Optimal. Leaf size=390 \[ \frac {a^2 (e x)^{3 n}}{3 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \text {Li}_2\left (-e^{2 i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n} \]
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Rubi [A] time = 0.40, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4208, 4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ -\frac {4 a b x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (3,-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (3,i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (2,-e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3719
Rule 4181
Rule 4184
Rule 4190
Rule 4204
Rule 4208
Rule 6589
Rubi steps
\begin {align*} \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx &=\frac {\left (x^{-3 n} (e x)^{3 n}\right ) \int x^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx}{e}\\ &=\frac {\left (x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,x^n\right )}{e n}\\ &=\frac {\left (x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \left (a^2 x^2+2 a b x^2 \sec (c+d x)+b^2 x^2 \sec ^2(c+d x)\right ) \, dx,x,x^n\right )}{e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}+\frac {\left (2 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,x^n\right )}{e n}+\frac {\left (b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x^2 \sec ^2(c+d x) \, dx,x,x^n\right )}{e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d e n}+\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d e n}-\frac {\left (2 b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \tan (c+d x) \, dx,x,x^n\right )}{d e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {\left (4 i a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}+\frac {\left (4 i a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}+\frac {\left (4 i b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,x^n\right )}{d e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{d^3 e n}-\frac {\left (2 b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}+\frac {\left (i b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \text {Li}_2\left (-e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}\\ \end {align*}
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Mathematica [F] time = 11.66, size = 0, normalized size = 0.00 \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx \]
Verification is Not applicable to the result.
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fricas [C] time = 1.28, size = 1028, normalized size = 2.64 \[ \text {result too large to display} \]
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 \[ \int {\left (b \sec \left (d x^{n} + c\right ) + a\right )}^{2} \left (e x\right )^{3 \, n - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.02, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{-1+3 n} \left (a +b \sec \left (c +d \,x^{n}\right )\right )^{2}\, dx \]
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 \[ \frac {\left (e x\right )^{3 \, n} a^{2}}{3 \, e n} + \frac {2 \, b^{2} e^{3 \, n} x^{2 \, n} \sin \left (2 \, d x^{n} + 2 \, c\right ) + 4 \, {\left (d e n \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e n \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} + 2 \, d e n \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e n\right )} \int \frac {a b d e^{3 \, n} x^{3 \, n} \cos \left (2 \, d x^{n} + 2 \, c\right ) \cos \left (d x^{n} + c\right ) + a b d e^{3 \, n} x^{3 \, n} \cos \left (d x^{n} + c\right ) + {\left (a b d e^{3 \, n} x^{3 \, n} \sin \left (d x^{n} + c\right ) - b^{2} e^{3 \, n} x^{2 \, n}\right )} \sin \left (2 \, d x^{n} + 2 \, c\right )}{d e x \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e x \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} + 2 \, d e x \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e x}\,{d x}}{d e n \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e n \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} + 2 \, d e n \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e n} \]
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 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x^n\right )}\right )}^2\,{\left (e\,x\right )}^{3\,n-1} \,d x \]
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 \[ \int \left (e x\right )^{3 n - 1} \left (a + b \sec {\left (c + d x^{n} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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