Optimal. Leaf size=94 \[ -\frac {x \cosh (a+b x)}{8 b}-\frac {x \cosh (3 a+3 b x)}{48 b}+\frac {x \cosh (5 a+5 b x)}{80 b}+\frac {\sinh (a+b x)}{8 b^2}+\frac {\sinh (3 a+3 b x)}{144 b^2}-\frac {\sinh (5 a+5 b x)}{400 b^2} \]
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Rubi [A]
time = 0.08, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5556, 3377,
2717} \begin {gather*} \frac {\sinh (a+b x)}{8 b^2}+\frac {\sinh (3 a+3 b x)}{144 b^2}-\frac {\sinh (5 a+5 b x)}{400 b^2}-\frac {x \cosh (a+b x)}{8 b}-\frac {x \cosh (3 a+3 b x)}{48 b}+\frac {x \cosh (5 a+5 b x)}{80 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 2717
Rule 3377
Rule 5556
Rubi steps
\begin {align*} \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx &=\int \left (-\frac {1}{8} x \sinh (a+b x)-\frac {1}{16} x \sinh (3 a+3 b x)+\frac {1}{16} x \sinh (5 a+5 b x)\right ) \, dx\\ &=-\left (\frac {1}{16} \int x \sinh (3 a+3 b x) \, dx\right )+\frac {1}{16} \int x \sinh (5 a+5 b x) \, dx-\frac {1}{8} \int x \sinh (a+b x) \, dx\\ &=-\frac {x \cosh (a+b x)}{8 b}-\frac {x \cosh (3 a+3 b x)}{48 b}+\frac {x \cosh (5 a+5 b x)}{80 b}-\frac {\int \cosh (5 a+5 b x) \, dx}{80 b}+\frac {\int \cosh (3 a+3 b x) \, dx}{48 b}+\frac {\int \cosh (a+b x) \, dx}{8 b}\\ &=-\frac {x \cosh (a+b x)}{8 b}-\frac {x \cosh (3 a+3 b x)}{48 b}+\frac {x \cosh (5 a+5 b x)}{80 b}+\frac {\sinh (a+b x)}{8 b^2}+\frac {\sinh (3 a+3 b x)}{144 b^2}-\frac {\sinh (5 a+5 b x)}{400 b^2}\\ \end {align*}
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Mathematica [A]
time = 0.13, size = 70, normalized size = 0.74 \begin {gather*} \frac {-450 b x \cosh (a+b x)-75 b x \cosh (3 (a+b x))+45 b x \cosh (5 (a+b x))+450 \sinh (a+b x)+25 \sinh (3 (a+b x))-9 \sinh (5 (a+b x))}{3600 b^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.08, size = 131, normalized size = 1.39
method | result | size |
risch | \(\frac {\left (5 b x -1\right ) {\mathrm e}^{5 b x +5 a}}{800 b^{2}}-\frac {\left (3 b x -1\right ) {\mathrm e}^{3 b x +3 a}}{288 b^{2}}-\frac {\left (b x -1\right ) {\mathrm e}^{b x +a}}{16 b^{2}}-\frac {\left (b x +1\right ) {\mathrm e}^{-b x -a}}{16 b^{2}}-\frac {\left (3 b x +1\right ) {\mathrm e}^{-3 b x -3 a}}{288 b^{2}}+\frac {\left (5 b x +1\right ) {\mathrm e}^{-5 b x -5 a}}{800 b^{2}}\) | \(117\) |
default | \(-\frac {\left (b x +a \right ) \cosh \left (b x +a \right )-\sinh \left (b x +a \right )-a \cosh \left (b x +a \right )}{8 b^{2}}-\frac {\left (3 b x +3 a \right ) \cosh \left (3 b x +3 a \right )-\sinh \left (3 b x +3 a \right )-3 a \cosh \left (3 b x +3 a \right )}{144 b^{2}}+\frac {\left (5 b x +5 a \right ) \cosh \left (5 b x +5 a \right )-\sinh \left (5 b x +5 a \right )-5 a \cosh \left (5 b x +5 a \right )}{400 b^{2}}\) | \(131\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 129, normalized size = 1.37 \begin {gather*} \frac {{\left (5 \, b x e^{\left (5 \, a\right )} - e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{800 \, b^{2}} - \frac {{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{288 \, b^{2}} - \frac {{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{2}} - \frac {{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{16 \, b^{2}} - \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{288 \, b^{2}} + \frac {{\left (5 \, b x + 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{800 \, b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 153, normalized size = 1.63 \begin {gather*} \frac {45 \, b x \cosh \left (b x + a\right )^{5} + 225 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 75 \, b x \cosh \left (b x + a\right )^{3} - 9 \, \sinh \left (b x + a\right )^{5} - 5 \, {\left (18 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )^{3} - 450 \, b x \cosh \left (b x + a\right ) + 225 \, {\left (2 \, b x \cosh \left (b x + a\right )^{3} - b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 15 \, {\left (3 \, \cosh \left (b x + a\right )^{4} - 5 \, \cosh \left (b x + a\right )^{2} - 30\right )} \sinh \left (b x + a\right )}{3600 \, b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.43, size = 112, normalized size = 1.19 \begin {gather*} \begin {cases} \frac {x \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 x \cosh ^{5}{\left (a + b x \right )}}{15 b} + \frac {26 \sinh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac {13 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {2 \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{15 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh ^{3}{\left (a \right )} \cosh ^{2}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 116, normalized size = 1.23 \begin {gather*} \frac {{\left (5 \, b x - 1\right )} e^{\left (5 \, b x + 5 \, a\right )}}{800 \, b^{2}} - \frac {{\left (3 \, b x - 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{288 \, b^{2}} - \frac {{\left (b x - 1\right )} e^{\left (b x + a\right )}}{16 \, b^{2}} - \frac {{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{16 \, b^{2}} - \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{288 \, b^{2}} + \frac {{\left (5 \, b x + 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{800 \, b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.14, size = 71, normalized size = 0.76 \begin {gather*} \frac {\frac {26\,\mathrm {sinh}\left (a+b\,x\right )}{225}-b\,\left (\frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{3}-\frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^5}{5}\right )+\frac {13\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{225}-\frac {{\mathrm {cosh}\left (a+b\,x\right )}^4\,\mathrm {sinh}\left (a+b\,x\right )}{25}}{b^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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