Optimal. Leaf size=347 \[ -\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e}+\frac {2 b^2 \text {Li}_3\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {Li}_3\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e} \]
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Rubi [A] time = 0.51, antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4741, 4519, 2190, 2531, 2282, 6589} \[ -\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}+\frac {2 b^2 \text {PolyLog}\left (3,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {PolyLog}\left (3,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 4519
Rule 4741
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d+e x} \, dx &=\operatorname {Subst}\left (\int \frac {(a+b x)^2 \cos (x)}{c d+e \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e}+\operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{c d-\sqrt {c^2 d^2-e^2}-i e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )+\operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{c d+\sqrt {c^2 d^2-e^2}-i e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-\frac {i e e^{i x}}{c d-\sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e}-\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-\frac {i e e^{i x}}{c d+\sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {i e e^{i x}}{c d-\sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {i e e^{i x}}{c d+\sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i e x}{c d-\sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i e x}{c d+\sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {Li}_3\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {Li}_3\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 332, normalized size = 0.96 \[ \frac {6 b \left (b \text {Li}_3\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )-i \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )\right )+6 b \left (b \text {Li}_3\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )-i \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )+3 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+3 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{b}}{3 e} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{e x + d}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{e x + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{e x +d}\, dx \]
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 \[ \frac {a^{2} \log \left (e x + d\right )}{e} + \int \frac {b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{e x + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{d+e\,x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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