Optimal. Leaf size=107 \[ -\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+4 a^3 x (B+i A)+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d} \]
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Rubi [A] time = 0.28, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3594, 3589, 3475, 3531} \[ -\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+4 a^3 x (B+i A)+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d} \]
Antiderivative was successfully verified.
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Rule 3475
Rule 3531
Rule 3589
Rule 3594
Rubi steps
\begin {align*} \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac {i a B (a+i a \tan (c+d x))^2}{2 d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x))^2 (2 a A+2 a (i A+2 B) \tan (c+d x)) \, dx\\ &=\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (2 a^2 A+2 a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) \left (2 a^3 A+8 a^3 (i A+B) \tan (c+d x)\right ) \, dx-\left (a^3 (3 A-4 i B)\right ) \int \tan (c+d x) \, dx\\ &=4 a^3 (i A+B) x+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\left (a^3 A\right ) \int \cot (c+d x) \, dx\\ &=4 a^3 (i A+B) x+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}\\ \end {align*}
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Mathematica [B] time = 8.48, size = 281, normalized size = 2.63 \[ \frac {a^3 \sec (c) \sec ^2(c+d x) (\cos (3 d x)+i \sin (3 d x)) \left (2 \cos (c) \left ((3 A-4 i B) \log \left (\cos ^2(c+d x)\right )+A \log \left (\sin ^2(c+d x)\right )+8 i A d x+8 B d x-2 i B\right )+\cos (c+2 d x) \left ((3 A-4 i B) \log \left (\cos ^2(c+d x)\right )+8 d x (B+i A)+A \log \left (\sin ^2(c+d x)\right )\right )-4 i A \sin (c+2 d x)+8 i A d x \cos (3 c+2 d x)+3 A \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+A \cos (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )+4 i A \sin (c)-12 B \sin (c+2 d x)+8 B d x \cos (3 c+2 d x)-4 i B \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+12 B \sin (c)\right )}{8 d (\cos (d x)+i \sin (d x))^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 172, normalized size = 1.61 \[ \frac {2 \, {\left (A - 4 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, {\left (A - 3 i \, B\right )} a^{3} + {\left ({\left (3 \, A - 4 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (3 \, A - 4 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (3 \, A - 4 i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.35, size = 265, normalized size = 2.48 \[ \frac {2 \, A a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, {\left (3 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, {\left (4 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 2 \, {\left (3 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 28 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A a^{3} - 12 i \, B a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.46, size = 135, normalized size = 1.26 \[ 4 i A x \,a^{3}-\frac {i a^{3} A \tan \left (d x +c \right )}{d}+\frac {4 i A \,a^{3} c}{d}-\frac {i a^{3} B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {4 i a^{3} B \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+4 a^{3} B x -\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {4 a^{3} B c}{d}+\frac {a^{3} A \ln \left (\sin \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.85, size = 89, normalized size = 0.83 \[ -\frac {i \, B a^{3} \tan \left (d x + c\right )^{2} - 2 \, {\left (d x + c\right )} {\left (4 i \, A + 4 \, B\right )} a^{3} + 4 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, A a^{3} \log \left (\tan \left (d x + c\right )\right ) - {\left (-2 i \, A - 6 \, B\right )} a^{3} \tan \left (d x + c\right )}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.15, size = 87, normalized size = 0.81 \[ \frac {A\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^3+a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\right )}{d}-\frac {4\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 3.19, size = 233, normalized size = 2.18 \[ \frac {A a^{3} \log {\left (\frac {- A a^{3} + 2 i B a^{3}}{A a^{3} e^{2 i c} - 2 i B a^{3} e^{2 i c}} + e^{2 i d x} \right )}}{d} + \frac {a^{3} \left (3 A - 4 i B\right ) \log {\left (e^{2 i d x} + \frac {- 2 A a^{3} + 2 i B a^{3} + a^{3} \left (3 A - 4 i B\right )}{A a^{3} e^{2 i c} - 2 i B a^{3} e^{2 i c}} \right )}}{d} + \frac {- 2 i A a^{3} - 6 B a^{3} + \left (- 2 i A a^{3} e^{2 i c} - 8 B a^{3} e^{2 i c}\right ) e^{2 i d x}}{- i d e^{4 i c} e^{4 i d x} - 2 i d e^{2 i c} e^{2 i d x} - i d} \]
Verification of antiderivative is not currently implemented for this CAS.
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