Optimal. Leaf size=66 \[ -\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac {5 i \tan ^{-1}\left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}+\frac {5 x}{64} \]
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Rubi [A] time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2664, 12, 2657} \[ -\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac {5 i \tan ^{-1}\left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}+\frac {5 x}{64} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2657
Rule 2664
Rubi steps
\begin {align*} \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx &=-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac {1}{16} \int -\frac {5}{5+3 i \sinh (c+d x)} \, dx\\ &=-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}+\frac {5}{16} \int \frac {1}{5+3 i \sinh (c+d x)} \, dx\\ &=\frac {5 x}{64}-\frac {5 i \tan ^{-1}\left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\\ \end {align*}
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Mathematica [B] time = 0.25, size = 183, normalized size = 2.77 \[ \frac {-\frac {120 \cosh (c+d x)}{3 \sinh (c+d x)-5 i}-25 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+25 \log (4 \sinh (c+d x)+5 \cosh (c+d x))-50 i \tan ^{-1}\left (\frac {2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-2 \sinh \left (\frac {1}{2} (c+d x)\right )}\right )+50 i \tan ^{-1}\left (\frac {2 \sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right )}{\sinh \left (\frac {1}{2} (c+d x)\right )+2 \cosh \left (\frac {1}{2} (c+d x)\right )}\right )+24 i}{640 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 102, normalized size = 1.55 \[ \frac {5 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - 5 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 40 i \, e^{\left (d x + c\right )} - 24}{192 \, d e^{\left (2 \, d x + 2 \, c\right )} - 640 i \, d e^{\left (d x + c\right )} - 192 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 65, normalized size = 0.98 \[ -\frac {\frac {8 \, {\left (5 i \, e^{\left (d x + c\right )} + 3\right )}}{3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3} - 5 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right ) + 5 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{64 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 134, normalized size = 2.03 \[ -\frac {9}{80 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}+\frac {3 i}{20 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64 d}-\frac {9}{80 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}-\frac {3 i}{20 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 64, normalized size = 0.97 \[ -\frac {5 i \, \arctan \left (\frac {3}{4} \, e^{\left (-d x - c\right )} + \frac {5}{4} i\right )}{32 \, d} - \frac {5 i \, e^{\left (-d x - c\right )} - 3}{-8 \, d {\left (-10 i \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.07, size = 102, normalized size = 1.55 \[ \frac {3}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )}-\frac {5\,\ln \left (-\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}+\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {5\,\ln \left (\frac {45\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}-\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,5{}\mathrm {i}}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotInvertible} \]
Verification of antiderivative is not currently implemented for this CAS.
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