Integrand size = 32, antiderivative size = 110 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(i A+B) x}{8 a^3}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3671, 3607, 3560, 8} \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (B+i A)}{8 a^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3} \]
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Rule 8
Rule 3560
Rule 3607
Rule 3671
Rubi steps \begin{align*} \text {integral}& = -\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{2 a^2} \\ & = -\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}-\frac {(i A+B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = -\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(i A+B) \int 1 \, dx}{8 a^3} \\ & = -\frac {(i A+B) x}{8 a^3}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.53 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {(\cos (3 (c+d x))-i \sin (3 (c+d x))) (3 (A+3 i B) \cos (c+d x)-2 (A+6 i A d x+B (i+6 d x)) \cos (3 (c+d x))+9 i A \sin (c+d x)-3 B \sin (c+d x)+2 i A \sin (3 (c+d x))-2 B \sin (3 (c+d x))+12 A d x \sin (3 (c+d x))-12 i B d x \sin (3 (c+d x)))}{96 a^3 d} \]
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Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16
method | result | size |
risch | \(-\frac {x B}{8 a^{3}}-\frac {i x A}{8 a^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}\) | \(128\) |
derivativedivides | \(-\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(158\) |
default | \(-\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(158\) |
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Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, {\left (i \, A + B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
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Time = 0.29 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.36 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (\left (- 512 A a^{6} d^{2} e^{6 i c} - 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 768 A a^{6} d^{2} e^{8 i c} + 768 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (1536 A a^{6} d^{2} e^{10 i c} + 1536 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- i A - B}{8 a^{3}} + \frac {\left (- i A e^{6 i c} - i A e^{4 i c} + i A e^{2 i c} + i A - B e^{6 i c} + B e^{4 i c} + B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- i A - B\right )}{8 a^{3}} \]
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Exception generated. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.56 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {11 \, A \tan \left (d x + c\right )^{3} - 11 i \, B \tan \left (d x + c\right )^{3} - 45 i \, A \tan \left (d x + c\right )^{2} - 45 \, B \tan \left (d x + c\right )^{2} - 69 \, A \tan \left (d x + c\right ) + 21 i \, B \tan \left (d x + c\right ) + 19 i \, A + 3 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]
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Time = 7.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.34 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{8\,a^3}-\frac {B\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{8\,a^3}+\frac {A\,3{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d} \]
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