Optimal. Leaf size=97 \[ -\frac {9 x}{4 a^2}-\frac {9 \cot (c+d x)}{4 a^2 d}-\frac {2 i \log (\sin (c+d x))}{a^2 d}+\frac {\cot (c+d x)}{a^2 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A]
time = 0.14, 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 = {3640, 3677,
3610, 3612, 3556} \begin {gather*} -\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} \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 ^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] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(276\) vs. \(2(97)=194\).
time = 1.98, size = 276, normalized size = 2.85 \begin {gather*} -\frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-32 d x+64 d x \cos ^2(c)-12 i \cos (2 d x)+32 i d x \cot (c)+8 i \cos (2 c-d x) \csc (c) \csc (c+d x)-8 i \cos (2 c+d x) \csc (c) \csc (c+d x)-32 \text {ArcTan}(\tan (d x)) (\cos (2 c)+i \sin (2 c))-36 i d x \sin (2 c)-\cos (4 d x) \sin (2 c)+16 \log \left (\sin ^2(c+d x)\right ) \sin (2 c)-12 \sin (2 d x)+i \sin (2 c) \sin (4 d x)-i \cos (2 c) \left (\cos (4 d x)+32 d x \cot (c)-i \left (36 d x+16 i \log \left (\sin ^2(c+d x)\right )+\sin (4 d x)\right )\right )-8 \csc (c) \csc (c+d x) \sin (2 c-d x)+8 \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 = 82, normalized size = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {i}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {17 i \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {5}{4 \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{\tan \left (d x +c \right )}-2 i \ln \left (\tan \left (d x +c \right )\right )-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\) | \(82\) |
default | \(\frac {\frac {i}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {17 i \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {5}{4 \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{\tan \left (d x +c \right )}-2 i \ln \left (\tan \left (d x +c \right )\right )-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\) | \(82\) |
risch | \(-\frac {17 x}{4 a^{2}}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {4 c}{a^{2} d}-\frac {2 i}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(96\) |
norman | \(\frac {-\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{d a}-\frac {1}{d a}-\frac {9 \left (\tan ^{4}\left (d x +c \right )\right )}{4 d a}-\frac {9 x \tan \left (d x +c \right )}{4 a}-\frac {9 x \left (\tan ^{3}\left (d x +c \right )\right )}{2 a}-\frac {9 x \left (\tan ^{5}\left (d x +c \right )\right )}{4 a}-\frac {15 \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}-\frac {3 i \tan \left (d x +c \right )}{2 d a}}{\tan \left (d x +c \right ) a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}-\frac {2 i \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\) | \(175\) |
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 = 114, normalized size = 1.18 \begin {gather*} -\frac {68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, {\left (17 \, d x - 11 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 32 \, {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.28, size = 180, normalized size = 1.86 \begin {gather*} \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}\: 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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.88, size = 109, normalized size = 1.12 \begin {gather*} -\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} \end {gather*}
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 \begin {gather*} -\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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