Optimal. Leaf size=83 \[ \frac {\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5}-\frac {a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac {\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac {a \cosh ^3(x)}{3 b^2}+\frac {\cosh ^4(x)}{4 b} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.11, antiderivative size = 83, 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 = {2668, 697} \[ \frac {\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac {a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac {\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5}-\frac {a \cosh ^3(x)}{3 b^2}+\frac {\cosh ^4(x)}{4 b} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 697
Rule 2668
Rubi steps
\begin {align*} \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{a+x} \, dx,x,b \cosh (x)\right )}{b^5}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a^3 \left (1-\frac {2 b^2}{a^2}\right )+\left (a^2-2 b^2\right ) x-a x^2+x^3+\frac {\left (a^2-b^2\right )^2}{a+x}\right ) \, dx,x,b \cosh (x)\right )}{b^5}\\ &=-\frac {a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac {\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac {a \cosh ^3(x)}{3 b^2}+\frac {\cosh ^4(x)}{4 b}+\frac {\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.12, size = 84, normalized size = 1.01 \[ \frac {-12 b^2 \left (3 b^2-2 a^2\right ) \cosh (2 x)-24 a b \left (4 a^2-7 b^2\right ) \cosh (x)+96 \left (a^2-b^2\right )^2 \log (a+b \cosh (x))-8 a b^3 \cosh (3 x)+3 b^4 \cosh (4 x)}{96 b^5} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 1.03, size = 866, normalized size = 10.43 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.13, size = 124, normalized size = 1.49 \[ \frac {3 \, b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 8 \, a b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 24 \, a^{2} b {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 48 \, b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 96 \, a^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} + 192 \, a b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.07, size = 599, normalized size = 7.22 \[ \frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {3}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {5}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {3}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {5}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a^{4}}{b^{4} \left (a -b \right )}-\frac {2 \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a^{3}}{b^{3} \left (a -b \right )}+\frac {2 \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a^{2}}{b^{2} \left (a -b \right )}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a}{b \left (a -b \right )}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a^{5}}{b^{5} \left (a -b \right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{a -b}+\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{3}}+\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}-\frac {a^{3}}{b^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a}{3 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{4}}{b^{5}}+\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {a}{3 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{4}}{b^{5}}+\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a^{3}}{b^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {3 a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.33, size = 178, normalized size = 2.14 \[ -\frac {{\left (8 \, a b^{2} e^{\left (-x\right )} - 3 \, b^{3} - 12 \, {\left (2 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 24 \, {\left (4 \, a^{3} - 7 \, a b^{2}\right )} e^{\left (-3 \, x\right )}\right )} e^{\left (4 \, x\right )}}{192 \, b^{4}} - \frac {8 \, a b^{2} e^{\left (-3 \, x\right )} - 3 \, b^{3} e^{\left (-4 \, x\right )} + 24 \, {\left (4 \, a^{3} - 7 \, a b^{2}\right )} e^{\left (-x\right )} - 12 \, {\left (2 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x}{b^{5}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 1.31, size = 169, normalized size = 2.04 \[ \frac {{\mathrm {e}}^{-4\,x}}{64\,b}+\frac {{\mathrm {e}}^{4\,x}}{64\,b}-\frac {x\,{\left (a^2-b^2\right )}^2}{b^5}+\frac {{\mathrm {e}}^{-x}\,\left (7\,a\,b^2-4\,a^3\right )}{8\,b^4}+\frac {\ln \left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}-\frac {a\,{\mathrm {e}}^{-3\,x}}{24\,b^2}-\frac {a\,{\mathrm {e}}^{3\,x}}{24\,b^2}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a^2-3\,b^2\right )}{16\,b^3}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a^2-3\,b^2\right )}{16\,b^3}+\frac {{\mathrm {e}}^x\,\left (7\,a\,b^2-4\,a^3\right )}{8\,b^4} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________