Optimal. Leaf size=439 \[ -\frac {4 b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}+\frac {4 b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}-\frac {8 b^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}+\frac {2 i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4}-\frac {2 i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4}-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {4 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4}+\frac {4 i b^4 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {Li}_4\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 i b^4 \text {Li}_4\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4} \]
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Rubi [A] time = 0.58, antiderivative size = 439, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {4805, 12, 4627, 4701, 4709, 4183, 2531, 6609, 2282, 6589, 2279, 2391} \[ -\frac {4 b^3 \text {PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}+\frac {4 b^3 \text {PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}+\frac {2 i b^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4}-\frac {2 i b^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4}+\frac {4 i b^4 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {PolyLog}\left (4,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 i b^4 \text {PolyLog}\left (4,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {8 b^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {4 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 4183
Rule 4627
Rule 4701
Rule 4709
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)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x^3 \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e^4}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {(2 b) \operatorname {Subst}\left (\int (a+b x)^3 \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{3 d e^4}+\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (4 i b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (4 i b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {\left (4 i b^4\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (4 i b^4\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {4 i b^4 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (4 i b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {\left (4 i b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {4 i b^4 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 i b^4 \text {Li}_4\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 i b^4 \text {Li}_4\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}\\ \end {align*}
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Mathematica [B] time = 11.39, size = 1274, normalized size = 2.90 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \arcsin \left (d x + c\right )^{4} + 4 \, a b^{3} \arcsin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} + 4 \, a^{3} b \arcsin \left (d x + c\right ) + a^{4}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.42, size = 1327, normalized size = 3.02 \[ \text {result too large to display} \]
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 \[ -\frac {a^{4}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac {b^{4} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{4} + 2 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )} \int \frac {2 \, {\left (b^{4} d x + b^{4} c\right )} \sqrt {d x + c + 1} \sqrt {-d x - c + 1} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} - 6 \, {\left (a b^{3} d^{2} x^{2} + 2 \, a b^{3} c d x + a b^{3} c^{2} - a b^{3}\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} - 9 \, {\left (a^{2} b^{2} d^{2} x^{2} + 2 \, a^{2} b^{2} c d x + a^{2} b^{2} c^{2} - a^{2} b^{2}\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} - 6 \, {\left (a^{3} b d^{2} x^{2} + 2 \, a^{3} b c d x + a^{3} b c^{2} - a^{3} b\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + {\left (15 \, c^{2} - 1\right )} d^{4} e^{4} x^{4} + 4 \, {\left (5 \, c^{3} - c\right )} d^{3} e^{4} x^{3} + 3 \, {\left (5 \, c^{4} - 2 \, c^{2}\right )} d^{2} e^{4} x^{2} + 2 \, {\left (3 \, c^{5} - 2 \, c^{3}\right )} d e^{4} x + {\left (c^{6} - c^{4}\right )} e^{4}}\,{d x}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]
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 \[ \frac {\int \frac {a^{4}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{4} \operatorname {asin}^{4}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {4 a b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {4 a^{3} b \operatorname {asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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