Integrand size = 24, antiderivative size = 116 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {x}{16 a^4}-\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{4 a d (a+i a \tan (c+d x))^3}-\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \]
[Out]
Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3621, 3607, 3560, 8} \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {x}{16 a^4}-\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{4 a d (a+i a \tan (c+d x))^3}-\frac {i}{8 d (a+i a \tan (c+d x))^4} \]
[In]
[Out]
Rule 8
Rule 3560
Rule 3607
Rule 3621
Rubi steps \begin{align*} \text {integral}& = -\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {a-2 i a \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{2 a^2} \\ & = -\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{4 a d (a+i a \tan (c+d x))^3}-\frac {\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2} \\ & = -\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{4 a d (a+i a \tan (c+d x))^3}-\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {\int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{4 a d (a+i a \tan (c+d x))^3}-\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\int 1 \, dx}{16 a^4} \\ & = -\frac {x}{16 a^4}-\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{4 a d (a+i a \tan (c+d x))^3}-\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 0.49 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.59 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {(\cos (4 (c+d x))-i \sin (4 (c+d x))) (-4 i+(i+8 d x) \cos (4 (c+d x))+(1+8 i d x) \sin (4 (c+d x)))}{128 a^4 d} \]
[In]
[Out]
Time = 0.40 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.38
method | result | size |
risch | \(-\frac {x}{16 a^{4}}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\) | \(44\) |
derivativedivides | \(-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {1}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {1}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\) | \(95\) |
default | \(-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {1}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {1}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\) | \(95\) |
norman | \(\frac {-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{d a}-\frac {x}{16 a}-\frac {9 \left (\tan ^{5}\left (d x +c \right )\right )}{16 a d}-\frac {\tan ^{7}\left (d x +c \right )}{16 a d}-\frac {x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}-\frac {3 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}-\frac {x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {\tan \left (d x +c \right )}{16 a d}+\frac {9 \left (\tan ^{3}\left (d x +c \right )\right )}{16 a d}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}\) | \(159\) |
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.37 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {{\left (8 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{128 \, a^{4} d} \]
[In]
[Out]
Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (128 i a^{4} d e^{8 i c} e^{- 4 i d x} - 32 i a^{4} d e^{4 i c} e^{- 8 i d x}\right ) e^{- 12 i c}}{4096 a^{8} d^{2}} & \text {for}\: a^{8} d^{2} e^{12 i c} \neq 0 \\x \left (\frac {\left (- e^{8 i c} + 2 e^{4 i c} - 1\right ) e^{- 8 i c}}{16 a^{4}} + \frac {1}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x}{16 a^{4}} \]
[In]
[Out]
Exception generated. \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
none
Time = 0.90 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {2 i \, \log \left (\tan \left (2 \, d x + 2 \, c\right ) + i\right )}{a^{4}} - \frac {2 i \, \log \left (\tan \left (2 \, d x + 2 \, c\right ) - i\right )}{a^{4}} + \frac {3 i \, \tan \left (2 \, d x + 2 \, c\right )^{2} - 6 \, \tan \left (2 \, d x + 2 \, c\right ) + 5 i}{a^{4} {\left (\tan \left (2 \, d x + 2 \, c\right ) - i\right )}^{2}}}{128 \, d} \]
[In]
[Out]
Time = 4.73 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {x}{16\,a^4}+\frac {\frac {\mathrm {tan}\left (c+d\,x\right )}{16\,a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{16\,a^4}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )} \]
[In]
[Out]