Optimal. Leaf size=155 \[ -\frac{3 x^2 \sinh (a+b x) \cosh ^3(a+b x)}{16 b^2}-\frac{9 x^2 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac{3 x \cosh ^4(a+b x)}{32 b^3}+\frac{9 x \cosh ^2(a+b x)}{32 b^3}-\frac{3 \sinh (a+b x) \cosh ^3(a+b x)}{128 b^4}-\frac{45 \sinh (a+b x) \cosh (a+b x)}{256 b^4}+\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{45 x}{256 b^3}-\frac{3 x^3}{32 b} \]
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Rubi [A] time = 0.141297, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {5373, 3311, 30, 2635, 8} \[ -\frac{3 x^2 \sinh (a+b x) \cosh ^3(a+b x)}{16 b^2}-\frac{9 x^2 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac{3 x \cosh ^4(a+b x)}{32 b^3}+\frac{9 x \cosh ^2(a+b x)}{32 b^3}-\frac{3 \sinh (a+b x) \cosh ^3(a+b x)}{128 b^4}-\frac{45 \sinh (a+b x) \cosh (a+b x)}{256 b^4}+\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{45 x}{256 b^3}-\frac{3 x^3}{32 b} \]
Antiderivative was successfully verified.
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Rule 5373
Rule 3311
Rule 30
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int x^3 \cosh ^3(a+b x) \sinh (a+b x) \, dx &=\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{3 \int x^2 \cosh ^4(a+b x) \, dx}{4 b}\\ &=\frac{3 x \cosh ^4(a+b x)}{32 b^3}+\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac{3 \int \cosh ^4(a+b x) \, dx}{32 b^3}-\frac{9 \int x^2 \cosh ^2(a+b x) \, dx}{16 b}\\ &=\frac{9 x \cosh ^2(a+b x)}{32 b^3}+\frac{3 x \cosh ^4(a+b x)}{32 b^3}+\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{3 \cosh ^3(a+b x) \sinh (a+b x)}{128 b^4}-\frac{3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac{9 \int \cosh ^2(a+b x) \, dx}{128 b^3}-\frac{9 \int \cosh ^2(a+b x) \, dx}{32 b^3}-\frac{9 \int x^2 \, dx}{32 b}\\ &=-\frac{3 x^3}{32 b}+\frac{9 x \cosh ^2(a+b x)}{32 b^3}+\frac{3 x \cosh ^4(a+b x)}{32 b^3}+\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{45 \cosh (a+b x) \sinh (a+b x)}{256 b^4}-\frac{9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{3 \cosh ^3(a+b x) \sinh (a+b x)}{128 b^4}-\frac{3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac{9 \int 1 \, dx}{256 b^3}-\frac{9 \int 1 \, dx}{64 b^3}\\ &=-\frac{45 x}{256 b^3}-\frac{3 x^3}{32 b}+\frac{9 x \cosh ^2(a+b x)}{32 b^3}+\frac{3 x \cosh ^4(a+b x)}{32 b^3}+\frac{x^3 \cosh ^4(a+b x)}{4 b}-\frac{45 \cosh (a+b x) \sinh (a+b x)}{256 b^4}-\frac{9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{3 \cosh ^3(a+b x) \sinh (a+b x)}{128 b^4}-\frac{3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}\\ \end{align*}
Mathematica [A] time = 0.599431, size = 91, normalized size = 0.59 \[ \frac{32 b x \left (2 b^2 x^2+3\right ) \cosh (2 (a+b x))+2 b x \left (8 b^2 x^2+3\right ) \cosh (4 (a+b x))-3 \sinh (2 (a+b x)) \left (\left (8 b^2 x^2+1\right ) \cosh (2 (a+b x))+32 b^2 x^2+16\right )}{512 b^4} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.009, size = 414, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.082, size = 231, normalized size = 1.49 \begin{align*} \frac{{\left (32 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 24 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 12 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{2048 \, b^{4}} + \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 )}}{64 \, 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 )}}{64 \, b^{4}} + \frac{{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.77661, size = 468, normalized size = 3.02 \begin{align*} \frac{{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{4} - 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} +{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \sinh \left (b x + a\right )^{4} + 16 \,{\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (16 \, b^{3} x^{3} + 3 \,{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} + 24 \, b x\right )} \sinh \left (b x + a\right )^{2} - 3 \,{\left ({\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} + 16 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{256 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 8.53415, size = 226, normalized size = 1.46 \begin{align*} \begin{cases} - \frac{3 x^{3} \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} + \frac{5 x^{3} \cosh ^{4}{\left (a + b x \right )}}{32 b} + \frac{9 x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{32 b^{2}} - \frac{15 x^{2} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b^{2}} - \frac{45 x \sinh ^{4}{\left (a + b x \right )}}{256 b^{3}} + \frac{9 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{128 b^{3}} + \frac{51 x \cosh ^{4}{\left (a + b x \right )}}{256 b^{3}} + \frac{45 \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{256 b^{4}} - \frac{51 \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{256 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh{\left (a \right )} \cosh ^{3}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15055, size = 196, normalized size = 1.26 \begin{align*} \frac{{\left (32 \, b^{3} x^{3} - 24 \, b^{2} x^{2} + 12 \, b x - 3\right )} e^{\left (4 \, b x + 4 \, a\right )}}{2048 \, 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 )}}{64 \, 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 )}}{64 \, b^{4}} + \frac{{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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