Optimal. Leaf size=259 \[ -\frac {5 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {5 i b c^3 \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {5 i b c^3 \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {b c}{6 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {b c^3}{3 d^2 \sqrt {1-c^2 x^2}}-\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d^2} \]
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Rubi [A] time = 0.31, antiderivative size = 285, normalized size of antiderivative = 1.10, number of steps used = 19, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4701, 4655, 4657, 4181, 2279, 2391, 261, 266, 51, 63, 208} \[ \frac {5 i b c^3 \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {5 i b c^3 \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {5 b c^3}{6 d^2 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d^2 x^2}+\frac {b c}{3 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d^2} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 208
Rule 261
Rule 266
Rule 2279
Rule 2391
Rule 4181
Rule 4655
Rule 4657
Rule 4701
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )^2} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}+\frac {1}{3} \left (5 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^2} \, dx+\frac {(b c) \int \frac {1}{x^3 \left (1-c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\left (5 c^4\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}+\frac {\left (5 b c^3\right ) \int \frac {1}{x \left (1-c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac {b c}{3 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 d^2}+\frac {\left (5 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}-\frac {\left (5 b c^5\right ) \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac {\left (5 c^4\right ) \int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{2 d}\\ &=-\frac {5 b c^3}{6 d^2 \sqrt {1-c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {\left (5 c^3\right ) \operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}+\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{4 d^2}+\frac {\left (5 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac {5 b c^3}{6 d^2 \sqrt {1-c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {5 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{2 d^2}-\frac {(5 b c) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{3 d^2}-\frac {\left (5 b c^3\right ) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}+\frac {\left (5 b c^3\right ) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}\\ &=-\frac {5 b c^3}{6 d^2 \sqrt {1-c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {5 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d^2}+\frac {\left (5 i b c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {\left (5 i b c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 d^2}\\ &=-\frac {5 b c^3}{6 d^2 \sqrt {1-c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac {5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {5 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d^2}+\frac {5 i b c^3 \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {5 i b c^3 \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}\\ \end {align*}
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Mathematica [A] time = 0.90, size = 426, normalized size = 1.64 \[ -\frac {15 a c^3 \log (1-c x)-15 a c^3 \log (c x+1)+\frac {24 a c^2}{x}+\frac {6 a c^4 x}{c^2 x^2-1}+\frac {4 a}{x^3}-30 i b c^3 \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+30 i b c^3 \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+\frac {3 b c^3 \sin ^{-1}(c x)}{c x-1}+\frac {3 b c^3 \sin ^{-1}(c x)}{c x+1}+15 i \pi b c^3 \sin ^{-1}(c x)-30 b c^3 \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c^3 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+30 b c^3 \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c^3 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+15 \pi b c^3 \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+15 \pi b c^3 \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\frac {2 b c \sqrt {1-c^2 x^2}}{x^2}+\frac {24 b c^2 \sin ^{-1}(c x)}{x}-\frac {3 b c^3 \sqrt {1-c^2 x^2}}{c x-1}+\frac {3 b c^3 \sqrt {1-c^2 x^2}}{c x+1}+26 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )+\frac {4 b \sin ^{-1}(c x)}{x^3}}{12 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{c^{4} d^{2} x^{8} - 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, x\right ) \]
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 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{2} x^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.33, size = 426, normalized size = 1.64 \[ -\frac {c^{3} a}{4 d^{2} \left (c x +1\right )}+\frac {5 c^{3} a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{3 d^{2} x^{3}}-\frac {2 c^{2} a}{d^{2} x}-\frac {c^{3} a}{4 d^{2} \left (c x -1\right )}-\frac {5 c^{3} a \ln \left (c x -1\right )}{4 d^{2}}-\frac {5 c^{4} b \arcsin \left (c x \right ) x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {c^{3} b \sqrt {-c^{2} x^{2}+1}}{3 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {5 c^{2} b \arcsin \left (c x \right )}{3 d^{2} x \left (c^{2} x^{2}-1\right )}+\frac {c b \sqrt {-c^{2} x^{2}+1}}{6 d^{2} x^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right )}{3 d^{2} x^{3} \left (c^{2} x^{2}-1\right )}-\frac {13 c^{3} b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{6 d^{2}}+\frac {13 c^{3} b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{6 d^{2}}-\frac {5 c^{3} b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {5 c^{3} b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {5 i c^{3} b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {5 i c^{3} b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}} \]
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 \[ \frac {1}{12} \, {\left (\frac {15 \, c^{3} \log \left (c x + 1\right )}{d^{2}} - \frac {15 \, c^{3} \log \left (c x - 1\right )}{d^{2}} - \frac {2 \, {\left (15 \, c^{4} x^{4} - 10 \, c^{2} x^{2} - 2\right )}}{c^{2} d^{2} x^{5} - d^{2} x^{3}}\right )} a + \frac {{\left (15 \, {\left (c^{5} x^{5} - c^{3} x^{3}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 15 \, {\left (c^{5} x^{5} - c^{3} x^{3}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (15 \, c^{4} x^{4} - 10 \, c^{2} x^{2} - 2\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (c^{2} d^{2} x^{5} - d^{2} x^{3}\right )} \int \frac {{\left (30 \, c^{5} x^{4} - 20 \, c^{3} x^{2} - 15 \, {\left (c^{6} x^{5} - c^{4} x^{3}\right )} \log \left (c x + 1\right ) + 15 \, {\left (c^{6} x^{5} - c^{4} x^{3}\right )} \log \left (-c x + 1\right ) - 4 \, c\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}\,{d x}\right )} b}{12 \, {\left (c^{2} d^{2} x^{5} - d^{2} x^{3}\right )}} \]
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 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4\,{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a}{c^{4} x^{8} - 2 c^{2} x^{6} + x^{4}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{4} x^{8} - 2 c^{2} x^{6} + x^{4}}\, dx}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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