Optimal. Leaf size=116 \[ e^x-\frac {\text {ArcTan}\left (e^x\right )}{2}+\frac {\text {ArcTan}\left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\text {ArcTan}\left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}} \]
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Rubi [A]
time = 0.05, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 12, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.500, Rules used = {2320, 396,
220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {1}{2} \text {ArcTan}\left (e^x\right )+\frac {\text {ArcTan}\left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\text {ArcTan}\left (\sqrt {2} e^x+1\right )}{2 \sqrt {2}}+e^x+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{4 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{4 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (e^x\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 210
Rule 212
Rule 217
Rule 218
Rule 220
Rule 396
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2320
Rubi steps
\begin {align*} \int e^x \coth (4 x) \, dx &=\text {Subst}\left (\int \frac {-1-x^8}{1-x^8} \, dx,x,e^x\right )\\ &=e^x-2 \text {Subst}\left (\int \frac {1}{1-x^8} \, dx,x,e^x\right )\\ &=e^x-\text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,e^x\right )\\ &=e^x-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=e^x-\frac {1}{2} \tan ^{-1}\left (e^x\right )-\frac {1}{2} \tanh ^{-1}\left (e^x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,e^x\right )+\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,e^x\right )}{4 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,e^x\right )}{4 \sqrt {2}}\\ &=e^x-\frac {1}{2} \tan ^{-1}\left (e^x\right )-\frac {1}{2} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^x\right )}{2 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^x\right )}{2 \sqrt {2}}\\ &=e^x-\frac {1}{2} \tan ^{-1}\left (e^x\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.01, size = 22, normalized size = 0.19 \begin {gather*} e^x-2 e^x \, _2F_1\left (\frac {1}{8},1;\frac {9}{8};e^{8 x}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.98, size = 56, normalized size = 0.48
method | result | size |
risch | \({\mathrm e}^{x}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{4}+\left (\munderset {\textit {\_R} =\RootOf \left (256 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-4 \textit {\_R} \right )\right )+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{4}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{4}\) | \(56\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 97, normalized size = 0.84 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 134, normalized size = 1.16 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int e^{x} \coth {\left (4 x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 98, normalized size = 0.84 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.33, size = 104, normalized size = 0.90 \begin {gather*} \frac {\ln \left (2-2\,{\mathrm {e}}^x\right )}{4}-\frac {\ln \left (-2\,{\mathrm {e}}^x-2\right )}{4}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}+{\mathrm {e}}^x-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}{2}\right )}{4}+\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}^2+2\right )}{8}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}{2}\right )}{4}-\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}^2+2\right )}{8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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