Optimal. Leaf size=68 \[ \frac {1}{8 a (1-\cosh (x))}+\frac {3}{4 a (\cosh (x)+1)}-\frac {1}{8 a (\cosh (x)+1)^2}+\frac {5 \log (1-\cosh (x))}{16 a}+\frac {11 \log (\cosh (x)+1)}{16 a} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.09, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3879, 88} \[ \frac {1}{8 a (1-\cosh (x))}+\frac {3}{4 a (\cosh (x)+1)}-\frac {1}{8 a (\cosh (x)+1)^2}+\frac {5 \log (1-\cosh (x))}{16 a}+\frac {11 \log (\cosh (x)+1)}{16 a} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 88
Rule 3879
Rubi steps
\begin {align*} \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx &=a^4 \operatorname {Subst}\left (\int \frac {x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\cosh (x)\right )\\ &=a^4 \operatorname {Subst}\left (\int \left (\frac {1}{8 a^5 (-1+x)^2}+\frac {5}{16 a^5 (-1+x)}+\frac {1}{4 a^5 (1+x)^3}-\frac {3}{4 a^5 (1+x)^2}+\frac {11}{16 a^5 (1+x)}\right ) \, dx,x,\cosh (x)\right )\\ &=\frac {1}{8 a (1-\cosh (x))}-\frac {1}{8 a (1+\cosh (x))^2}+\frac {3}{4 a (1+\cosh (x))}+\frac {5 \log (1-\cosh (x))}{16 a}+\frac {11 \log (1+\cosh (x))}{16 a}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.19, size = 66, normalized size = 0.97 \[ \frac {\text {sech}(x) \left (-2 \coth ^2\left (\frac {x}{2}\right )-\text {sech}^2\left (\frac {x}{2}\right )+4 \cosh ^2\left (\frac {x}{2}\right ) \left (5 \log \left (\sinh \left (\frac {x}{2}\right )\right )+11 \log \left (\cosh \left (\frac {x}{2}\right )\right )\right )+12\right )}{16 a (\text {sech}(x)+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.41, size = 773, normalized size = 11.37 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.13, size = 94, normalized size = 1.38 \[ \frac {11 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, a} + \frac {5 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, a} - \frac {5 \, e^{\left (-x\right )} + 5 \, e^{x} - 6}{16 \, a {\left (e^{\left (-x\right )} + e^{x} - 2\right )}} - \frac {33 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 84 \, e^{\left (-x\right )} + 84 \, e^{x} + 52}{32 \, a {\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.16, size = 69, normalized size = 1.01 \[ -\frac {\tanh ^{4}\left (\frac {x}{2}\right )}{32 a}-\frac {5 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{16 a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {1}{16 a \tanh \left (\frac {x}{2}\right )^{2}}+\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{8 a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.48, size = 108, normalized size = 1.59 \[ \frac {x}{a} + \frac {5 \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} - 14 \, e^{\left (-3 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )}}{4 \, {\left (2 \, a e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} - 4 \, a e^{\left (-3 \, x\right )} - a e^{\left (-4 \, x\right )} + 2 \, a e^{\left (-5 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} + \frac {11 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a} + \frac {5 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 1.43, size = 160, normalized size = 2.35 \[ \frac {\ln \left (9\,{\mathrm {e}}^{2\,x}-9\right )}{a}-\frac {x}{a}-\frac {1}{2\,\left (a+4\,a\,{\mathrm {e}}^x+6\,a\,{\mathrm {e}}^{2\,x}+4\,a\,{\mathrm {e}}^{3\,x}+a\,{\mathrm {e}}^{4\,x}\right )}+\frac {1}{a+3\,a\,{\mathrm {e}}^x+3\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{3\,x}}-\frac {1}{4\,\left (a-2\,a\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}\right )}-\frac {2}{a+2\,a\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}}+\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {-a^2}}{a}\right )}{4\,\sqrt {-a^2}}+\frac {3}{2\,\left (a+a\,{\mathrm {e}}^x\right )}+\frac {1}{4\,\left (a-a\,{\mathrm {e}}^x\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\coth ^{3}{\relax (x )}}{\operatorname {sech}{\relax (x )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________