Integrand size = 34, antiderivative size = 129 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {3 (i A-B) x}{2 a}-\frac {(A+2 i B) \log (\cos (c+d x))}{a d}-\frac {3 (i A-B) \tan (c+d x)}{2 a d}-\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}+\frac {(i A-B) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))} \]
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Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3676, 3609, 3606, 3556} \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {(-B+i A) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}-\frac {3 (-B+i A) \tan (c+d x)}{2 a d}-\frac {(A+2 i B) \log (\cos (c+d x))}{a d}+\frac {3 x (-B+i A)}{2 a} \]
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Rule 3556
Rule 3606
Rule 3609
Rule 3676
Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \tan ^2(c+d x) (3 a (i A-B)+2 a (A+2 i B) \tan (c+d x)) \, dx}{2 a^2} \\ & = -\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}+\frac {(i A-B) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \tan (c+d x) (-2 a (A+2 i B)+3 a (i A-B) \tan (c+d x)) \, dx}{2 a^2} \\ & = \frac {3 (i A-B) x}{2 a}-\frac {3 (i A-B) \tan (c+d x)}{2 a d}-\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}+\frac {(i A-B) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(A+2 i B) \int \tan (c+d x) \, dx}{a} \\ & = \frac {3 (i A-B) x}{2 a}-\frac {(A+2 i B) \log (\cos (c+d x))}{a d}-\frac {3 (i A-B) \tan (c+d x)}{2 a d}-\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}+\frac {(i A-B) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))} \\ \end{align*}
Time = 2.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\frac {2 (2 A+i B) \tan ^2(c+d x)}{a+i a \tan (c+d x)}+\frac {2 B \tan ^3(c+d x)}{a+i a \tan (c+d x)}+\frac {(5 A+7 i B) \log (i-\tan (c+d x))-(A-i B) \log (i+\tan (c+d x))+\frac {6 (-i A+B)}{-i+\tan (c+d x)}}{a}}{4 d} \]
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Time = 0.10 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.13
method | result | size |
norman | \(\frac {\frac {\left (-i A +B \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{a d}-\frac {3 \left (-i A +B \right ) x}{2 a}+\frac {2 i B +A}{2 a d}+\frac {3 \left (-i A +B \right ) \tan \left (d x +c \right )}{2 a d}-\frac {3 \left (-i A +B \right ) x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {i B \left (\tan ^{4}\left (d x +c \right )\right )}{2 a d}}{1+\tan ^{2}\left (d x +c \right )}+\frac {\left (2 i B +A \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a d}\) | \(146\) |
risch | \(-\frac {7 x B}{2 a}+\frac {5 i x A}{2 a}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a d}-\frac {4 B c}{a d}+\frac {2 i A c}{a d}+\frac {2 i \left (-i A \,{\mathrm e}^{2 i \left (d x +c \right )}-i A +B \right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{a d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{a d}\) | \(159\) |
derivativedivides | \(\frac {B \tan \left (d x +c \right )}{d a}-\frac {i B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i A \tan \left (d x +c \right )}{d a}+\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}+\frac {3 i A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d a}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )}\) | \(165\) |
default | \(\frac {B \tan \left (d x +c \right )}{d a}-\frac {i B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i A \tan \left (d x +c \right )}{d a}+\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}+\frac {3 i A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d a}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )}\) | \(165\) |
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Time = 0.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.44 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (-5 i \, A + 7 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (4 \, {\left (-5 i \, A + 7 \, B\right )} d x - 9 \, A - i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left ({\left (-5 i \, A + 7 \, B\right )} d x - 5 \, A - 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left ({\left (A + 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (A + 2 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - A - i \, B}{4 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
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Time = 0.41 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.52 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {2 A e^{2 i c} e^{2 i d x} + 2 A + 2 i B}{a d e^{4 i c} e^{4 i d x} + 2 a d e^{2 i c} e^{2 i d x} + a d} + \begin {cases} \frac {\left (A + i B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (- \frac {5 i A - 7 B}{2 a} + \frac {\left (5 i A e^{2 i c} - i A - 7 B e^{2 i c} + B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x \left (5 i A - 7 B\right )}{2 a} - \frac {\left (A + 2 i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \]
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Exception generated. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.61 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\frac {{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a} - \frac {{\left (5 \, A + 7 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {2 \, {\left (i \, B a \tan \left (d x + c\right )^{2} + 2 i \, A a \tan \left (d x + c\right ) - 2 \, B a \tan \left (d x + c\right )\right )}}{a^{2}} + \frac {5 \, A \tan \left (d x + c\right ) + 7 i \, B \tan \left (d x + c\right ) - 3 i \, A + 5 \, B}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]
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Time = 7.48 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\frac {A}{2\,a}-\frac {A+B\,1{}\mathrm {i}}{2\,a}+\frac {A+B\,2{}\mathrm {i}}{2\,a}}{d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a}+\frac {A\,1{}\mathrm {i}}{a}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (5\,A+B\,7{}\mathrm {i}\right )}{4\,a\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a\,d} \]
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