Optimal. Leaf size=124 \[ -\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {b \sqrt {1-c^2 x^2}}{c^3 d} \]
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Rubi [A] time = 0.14, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4715, 4657, 4181, 2279, 2391, 261} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {b \sqrt {1-c^2 x^2}}{c^3 d} \]
Antiderivative was successfully verified.
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Rule 261
Rule 2279
Rule 2391
Rule 4181
Rule 4657
Rule 4715
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{c^2}+\frac {b \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{c d}\\ &=-\frac {b \sqrt {1-c^2 x^2}}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}+\frac {\operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d}\\ &=-\frac {b \sqrt {1-c^2 x^2}}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {b \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d}+\frac {b \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d}\\ &=-\frac {b \sqrt {1-c^2 x^2}}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^3 d}\\ &=-\frac {b \sqrt {1-c^2 x^2}}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 238, normalized size = 1.92 \[ -\frac {2 a c x+a \log (1-c x)-a \log (c x+1)+2 b \sqrt {1-c^2 x^2}-2 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+2 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+2 b c x \sin ^{-1}(c x)+i \pi b \sin ^{-1}(c x)-2 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+2 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{2 c^3 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{2} \arcsin \left (c x\right ) + a x^{2}}{c^{2} d x^{2} - d}, x\right ) \]
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 \[ \int -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{c^{2} d x^{2} - d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 218, normalized size = 1.76 \[ -\frac {a x}{c^{2} d}-\frac {a \ln \left (c x -1\right )}{2 c^{3} d}+\frac {a \ln \left (c x +1\right )}{2 c^{3} d}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d}-\frac {b \arcsin \left (c x \right ) x}{c^{2} d}-\frac {b \sqrt {-c^{2} x^{2}+1}}{c^{3} d} \]
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 \[ -\frac {1}{2} \, a {\left (\frac {2 \, x}{c^{2} d} - \frac {\log \left (c x + 1\right )}{c^{3} d} + \frac {\log \left (c x - 1\right )}{c^{3} d}\right )} + \frac {-{\left (c^{3} d {\left (\frac {2 \, \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{3} d} + \int -\frac {\sqrt {c x + 1} \sqrt {-c x + 1} {\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{c^{4} d x^{2} - c^{2} d}\,{d x}\right )} + 2 \, c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) + \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right )\right )} b}{2 \, c^{3} d} \]
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 \[ \int \frac {x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{d-c^2\,d\,x^2} \,d x \]
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 \[ - \frac {\int \frac {a x^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{2} \operatorname {asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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