Integrand size = 26, antiderivative size = 145 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {(A-i B) x}{16 a^4}+\frac {i A-B}{8 d (a+i a \tan (c+d x))^4}+\frac {i A+B}{12 a d (a+i a \tan (c+d x))^3}+\frac {i A+B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i A+B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \]
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Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3607, 3560, 8} \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {B+i A}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {x (A-i B)}{16 a^4}+\frac {B+i A}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {-B+i A}{8 d (a+i a \tan (c+d x))^4}+\frac {B+i A}{12 a d (a+i a \tan (c+d x))^3} \]
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Rule 8
Rule 3560
Rule 3607
Rubi steps \begin{align*} \text {integral}& = \frac {i A-B}{8 d (a+i a \tan (c+d x))^4}+\frac {(A-i B) \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx}{2 a} \\ & = \frac {i A-B}{8 d (a+i a \tan (c+d x))^4}+\frac {i A+B}{12 a 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}{4 a^2} \\ & = \frac {i A-B}{8 d (a+i a \tan (c+d x))^4}+\frac {i A+B}{12 a d (a+i a \tan (c+d x))^3}+\frac {i A+B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {(A-i B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {i A-B}{8 d (a+i a \tan (c+d x))^4}+\frac {i A+B}{12 a d (a+i a \tan (c+d x))^3}+\frac {i A+B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i A+B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {(A-i B) \int 1 \, dx}{16 a^4} \\ & = \frac {(A-i B) x}{16 a^4}+\frac {i A-B}{8 d (a+i a \tan (c+d x))^4}+\frac {i A+B}{12 a d (a+i a \tan (c+d x))^3}+\frac {i A+B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i A+B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^3(c+d x) (18 i A \cos (c+d x)+2 (7 i A+4 B) \cos (3 (c+d x))-(A-i B) (5 \sin (c+d x)+11 \sin (3 (c+d x)))+6 (A-i B) \arctan (\tan (c+d x)) \sec (c+d x) (\cos (4 (c+d x))+i \sin (4 (c+d x))))}{96 a^4 d (-i+\tan (c+d x))^4} \]
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Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01
method | result | size |
risch | \(-\frac {i x B}{16 a^{4}}+\frac {x A}{16 a^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{16 d \,a^{4}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{8 d \,a^{4}}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )} A}{32 d \,a^{4}}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} B}{48 d \,a^{4}}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} A}{24 d \,a^{4}}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )} B}{128 d \,a^{4}}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )} A}{128 d \,a^{4}}\) | \(147\) |
derivativedivides | \(-\frac {i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\) | \(199\) |
default | \(-\frac {i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\) | \(199\) |
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Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (24 \, {\left (A - i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 24 \, {\left (-2 i \, A - B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, A e^{\left (4 i \, d x + 4 i \, c\right )} - 8 \, {\left (-2 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]
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Time = 0.35 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.06 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (294912 i A a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + \left (24576 i A a^{12} d^{3} e^{12 i c} - 24576 B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (131072 i A a^{12} d^{3} e^{14 i c} - 65536 B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (393216 i A a^{12} d^{3} e^{18 i c} + 196608 B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {A - i B}{16 a^{4}} + \frac {\left (A e^{8 i c} + 4 A e^{6 i c} + 6 A e^{4 i c} + 4 A e^{2 i c} + A - i B e^{8 i c} - 2 i B e^{6 i c} + 2 i B e^{2 i c} + i B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (A - i B\right )}{16 a^{4}} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.61 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.06 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {12 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {25 i \, A \tan \left (d x + c\right )^{4} + 25 \, B \tan \left (d x + c\right )^{4} + 124 \, A \tan \left (d x + c\right )^{3} - 124 i \, B \tan \left (d x + c\right )^{3} - 246 i \, A \tan \left (d x + c\right )^{2} - 246 \, B \tan \left (d x + c\right )^{2} - 252 \, A \tan \left (d x + c\right ) + 252 i \, B \tan \left (d x + c\right ) + 153 i \, A + 57 \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
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Time = 7.67 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {B}{12\,a^4}+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {A}{16\,a^4}-\frac {B\,1{}\mathrm {i}}{16\,a^4}\right )+\frac {A\,1{}\mathrm {i}}{3\,a^4}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B}{4\,a^4}+\frac {A\,1{}\mathrm {i}}{4\,a^4}\right )-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {19\,A}{48\,a^4}-\frac {B\,19{}\mathrm {i}}{48\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}-\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^4} \]
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