Integrand size = 8, antiderivative size = 116 \[ \int e^x \coth (4 x) \, dx=e^x-\frac {\arctan \left (e^x\right )}{2}+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (e^x\right )}{2}+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}} \]
[Out]
Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.500, Rules used = {2320, 396, 220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \[ \int e^x \coth (4 x) \, dx=-\frac {1}{2} \arctan \left (e^x\right )+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (e^x\right )}{2}+e^x+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{4 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{4 \sqrt {2}} \]
[In]
[Out]
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*} \text {integral}& = \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 {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2}-\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 {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2}+\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 {\arctan \left (e^x\right )}{2}+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (e^x\right )}{2}+\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*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.19 \[ \int e^x \coth (4 x) \, dx=e^x-2 e^x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},1,\frac {9}{8},e^{8 x}\right ) \]
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.48
method | result | size |
risch | \({\mathrm e}^{x}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{4}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{4}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{4}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-4 \textit {\_R} \right )\right )\) | \(56\) |
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int e^x \coth (4 x) \, dx=-\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) - \frac {1}{2} \, \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + \cosh \left (x\right ) - \frac {1}{4} \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{4} \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + \sinh \left (x\right ) \]
[In]
[Out]
\[ \int e^x \coth (4 x) \, dx=\int e^{x} \coth {\left (4 x \right )}\, dx \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int e^x \coth (4 x) \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84 \[ \int e^x \coth (4 x) \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
[In]
[Out]
Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int e^x \coth (4 x) \, dx=\frac {\ln \left (2-2\,{\mathrm {e}}^x\right )}{4}-\frac {\ln \left (-2\,{\mathrm {e}}^x-2\right )}{4}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}+{\mathrm {e}}^x-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}{2}\right )}{4}+\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}^2+2\right )}{8}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}{2}\right )}{4}-\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}^2+2\right )}{8} \]
[In]
[Out]