Optimal. Leaf size=157 \[ \frac{x^3 \sqrt{1-a^2 x^2}}{32 a^2}+\frac{15 x \sqrt{1-a^2 x^2}}{64 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{4 a^2}+\frac{3 x^2 \sin ^{-1}(a x)}{8 a^3}-\frac{3 x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{8 a^4}+\frac{\sin ^{-1}(a x)^3}{8 a^5}-\frac{15 \sin ^{-1}(a x)}{64 a^5}+\frac{x^4 \sin ^{-1}(a x)}{8 a} \]
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Rubi [A] time = 0.270624, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4707, 4641, 4627, 321, 216} \[ \frac{x^3 \sqrt{1-a^2 x^2}}{32 a^2}+\frac{15 x \sqrt{1-a^2 x^2}}{64 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{4 a^2}+\frac{3 x^2 \sin ^{-1}(a x)}{8 a^3}-\frac{3 x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{8 a^4}+\frac{\sin ^{-1}(a x)^3}{8 a^5}-\frac{15 \sin ^{-1}(a x)}{64 a^5}+\frac{x^4 \sin ^{-1}(a x)}{8 a} \]
Antiderivative was successfully verified.
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Rule 4707
Rule 4641
Rule 4627
Rule 321
Rule 216
Rubi steps
\begin{align*} \int \frac{x^4 \sin ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx &=-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{4 a^2}+\frac{3 \int \frac{x^2 \sin ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{4 a^2}+\frac{\int x^3 \sin ^{-1}(a x) \, dx}{2 a}\\ &=\frac{x^4 \sin ^{-1}(a x)}{8 a}-\frac{3 x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-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{\sin ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{8 a^4}+\frac{3 \int x \sin ^{-1}(a x) \, dx}{4 a^3}\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{32 a^2}+\frac{3 x^2 \sin ^{-1}(a x)}{8 a^3}+\frac{x^4 \sin ^{-1}(a x)}{8 a}-\frac{3 x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{4 a^2}+\frac{\sin ^{-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 \sin ^{-1}(a x)}{8 a^3}+\frac{x^4 \sin ^{-1}(a x)}{8 a}-\frac{3 x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{4 a^2}+\frac{\sin ^{-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 \sin ^{-1}(a x)}{64 a^5}+\frac{3 x^2 \sin ^{-1}(a x)}{8 a^3}+\frac{x^4 \sin ^{-1}(a x)}{8 a}-\frac{3 x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{8 a^4}-\frac{x^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{4 a^2}+\frac{\sin ^{-1}(a x)^3}{8 a^5}\\ \end{align*}
Mathematica [A] time = 0.0516752, size = 100, normalized size = 0.64 \[ \frac{a x \sqrt{1-a^2 x^2} \left (2 a^2 x^2+15\right )-8 a x \sqrt{1-a^2 x^2} \left (2 a^2 x^2+3\right ) \sin ^{-1}(a x)^2+\left (8 a^4 x^4+24 a^2 x^2-15\right ) \sin ^{-1}(a x)+8 \sin ^{-1}(a x)^3}{64 a^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.063, size = 129, normalized size = 0.8 \begin{align*}{\frac{1}{64\,{a}^{5}} \left ( -16\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}\sqrt{-{a}^{2}{x}^{2}+1}{x}^{3}{a}^{3}+8\,{a}^{4}{x}^{4}\arcsin \left ( ax \right ) +2\,{a}^{3}{x}^{3}\sqrt{-{a}^{2}{x}^{2}+1}-24\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}\sqrt{-{a}^{2}{x}^{2}+1}xa+24\,{a}^{2}{x}^{2}\arcsin \left ( ax \right ) +8\, \left ( \arcsin \left ( ax \right ) \right ) ^{3}+15\,ax\sqrt{-{a}^{2}{x}^{2}+1}-15\,\arcsin \left ( ax \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \arcsin \left (a x\right )^{2}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.7379, size = 205, normalized size = 1.31 \begin{align*} \frac{8 \, \arcsin \left (a x\right )^{3} +{\left (8 \, a^{4} x^{4} + 24 \, a^{2} x^{2} - 15\right )} \arcsin \left (a x\right ) +{\left (2 \, a^{3} x^{3} - 8 \,{\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \arcsin \left (a x\right )^{2} + 15 \, a x\right )} \sqrt{-a^{2} x^{2} + 1}}{64 \, a^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 4.83173, size = 146, normalized size = 0.93 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{asin}{\left (a x \right )}}{8 a} - \frac{x^{3} \sqrt{- a^{2} x^{2} + 1} \operatorname{asin}^{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{asin}{\left (a x \right )}}{8 a^{3}} - \frac{3 x \sqrt{- a^{2} x^{2} + 1} \operatorname{asin}^{2}{\left (a x \right )}}{8 a^{4}} + \frac{15 x \sqrt{- a^{2} x^{2} + 1}}{64 a^{4}} + \frac{\operatorname{asin}^{3}{\left (a x \right )}}{8 a^{5}} - \frac{15 \operatorname{asin}{\left (a x \right )}}{64 a^{5}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.2664, size = 193, normalized size = 1.23 \begin{align*} \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x \arcsin \left (a x\right )^{2}}{4 \, a^{4}} - \frac{5 \, \sqrt{-a^{2} x^{2} + 1} x \arcsin \left (a x\right )^{2}}{8 \, a^{4}} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{32 \, a^{4}} + \frac{{\left (a^{2} x^{2} - 1\right )}^{2} \arcsin \left (a x\right )}{8 \, a^{5}} + \frac{\arcsin \left (a x\right )^{3}}{8 \, a^{5}} + \frac{17 \, \sqrt{-a^{2} x^{2} + 1} x}{64 \, a^{4}} + \frac{5 \,{\left (a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )}{8 \, a^{5}} + \frac{17 \, \arcsin \left (a x\right )}{64 \, a^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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