Optimal. Leaf size=115 \[ -24 a b^2 x-\frac {48 b^3 \sqrt {2 d x^2-d^2 x^4}}{d x}+24 b^3 x \text {ArcSin}\left (1-d x^2\right )+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^3 \]
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Rubi [A]
time = 0.04, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4898, 4924, 12,
1602} \begin {gather*} \frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^3-24 a b^2 x+24 b^3 x \text {ArcSin}\left (1-d x^2\right )-\frac {48 b^3 \sqrt {2 d x^2-d^2 x^4}}{d x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 1602
Rule 4898
Rule 4924
Rubi steps
\begin {align*} \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3 \, dx &=\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3-\left (24 b^2\right ) \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right ) \, dx\\ &=-24 a b^2 x+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3+\left (24 b^3\right ) \int \sin ^{-1}\left (1-d x^2\right ) \, dx\\ &=-24 a b^2 x+24 b^3 x \sin ^{-1}\left (1-d x^2\right )+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3-\left (24 b^3\right ) \int -\frac {2 d x^2}{\sqrt {2 d x^2-d^2 x^4}} \, dx\\ &=-24 a b^2 x+24 b^3 x \sin ^{-1}\left (1-d x^2\right )+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3+\left (48 b^3 d\right ) \int \frac {x^2}{\sqrt {2 d x^2-d^2 x^4}} \, dx\\ &=-24 a b^2 x-\frac {48 b^3 \sqrt {2 d x^2-d^2 x^4}}{d x}+24 b^3 x \sin ^{-1}\left (1-d x^2\right )+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 166, normalized size = 1.44 \begin {gather*} \frac {a \left (a^2-24 b^2\right ) d x^2+6 b \left (a^2-8 b^2\right ) \sqrt {d x^2 \left (2-d x^2\right )}-3 b \left (a^2 d x^2-8 b^2 d x^2+4 a b \sqrt {-d x^2 \left (-2+d x^2\right )}\right ) \text {ArcSin}\left (1-d x^2\right )+3 b^2 \left (a d x^2+2 b \sqrt {-d x^2 \left (-2+d x^2\right )}\right ) \text {ArcSin}\left (1-d x^2\right )^2-b^3 d x^2 \text {ArcSin}\left (1-d x^2\right )^3}{d x} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{3}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.30, size = 144, normalized size = 1.25 \begin {gather*} \frac {b^{3} d x^{2} \arcsin \left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} d x^{2} \arcsin \left (d x^{2} - 1\right )^{2} + 3 \, {\left (a^{2} b - 8 \, b^{3}\right )} d x^{2} \arcsin \left (d x^{2} - 1\right ) + {\left (a^{3} - 24 \, a b^{2}\right )} d x^{2} + 6 \, \sqrt {-d^{2} x^{4} + 2 \, d x^{2}} {\left (b^{3} \arcsin \left (d x^{2} - 1\right )^{2} + 2 \, a b^{2} \arcsin \left (d x^{2} - 1\right ) + a^{2} b - 8 \, b^{3}\right )}}{d x} \end {gather*}
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 \begin {gather*} \int \left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{3}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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