Optimal. Leaf size=82 \[ \sinh (x)-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \sinh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \sinh (x)\right ) \]
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Rubi [A] time = 0.187413, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {1676, 1166, 203} \[ \sinh (x)-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \sinh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \sinh (x)\right ) \]
Antiderivative was successfully verified.
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Rule 1676
Rule 1166
Rule 203
Rubi steps
\begin{align*} \int \coth (5 x) \sinh (x) \, dx &=\operatorname{Subst}\left (\int \frac{1+12 x^2+16 x^4}{5+20 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (1-\frac{4 \left (1+2 x^2\right )}{5+20 x^2+16 x^4}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-4 \operatorname{Subst}\left (\int \frac{1+2 x^2}{5+20 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac{1}{5} \left (4 \left (5-\sqrt{5}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{10-2 \sqrt{5}+16 x^2} \, dx,x,\sinh (x)\right )-\frac{1}{5} \left (4 \left (5+\sqrt{5}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{10+2 \sqrt{5}+16 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \sinh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \sinh (x)\right )+\sinh (x)\\ \end{align*}
Mathematica [A] time = 0.223715, size = 76, normalized size = 0.93 \[ \frac{1}{10} \left (10 \sinh (x)-\sqrt{10-2 \sqrt{5}} \tan ^{-1}\left (\sqrt{2+\frac{2}{\sqrt{5}}} \sinh (x)\right )-\sqrt{2 \left (5+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \sinh (x)\right )\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.155, size = 246, normalized size = 3. \begin{align*}{\frac{\sqrt{5}}{10\,\sqrt{5-2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5-2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{1}{2\,\sqrt{5-2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5-2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{\sqrt{5}}{10\,\sqrt{5+2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5+2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{1}{2\,\sqrt{5+2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5+2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }+{\frac{\sqrt{5}}{2\,\sqrt{25-10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25-10\,\sqrt{5}}}} \right ) }-{\frac{3}{2\,\sqrt{25-10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25-10\,\sqrt{5}}}} \right ) }-{\frac{\sqrt{5}}{2\,\sqrt{25+10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25+10\,\sqrt{5}}}} \right ) }-{\frac{3}{2\,\sqrt{25+10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25+10\,\sqrt{5}}}} \right ) }- \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}- \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )} - \frac{1}{2} \, \int \frac{e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x}}{e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1}\,{d x} - \frac{1}{2} \, \int \frac{e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + e^{x}}{e^{\left (4 \, x\right )} - e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - e^{x} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.38223, size = 699, normalized size = 8.52 \begin{align*} -\frac{1}{10} \,{\left (2 \, \sqrt{2} \sqrt{\sqrt{5} + 5} \arctan \left (\frac{1}{40} \,{\left (\sqrt{2 \,{\left (\sqrt{5} + 1\right )} e^{\left (2 \, x\right )} + 4 \, e^{\left (4 \, x\right )} + 4}{\left (\sqrt{5} \sqrt{2} - 5 \, \sqrt{2}\right )} \sqrt{\sqrt{5} + 5} - 2 \,{\left ({\left (\sqrt{5} \sqrt{2} - 5 \, \sqrt{2}\right )} e^{\left (2 \, x\right )} - \sqrt{5} \sqrt{2} + 5 \, \sqrt{2}\right )} \sqrt{\sqrt{5} + 5}\right )} e^{\left (-x\right )}\right ) e^{x} - 2 \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \arctan \left (\frac{1}{40} \,{\left (\sqrt{-2 \,{\left (\sqrt{5} - 1\right )} e^{\left (2 \, x\right )} + 4 \, e^{\left (4 \, x\right )} + 4}{\left (\sqrt{5} \sqrt{2} + 5 \, \sqrt{2}\right )} \sqrt{-\sqrt{5} + 5} - 2 \,{\left ({\left (\sqrt{5} \sqrt{2} + 5 \, \sqrt{2}\right )} e^{\left (2 \, x\right )} - \sqrt{5} \sqrt{2} - 5 \, \sqrt{2}\right )} \sqrt{-\sqrt{5} + 5}\right )} e^{\left (-x\right )}\right ) e^{x} - 5 \, e^{\left (2 \, x\right )} + 5\right )} e^{\left (-x\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (x \right )} \coth{\left (5 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.25231, size = 101, normalized size = 1.23 \begin{align*} -\frac{1}{10} \, \sqrt{2 \, \sqrt{5} + 10} \arctan \left (-\frac{e^{\left (-x\right )} - e^{x}}{\sqrt{\frac{1}{2} \, \sqrt{5} + \frac{5}{2}}}\right ) - \frac{1}{10} \, \sqrt{-2 \, \sqrt{5} + 10} \arctan \left (-\frac{e^{\left (-x\right )} - e^{x}}{\sqrt{-\frac{1}{2} \, \sqrt{5} + \frac{5}{2}}}\right ) - \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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