Optimal. Leaf size=126 \[ -\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}+\frac {b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e} \]
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Rubi [A] time = 0.18, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4805, 12, 4625, 3717, 2190, 2531, 2282, 6589} \[ -\frac {i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}+\frac {b^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2282
Rule 2531
Rule 3717
Rule 4625
Rule 4805
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {i b \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {i b \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {i b \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 170, normalized size = 1.35 \[ \frac {a^2 \log (c+d x)-i a b \left (\sin ^{-1}(c+d x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )\right )+2 a b \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )+b^2 \left (i \sin ^{-1}(c+d x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+\frac {1}{3} i \sin ^{-1}(c+d x)^3+\sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac {i \pi ^3}{24}\right )}{d e} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{d e x + c e}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{d e x + c e}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 455, normalized size = 3.61 \[ \frac {a^{2} \ln \left (d x +c \right )}{d e}-\frac {i b^{2} \arcsin \left (d x +c \right )^{3}}{3 d e}+\frac {b^{2} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {2 i b^{2} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {2 b^{2} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {b^{2} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {2 i b^{2} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {2 b^{2} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {i a b \arcsin \left (d x +c \right )^{2}}{d e}+\frac {2 a b \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {2 a b \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {2 i a b \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {2 i a b \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e} \]
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 \[ \frac {a^{2} \log \left (d e x + c e\right )}{d e} + \int \frac {b^{2} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 2 \, a b \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d e x + c e}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{c\,e+d\,e\,x} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c + d x}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c + d x \right )}}{c + d x}\, dx}{e} \]
Verification of antiderivative is not currently implemented for this CAS.
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