Optimal. Leaf size=454 \[ -\frac {i d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}} \]
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Rubi [A] time = 0.66, antiderivative size = 454, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {4673, 4763, 4651, 4675, 3719, 2190, 2279, 2391, 4677, 4657, 4181} \[ -\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 3719
Rule 4181
Rule 4651
Rule 4657
Rule 4673
Rule 4675
Rule 4677
Rule 4763
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {(d+c d x) \left (a+b \sin ^{-1}(c x)\right )^2}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{(d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {\left (1-c^2 x^2\right )^{3/2} \int \left (\frac {d \left (a+b \sin ^{-1}(c x)\right )^2}{\left (1-c^2 x^2\right )^{3/2}}+\frac {c d x \left (a+b \sin ^{-1}(c x)\right )^2}{\left (1-c^2 x^2\right )^{3/2}}\right ) \, dx}{(d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {\left (d \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{(d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {\left (c d \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{(d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 b d \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 b c d \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{(d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 b d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 b d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {\left (4 i b d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {\left (2 b^2 d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 b^2 d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 i b^2 d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {\left (2 i b^2 d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {\left (2 b^2 d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {\left (i b^2 d \left (1-c^2 x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}\\ &=\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}\\ \end {align*}
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Mathematica [A] time = 1.81, size = 221, normalized size = 0.49 \[ -\frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (2 b \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \left (a \tan \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )+2 b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )\right )+a \left (a c x+a+4 b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )\right )-4 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2 \left (\tan \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )-i\right )\right )}{c d e^2 (c x-1) (c x+1)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{c^{3} d e^{2} x^{3} - c^{2} d e^{2} x^{2} - c d e^{2} x + d e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}\, \left (-c e x +e \right )^{\frac {3}{2}}}\, dx \]
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 \[ -\frac {2 \, \sqrt {-c^{2} d e x^{2} + d e} a b \arcsin \left (c x\right )}{c^{2} d e^{2} x - c d e^{2}} - \frac {\frac {b^{2} \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2}}{\sqrt {c x + 1} {\left (c x - 1\right )} \sqrt {-c x + 1}}\,{d x}}{e}}{\sqrt {d} \sqrt {e}} - \frac {\sqrt {-c^{2} d e x^{2} + d e} a^{2}}{c^{2} d e^{2} x - c d e^{2}} + \frac {2 \, a b \log \left (c x - 1\right )}{c \sqrt {d} e^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{\sqrt {d+c\,d\,x}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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