Integrand size = 26, antiderivative size = 112 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {(A-i B) x}{8 a^3}+\frac {i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac {i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac {i A+B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3607, 3560, 8} \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {B+i A}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x (A-i B)}{8 a^3}+\frac {-B+i A}{6 d (a+i a \tan (c+d x))^3}+\frac {B+i A}{8 a d (a+i a \tan (c+d x))^2} \]
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Rule 8
Rule 3560
Rule 3607
Rubi steps \begin{align*} \text {integral}& = \frac {i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac {(A-i B) \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{2 a} \\ & = \frac {i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac {i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac {(A-i B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac {i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac {i A+B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(A-i B) \int 1 \, dx}{8 a^3} \\ & = \frac {(A-i B) x}{8 a^3}+\frac {i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac {i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac {i A+B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-10 A+2 i B+3 (i A+B) \arctan (\tan (c+d x)) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+(-9 i A-9 B) \tan (c+d x)+3 (A-i B) \tan ^2(c+d x)}{24 a^3 d (-i+\tan (c+d x))^3} \]
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Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14
method | result | size |
risch | \(-\frac {i x B}{8 a^{3}}+\frac {x A}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}\) | \(128\) |
derivativedivides | \(\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}\) | \(158\) |
default | \(\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}\) | \(158\) |
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Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (12 \, {\left (A - i \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (-3 i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (-3 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
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Time = 0.27 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.30 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (\left (512 i A a^{6} d^{2} e^{6 i c} - 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (2304 i A a^{6} d^{2} e^{8 i c} - 768 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (4608 i A a^{6} d^{2} e^{10 i c} + 1536 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {A - i B}{8 a^{3}} + \frac {\left (A e^{6 i c} + 3 A e^{4 i c} + 3 A e^{2 i c} + A - i B e^{6 i c} - i B e^{4 i c} + i B e^{2 i c} + i B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (A - i B\right )}{8 a^{3}} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.54 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {6 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {-11 i \, A \tan \left (d x + c\right )^{3} - 11 \, B \tan \left (d x + c\right )^{3} - 45 \, A \tan \left (d x + c\right )^{2} + 45 i \, B \tan \left (d x + c\right )^{2} + 69 i \, A \tan \left (d x + c\right ) + 69 \, B \tan \left (d x + c\right ) + 51 \, A - 19 i \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]
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Time = 7.61 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B}{8\,a^3}+\frac {A\,1{}\mathrm {i}}{8\,a^3}\right )-\frac {A\,5{}\mathrm {i}}{12\,a^3}-\frac {B}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {3\,A}{8\,a^3}-\frac {B\,3{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3} \]
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