Optimal. Leaf size=297 \[ \frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c^3 d}-\frac {x (a+b \text {ArcSin}(c x))^2}{c^4 d}-\frac {x^3 (a+b \text {ArcSin}(c x))^2}{3 c^2 d}-\frac {2 i (a+b \text {ArcSin}(c x))^2 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{c^5 d}+\frac {2 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{c^5 d}-\frac {2 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{c^5 d}-\frac {2 b^2 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )}{c^5 d}+\frac {2 b^2 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )}{c^5 d} \]
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Rubi [A]
time = 0.38, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps
used = 16, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4795, 4749,
4266, 2611, 2320, 6724, 4767, 8, 30} \begin {gather*} -\frac {2 i \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{c^5 d}+\frac {2 i b \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^5 d}-\frac {2 i b \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^5 d}-\frac {x (a+b \text {ArcSin}(c x))^2}{c^4 d}-\frac {x^3 (a+b \text {ArcSin}(c x))^2}{3 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c^3 d}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \text {ArcSin}(c x)}\right )}{c^5 d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \text {ArcSin}(c x)}\right )}{c^5 d}+\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2320
Rule 2611
Rule 4266
Rule 4749
Rule 4767
Rule 4795
Rule 6724
Rubi steps
\begin {align*} \int \frac {x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^2}+\frac {(2 b) \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{3 c d}\\ &=-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^4}+\frac {(4 b) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{9 c^3 d}+\frac {(2 b) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{c^3 d}+\frac {\left (2 b^2\right ) \int x^2 \, dx}{9 c^2 d}\\ &=\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\text {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac {\left (4 b^2\right ) \int 1 \, dx}{9 c^4 d}+\frac {\left (2 b^2\right ) \int 1 \, dx}{c^4 d}\\ &=\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 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^5 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^5 d}\\ &=\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 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^5 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^5 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^5 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^5 d}\\ &=\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 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^5 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^5 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^5 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^5 d}\\ &=\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 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^5 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^5 d}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ \end {align*}
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Mathematica [A]
time = 0.53, size = 508, normalized size = 1.71 \begin {gather*} -\frac {108 a^2 c x-270 b^2 c x+36 a^2 c^3 x^3+264 a b \sqrt {1-c^2 x^2}+24 a b c^2 x^2 \sqrt {1-c^2 x^2}+108 i a b \pi \text {ArcSin}(c x)+216 a b c x \text {ArcSin}(c x)+72 a b c^3 x^3 \text {ArcSin}(c x)+270 b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x)+135 b^2 c x \text {ArcSin}(c x)^2-6 b^2 \text {ArcSin}(c x) \cos (3 \text {ArcSin}(c x))-108 a b \pi \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-216 a b \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-108 b^2 \text {ArcSin}(c x)^2 \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-108 a b \pi \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+216 a b \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+108 b^2 \text {ArcSin}(c x)^2 \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+54 a^2 \log (1-c x)-54 a^2 \log (1+c x)+108 a b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+108 a b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-216 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+216 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )+216 b^2 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )-216 b^2 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )+2 b^2 \sin (3 \text {ArcSin}(c x))-9 b^2 \text {ArcSin}(c x)^2 \sin (3 \text {ArcSin}(c x))}{108 c^5 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (a +b \arcsin \left (c x \right )\right )^{2}}{-c^{2} d \,x^{2}+d}\, dx\]
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 \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} x^{4}}{c^{2} x^{2} - 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {asin}^{2}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx + \int \frac {2 a b x^{4} \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 = -0.00 \begin {gather*} \int \frac {x^4\,{\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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