Integrand size = 32, antiderivative size = 143 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {(i A+B) x}{16 a^4}-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \]
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Time = 0.23 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3671, 3607, 3560, 8} \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {x (B+i A)}{16 a^4}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4} \]
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Rule 8
Rule 3560
Rule 3607
Rule 3671
Rubi steps \begin{align*} \text {integral}& = -\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}-\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{2 a^2} \\ & = -\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}-\frac {(i A+B) \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2} \\ & = -\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {(i A+B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {(i A+B) \int 1 \, dx}{16 a^4} \\ & = -\frac {(i A+B) x}{16 a^4}-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 0.79 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (12 i B+16 A \cos (2 (c+d x))-3 (A+8 i A d x+B (i+8 d x)) \cos (4 (c+d x))+32 i A \sin (2 (c+d x))+3 i A \sin (4 (c+d x))-3 B \sin (4 (c+d x))+24 A d x \sin (4 (c+d x))-24 i B d x \sin (4 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \]
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Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.76
method | result | size |
risch | \(-\frac {x B}{16 a^{4}}-\frac {i x A}{16 a^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 d \,a^{4}}+\frac {i B \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 d \,a^{4}}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 d \,a^{4}}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )} B}{128 d \,a^{4}}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )} A}{128 d \,a^{4}}\) | \(109\) |
derivativedivides | \(-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}\) | \(199\) |
default | \(-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}\) | \(199\) |
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Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.55 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {{\left (24 \, {\left (i \, A + B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 24 \, A e^{\left (6 i \, d x + 6 i \, c\right )} - 12 i \, B e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, A e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]
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Time = 0.31 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.72 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (196608 A a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 65536 A a^{12} d^{3} e^{14 i c} e^{- 6 i d x} + 98304 i B a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + \left (- 24576 A a^{12} d^{3} e^{12 i c} - 24576 i B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {- i A - B}{16 a^{4}} + \frac {\left (- i A e^{8 i c} - 2 i A e^{6 i c} + 2 i A e^{2 i c} + i A - B e^{8 i c} + 2 B e^{4 i c} - B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- i A - B\right )}{16 a^{4}} \]
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Exception generated. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.64 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.04 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {12 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {12 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} + \frac {25 \, A \tan \left (d x + c\right )^{4} - 25 i \, B \tan \left (d x + c\right )^{4} - 124 i \, A \tan \left (d x + c\right )^{3} - 124 \, B \tan \left (d x + c\right )^{3} - 246 \, A \tan \left (d x + c\right )^{2} + 246 i \, B \tan \left (d x + c\right )^{2} + 252 i \, A \tan \left (d x + c\right ) + 124 \, B \tan \left (d x + c\right ) + 57 \, A - 25 i \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
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Time = 8.42 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.20 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{4\,a^4}-\frac {B\,1{}\mathrm {i}}{4\,a^4}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B}{16\,a^4}+\frac {A\,1{}\mathrm {i}}{16\,a^4}\right )-\frac {A}{12\,a^4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B}{16\,a^4}+\frac {A\,19{}\mathrm {i}}{48\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{32\,a^4\,d} \]
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