Optimal. Leaf size=155 \[ \frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}-\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}+\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \]
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Rubi [A] time = 0.41, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5653, 5717, 5779, 3308, 2180, 2204, 2205} \[ \frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}-\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}+\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \]
Antiderivative was successfully verified.
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Rule 2180
Rule 2204
Rule 2205
Rule 3308
Rule 5653
Rule 5717
Rule 5779
Rubi steps
\begin {align*} \int \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \, dx &=x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {1}{2} (5 b c) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {1}{4} \left (15 b^2\right ) \int \sqrt {a+b \sinh ^{-1}(c x)} \, dx\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {1}{8} \left (15 b^3 c\right ) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c}\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}-\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c}\\ &=\frac {15}{4} b^2 x \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}-\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c}\\ \end {align*}
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Mathematica [A] time = 3.65, size = 282, normalized size = 1.82 \[ \frac {\sqrt {b} e^{-\frac {a}{b}} \left (-\left (\sqrt {\pi } \left (4 a^2-15 b^2\right ) e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )\right )+\sqrt {\pi } \left (4 a^2-15 b^2\right ) \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )+\frac {4 \sqrt {b} \left (-2 a^2 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )-2 a^2 \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )+e^{a/b} \left (a+b \sinh ^{-1}(c x)\right ) \left (5 \left (3 b c x-2 a \sqrt {c^2 x^2+1}\right )+2 \sinh ^{-1}(c x) \left (4 a c x-5 b \sqrt {c^2 x^2+1}\right )+4 b c x \sinh ^{-1}(c x)^2\right )\right )}{\sqrt {a+b \sinh ^{-1}(c x)}}\right )}{16 c} \]
Warning: Unable to verify antiderivative.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.37, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsinh \left (c x \right )\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]
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 \[ \int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{5/2} \,d x \]
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 \[ \int \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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