Optimal. Leaf size=128 \[ -\frac {i \text {ArcSin}(a+b x)^3}{b}+\frac {(a+b x) \text {ArcSin}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {3 \text {ArcSin}(a+b x)^2 \log \left (1+e^{2 i \text {ArcSin}(a+b x)}\right )}{b}-\frac {3 i \text {ArcSin}(a+b x) \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(a+b x)}\right )}{b}+\frac {3 \text {PolyLog}\left (3,-e^{2 i \text {ArcSin}(a+b x)}\right )}{2 b} \]
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Rubi [A]
time = 0.14, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4891, 4745,
4765, 3800, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {3 i \text {ArcSin}(a+b x) \text {Li}_2\left (-e^{2 i \text {ArcSin}(a+b x)}\right )}{b}+\frac {3 \text {Li}_3\left (-e^{2 i \text {ArcSin}(a+b x)}\right )}{2 b}+\frac {(a+b x) \text {ArcSin}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}-\frac {i \text {ArcSin}(a+b x)^3}{b}+\frac {3 \text {ArcSin}(a+b x)^2 \log \left (1+e^{2 i \text {ArcSin}(a+b x)}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 4745
Rule 4765
Rule 4891
Rule 6724
Rubi steps
\begin {align*} \int \frac {\sin ^{-1}(a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin ^{-1}(x)^3}{\left (1-x^2\right )^{3/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}-\frac {3 \text {Subst}\left (\int \frac {x \sin ^{-1}(x)^2}{1-x^2} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}-\frac {3 \text {Subst}\left (\int x^2 \tan (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {i \sin ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {(6 i) \text {Subst}\left (\int \frac {e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {i \sin ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {i \sin ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac {3 i \sin ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac {(3 i) \text {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {i \sin ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac {3 i \sin ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac {3 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}\\ &=-\frac {i \sin ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac {3 i \sin ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac {3 \text {Li}_3\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}\\ \end {align*}
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Mathematica [A]
time = 0.39, size = 144, normalized size = 1.12 \begin {gather*} \frac {2 \text {ArcSin}(a+b x)^2 \left (\frac {\left (a+b x-i \sqrt {1-a^2-2 a b x-b^2 x^2}\right ) \text {ArcSin}(a+b x)}{\sqrt {1-a^2-2 a b x-b^2 x^2}}+3 \log \left (1+e^{2 i \text {ArcSin}(a+b x)}\right )\right )-6 i \text {ArcSin}(a+b x) \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(a+b x)}\right )+3 \text {PolyLog}\left (3,-e^{2 i \text {ArcSin}(a+b x)}\right )}{2 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.90, size = 229, normalized size = 1.79
method | result | size |
default | \(\frac {\left (-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x +i b^{2} x^{2}-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +2 i a b x +i a^{2}-i\right ) \arcsin \left (b x +a \right )^{3}}{b \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}-\frac {4 i \arcsin \left (b x +a \right )^{3}-6 \arcsin \left (b x +a \right )^{2} \ln \left (1+\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )+6 i \arcsin \left (b x +a \right ) \polylog \left (2, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )-3 \polylog \left (3, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{2 b}\) | \(229\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.62, size = 137, normalized size = 1.07 \begin {gather*} \frac {3}{2} \, b {\left (\frac {\log \left (b x + a + 1\right )}{b^{2}} + \frac {\log \left (b x + a - 1\right )}{b^{2}}\right )} \arcsin \left (b x + a\right )^{2} + {\left (\frac {b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {a b}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}\right )} \arcsin \left (b x + a\right )^{3} \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*} \int \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}\, dx \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 {{\mathrm {asin}\left (a+b\,x\right )}^3}{{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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