Optimal. Leaf size=429 \[ \frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c (a+b \text {ArcSin}(c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c (a+b \text {ArcSin}(c x))}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {(a+b \text {ArcSin}(c x))^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x (a+b \text {ArcSin}(c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x (a+b \text {ArcSin}(c x))^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c (a+b \text {ArcSin}(c x))^2 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{4 d^3}-\frac {4 b c (a+b \text {ArcSin}(c x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )}{d^3}+\frac {11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac {2 i b^2 c \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )}{d^3}+\frac {15 i b c (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{4 d^3}-\frac {15 i b c (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{4 d^3}-\frac {2 i b^2 c \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )}{d^3}-\frac {15 b^2 c \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )}{4 d^3}+\frac {15 b^2 c \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )}{4 d^3} \]
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Rubi [A]
time = 0.53, antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps
used = 27, number of rules used = 15, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used =
{4789, 4747, 4749, 4266, 2611, 2320, 6724, 4767, 212, 205, 4793, 4803, 4268, 2317, 2438}
\begin {gather*} -\frac {15 i c \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{4 d^3}-\frac {7 b c (a+b \text {ArcSin}(c x))}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {b c (a+b \text {ArcSin}(c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {15 c^2 x (a+b \text {ArcSin}(c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {5 c^2 x (a+b \text {ArcSin}(c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {(a+b \text {ArcSin}(c x))^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {15 i b c \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{4 d^3}-\frac {15 i b c \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{4 d^3}-\frac {4 b c \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d^3}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right )}{d^3}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right )}{d^3}-\frac {15 b^2 c \text {Li}_3\left (-i e^{i \text {ArcSin}(c x)}\right )}{4 d^3}+\frac {15 b^2 c \text {Li}_3\left (i e^{i \text {ArcSin}(c x)}\right )}{4 d^3}+\frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac {11 b^2 c \tanh ^{-1}(c x)}{6 d^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 212
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 4266
Rule 4268
Rule 4747
Rule 4749
Rule 4767
Rule 4789
Rule 4793
Rule 4803
Rule 6724
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 )^3} \, dx &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\left (5 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx+\frac {(2 b c) \int \frac {a+b \sin ^{-1}(c x)}{x \left (1-c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=\frac {2 b c \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(2 b c) \int \frac {a+b \sin ^{-1}(c x)}{x \left (1-c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac {\left (2 b^2 c^2\right ) \int \frac {1}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^3}-\frac {\left (5 b c^3\right ) \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 {\left (15 c^2\right ) \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^2 c^2 x}{3 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \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^3}-\frac {\left (b^2 c^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{3 d^3}+\frac {\left (5 b^2 c^2\right ) \int \frac {1}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}-\frac {\left (2 b^2 c^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{d^3}-\frac {\left (15 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}{4 d^3}+\frac {\left (15 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {7 b^2 c \tanh ^{-1}(c x)}{3 d^3}+\frac {(15 c) \text {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}+\frac {(2 b c) \text {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac {\left (5 b^2 c^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{12 d^3}+\frac {\left (15 b^2 c^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{4 d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac {11 b^2 c \tanh ^{-1}(c x)}{6 d^3}-\frac {(15 b c) \text {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}+\frac {(15 b c) \text {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}-\frac {\left (2 b^2 c\right ) \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac {\left (2 b^2 c\right ) \text {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac {11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac {15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac {\left (2 i b^2 c\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac {\left (2 i b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac {\left (15 i b^2 c\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}+\frac {\left (15 i b^2 c\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac {11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac {15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac {\left (15 b^2 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac {\left (15 b^2 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt {1-c^2 x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac {11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac {15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac {15 b^2 c \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac {15 b^2 c \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(1351\) vs. \(2(429)=858\).
time = 9.97, size = 1351, normalized size = 3.15 \begin {gather*} \text {Too large to display} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1073 vs. \(2 (458 ) = 916\).
time = 0.42, size = 1074, normalized size = 2.50 Too large to display
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} - \frac {\int \frac {a^{2}}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c x \right )}}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx}{d^{3}} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x^2\,{\left (d-c^2\,d\,x^2\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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