Integrand size = 32, antiderivative size = 162 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {(15 i A-B) x}{16 a^4}+\frac {A \log (\sin (c+d x))}{a^4 d}+\frac {7 A+i B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {15 A+i B}{16 a^4 d (1+i \tan (c+d x))}+\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3} \]
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Time = 0.55 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3677, 3612, 3556} \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {15 A+i B}{16 a^4 d (1+i \tan (c+d x))}+\frac {7 A+i B}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {x (-B+15 i A)}{16 a^4}+\frac {A \log (\sin (c+d x))}{a^4 d}+\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3} \]
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Rule 3556
Rule 3612
Rule 3677
Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {\cot (c+d x) (8 a A-4 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2} \\ & = \frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot (c+d x) \left (48 a^2 A-12 a^2 (3 i A-B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4} \\ & = \frac {7 A+i B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot (c+d x) \left (192 a^3 A-24 a^3 (7 i A-B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6} \\ & = \frac {7 A+i B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {15 A+i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (384 a^4 A-24 a^4 (15 i A-B) \tan (c+d x)\right ) \, dx}{384 a^8} \\ & = -\frac {(15 i A-B) x}{16 a^4}+\frac {7 A+i B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {15 A+i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {A \int \cot (c+d x) \, dx}{a^4} \\ & = -\frac {(15 i A-B) x}{16 a^4}+\frac {A \log (\sin (c+d x))}{a^4 d}+\frac {7 A+i B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {3 A+i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {15 A+i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.72 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {-3 (31 A+i B) \log (i-\tan (c+d x))+96 A \log (\tan (c+d x))-3 (A-i B) \log (i+\tan (c+d x))+\frac {12 (A+i B)}{(-i+\tan (c+d x))^4}+\frac {24 i A-8 B}{(-i+\tan (c+d x))^3}-\frac {6 (7 A+i B)}{(-i+\tan (c+d x))^2}+\frac {6 (-15 i A+B)}{-i+\tan (c+d x)}}{96 a^4 d} \]
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Time = 0.20 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.21
method | result | size |
risch | \(\frac {x B}{16 a^{4}}-\frac {31 i x A}{16 a^{4}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 d \,a^{4}}+\frac {13 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 d \,a^{4}}+\frac {3 i B \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 d \,a^{4}}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{4 d \,a^{4}}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{24 d \,a^{4}}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{16 d \,a^{4}}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )} B}{128 d \,a^{4}}+\frac {{\mathrm e}^{-8 i \left (d x +c \right )} A}{128 d \,a^{4}}-\frac {2 i A c}{a^{4} d}+\frac {A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\) | \(196\) |
derivativedivides | \(-\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}-\frac {15 i A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {7 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {15 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {i A}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{4}}\) | \(234\) |
default | \(-\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}-\frac {15 i A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {7 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {15 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {i A}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{4}}\) | \(234\) |
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Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {{\left (24 \, {\left (31 i \, A - B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 384 \, A e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 24 \, {\left (13 \, A + 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 12 \, {\left (8 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 \, {\left (3 \, A + 2 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]
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Time = 0.45 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.22 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {A \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{4} d} + \begin {cases} \frac {\left (\left (24576 A a^{12} d^{3} e^{12 i c} + 24576 i B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (196608 A a^{12} d^{3} e^{14 i c} + 131072 i B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (786432 A a^{12} d^{3} e^{16 i c} + 294912 i B a^{12} d^{3} e^{16 i c}\right ) e^{- 4 i d x} + \left (2555904 A a^{12} d^{3} e^{18 i c} + 393216 i B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {- 31 i A + B}{16 a^{4}} + \frac {\left (- 31 i A e^{8 i c} - 26 i A e^{6 i c} - 16 i A e^{4 i c} - 6 i A e^{2 i c} - i A + B e^{8 i c} + 4 B e^{6 i c} + 6 B e^{4 i c} + 4 B e^{2 i c} + B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 31 i A + B\right )}{16 a^{4}} \]
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Exception generated. \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.91 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac {12 \, {\left (31 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {384 \, A \log \left (\tan \left (d x + c\right )\right )}{a^{4}} - \frac {775 \, A \tan \left (d x + c\right )^{4} + 25 i \, B \tan \left (d x + c\right )^{4} - 3460 i \, A \tan \left (d x + c\right )^{3} + 124 \, B \tan \left (d x + c\right )^{3} - 5898 \, A \tan \left (d x + c\right )^{2} - 246 i \, B \tan \left (d x + c\right )^{2} + 4612 i \, A \tan \left (d x + c\right ) - 252 \, B \tan \left (d x + c\right ) + 1447 \, A + 153 i \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
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Time = 7.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.21 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {7\,A}{4\,a^4}-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {B}{16\,a^4}+\frac {A\,15{}\mathrm {i}}{16\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {13\,A}{4\,a^4}+\frac {B\,1{}\mathrm {i}}{4\,a^4}\right )+\frac {B\,1{}\mathrm {i}}{3\,a^4}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {19\,B}{48\,a^4}+\frac {A\,63{}\mathrm {i}}{16\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}+\frac {A\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (31\,A+B\,1{}\mathrm {i}\right )}{32\,a^4\,d} \]
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