Optimal. Leaf size=87 \[ \sinh (x)-\frac{1}{5} \tan ^{-1}(\sinh (x))-\frac{1}{5} \sqrt{\frac{1}{2} \left (3+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{3+\sqrt{5}}} \sinh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (3-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{2 \left (3+\sqrt{5}\right )} \sinh (x)\right ) \]
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Rubi [A] time = 0.290608, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6742, 2073, 203, 1166} \[ \sinh (x)-\frac{1}{5} \tan ^{-1}(\sinh (x))-\frac{1}{5} \sqrt{\frac{1}{2} \left (3+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{3+\sqrt{5}}} \sinh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (3-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{2 \left (3+\sqrt{5}\right )} \sinh (x)\right ) \]
Antiderivative was successfully verified.
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Rule 6742
Rule 2073
Rule 203
Rule 1166
Rubi steps
\begin{align*} \int \sinh (x) \tanh (5 x) \, dx &=-\operatorname{Subst}\left (\int \frac{x^2 \left (-5-20 x^2-16 x^4\right )}{1+13 x^2+28 x^4+16 x^6} \, dx,x,\sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-1+\frac{1+8 x^2+8 x^4}{1+13 x^2+28 x^4+16 x^6}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\operatorname{Subst}\left (\int \frac{1+8 x^2+8 x^4}{1+13 x^2+28 x^4+16 x^6} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\operatorname{Subst}\left (\int \left (\frac{1}{5 \left (1+x^2\right )}+\frac{4 \left (1+6 x^2\right )}{5 \left (1+12 x^2+16 x^4\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )-\frac{4}{5} \operatorname{Subst}\left (\int \frac{1+6 x^2}{1+12 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{5} \tan ^{-1}(\sinh (x))+\sinh (x)-\frac{1}{5} \left (4 \left (3-\sqrt{5}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{6-2 \sqrt{5}+16 x^2} \, dx,x,\sinh (x)\right )-\frac{1}{5} \left (4 \left (3+\sqrt{5}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{6+2 \sqrt{5}+16 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{5} \tan ^{-1}(\sinh (x))-\frac{1}{5} \sqrt{\frac{1}{2} \left (3+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{3+\sqrt{5}}} \sinh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (3-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{2 \left (3+\sqrt{5}\right )} \sinh (x)\right )+\sinh (x)\\ \end{align*}
Mathematica [A] time = 0.187241, size = 81, normalized size = 0.93 \[ \frac{1}{10} \left (10 \sinh (x)-2 \tan ^{-1}(\sinh (x))-\sqrt{2 \left (3+\sqrt{5}\right )} \tan ^{-1}\left (2 \sqrt{\frac{2}{3+\sqrt{5}}} \sinh (x)\right )-\sqrt{6-2 \sqrt{5}} \tan ^{-1}\left (\sqrt{2 \left (3+\sqrt{5}\right )} \sinh (x)\right )\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.089, size = 60, normalized size = 0.7 \begin{align*}{\frac{{{\rm e}^{x}}}{2}}-{\frac{{{\rm e}^{-x}}}{2}}+{\frac{i}{5}}\ln \left ({{\rm e}^{x}}-i \right ) -{\frac{i}{5}}\ln \left ({{\rm e}^{x}}+i \right ) +\sum _{{\it \_R}={\it RootOf} \left ( 10000\,{{\it \_Z}}^{4}+300\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( -10\,{\it \_R}\,{{\rm e}^{x}}+{{\rm e}^{2\,x}}-1 \right ) \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{2}{5} \, \arctan \left (e^{x}\right ) - \frac{1}{2} \, \int \frac{2 \,{\left (3 \, e^{\left (7 \, x\right )} - e^{\left (5 \, x\right )} - e^{\left (3 \, x\right )} + 3 \, e^{x}\right )}}{5 \,{\left (e^{\left (8 \, x\right )} - e^{\left (6 \, x\right )} + e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.4281, size = 724, normalized size = 8.32 \begin{align*} -\frac{1}{10} \,{\left (2 \, \sqrt{2} \sqrt{\sqrt{5} + 3} \arctan \left (\frac{1}{8} \,{\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} - 3 \, \sqrt{2}\right )} \sqrt{\sqrt{5} + 3} - 2 \,{\left ({\left (\sqrt{5} \sqrt{2} - 3 \, \sqrt{2}\right )} e^{\left (2 \, x\right )} - \sqrt{5} \sqrt{2} + 3 \, \sqrt{2}\right )} \sqrt{\sqrt{5} + 3}\right )} e^{\left (-x\right )}\right ) e^{x} - 2 \, \sqrt{2} \sqrt{-\sqrt{5} + 3} \arctan \left (\frac{1}{8} \,{\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} + 3 \, \sqrt{2}\right )} \sqrt{-\sqrt{5} + 3} - 2 \,{\left ({\left (\sqrt{5} \sqrt{2} + 3 \, \sqrt{2}\right )} e^{\left (2 \, x\right )} - \sqrt{5} \sqrt{2} - 3 \, \sqrt{2}\right )} \sqrt{-\sqrt{5} + 3}\right )} e^{\left (-x\right )}\right ) e^{x} + 4 \, \arctan \left (e^{x}\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 )} \tanh{\left (5 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.24147, size = 109, normalized size = 1.25 \begin{align*} -\frac{1}{10} \, \pi - \frac{1}{10} \,{\left (\sqrt{5} + 1\right )} \arctan \left (-\frac{2 \,{\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt{5} + 1}\right ) - \frac{1}{10} \,{\left (\sqrt{5} - 1\right )} \arctan \left (-\frac{2 \,{\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt{5} - 1}\right ) - \frac{1}{5} \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\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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