Optimal. Leaf size=104 \[ -\frac {2}{15} \sqrt {\pi } e^{-a} b^{5/2} \text {erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {2}{15} \sqrt {\pi } e^a b^{5/2} \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {4}{15} b^2 x \sinh \left (a+\frac {b}{x^2}\right )+\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )+\frac {2}{15} b x^3 \cosh \left (a+\frac {b}{x^2}\right ) \]
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Rubi [A] time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5346, 5326, 5327, 5299, 2204, 2205} \[ -\frac {2}{15} \sqrt {\pi } e^{-a} b^{5/2} \text {Erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {2}{15} \sqrt {\pi } e^a b^{5/2} \text {Erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {4}{15} b^2 x \sinh \left (a+\frac {b}{x^2}\right )+\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )+\frac {2}{15} b x^3 \cosh \left (a+\frac {b}{x^2}\right ) \]
Antiderivative was successfully verified.
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Rule 2204
Rule 2205
Rule 5299
Rule 5326
Rule 5327
Rule 5346
Rubi steps
\begin {align*} \int x^4 \sinh \left (a+\frac {b}{x^2}\right ) \, dx &=-\operatorname {Subst}\left (\int \frac {\sinh \left (a+b x^2\right )}{x^6} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{5} (2 b) \operatorname {Subst}\left (\int \frac {\cosh \left (a+b x^2\right )}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{15} b x^3 \cosh \left (a+\frac {b}{x^2}\right )+\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{15} \left (4 b^2\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (a+b x^2\right )}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{15} b x^3 \cosh \left (a+\frac {b}{x^2}\right )+\frac {4}{15} b^2 x \sinh \left (a+\frac {b}{x^2}\right )+\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{15} \left (8 b^3\right ) \operatorname {Subst}\left (\int \cosh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{15} b x^3 \cosh \left (a+\frac {b}{x^2}\right )+\frac {4}{15} b^2 x \sinh \left (a+\frac {b}{x^2}\right )+\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{15} \left (4 b^3\right ) \operatorname {Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac {1}{x}\right )-\frac {1}{15} \left (4 b^3\right ) \operatorname {Subst}\left (\int e^{a+b x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{15} b x^3 \cosh \left (a+\frac {b}{x^2}\right )-\frac {2}{15} b^{5/2} e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {2}{15} b^{5/2} e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {4}{15} b^2 x \sinh \left (a+\frac {b}{x^2}\right )+\frac {1}{5} x^5 \sinh \left (a+\frac {b}{x^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.11, size = 102, normalized size = 0.98 \[ \frac {1}{15} \left (2 \sqrt {\pi } b^{5/2} (\sinh (a)-\cosh (a)) \text {erf}\left (\frac {\sqrt {b}}{x}\right )-2 \sqrt {\pi } b^{5/2} (\sinh (a)+\cosh (a)) \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+4 b^2 x \sinh \left (a+\frac {b}{x^2}\right )+3 x^5 \sinh \left (a+\frac {b}{x^2}\right )+2 b x^3 \cosh \left (a+\frac {b}{x^2}\right )\right ) \]
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 323, normalized size = 3.11 \[ -\frac {3 \, x^{5} - 2 \, b x^{3} + 4 \, b^{2} x - {\left (3 \, x^{5} + 2 \, b x^{3} + 4 \, b^{2} x\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 4 \, \sqrt {\pi } {\left (b^{2} \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b^{2} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (b^{2} \cosh \relax (a) + b^{2} \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) + 4 \, \sqrt {\pi } {\left (b^{2} \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - b^{2} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (b^{2} \cosh \relax (a) - b^{2} \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) - 2 \, {\left (3 \, x^{5} + 2 \, b x^{3} + 4 \, b^{2} x\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - {\left (3 \, x^{5} + 2 \, b x^{3} + 4 \, b^{2} x\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2}}{30 \, {\left (\cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \sinh \left (a + \frac {b}{x^{2}}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 138, normalized size = 1.33 \[ -\frac {{\mathrm e}^{-a} x^{5} {\mathrm e}^{-\frac {b}{x^{2}}}}{10}+\frac {{\mathrm e}^{-a} b \,x^{3} {\mathrm e}^{-\frac {b}{x^{2}}}}{15}-\frac {2 \,{\mathrm e}^{-a} b^{\frac {5}{2}} \sqrt {\pi }\, \erf \left (\frac {\sqrt {b}}{x}\right )}{15}-\frac {2 \,{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}} b^{2} x}{15}+\frac {{\mathrm e}^{a} x^{5} {\mathrm e}^{\frac {b}{x^{2}}}}{10}+\frac {{\mathrm e}^{a} b \,x^{3} {\mathrm e}^{\frac {b}{x^{2}}}}{15}+\frac {2 \,{\mathrm e}^{a} b^{2} {\mathrm e}^{\frac {b}{x^{2}}} x}{15}-\frac {2 \,{\mathrm e}^{a} b^{3} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b}}{x}\right )}{15 \sqrt {-b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.34, size = 62, normalized size = 0.60 \[ \frac {1}{5} \, x^{5} \sinh \left (a + \frac {b}{x^{2}}\right ) + \frac {1}{10} \, {\left (x^{3} \left (\frac {b}{x^{2}}\right )^{\frac {3}{2}} e^{\left (-a\right )} \Gamma \left (-\frac {3}{2}, \frac {b}{x^{2}}\right ) + x^{3} \left (-\frac {b}{x^{2}}\right )^{\frac {3}{2}} e^{a} \Gamma \left (-\frac {3}{2}, -\frac {b}{x^{2}}\right )\right )} b \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\mathrm {sinh}\left (a+\frac {b}{x^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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