Optimal. Leaf size=230 \[ -\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d^2}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d^2}-\frac {i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c d^2}-\frac {b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c d^2} \]
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Rubi [A] time = 0.24, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4655, 4657, 4181, 2531, 2282, 6589, 4677, 206} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d^2}-\frac {i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d^2}-\frac {b^2 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c d^2} \]
Antiderivative was successfully verified.
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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 )^2} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {(b c) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{d^2}+\frac {\int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{2 d}\\ &=-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}+\frac {b^2 \int \frac {1}{1-c^2 x^2} \, dx}{d^2}+\frac {\operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 c d^2}\\ &=-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c d^2}-\frac {b \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d^2}+\frac {b \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d^2}\\ &=-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c d^2}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d^2}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d^2}\\ &=-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c d^2}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d^2}\\ &=-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c d^2}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d^2}+\frac {b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c d^2}\\ \end {align*}
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Mathematica [A] time = 2.62, size = 359, normalized size = 1.56 \[ \frac {-\frac {2 a^2 x}{c^2 x^2-1}-\frac {a^2 \log (1-c x)}{c}+\frac {a^2 \log (c x+1)}{c}+\frac {2 a b \left (\frac {2 \left (c^2 x^2+\sqrt {1-c^2 x^2}+\sin ^{-1}(c x) \left (\left (c^2 x^2-1\right ) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+\left (1-c^2 x^2\right ) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-c x\right )-1\right )}{c^2 x^2-1}+2 i \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-2 i \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )\right )}{c}+\frac {4 b^2 \left (\frac {c x \sin ^{-1}(c x)^2}{2-2 c^2 x^2}-\frac {\sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+i \sin ^{-1}(c x) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-i \sin ^{-1}(c x) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )-\text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )+\text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )+\tanh ^{-1}(c x)-i \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )\right )}{c}}{4 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.18, size = 593, normalized size = 2.58 \[ -\frac {a^{2}}{4 c \,d^{2} \left (c x +1\right )}+\frac {a^{2} \ln \left (c x +1\right )}{4 c \,d^{2}}-\frac {a^{2}}{4 c \,d^{2} \left (c x -1\right )}-\frac {a^{2} \ln \left (c x -1\right )}{4 c \,d^{2}}-\frac {b^{2} \arcsin \left (c x \right )^{2} x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{c \,d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c \,d^{2}}-\frac {i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}+\frac {b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c \,d^{2}}-\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}-\frac {b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}-\frac {2 i b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c \,d^{2}}-\frac {a b \arcsin \left (c x \right ) x}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {a b \sqrt {-c^{2} x^{2}+1}}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}+\frac {a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}+\frac {i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}}+\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c \,d^{2}} \]
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 \[ -\frac {1}{4} \, a^{2} {\left (\frac {2 \, x}{c^{2} d^{2} x^{2} - d^{2}} - \frac {\log \left (c x + 1\right )}{c d^{2}} + \frac {\log \left (c x - 1\right )}{c d^{2}}\right )} - \frac {2 \, b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} - {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) + {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, {\left (c^{3} d^{2} x^{2} - c d^{2}\right )} \int \frac {4 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (2 \, b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) + {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}\,{d x}}{4 \, {\left (c^{3} d^{2} x^{2} - c d^{2}\right )}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]
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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