Optimal. Leaf size=167 \[ \frac {3 b^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}-\frac {3 i b^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e^3} \]
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Rubi [A] time = 0.25, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4805, 12, 4627, 4681, 4625, 3717, 2190, 2279, 2391} \[ -\frac {3 i b^3 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e^3}+\frac {3 b^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2279
Rule 2391
Rule 3717
Rule 4625
Rule 4627
Rule 4681
Rule 4805
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{(c e+d e x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x^2 \sqrt {1-x^2}} \, dx,x,c+d x\right )}{2 d e^3}\\ &=-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}-\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac {\left (3 i b^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e^3}\\ &=-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac {3 i b^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e^3}\\ \end {align*}
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Mathematica [A] time = 0.89, size = 248, normalized size = 1.49 \[ -\frac {a \left (a \left (a+3 b (c+d x) \sqrt {-c^2-2 c d x-d^2 x^2+1}\right )-6 b^2 (c+d x)^2 \log (c+d x)\right )+3 b^2 \sin ^{-1}(c+d x)^2 \left (a+b (c+d x) \left (\sqrt {-c^2-2 c d x-d^2 x^2+1}+i c+i d x\right )\right )+3 b \sin ^{-1}(c+d x) \left (a \left (a+2 b (c+d x) \sqrt {-c^2-2 c d x-d^2 x^2+1}\right )-2 b^2 (c+d x)^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )\right )+3 i b^3 (c+d x)^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )+b^3 \sin ^{-1}(c+d x)^3}{2 d e^3 (c+d x)^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.29, size = 403, normalized size = 2.41 \[ -\frac {a^{3}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {3 i b^{3} \arcsin \left (d x +c \right )^{2}}{2 d \,e^{3}}-\frac {3 b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{2 d \,e^{3} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {3 b^{3} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {3 b^{3} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {3 i b^{3} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {3 i b^{3} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right )^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}+\frac {3 a \,b^{2} \ln \left (d x +c \right )}{d \,e^{3}}-\frac {3 a^{2} b \arcsin \left (d x +c \right )}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {3 a^{2} b \sqrt {1-\left (d x +c \right )^{2}}}{2 d \,e^{3} \left (d x +c \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \]
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 \[ \frac {\int \frac {a^{3}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a^{2} b \operatorname {asin}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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