Optimal. Leaf size=204 \[ -\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}+\frac {3 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac {3 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac {5 b}{8 c^5 d^3 \sqrt {1-c^2 x^2}}-\frac {b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}} \]
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Rubi [A] time = 0.24, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4703, 4657, 4181, 2279, 2391, 261, 266, 43} \[ \frac {3 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac {3 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {5 b}{8 c^5 d^3 \sqrt {1-c^2 x^2}}-\frac {b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 43
Rule 261
Rule 266
Rule 2279
Rule 2391
Rule 4181
Rule 4657
Rule 4703
Rubi steps
\begin {align*} \int \frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {b \int \frac {x^3}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}-\frac {3 \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^2} \, dx}{4 c^2 d}\\ &=\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}+\frac {(3 b) \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 c^3 d^3}-\frac {b \operatorname {Subst}\left (\int \frac {x}{\left (1-c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{8 c d^3}+\frac {3 \int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 c^4 d^2}\\ &=\frac {3 b}{8 c^5 d^3 \sqrt {1-c^2 x^2}}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}+\frac {3 \operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^5 d^3}-\frac {b \operatorname {Subst}\left (\int \left (\frac {1}{c^2 \left (1-c^2 x\right )^{5/2}}-\frac {1}{c^2 \left (1-c^2 x\right )^{3/2}}\right ) \, dx,x,x^2\right )}{8 c d^3}\\ &=-\frac {b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b}{8 c^5 d^3 \sqrt {1-c^2 x^2}}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {(3 b) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^5 d^3}+\frac {(3 b) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^5 d^3}\\ &=-\frac {b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b}{8 c^5 d^3 \sqrt {1-c^2 x^2}}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {(3 i b) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac {(3 i b) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}\\ &=-\frac {b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b}{8 c^5 d^3 \sqrt {1-c^2 x^2}}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac {3 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}\\ \end {align*}
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Mathematica [B] time = 0.94, size = 445, normalized size = 2.18 \[ \frac {\frac {30 a c x}{c^2 x^2-1}+\frac {12 a c x}{\left (c^2 x^2-1\right )^2}-9 a \log (1-c x)+9 a \log (c x+1)-\frac {15 b \sqrt {1-c^2 x^2}}{c x-1}+\frac {15 b \sqrt {1-c^2 x^2}}{c x+1}+\frac {b c x \sqrt {1-c^2 x^2}}{(c x-1)^2}-\frac {2 b \sqrt {1-c^2 x^2}}{(c x-1)^2}-\frac {b c x \sqrt {1-c^2 x^2}}{(c x+1)^2}-\frac {2 b \sqrt {1-c^2 x^2}}{(c x+1)^2}+18 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-18 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+\frac {15 b \sin ^{-1}(c x)}{c x-1}+\frac {15 b \sin ^{-1}(c x)}{c x+1}+\frac {3 b \sin ^{-1}(c x)}{(c x-1)^2}-\frac {3 b \sin ^{-1}(c x)}{(c x+1)^2}-9 i \pi b \sin ^{-1}(c x)+18 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+9 \pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-18 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+9 \pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-9 \pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-9 \pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{48 c^5 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{4} \arcsin \left (c x\right ) + a x^{4}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]
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 \[ \int -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} - d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.55, size = 389, normalized size = 1.91 \[ -\frac {a}{16 c^{5} d^{3} \left (c x +1\right )^{2}}+\frac {5 a}{16 c^{5} d^{3} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{16 c^{5} d^{3}}+\frac {a}{16 c^{5} d^{3} \left (c x -1\right )^{2}}+\frac {5 a}{16 c^{5} d^{3} \left (c x -1\right )}-\frac {3 a \ln \left (c x -1\right )}{16 c^{5} d^{3}}+\frac {5 b \arcsin \left (c x \right ) x^{3}}{8 c^{2} d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {5 b \,x^{2} \sqrt {-c^{2} x^{2}+1}}{8 c^{3} d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 b \arcsin \left (c x \right ) x}{8 c^{4} d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {13 b \sqrt {-c^{2} x^{2}+1}}{24 c^{5} d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c^{5} d^{3}}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c^{5} d^{3}}+\frac {3 i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c^{5} d^{3}}-\frac {3 i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c^{5} d^{3}} \]
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 \[ \frac {1}{16} \, a {\left (\frac {2 \, {\left (5 \, c^{2} x^{3} - 3 \, x\right )}}{c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}} + \frac {3 \, \log \left (c x + 1\right )}{c^{5} d^{3}} - \frac {3 \, \log \left (c x - 1\right )}{c^{5} d^{3}}\right )} + \frac {{\left (3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) + 2 \, {\left (5 \, c^{3} x^{3} - 3 \, c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (c^{9} d^{3} x^{4} - 2 \, c^{7} d^{3} x^{2} + c^{5} d^{3}\right )} \int \frac {{\left (10 \, c^{3} x^{3} - 6 \, c x + 3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \log \left (c x + 1\right ) - 3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{10} d^{3} x^{6} - 3 \, c^{8} d^{3} x^{4} + 3 \, c^{6} d^{3} x^{2} - c^{4} d^{3}}\,{d x}\right )} b}{16 \, {\left (c^{9} d^{3} x^{4} - 2 \, c^{7} d^{3} x^{2} + c^{5} d^{3}\right )}} \]
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 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]
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 \[ - \frac {\int \frac {a x^{4}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{4} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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