Optimal. Leaf size=113 \[ e^x+\frac{2 e^x}{3 \left (e^{6 x}+1\right )}+\frac{\log \left (-\sqrt{3} e^x+e^{2 x}+1\right )}{6 \sqrt{3}}-\frac{\log \left (\sqrt{3} e^x+e^{2 x}+1\right )}{6 \sqrt{3}}-\frac{2}{9} \tan ^{-1}\left (e^x\right )+\frac{1}{9} \tan ^{-1}\left (\sqrt{3}-2 e^x\right )-\frac{1}{9} \tan ^{-1}\left (2 e^x+\sqrt{3}\right ) \]
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Rubi [A] time = 0.212323, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {2282, 390, 288, 209, 634, 618, 204, 628, 203} \[ e^x+\frac{2 e^x}{3 \left (e^{6 x}+1\right )}+\frac{\log \left (-\sqrt{3} e^x+e^{2 x}+1\right )}{6 \sqrt{3}}-\frac{\log \left (\sqrt{3} e^x+e^{2 x}+1\right )}{6 \sqrt{3}}-\frac{2}{9} \tan ^{-1}\left (e^x\right )+\frac{1}{9} \tan ^{-1}\left (\sqrt{3}-2 e^x\right )-\frac{1}{9} \tan ^{-1}\left (2 e^x+\sqrt{3}\right ) \]
Antiderivative was successfully verified.
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Rule 2282
Rule 390
Rule 288
Rule 209
Rule 634
Rule 618
Rule 204
Rule 628
Rule 203
Rubi steps
\begin{align*} \int e^x \tanh ^2(3 x) \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^6\right )^2}{\left (1+x^6\right )^2} \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \left (1-\frac{4 x^6}{\left (1+x^6\right )^2}\right ) \, dx,x,e^x\right )\\ &=e^x-4 \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^6\right )^2} \, dx,x,e^x\right )\\ &=e^x+\frac{2 e^x}{3 \left (1+e^{6 x}\right )}-\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{1+x^6} \, dx,x,e^x\right )\\ &=e^x+\frac{2 e^x}{3 \left (1+e^{6 x}\right )}-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^x\right )-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{3} x}{2}}{1-\sqrt{3} x+x^2} \, dx,x,e^x\right )-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{3} x}{2}}{1+\sqrt{3} x+x^2} \, dx,x,e^x\right )\\ &=e^x+\frac{2 e^x}{3 \left (1+e^{6 x}\right )}-\frac{2}{9} \tan ^{-1}\left (e^x\right )-\frac{1}{18} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{3} x+x^2} \, dx,x,e^x\right )-\frac{1}{18} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{3} x+x^2} \, dx,x,e^x\right )+\frac{\operatorname{Subst}\left (\int \frac{-\sqrt{3}+2 x}{1-\sqrt{3} x+x^2} \, dx,x,e^x\right )}{6 \sqrt{3}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{3}+2 x}{1+\sqrt{3} x+x^2} \, dx,x,e^x\right )}{6 \sqrt{3}}\\ &=e^x+\frac{2 e^x}{3 \left (1+e^{6 x}\right )}-\frac{2}{9} \tan ^{-1}\left (e^x\right )+\frac{\log \left (1-\sqrt{3} e^x+e^{2 x}\right )}{6 \sqrt{3}}-\frac{\log \left (1+\sqrt{3} e^x+e^{2 x}\right )}{6 \sqrt{3}}+\frac{1}{9} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 e^x\right )+\frac{1}{9} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 e^x\right )\\ &=e^x+\frac{2 e^x}{3 \left (1+e^{6 x}\right )}-\frac{2}{9} \tan ^{-1}\left (e^x\right )+\frac{1}{9} \tan ^{-1}\left (\sqrt{3}-2 e^x\right )-\frac{1}{9} \tan ^{-1}\left (\sqrt{3}+2 e^x\right )+\frac{\log \left (1-\sqrt{3} e^x+e^{2 x}\right )}{6 \sqrt{3}}-\frac{\log \left (1+\sqrt{3} e^x+e^{2 x}\right )}{6 \sqrt{3}}\\ \end{align*}
Mathematica [C] time = 0.0759737, size = 97, normalized size = 0.86 \[ -\frac{1}{9} \text{RootSum}\left [\text{$\#$1}^4-\text{$\#$1}^2+1\& ,\frac{\text{$\#$1}^2 x-\text{$\#$1}^2 \log \left (e^x-\text{$\#$1}\right )+2 \log \left (e^x-\text{$\#$1}\right )-2 x}{2 \text{$\#$1}^3-\text{$\#$1}}\& \right ]+e^x+\frac{2 e^x}{3 \left (e^{6 x}+1\right )}-\frac{2}{9} \tan ^{-1}\left (e^x\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.106, size = 59, normalized size = 0.5 \begin{align*}{{\rm e}^{x}}+{\frac{2\,{{\rm e}^{x}}}{3+3\,{{\rm e}^{6\,x}}}}+\sum _{{\it \_R}={\it RootOf} \left ( 6561\,{{\it \_Z}}^{4}-81\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}-9\,{\it \_R} \right ) +{\frac{i}{9}}\ln \left ({{\rm e}^{x}}-i \right ) -{\frac{i}{9}}\ln \left ({{\rm e}^{x}}+i \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.57048, size = 109, normalized size = 0.96 \begin{align*} -\frac{1}{18} \, \sqrt{3} \log \left (\sqrt{3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{18} \, \sqrt{3} \log \left (-\sqrt{3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{2 \, e^{x}}{3 \,{\left (e^{\left (6 \, x\right )} + 1\right )}} - \frac{1}{9} \, \arctan \left (\sqrt{3} + 2 \, e^{x}\right ) - \frac{1}{9} \, \arctan \left (-\sqrt{3} + 2 \, e^{x}\right ) - \frac{2}{9} \, \arctan \left (e^{x}\right ) + e^{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.31428, size = 490, normalized size = 4.34 \begin{align*} \frac{4 \,{\left (e^{\left (6 \, x\right )} + 1\right )} \arctan \left (\sqrt{3} + \sqrt{-4 \, \sqrt{3} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} - 2 \, e^{x}\right ) + 4 \,{\left (e^{\left (6 \, x\right )} + 1\right )} \arctan \left (-\sqrt{3} + 2 \, \sqrt{\sqrt{3} e^{x} + e^{\left (2 \, x\right )} + 1} - 2 \, e^{x}\right ) - 4 \,{\left (e^{\left (6 \, x\right )} + 1\right )} \arctan \left (e^{x}\right ) -{\left (\sqrt{3} e^{\left (6 \, x\right )} + \sqrt{3}\right )} \log \left (4 \, \sqrt{3} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) +{\left (\sqrt{3} e^{\left (6 \, x\right )} + \sqrt{3}\right )} \log \left (-4 \, \sqrt{3} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 18 \, e^{\left (7 \, x\right )} + 30 \, e^{x}}{18 \,{\left (e^{\left (6 \, x\right )} + 1\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 e^{x} \tanh ^{2}{\left (3 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \tanh \left (3 \, x\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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