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 = 0.0346976, antiderivative size = 66, normalized size of antiderivative = 1., 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 2664
Rule 12
Rule 2657
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*}
Mathematica [B] time = 0.228439, 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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Maple [B] time = 0.043, size = 134, normalized size = 2. \begin{align*} -{\frac{9}{80\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-1}}-{\frac{{\frac{3\,i}{20}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-1}}+{\frac{5}{64\,d}\ln \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) }-{\frac{9}{80\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-1}}+{\frac{{\frac{3\,i}{20}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-1}}-{\frac{5}{64\,d}\ln \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.78205, size = 86, normalized size = 1.3 \begin{align*} -\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.08619, size = 298, normalized size = 4.52 \begin{align*} \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}{64 \,{\left (3 \, d e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, d e^{\left (d x + c\right )} - 3 \, d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 1.11671, size = 88, normalized size = 1.33 \begin{align*} \frac{- \frac{5 i e^{- c} e^{d x}}{24 d} - \frac{e^{- 2 c}}{8 d}}{e^{2 d x} - \frac{10 i e^{- c} e^{d x}}{3} - e^{- 2 c}} + \frac{- \frac{5 \log{\left (e^{d x} - 3 i e^{- c} \right )}}{64} + \frac{5 \log{\left (e^{d x} - \frac{i e^{- c}}{3} \right )}}{64}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.24642, size = 93, normalized size = 1.41 \begin{align*} \frac{5 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right )}{64 \, d} - \frac{5 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{64 \, d} + \frac{-5 i \, e^{\left (d x + c\right )} - 3}{8 \, d{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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