Optimal. Leaf size=194 \[ -\frac {a x}{8 \left (a^2-b^2\right )}-\frac {a b^2 x}{2 \left (a^2-b^2\right )^2}+\frac {a^2 b \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}-\frac {b \cosh ^4(x)}{4 \left (a^2-b^2\right )}+\frac {a \sinh (x) \cosh ^3(x)}{4 \left (a^2-b^2\right )}-\frac {a \sinh (x) \cosh (x)}{8 \left (a^2-b^2\right )}-\frac {a b^2 \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )^2}-\frac {a^2 b^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3}+\frac {a^3 b^2 x}{\left (a^2-b^2\right )^3} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.34, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3109, 2565, 30, 2568, 2635, 8, 2564, 3098, 3133} \[ -\frac {a x}{8 \left (a^2-b^2\right )}-\frac {a b^2 x}{2 \left (a^2-b^2\right )^2}+\frac {a^3 b^2 x}{\left (a^2-b^2\right )^3}+\frac {a^2 b \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}-\frac {b \cosh ^4(x)}{4 \left (a^2-b^2\right )}+\frac {a \sinh (x) \cosh ^3(x)}{4 \left (a^2-b^2\right )}-\frac {a \sinh (x) \cosh (x)}{8 \left (a^2-b^2\right )}-\frac {a b^2 \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )^2}-\frac {a^2 b^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 8
Rule 30
Rule 2564
Rule 2565
Rule 2568
Rule 2635
Rule 3098
Rule 3109
Rule 3133
Rubi steps
\begin {align*} \int \frac {\cosh ^3(x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac {a \int \cosh ^2(x) \sinh ^2(x) \, dx}{a^2-b^2}-\frac {b \int \cosh ^3(x) \sinh (x) \, dx}{a^2-b^2}+\frac {(a b) \int \frac {\cosh ^2(x) \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=\frac {a \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}+\frac {\left (a^2 b\right ) \int \cosh (x) \sinh (x) \, dx}{\left (a^2-b^2\right )^2}-\frac {\left (a b^2\right ) \int \cosh ^2(x) \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {\cosh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {a \int \cosh ^2(x) \, dx}{4 \left (a^2-b^2\right )}-\frac {b \operatorname {Subst}\left (\int x^3 \, dx,x,\cosh (x)\right )}{a^2-b^2}\\ &=\frac {a^3 b^2 x}{\left (a^2-b^2\right )^3}-\frac {b \cosh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {a b^2 \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )^2}-\frac {a \cosh (x) \sinh (x)}{8 \left (a^2-b^2\right )}+\frac {a \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}-\frac {\left (i a^2 b^3\right ) \int \frac {-i b \cosh (x)-i a \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^3}-\frac {\left (a^2 b\right ) \operatorname {Subst}(\int x \, dx,x,i \sinh (x))}{\left (a^2-b^2\right )^2}-\frac {\left (a b^2\right ) \int 1 \, dx}{2 \left (a^2-b^2\right )^2}-\frac {a \int 1 \, dx}{8 \left (a^2-b^2\right )}\\ &=\frac {a^3 b^2 x}{\left (a^2-b^2\right )^3}-\frac {a b^2 x}{2 \left (a^2-b^2\right )^2}-\frac {a x}{8 \left (a^2-b^2\right )}-\frac {b \cosh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {a^2 b^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3}-\frac {a b^2 \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )^2}-\frac {a \cosh (x) \sinh (x)}{8 \left (a^2-b^2\right )}+\frac {a \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}+\frac {a^2 b \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.61, size = 126, normalized size = 0.65 \[ \frac {4 b \left (a^4-b^4\right ) \cosh (2 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)+a \left (8 b^2 \left (b^2-a^2\right ) \sinh (2 x)+\left (a^2-b^2\right )^2 \sinh (4 x)-4 \left (x \left (a^4-6 a^2 b^2-3 b^4\right )+8 a b^3 \log (a \cosh (x)+b \sinh (x))\right )\right )}{32 (a-b)^3 (a+b)^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.61, size = 1162, normalized size = 5.99 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.12, size = 202, normalized size = 1.04 \[ -\frac {a^{2} b^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (a^{2} - 3 \, a b\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {{\left (6 \, a^{2} e^{\left (4 \, x\right )} - 18 \, a b e^{\left (4 \, x\right )} + 4 \, a b e^{\left (2 \, x\right )} - 4 \, b^{2} e^{\left (2 \, x\right )} - a^{2} + 2 \, a b - b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 4 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.22, size = 322, normalized size = 1.66 \[ \frac {2}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {8}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}+\frac {3 a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a^{2} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}+\frac {3 b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a}{8 \left (a +b \right )^{3}}-\frac {a^{2} b^{3} \ln \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {2}{\left (8 a -8 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {8}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (16 a -16 b \right )}+\frac {a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {3 a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a^{2} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}+\frac {3 b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a}{8 \left (a -b \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.33, size = 150, normalized size = 0.77 \[ -\frac {a^{2} b^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (a^{2} + 3 \, a b\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac {{\left (4 \, b e^{\left (-2 \, x\right )} + a + b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {4 \, b e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 1.82, size = 129, normalized size = 0.66 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {x\,\left (3\,a\,b-a^2\right )}{8\,{\left (a-b\right )}^3}+\frac {b\,{\mathrm {e}}^{2\,x}}{16\,{\left (a+b\right )}^2}+\frac {b\,{\mathrm {e}}^{-2\,x}}{16\,{\left (a-b\right )}^2}-\frac {a^2\,b^3\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6} \]
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]
________________________________________________________________________________________