Optimal. Leaf size=190 \[ \frac {6 i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac {6 i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {6 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2}-\frac {6 b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2} \]
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Rubi [A] time = 0.25, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4805, 12, 4627, 4709, 4183, 2531, 2282, 6589} \[ \frac {6 i b^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac {6 i b^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac {6 b^3 \text {PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 b^3 \text {PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {6 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2282
Rule 2531
Rule 4183
Rule 4627
Rule 4709
Rule 4805
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}
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Mathematica [A] time = 1.03, size = 342, normalized size = 1.80 \[ -\frac {\frac {a^3}{c+d x}+3 a^2 b \log \left (\sqrt {-c^2-2 c d x-d^2 x^2+1}+1\right )-3 a^2 b \log (c+d x)+\frac {3 a^2 b \sin ^{-1}(c+d x)}{c+d x}-6 i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )+6 i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )+\frac {3 a b^2 \sin ^{-1}(c+d x)^2}{c+d x}-6 a b^2 \sin ^{-1}(c+d x) \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )+6 a b^2 \sin ^{-1}(c+d x) \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )+6 b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )-6 b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )+\frac {b^3 \sin ^{-1}(c+d x)^3}{c+d x}-3 b^3 \sin ^{-1}(c+d x)^2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )+3 b^3 \sin ^{-1}(c+d x)^2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )}{d e^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.24, size = 532, normalized size = 2.80 \[ -\frac {a^{3}}{d \,e^{2} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{d \,e^{2} \left (d x +c \right )}-\frac {3 b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {6 i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {6 b^{3} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {3 b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {6 i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {6 b^{3} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right )^{2}}{d \,e^{2} \left (d x +c \right )}+\frac {6 a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {6 a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {6 i a \,b^{2} \dilog \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {6 i a \,b^{2} \dilog \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {3 a^{2} b \arcsin \left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}-\frac {3 a^{2} b \arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d \,e^{2}} \]
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 \[ -3 \, a^{2} b {\left (\frac {\arcsin \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}} + \frac {\log \left (\frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{{\left | d^{2} e^{2} x + c d e^{2} \right |}} + \frac {2}{{\left | d^{2} e^{2} x + c d e^{2} \right |}}\right )}{d e^{2}}\right )} - \frac {a^{3}}{d^{2} e^{2} x + c d e^{2}} - \frac {b^{3} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} - \frac {3}{2} \, {\left (a b^{2} {\left (\frac {2 \, c^{2}}{d^{4} e^{2} x + c d^{3} e^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3} e^{2}} + \frac {4 \, c \log \left (d x + c\right )}{d^{3} e^{2}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3} e^{2}}\right )} d^{2} \arctan \left (\frac {d x}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}} + \frac {c}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}}\right )^{2} - 2 \, a b^{2} c d {\left (\frac {2 \, c}{d^{3} e^{2} x + c d^{2} e^{2}} - \frac {{\left (c + 1\right )} \log \left (d x + c + 1\right )}{d^{2} e^{2}} + \frac {{\left (c - 1\right )} \log \left (d x + c - 1\right )}{d^{2} e^{2}} + \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}}\right )} \arctan \left (\frac {d x}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}} + \frac {c}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}}\right )^{2} + a b^{2} c^{2} {\left (\frac {2}{d^{2} e^{2} x + c d e^{2}} - \frac {\log \left (d x + c + 1\right )}{d e^{2}} + \frac {\log \left (d x + c - 1\right )}{d e^{2}}\right )} \arctan \left (\frac {d x}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}} + \frac {c}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}}\right )^{2} - a b^{2} {\left (\frac {2}{d^{2} e^{2} x + c d e^{2}} - \frac {\log \left (d x + c + 1\right )}{d e^{2}} + \frac {\log \left (d x + c - 1\right )}{d e^{2}}\right )} \arctan \left (\frac {d x}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}} + \frac {c}{\sqrt {d x + c + 1} \sqrt {-d x - c + 1}}\right )^{2}\right )} {\left (d^{2} e^{2} x + c d e^{2}\right )}}{d^{2} e^{2} x + c d e^{2}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {asin}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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