Optimal. Leaf size=94 \[ -\frac {3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}-\frac {3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 x}{8 b^3}+\frac {x^3}{4 b} \]
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Rubi [A] time = 0.07, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5372, 3311, 30, 2635, 8} \[ -\frac {3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}-\frac {3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 x}{8 b^3}+\frac {x^3}{4 b} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2635
Rule 3311
Rule 5372
Rubi steps
\begin {align*} \int x^3 \cosh (a+b x) \sinh (a+b x) \, dx &=\frac {x^3 \sinh ^2(a+b x)}{2 b}-\frac {3 \int x^2 \sinh ^2(a+b x) \, dx}{2 b}\\ &=-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}-\frac {3 \int \sinh ^2(a+b x) \, dx}{4 b^3}+\frac {3 \int x^2 \, dx}{4 b}\\ &=\frac {x^3}{4 b}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 \int 1 \, dx}{8 b^3}\\ &=\frac {3 x}{8 b^3}+\frac {x^3}{4 b}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 50, normalized size = 0.53 \[ \frac {\left (4 b^3 x^3+6 b x\right ) \cosh (2 (a+b x))-3 \left (2 b^2 x^2+1\right ) \sinh (2 (a+b x))}{16 b^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.39, size = 74, normalized size = 0.79 \[ \frac {{\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} - 3 \, {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \sinh \left (b x + a\right )^{2}}{8 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 73, normalized size = 0.78 \[ \frac {{\left (4 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, b x - 3\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 203, normalized size = 2.16 \[ \frac {\frac {\left (b x +a \right )^{3} \left (\cosh ^{2}\left (b x +a \right )\right )}{2}-\frac {3 \left (b x +a \right )^{2} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4}-\frac {\left (b x +a \right )^{3}}{4}+\frac {3 \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{4}-\frac {3 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}-\frac {3 b x}{8}-\frac {3 a}{8}-3 a \left (\frac {\left (b x +a \right )^{2} \left (\cosh ^{2}\left (b x +a \right )\right )}{2}-\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2}}{4}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{4}\right )+3 a^{2} \left (\frac {\left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{2}-\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4}-\frac {b x}{4}-\frac {a}{4}\right )-\frac {\left (\cosh ^{2}\left (b x +a \right )\right ) a^{3}}{2}}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 86, normalized size = 0.91 \[ \frac {{\left (4 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.50, size = 64, normalized size = 0.68 \[ -\frac {\frac {3\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{2}-2\,b^3\,x^3\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )+3\,b^2\,x^2\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )-3\,b\,x\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{8\,b^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.73, size = 119, normalized size = 1.27 \[ \begin {cases} \frac {x^{3} \sinh ^{2}{\left (a + b x \right )}}{4 b} + \frac {x^{3} \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac {3 x^{2} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b^{2}} + \frac {3 x \sinh ^{2}{\left (a + b x \right )}}{8 b^{3}} + \frac {3 x \cosh ^{2}{\left (a + b x \right )}}{8 b^{3}} - \frac {3 \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh {\relax (a )} \cosh {\relax (a )}}{4} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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