Optimal. Leaf size=324 \[ -\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 i b c \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {3 i b c \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {3 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^2}-\frac {4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {3 b^2 c \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {3 b^2 c \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {b^2 c \tanh ^{-1}(c x)}{d^2} \]
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Rubi [A] time = 0.56, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 14, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.518, Rules used = {4701, 4655, 4657, 4181, 2531, 2282, 6589, 4677, 206, 4705, 4709, 4183, 2279, 2391} \[ \frac {3 i b c \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {3 i b c \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}+\frac {2 i b^2 c \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {2 i b^2 c \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {3 b^2 c \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {3 b^2 c \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}-\frac {3 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^2}-\frac {4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}+\frac {b^2 c \tanh ^{-1}(c x)}{d^2} \]
Antiderivative was successfully verified.
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Rule 206
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 4181
Rule 4183
Rule 4655
Rule 4657
Rule 4677
Rule 4701
Rule 4705
Rule 4709
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x^2 \left (d-c^2 d x^2\right )^2} \, dx &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\left (3 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx+\frac {(2 b c) \int \frac {a+b \sin ^{-1}(c x)}{x \left (1-c^2 x^2\right )^{3/2}} \, dx}{d^2}\\ &=\frac {2 b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}+\frac {(2 b c) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{d^2}-\frac {\left (2 b^2 c^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{d^2}-\frac {\left (3 b c^3\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{d^2}+\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{2 d}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {2 b^2 c \tanh ^{-1}(c x)}{d^2}+\frac {(3 c) \operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}+\frac {(2 b c) \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac {\left (3 b^2 c^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {3 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {b^2 c \tanh ^{-1}(c x)}{d^2}-\frac {(3 b c) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac {(3 b c) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}-\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {3 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {b^2 c \tanh ^{-1}(c x)}{d^2}+\frac {3 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {3 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {\left (2 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {\left (2 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {\left (3 i b^2 c\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac {\left (3 i b^2 c\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {3 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {b^2 c \tanh ^{-1}(c x)}{d^2}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {3 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {3 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {\left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {\left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^2}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {3 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {b^2 c \tanh ^{-1}(c x)}{d^2}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {3 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {3 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac {3 b^2 c \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{d^2}+\frac {3 b^2 c \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{d^2}\\ \end {align*}
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Mathematica [B] time = 9.77, size = 1059, normalized size = 3.27 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{4} d^{2} x^{6} - 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} - d\right )}^{2} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.44, size = 778, normalized size = 2.40 \[ -\frac {a^{2}}{d^{2} x}-\frac {3 i c a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}+\frac {3 i c \,b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}+\frac {b^{2} \arcsin \left (c x \right )^{2}}{d^{2} \left (c^{2} x^{2}-1\right ) x}-\frac {c \,a^{2}}{4 d^{2} \left (c x +1\right )}-\frac {c \,a^{2}}{4 d^{2} \left (c x -1\right )}+\frac {3 c \,a^{2} \ln \left (c x +1\right )}{4 d^{2}}-\frac {3 c \,a^{2} \ln \left (c x -1\right )}{4 d^{2}}-\frac {3 b^{2} c \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}+\frac {3 b^{2} c \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}+\frac {c \,b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {3 i c \,b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}+\frac {3 i c a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}-\frac {3 b^{2} \arcsin \left (c x \right )^{2} c^{2} x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \arcsin \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right ) x}-\frac {3 c \,b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {2 c a b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 c a b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d^{2}}-\frac {2 i c \,b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 c \,b^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {3 c \,b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {2 i c \,b^{2} \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 i c \,b^{2} \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {3 c a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}-\frac {3 a b \arcsin \left (c x \right ) c^{2} x}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {c a b \sqrt {-c^{2} x^{2}+1}}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {3 c a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{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 \[ -\frac {1}{4} \, a^{2} {\left (\frac {2 \, {\left (3 \, c^{2} x^{2} - 2\right )}}{c^{2} d^{2} x^{3} - d^{2} x} - \frac {3 \, c \log \left (c x + 1\right )}{d^{2}} + \frac {3 \, c \log \left (c x - 1\right )}{d^{2}}\right )} + \frac {3 \, {\left (b^{2} c^{3} x^{3} - b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) - 3 \, {\left (b^{2} c^{3} x^{3} - b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, {\left (3 \, b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, {\left (c^{2} d^{2} x^{3} - d^{2} x\right )} \int \frac {4 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (3 \, {\left (b^{2} c^{4} x^{4} - b^{2} c^{2} x^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, {\left (b^{2} c^{4} x^{4} - b^{2} c^{2} x^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (3 \, b^{2} c^{3} x^{3} - 2 \, b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{4} d^{2} x^{6} - 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}\,{d x}}{4 \, {\left (c^{2} d^{2} x^{3} - d^{2} x\right )}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x^2\,{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c^{4} x^{6} - 2 c^{2} x^{4} + x^{2}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{c^{4} x^{6} - 2 c^{2} x^{4} + x^{2}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c x \right )}}{c^{4} x^{6} - 2 c^{2} x^{4} + x^{2}}\, dx}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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