Optimal. Leaf size=401 \[ \frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )}-\frac {b^2 c^3 d \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b^2 c^3 d \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}} \]
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Rubi [A] time = 0.64, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4743, 4773, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {b^2 c^3 d \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b^2 c^3 d \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )} \]
Antiderivative was successfully verified.
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Rule 31
Rule 2190
Rule 2264
Rule 2279
Rule 2391
Rule 2668
Rule 3323
Rule 3324
Rule 4743
Rule 4773
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{(d+e x)^3} \, dx &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac {(b c) \int \frac {a+b \sin ^{-1}(c x)}{(d+e x)^2 \sqrt {1-c^2 x^2}} \, dx}{e}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac {\left (b c^2\right ) \operatorname {Subst}\left (\int \frac {a+b x}{(c d+e \sin (x))^2} \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{c d+e \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d^2-e^2}+\frac {\left (b c^3 d\right ) \operatorname {Subst}\left (\int \frac {a+b x}{c d+e \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{e \left (c^2 d^2-e^2\right )}\\ &=\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{c d+x} \, dx,x,c e x\right )}{e \left (c^2 d^2-e^2\right )}+\frac {\left (2 b c^3 d\right ) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)}{i e+2 c d e^{i x}-i e e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e \left (c^2 d^2-e^2\right )}\\ &=\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )}-\frac {\left (2 i b c^3 d\right ) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)}{2 c d-2 \sqrt {c^2 d^2-e^2}-2 i e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right )^{3/2}}+\frac {\left (2 i b c^3 d\right ) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)}{2 c d+2 \sqrt {c^2 d^2-e^2}-2 i e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right )^{3/2}}\\ &=\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )}+\frac {\left (i b^2 c^3 d\right ) \operatorname {Subst}\left (\int \log \left (1-\frac {2 i e e^{i x}}{2 c d-2 \sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {\left (i b^2 c^3 d\right ) \operatorname {Subst}\left (\int \log \left (1-\frac {2 i e e^{i x}}{2 c d+2 \sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}\\ &=\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )}+\frac {\left (b^2 c^3 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i e x}{2 c d-2 \sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {\left (b^2 c^3 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i e x}{2 c d+2 \sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}\\ &=\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )}-\frac {b^2 c^3 d \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b^2 c^3 d \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}\\ \end {align*}
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Mathematica [A] time = 1.07, size = 315, normalized size = 0.79 \[ \frac {\frac {2 b c e \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}+\frac {2 b c^3 d \left (-i \left (a+b \sin ^{-1}(c x)\right ) \left (\log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )-\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )\right )-b \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )+b \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )}{\left (c^2 d^2-e^2\right )^{3/2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{(d+e x)^2}-\frac {2 b^2 c^2 \log (d+e x)}{c^2 d^2-e^2}}{2 e} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.32, size = 1173, normalized size = 2.93 \[ -\frac {c^{2} a^{2}}{2 \left (c e x +d c \right )^{2} e}-\frac {i c^{3} b^{2} \sqrt {-c^{2} d^{2}+e^{2}}\, d \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e \left (c^{2} d^{2}-e^{2}\right )^{2}}-\frac {i c^{4} b^{2} \arcsin \left (c x \right ) d^{2}}{\left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right ) e}+\frac {i c^{3} b^{2} \sqrt {-c^{2} d^{2}+e^{2}}\, d \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e \left (c^{2} d^{2}-e^{2}\right )^{2}}-\frac {c^{4} b^{2} \arcsin \left (c x \right )^{2} d^{2}}{2 \left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right ) e}+\frac {c^{3} b^{2} \arcsin \left (c x \right ) e \sqrt {-c^{2} x^{2}+1}\, x}{\left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right )}+\frac {c^{3} b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, d}{\left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right )}+\frac {c^{2} b^{2} \arcsin \left (c x \right )^{2} e}{2 \left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right )}+\frac {2 c^{2} b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{e \left (c^{2} d^{2}-e^{2}\right )}-\frac {c^{2} b^{2} \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2} e +2 i d c \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )-e \right )}{e \left (c^{2} d^{2}-e^{2}\right )}-\frac {c^{3} b^{2} \sqrt {-c^{2} d^{2}+e^{2}}\, d \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e \left (c^{2} d^{2}-e^{2}\right )^{2}}+\frac {c^{3} b^{2} \sqrt {-c^{2} d^{2}+e^{2}}\, d \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e \left (c^{2} d^{2}-e^{2}\right )^{2}}-\frac {2 i c^{4} b^{2} \arcsin \left (c x \right ) x d}{\left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right )}-\frac {i c^{4} b^{2} \arcsin \left (c x \right ) e \,x^{2}}{\left (c e x +d c \right )^{2} \left (c^{2} d^{2}-e^{2}\right )}-\frac {c^{2} a b \arcsin \left (c x \right )}{\left (c e x +d c \right )^{2} e}+\frac {c^{2} a b \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}{e \left (c^{2} d^{2}-e^{2}\right ) \left (c x +\frac {d c}{e}\right )}-\frac {c^{3} a b d \ln \left (\frac {-\frac {2 \left (c^{2} d^{2}-e^{2}\right )}{e^{2}}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}+2 \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}{c x +\frac {d c}{e}}\right )}{e^{2} \left (c^{2} d^{2}-e^{2}\right ) \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+e\,x\right )}^3} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d + e x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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