte. No podía encontrar a mí mismo, así que ¿podrías decirme donde me equivocaba ? $$\int \sqrt{\tan(x)}dx$$ $$u=\sqrt{\tan(x)}\qquad du = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}dx$$ $$du = \frac{u^4+1}{2u}dx\qquad dx = \frac{2u}{u^4+1}du$$ $$4\int \frac{u^2}{u^4+1}du$$ $$\frac{u^2}{u^4+1} = \frac{\frac{1}{2\sqrt{2}}u}{u^2-\sqrt{2}u+1}-\frac{\frac{1}{2\sqrt{2}}u}{u^2+\sqrt{2}u+1}$$ $$\frac{1}{2\sqrt{2}}\int\left(\frac{u}{u^2-\sqrt{2}u+1}-\frac{u}{u^2+\sqrt{2}u+1}\right)du$$ $$\frac{1}{2\sqrt{2}}\int\left(\frac{u}{u^2-\sqrt{2}u+\frac{1}{2}+\frac{1}{2}}-\frac{u}{u^2+\sqrt{2}u+\frac{1}{2}+\frac{1}{2}}\right)du$$ $$\frac{1}{2\sqrt{2}}\int\left(\frac{u_1}{\left(u_1-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\right)du_1-\frac{1}{2\sqrt{2}}\int\left(\frac{u_2}{\left(u_2+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\right)du_2$$ $$u_1-\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\tan(\theta_1)\qquad du_1 = \frac{1}{\sqrt{2}}\sec^2(\theta_1)d\theta_1$$ $$u_2+\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\tan(\theta_2)\qquad du_2 = \frac{1}{\sqrt{2}}\sec^2(\theta_2)d\theta_2$$ $$\frac{1}{2\sqrt{2}}\int\left(\frac{\frac{\sqrt{2}}{2} + \frac{1}{\sqrt{2}}\tan(\theta_1)}{\frac{1}{2}(1+\tan^2(\theta_1))}\right)\frac{1}{\sqrt{2}}\sec^2(\theta_1)d\theta_1-\frac{1}{2\sqrt{2}}\int\left(\frac{\frac{-\sqrt{2}}{2}+\frac{1}{\sqrt{2}}\tan(\theta_2)}{\frac{1}{2}(1+\tan^2(\theta_2))}\right)\frac{1}{\sqrt{2}}\sec^2(\theta_2)d\theta_2$$ $$\frac{1}{4}\int\left(\frac{\frac{\sqrt{2}}{2} + \frac{1}{\sqrt{2}}\tan(\theta_1)}{\frac{1}{2}\sec^2(\theta_1)}\right)\sec^2(\theta_1)d\theta_1-\frac{1}{4}\int\left(\frac{\frac{-\sqrt{2}}{2}+\frac{1}{\sqrt{2}}\tan(\theta_2)}{\frac{1}{2}\sec^2(\theta_2)}\right)\sec^2(\theta_2)d\theta_2$$ $$\frac{1}{2}\int\left(\frac{\sqrt{2}}{2} + \frac{1}{\sqrt{2}}\tan(\theta_1)\right)d\theta_1-\frac{1}{2}\int\left(\frac{-\sqrt{2}}{2}+\frac{1}{\sqrt{2}}\tan(\theta_2)\right)d\theta_2$$ $$\frac{1}{2}\left(\frac{\sqrt{2}}{2}\theta_1 + \frac{1}{\sqrt{2}}\ln|\sec(\theta_1)|\right)-\frac{1}{2}\left(-\frac{\sqrt{2}}{2}\theta_2 + \frac{1}{\sqrt{2}}\ln|\sec(\theta_2)|\right)$$ $$\frac{1}{2}\left(\frac{\sqrt{2}}{2}\tan^{-1}(u\sqrt{2}-1) + \frac{1}{\sqrt{2}}\ln|(u\sqrt{2}-1)^2+1|\right)-\frac{1}{2}\left(-\frac{\sqrt{2}}{2}\tan^{-1}(u\sqrt{2}+1) + \frac{1}{\sqrt{2}}\ln|(u\sqrt{2}+1)^2+1|\right)$$ La anterior es una línea entera. $$\frac{1}{2}\left[\frac{\sqrt{2}}{2}\tan^{-1}\left(\frac{u\sqrt{2}}{2-2u^2}\right)+\frac{1}{\sqrt{2}}\ln\left|\frac{(u\sqrt{2}-1)^2+1}{(u\sqrt{2}+1)^2+1}\right|\right]$$ Aquí he utilizado la identidad $$\arctan(x) + \arctan(y) = \left(\frac{x+y}{1-xy}\right)$$
$$\frac{\sqrt{2}}{4}\tan^{-1}\left(\frac{u\sqrt{2}}{2-2u^2}\right)+\frac{1}{2\sqrt{2}}\ln\left|\frac{(u\sqrt{2}-1)^2+1}{(u\sqrt{2}+1)^2+1}\right|$$ $$\frac{\sqrt{2}}{4}\tan^{-1}\left(\frac{\sqrt{2\tan(x)}}{2-2\tan(x)}\right)+\frac{1}{2\sqrt{2}}\ln\left|\frac{\tan(x)-\sqrt{2\tan(x)}+1}{\tan(x)+\sqrt{2\tan(x)}+1}\right|$$
