Optimal. Leaf size=97 \[ -\frac {9 \cot (c+d x)}{4 a^2 d}-\frac {2 i \log (\sin (c+d x))}{a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {9 x}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.20, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3559, 3596, 3529, 3531, 3475} \[ -\frac {9 \cot (c+d x)}{4 a^2 d}-\frac {2 i \log (\sin (c+d x))}{a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {9 x}{4 a^2}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3475
Rule 3529
Rule 3531
Rule 3559
Rule 3596
Rubi steps
\begin {align*} \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^2(c+d x) (5 a-3 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot ^2(c+d x) \left (18 a^2-16 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac {9 \cot (c+d x)}{4 a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \left (-16 i a^2-18 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac {9 x}{4 a^2}-\frac {9 \cot (c+d x)}{4 a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \int \cot (c+d x) \, dx}{a^2}\\ &=-\frac {9 x}{4 a^2}-\frac {9 \cot (c+d x)}{4 a^2 d}-\frac {2 i \log (\sin (c+d x))}{a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 1.82, size = 276, normalized size = 2.85 \[ -\frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-36 i d x \sin (2 c)+i \sin (2 c) \sin (4 d x)+64 d x \cos ^2(c)+32 i d x \cot (c)+16 \sin (2 c) \log \left (\sin ^2(c+d x)\right )-\sin (2 c) \cos (4 d x)+8 i \csc (c) \cos (2 c-d x) \csc (c+d x)-8 i \csc (c) \cos (2 c+d x) \csc (c+d x)-8 \csc (c) \sin (2 c-d x) \csc (c+d x)+8 \csc (c) \sin (2 c+d x) \csc (c+d x)-32 (\cos (2 c)+i \sin (2 c)) \tan ^{-1}(\tan (d x))-i \cos (2 c) \left (32 d x \cot (c)-i \left (16 i \log \left (\sin ^2(c+d x)\right )+36 d x+\sin (4 d x)\right )+\cos (4 d x)\right )-32 d x-12 \sin (2 d x)-12 i \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 114, normalized size = 1.18 \[ -\frac {68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (68 \, d x - 44 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (-32 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 32 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.94, size = 109, normalized size = 1.12 \[ -\frac {\frac {32 i \, \log \left (i \, \tan \left (d x + c\right )\right )}{a^{2}} - \frac {34 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {2 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {16 \, {\left (-2 i \, \tan \left (d x + c\right ) + 1\right )}}{a^{2} \tan \left (d x + c\right )} + \frac {51 i \, \tan \left (d x + c\right )^{2} + 122 \, \tan \left (d x + c\right ) - 75 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.40, size = 111, normalized size = 1.14 \[ -\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}-\frac {1}{d \,a^{2} \tan \left (d x +c \right )}-\frac {2 i \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {17 i \ln \left (\tan \left (d x +c \right )-i\right )}{8 d \,a^{2}}-\frac {5}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.96, size = 125, normalized size = 1.29 \[ -\frac {\frac {7\,\mathrm {tan}\left (c+d\,x\right )}{2\,a^2}-\frac {1{}\mathrm {i}}{a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,9{}\mathrm {i}}{4\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^2-\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,17{}\mathrm {i}}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,2{}\mathrm {i}}{a^2\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.51, size = 182, normalized size = 1.88 \[ \begin {cases} \frac {\left (- 48 i a^{2} d e^{4 i c} e^{- 2 i d x} - 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 17 e^{4 i c} - 6 e^{2 i c} - 1\right ) e^{- 4 i c}}{4 a^{2}} + \frac {17}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{2} d e^{2 i c} e^{2 i d x} - a^{2} d} - \frac {17 x}{4 a^{2}} - \frac {2 i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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