Optimal. Leaf size=122 \[ \frac {15 i x}{4 a^2}+\frac {15 i \cot (c+d x)}{4 a^2 d}-\frac {2 \cot ^2(c+d x)}{a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A]
time = 0.16, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3640, 3677,
3610, 3612, 3556} \begin {gather*} -\frac {2 \cot ^2(c+d x)}{a^2 d}+\frac {15 i \cot (c+d x)}{4 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {15 i x}{4 a^2}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 3556
Rule 3610
Rule 3612
Rule 3640
Rule 3677
Rubi steps
\begin {align*} \int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^3(c+d x) (6 a-4 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot ^3(c+d x) \left (32 a^2-30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac {2 \cot ^2(c+d x)}{a^2 d}+\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot ^2(c+d x) \left (-30 i a^2-32 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac {15 i \cot (c+d x)}{4 a^2 d}-\frac {2 \cot ^2(c+d x)}{a^2 d}+\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \left (-32 a^2+30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac {15 i x}{4 a^2}+\frac {15 i \cot (c+d x)}{4 a^2 d}-\frac {2 \cot ^2(c+d x)}{a^2 d}+\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {4 \int \cot (c+d x) \, dx}{a^2}\\ &=\frac {15 i x}{4 a^2}+\frac {15 i \cot (c+d x)}{4 a^2 d}-\frac {2 \cot ^2(c+d x)}{a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {5 \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {\cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(319\) vs. \(2(122)=244\).
time = 1.62, size = 319, normalized size = 2.61 \begin {gather*} \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-64 i d x+128 i d x \cos ^2(c)-60 i d x \cos (2 c)+16 \cos (2 d x)+\cos (2 c) \cos (4 d x)-64 d x \cot (c)+64 d x \cos (2 c) \cot (c)-16 \cos (2 c-d x) \csc (c) \csc (c+d x)+16 \cos (2 c+d x) \csc (c) \csc (c+d x)+8 \cos (2 c) \csc ^2(c+d x)+32 \cos (2 c) \log \left (\sin ^2(c+d x)\right )+60 d x \sin (2 c)-i \cos (4 d x) \sin (2 c)+8 i \csc ^2(c+d x) \sin (2 c)+32 i \log \left (\sin ^2(c+d x)\right ) \sin (2 c)+64 \text {ArcTan}(\tan (d x)) (-i \cos (2 c)+\sin (2 c))-16 i \sin (2 d x)-i \cos (2 c) \sin (4 d x)-\sin (2 c) \sin (4 d x)-16 i \csc (c) \csc (c+d x) \sin (2 c-d x)+16 i \csc (c) \csc (c+d x) \sin (2 c+d x)\right )}{16 a^2 d (-i+\tan (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.30, size = 90, normalized size = 0.74
method | result | size |
derivativedivides | \(\frac {\frac {7 i}{4 \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {1}{2 \tan \left (d x +c \right )^{2}}+\frac {2 i}{\tan \left (d x +c \right )}-4 \ln \left (\tan \left (d x +c \right )\right )+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\) | \(90\) |
default | \(\frac {\frac {7 i}{4 \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {1}{2 \tan \left (d x +c \right )^{2}}+\frac {2 i}{\tan \left (d x +c \right )}-4 \ln \left (\tan \left (d x +c \right )\right )+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\) | \(90\) |
risch | \(\frac {31 i x}{4 a^{2}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {8 i c}{a^{2} d}-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-2\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(105\) |
norman | \(\frac {-\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{d a}-\frac {1}{2 d a}-\frac {2 \left (\tan ^{4}\left (d x +c \right )\right )}{d a}+\frac {15 i x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}+\frac {15 i x \left (\tan ^{4}\left (d x +c \right )\right )}{2 a}+\frac {15 i x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}+\frac {2 i \tan \left (d x +c \right )}{d a}+\frac {25 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}+\frac {15 i \left (\tan ^{5}\left (d x +c \right )\right )}{4 d a}}{\tan \left (d x +c \right )^{2} a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {4 \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\) | \(195\) |
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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 151, normalized size = 1.24 \begin {gather*} \frac {124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \, {\left (31 i \, d x + 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 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 ) - 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, 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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.46, size = 216, normalized size = 1.77 \begin {gather*} \frac {- 2 e^{2 i c} e^{2 i d x} + 4}{a^{2} d e^{4 i c} e^{4 i d x} - 2 a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \begin {cases} \frac {\left (- 16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (31 i e^{4 i c} + 8 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} - \frac {31 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {31 i x}{4 a^{2}} - \frac {4 \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.61, size = 109, normalized size = 0.89 \begin {gather*} \frac {\frac {4 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {124 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {128 \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}} + \frac {3 \, \tan \left (d x + c\right )^{4} + 114 i \, \tan \left (d x + c\right )^{3} + 173 \, \tan \left (d x + c\right )^{2} - 32 i \, \tan \left (d x + c\right ) + 16}{{\left (\tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right )\right )}^{2} a^{2}}}{32 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.01, size = 135, normalized size = 1.11 \begin {gather*} \frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}+\frac {\frac {\mathrm {tan}\left (c+d\,x\right )}{a^2}-\frac {15\,{\mathrm {tan}\left (c+d\,x\right )}^3}{4\,a^2}+\frac {1{}\mathrm {i}}{2\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,11{}\mathrm {i}}{2\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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