Optimal. Leaf size=134 \[ e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {\text {ArcTan}\left (e^x\right )}{8}+\frac {\text {ArcTan}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\text {ArcTan}\left (1+\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {1}{8} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}} \]
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Rubi [A]
time = 0.07, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps
used = 17, number of rules used = 13, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.300, Rules used = {2320, 398,
294, 220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {1}{8} \text {ArcTan}\left (e^x\right )+\frac {\text {ArcTan}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\text {ArcTan}\left (\sqrt {2} e^x+1\right )}{8 \sqrt {2}}+e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {1}{8} \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 294
Rule 398
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2320
Rubi steps
\begin {align*} \int e^x \coth ^2(4 x) \, dx &=\text {Subst}\left (\int \frac {\left (1+x^8\right )^2}{\left (1-x^8\right )^2} \, dx,x,e^x\right )\\ &=\text {Subst}\left (\int \left (1+\frac {4 x^8}{\left (1-x^8\right )^2}\right ) \, dx,x,e^x\right )\\ &=e^x+4 \text {Subst}\left (\int \frac {x^8}{\left (1-x^8\right )^2} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^8} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^x\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right )-\frac {1}{16} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^x\right )-\frac {1}{16} \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 )}{16 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,e^x\right )}{16 \sqrt {2}}\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^x\right )}{8 \sqrt {2}}\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {1}{8} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \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 = 1.43, size = 113, normalized size = 0.84 \begin {gather*} \frac {e^{-15 x} \left (-44217-80225 e^{8 x}-15127 e^{16 x}+9361 e^{24 x}+9 \left (4913+8368 e^{8 x}+1486 e^{16 x}-1456 e^{24 x}+e^{32 x}\right ) \, _2F_1\left (\frac {1}{8},1;\frac {9}{8};e^{8 x}\right )\right )}{9216}+\frac {64 e^{9 x} \left (1+e^{8 x}\right )^2 \, _4F_3\left (\frac {9}{8},2,2,2;1,1,\frac {33}{8};e^{8 x}\right )}{3825} \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 = 1.02, size = 68, normalized size = 0.51
method | result | size |
risch | \({\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{2 \left ({\mathrm e}^{8 x}-1\right )}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{16}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{16}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{16}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{16}+\left (\munderset {\textit {\_R} =\RootOf \left (65536 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-16 \textit {\_R} \right )\right )\) | \(68\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 109, normalized size = 0.81 \begin {gather*} -\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left (e^{x} - 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 213 vs.
\(2 (90) = 180\).
time = 0.37, size = 213, normalized size = 1.59 \begin {gather*} \frac {4 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 4 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \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 ) - 4 \, {\left (e^{\left (8 \, x\right )} - 1\right )} \arctan \left (e^{x}\right ) - {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) - 2 \, {\left (e^{\left (8 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) + 2 \, {\left (e^{\left (8 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) + 32 \, e^{\left (9 \, x\right )} - 48 \, e^{x}}{32 \, {\left (e^{\left (8 \, x\right )} - 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 ^{2}{\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 = 110, normalized size = 0.82 \begin {gather*} -\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \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 = 1.39, size = 122, normalized size = 0.91 \begin {gather*} \frac {\ln \left (\frac {1}{2}-\frac {{\mathrm {e}}^x}{2}\right )}{16}-\frac {\ln \left (-\frac {{\mathrm {e}}^x}{2}-\frac {1}{2}\right )}{16}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{8}+{\mathrm {e}}^x-\frac {{\mathrm {e}}^x}{2\,\left ({\mathrm {e}}^{8\,x}-1\right )}-\frac {\sqrt {2}\,\mathrm {atan}\left (2\,\sqrt {2}\,\left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2}}{4}\right )\right )}{16}-\frac {\sqrt {2}\,\mathrm {atan}\left (2\,\sqrt {2}\,\left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2}}{4}\right )\right )}{16}+\frac {\sqrt {2}\,\ln \left ({\left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2}}{4}\right )}^2+\frac {1}{8}\right )}{32}-\frac {\sqrt {2}\,\ln \left ({\left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2}}{4}\right )}^2+\frac {1}{8}\right )}{32} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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