Optimal. Leaf size=156 \[ -\frac {2 i (a+b \text {ArcSin}(c x))^2 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{c d}+\frac {2 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{c d}-\frac {2 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{c d}-\frac {2 b^2 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )}{c d}+\frac {2 b^2 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )}{c d} \]
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Rubi [A]
time = 0.09, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4749, 4266,
2611, 2320, 6724} \begin {gather*} -\frac {2 i \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{c d}+\frac {2 i b \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c d}-\frac {2 i b \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c d}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \text {ArcSin}(c x)}\right )}{c d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \text {ArcSin}(c x)}\right )}{c d} \end {gather*}
Antiderivative was successfully verified.
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Rule 2320
Rule 2611
Rule 4266
Rule 4749
Rule 6724
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=\frac {\text {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}+\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}+\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d}\\ &=-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}\\ \end {align*}
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Mathematica [A]
time = 0.26, size = 207, normalized size = 1.33 \begin {gather*} \frac {-4 i b^2 \text {ArcSin}(c x)^2 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )+4 a b \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-4 a b \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )-a^2 \log (1-c x)+a^2 \log (1+c x)+4 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )-4 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )-4 b^2 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )+4 b^2 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )}{2 c d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.16, size = 375, normalized size = 2.40
method | result | size |
derivativedivides | \(\frac {\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i a b \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i a b \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i a^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}}{c}\) | \(375\) |
default | \(\frac {\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i a b \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i a b \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i a^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}}{c}\) | \(375\) |
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 [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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \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 \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{d-c^2\,d\,x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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