Optimal. Leaf size=332 \[ -\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac {3 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{4 c d^3}+\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {3 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {3 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3} \]
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Rubi [A] time = 0.35, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4655, 4657, 4181, 2531, 2282, 6589, 4677, 206, 199} \[ \frac {3 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac {3 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac {3 b^2 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {3 b^2 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{4 c d^3}+\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3} \]
Antiderivative was successfully verified.
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Rule 199
Rule 206
Rule 2282
Rule 2531
Rule 4181
Rule 4655
Rule 4657
Rule 4677
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {(b c) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac {3 \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b^2 \int \frac {1}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}-\frac {(3 b c) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{4 d^3}+\frac {3 \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b^2 \int \frac {1}{1-c^2 x^2} \, dx}{12 d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{4 d^3}+\frac {3 \operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c d^3}\\ &=\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3}-\frac {(3 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}+\frac {(3 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}\\ &=\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac {3 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {3 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}+\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}\\ &=\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac {3 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {3 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}\\ &=\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac {3 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {3 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac {3 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac {3 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}\\ \end {align*}
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Mathematica [A] time = 6.02, size = 556, normalized size = 1.67 \[ \frac {-\frac {36 a^2 x}{c^2 x^2-1}+\frac {24 a^2 x}{\left (c^2 x^2-1\right )^2}-\frac {18 a^2 \log (1-c x)}{c}+\frac {18 a^2 \log (c x+1)}{c}+\frac {a b \left (-72 i \left (c^2 x^2-1\right )^2 \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )-70 \sqrt {1-c^2 x^2}+40 \cos \left (2 \sin ^{-1}(c x)\right )-18 \cos \left (3 \sin ^{-1}(c x)\right )+10 \cos \left (4 \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \left (22 c x+6 \sin \left (3 \sin ^{-1}(c x)\right )+9 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-9 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+12 \left (\log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\log \left (1+i e^{i \sin ^{-1}(c x)}\right )\right ) \cos \left (2 \sin ^{-1}(c x)\right )+3 \left (\log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\log \left (1+i e^{i \sin ^{-1}(c x)}\right )\right ) \cos \left (4 \sin ^{-1}(c x)\right )\right )+30\right )}{c \left (c^2 x^2-1\right )^2}+\frac {72 i a b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {4 b^2 \left (\frac {2 c x}{c^2 x^2-1}+\frac {9 c x \sin ^{-1}(c x)^2}{c^2 x^2-1}-\frac {6 c x \sin ^{-1}(c x)^2}{\left (c^2 x^2-1\right )^2}+\frac {18 \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\frac {4 \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{3/2}}-18 i \sin ^{-1}(c x) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+18 i \sin ^{-1}(c x) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+18 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )-18 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )-20 \tanh ^{-1}(c x)+18 i \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )\right )}{c}}{96 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 3.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]
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 \[ \int -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} - d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.28, size = 890, normalized size = 2.68 \[ \frac {3 b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c \,d^{3}}-\frac {3 b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c \,d^{3}}-\frac {5 i b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 c \,d^{3}}+\frac {5 b^{2} \arcsin \left (c x \right )^{2} x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {c^{2} b^{2} x^{3}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 c^{2} a b \arcsin \left (c x \right ) x^{3}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 c a b \,x^{2} \sqrt {-c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 c \,b^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x^{2}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}+\frac {3 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}+\frac {b^{2} x}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 a^{2} \ln \left (c x +1\right )}{16 c \,d^{3}}-\frac {3 a^{2} \ln \left (c x -1\right )}{16 c \,d^{3}}-\frac {3 a^{2}}{16 c \,d^{3} \left (c x -1\right )}-\frac {a^{2}}{16 c \,d^{3} \left (c x +1\right )^{2}}-\frac {3 a^{2}}{16 c \,d^{3} \left (c x +1\right )}+\frac {a^{2}}{16 c \,d^{3} \left (c x -1\right )^{2}}+\frac {5 a b \arcsin \left (c x \right ) x}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 a b \sqrt {-c^{2} x^{2}+1}}{12 c \,d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 b^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{12 c \,d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 c^{2} b^{2} \arcsin \left (c x \right )^{2} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}+\frac {3 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}-\frac {3 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}+\frac {3 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}+\frac {3 i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,d^{3}}-\frac {3 i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 c \,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 \[ -\frac {1}{16} \, a^{2} {\left (\frac {2 \, {\left (3 \, c^{2} x^{3} - 5 \, x\right )}}{c^{4} d^{3} x^{4} - 2 \, c^{2} d^{3} x^{2} + d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c d^{3}}\right )} + \frac {3 \, {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) - 3 \, {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, {\left (3 \, b^{2} c^{3} x^{3} - 5 \, b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} - 2 \, {\left (c^{5} d^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )} \int \frac {16 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (3 \, {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (3 \, b^{2} c^{3} x^{3} - 5 \, b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}\,{d x}}{16 \, {\left (c^{5} d^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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