Optimal. Leaf size=202 \[ \frac {3 i b^3 \text {Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}+\frac {3 b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}-\frac {2 i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^4}{d e}-\frac {3 b^4 \text {Li}_5\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e} \]
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Rubi [A] time = 0.23, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4805, 12, 4625, 3717, 2190, 2531, 6609, 2282, 6589} \[ \frac {3 b^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}+\frac {3 i b^3 \text {PolyLog}\left (4,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}-\frac {2 i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e}-\frac {3 b^4 \text {PolyLog}\left (5,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^4}{d e} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2282
Rule 2531
Rule 3717
Rule 4625
Rule 4805
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{c e+d e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^4 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^4}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {(4 b) \operatorname {Subst}\left (\int (a+b x)^3 \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {\left (3 i b^4\right ) \operatorname {Subst}\left (\int \text {Li}_4\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {\left (3 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {3 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {3 b^4 \text {Li}_5\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ \end {align*}
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Mathematica [B] time = 0.47, size = 439, normalized size = 2.17 \[ \frac {16 a^4 \log (c+d x)+64 a^3 b \left (\sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )-\frac {1}{2} i \left (\sin ^{-1}(c+d x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )\right )\right )+4 a^2 b^2 \left (24 i \sin ^{-1}(c+d x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+12 \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+8 i \sin ^{-1}(c+d x)^3+24 \sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-i \pi ^3\right )-i a b^3 \left (-96 \sin ^{-1}(c+d x)^2 \text {Li}_2\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+96 i \sin ^{-1}(c+d x) \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+48 \text {Li}_4\left (e^{-2 i \sin ^{-1}(c+d x)}\right )-16 \sin ^{-1}(c+d x)^4+64 i \sin ^{-1}(c+d x)^3 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )+\pi ^4\right )+16 b^4 \left (2 i \sin ^{-1}(c+d x)^3 \text {Li}_2\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+3 \sin ^{-1}(c+d x)^2 \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )-3 i \sin ^{-1}(c+d x) \text {Li}_4\left (e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac {3}{2} \text {Li}_5\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+\frac {1}{5} i \sin ^{-1}(c+d x)^5+\sin ^{-1}(c+d x)^4 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac {i \pi ^5}{160}\right )}{16 d e} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \arcsin \left (d x + c\right )^{4} + 4 \, a b^{3} \arcsin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} + 4 \, a^{3} b \arcsin \left (d x + c\right ) + a^{4}}{d e x + c e}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{4}}{d e x + c e}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 1295, normalized size = 6.41 \[ \text {result too large to display} \]
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 \[ \frac {a^{4} \log \left (d e x + c e\right )}{d e} + \int \frac {b^{4} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{4} + 4 \, a b^{3} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} + 6 \, a^{2} b^{2} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 4 \, a^{3} b \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d e x + c e}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^4}{c\,e+d\,e\,x} \,d x \]
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 \[ \frac {\int \frac {a^{4}}{c + d x}\, dx + \int \frac {b^{4} \operatorname {asin}^{4}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {4 a b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {4 a^{3} b \operatorname {asin}{\left (c + d x \right )}}{c + d x}\, dx}{e} \]
Verification of antiderivative is not currently implemented for this CAS.
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