Optimal. Leaf size=198 \[ -\frac {6 i b^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3}+\frac {6 b^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {3 b^4 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3} \]
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Rubi [A] time = 0.32, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4805, 12, 4627, 4681, 4625, 3717, 2190, 2531, 2282, 6589} \[ -\frac {6 i b^3 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3}+\frac {3 b^4 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac {6 b^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2282
Rule 2531
Rule 3717
Rule 4625
Rule 4627
Rule 4681
Rule 4805
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{(c e+d e x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x^2 \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}-\frac {\left (12 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac {\left (12 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac {6 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac {\left (6 i b^4\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac {6 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac {\left (3 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}\\ &=-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac {6 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac {3 b^4 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}\\ \end {align*}
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Mathematica [A] time = 1.45, size = 385, normalized size = 1.94 \[ \frac {-\frac {2 a^4}{(c+d x)^2}-\frac {8 a^3 b \sqrt {1-(c+d x)^2}}{c+d x}-\frac {8 a^3 b \sin ^{-1}(c+d x)}{(c+d x)^2}+24 a^2 b^2 \left (\log (c+d x)-\frac {\sin ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac {\sqrt {1-(c+d x)^2} \sin ^{-1}(c+d x)}{c+d x}\right )+8 a b^3 \left (-3 i \left (\sin ^{-1}(c+d x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )\right )-\frac {\sin ^{-1}(c+d x)^3}{(c+d x)^2}-\frac {3 \sqrt {1-(c+d x)^2} \sin ^{-1}(c+d x)^2}{c+d x}+6 \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )\right )+b^4 \left (24 i \sin ^{-1}(c+d x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+12 \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac {8 \sqrt {1-(c+d x)^2} \sin ^{-1}(c+d x)^3}{c+d x}+8 i \sin ^{-1}(c+d x)^3+24 \sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-i \pi ^3\right )-\frac {2 b^4 \sin ^{-1}(c+d x)^4}{(c+d x)^2}}{4 d e^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \arcsin \left (d x + c\right )^{4} + 4 \, a b^{3} \arcsin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} + 4 \, a^{3} b \arcsin \left (d x + c\right ) + a^{4}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.26, size = 747, normalized size = 3.77 \[ -\frac {a^{4}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {12 i a \,b^{3} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {2 b^{4} \arcsin \left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {b^{4} \arcsin \left (d x +c \right )^{4}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {6 b^{4} \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^{3}}-\frac {12 i b^{4} \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^{3}}+\frac {12 b^{4} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {6 b^{4} \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^{3}}-\frac {12 i b^{4} \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^{3}}+\frac {12 b^{4} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {12 i a \,b^{3} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {6 a \,b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {2 a \,b^{3} \arcsin \left (d x +c \right )^{3}}{d \,e^{3} \left (d x +c \right )^{2}}+\frac {12 a \,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 {12 a \,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 {2 i b^{4} \arcsin \left (d x +c \right )^{3}}{d \,e^{3}}-\frac {6 i a \,b^{3} \arcsin \left (d x +c \right )^{2}}{d \,e^{3}}-\frac {3 a^{2} b^{2} \arcsin \left (d x +c \right )^{2}}{d \,e^{3} \left (d x +c \right )^{2}}-\frac {6 a^{2} 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 {6 a^{2} b^{2} \ln \left (d x +c \right )}{d \,e^{3}}-\frac {2 a^{3} b \arcsin \left (d x +c \right )}{d \,e^{3} \left (d x +c \right )^{2}}-\frac {2 a^{3} b \sqrt {1-\left (d x +c \right )^{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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \]
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 \[ \frac {\int \frac {a^{4}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{4} \operatorname {asin}^{4}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {4 a b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {4 a^{3} b \operatorname {asin}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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