Optimal. Leaf size=153 \[ \frac {\sinh ^{-1}(a x)^3}{8 a^5}+\frac {15 \sinh ^{-1}(a x)}{64 a^5}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}+\frac {x^3 \sqrt {a^2 x^2+1}}{32 a^2}+\frac {x^3 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{4 a^2}-\frac {15 x \sqrt {a^2 x^2+1}}{64 a^4}-\frac {3 x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{8 a^4}-\frac {x^4 \sinh ^{-1}(a x)}{8 a} \]
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Rubi [A] time = 0.29, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5758, 5675, 5661, 321, 215} \[ \frac {x^3 \sqrt {a^2 x^2+1}}{32 a^2}-\frac {15 x \sqrt {a^2 x^2+1}}{64 a^4}+\frac {x^3 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {3 x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}+\frac {15 \sinh ^{-1}(a x)}{64 a^5}-\frac {x^4 \sinh ^{-1}(a x)}{8 a} \]
Antiderivative was successfully verified.
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Rule 215
Rule 321
Rule 5661
Rule 5675
Rule 5758
Rubi steps
\begin {align*} \int \frac {x^4 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx &=\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}-\frac {3 \int \frac {x^2 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2}-\frac {\int x^3 \sinh ^{-1}(a x) \, dx}{2 a}\\ &=-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {1}{8} \int \frac {x^4}{\sqrt {1+a^2 x^2}} \, dx+\frac {3 \int \frac {\sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{8 a^4}+\frac {3 \int x \sinh ^{-1}(a x) \, dx}{4 a^3}\\ &=\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}-\frac {3 \int \frac {x^2}{\sqrt {1+a^2 x^2}} \, dx}{32 a^2}-\frac {3 \int \frac {x^2}{\sqrt {1+a^2 x^2}} \, dx}{8 a^2}\\ &=-\frac {15 x \sqrt {1+a^2 x^2}}{64 a^4}+\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}+\frac {3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{64 a^4}+\frac {3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{16 a^4}\\ &=-\frac {15 x \sqrt {1+a^2 x^2}}{64 a^4}+\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {15 \sinh ^{-1}(a x)}{64 a^5}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}\\ \end {align*}
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Mathematica [A] time = 0.08, size = 98, normalized size = 0.64 \[ \frac {a x \sqrt {a^2 x^2+1} \left (2 a^2 x^2-15\right )+8 a x \sqrt {a^2 x^2+1} \left (2 a^2 x^2-3\right ) \sinh ^{-1}(a x)^2+\left (-8 a^4 x^4+24 a^2 x^2+15\right ) \sinh ^{-1}(a x)+8 \sinh ^{-1}(a x)^3}{64 a^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 131, normalized size = 0.86 \[ \frac {8 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 8 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} - {\left (8 \, a^{4} x^{4} - 24 \, a^{2} x^{2} - 15\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (2 \, a^{3} x^{3} - 15 \, a x\right )} \sqrt {a^{2} x^{2} + 1}}{64 \, a^{5}} \]
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 \[ \int \frac {x^{4} \operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 125, normalized size = 0.82 \[ \frac {16 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}-8 \arcsinh \left (a x \right ) x^{4} a^{4}+2 \sqrt {a^{2} x^{2}+1}\, x^{3} a^{3}-24 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a x +24 \arcsinh \left (a x \right ) x^{2} a^{2}+8 \arcsinh \left (a x \right )^{3}-15 \sqrt {a^{2} x^{2}+1}\, x a +15 \arcsinh \left (a x \right )}{64 a^{5}} \]
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 \[ \int \frac {x^{4} \operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]
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 \[ \int \frac {x^4\,{\mathrm {asinh}\left (a\,x\right )}^2}{\sqrt {a^2\,x^2+1}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.68, size = 146, normalized size = 0.95 \[ \begin {cases} - \frac {x^{4} \operatorname {asinh}{\left (a x \right )}}{8 a} + \frac {x^{3} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{4 a^{2}} + \frac {x^{3} \sqrt {a^{2} x^{2} + 1}}{32 a^{2}} + \frac {3 x^{2} \operatorname {asinh}{\left (a x \right )}}{8 a^{3}} - \frac {3 x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{8 a^{4}} - \frac {15 x \sqrt {a^{2} x^{2} + 1}}{64 a^{4}} + \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{8 a^{5}} + \frac {15 \operatorname {asinh}{\left (a x \right )}}{64 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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