Optimal. Leaf size=242 \[ \frac {a^2 x^6}{6}-\frac {2 a b \text {Li}_3\left (-i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {2 a b \text {Li}_3\left (i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {2 i a b x^2 \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {i b^2 \text {Li}_2\left (-e^{2 i \left (d x^2+c\right )}\right )}{2 d^3}+\frac {b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac {i b^2 x^4}{2 d} \]
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Rubi [A] time = 0.37, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac {2 i a b x^2 \text {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 a b \text {PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {2 a b \text {PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac {i b^2 \text {PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}+\frac {a^2 x^6}{6}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac {i b^2 x^4}{2 d} \]
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 6589
Rubi steps
\begin {align*} \int x^5 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac {1}{2} \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^2\right )\\ &=\frac {a^2 x^6}{6}+(a b) \operatorname {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,x^2\right )+\frac {1}{2} b^2 \operatorname {Subst}\left (\int x^2 \sec ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac {a^2 x^6}{6}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac {(2 a b) \operatorname {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac {(2 a b) \operatorname {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}-\frac {b^2 \operatorname {Subst}\left (\int x \tan (c+d x) \, dx,x,x^2\right )}{d}\\ &=-\frac {i b^2 x^4}{2 d}+\frac {a^2 x^6}{6}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {2 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac {(2 i a b) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}+\frac {(2 i a b) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,x^2\right )}{d}\\ &=-\frac {i b^2 x^4}{2 d}+\frac {a^2 x^6}{6}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac {(2 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {(2 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac {b^2 \operatorname {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}\\ &=-\frac {i b^2 x^4}{2 d}+\frac {a^2 x^6}{6}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 a b \text {Li}_3\left (-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {2 a b \text {Li}_3\left (i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}\\ &=-\frac {i b^2 x^4}{2 d}+\frac {a^2 x^6}{6}-\frac {2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b^2 \text {Li}_2\left (-e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}-\frac {2 a b \text {Li}_3\left (-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {2 a b \text {Li}_3\left (i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b^2 x^4 \tan \left (c+d x^2\right )}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.61, size = 229, normalized size = 0.95 \[ \frac {a^2 d^3 x^6-12 i a b d^2 x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )+12 i a b d x^2 \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )-12 i a b d x^2 \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )-12 a b \text {Li}_3\left (-i e^{i \left (d x^2+c\right )}\right )+12 a b \text {Li}_3\left (i e^{i \left (d x^2+c\right )}\right )+3 b^2 d^2 x^4 \tan \left (c+d x^2\right )-3 i b^2 \text {Li}_2\left (-e^{2 i \left (d x^2+c\right )}\right )+6 b^2 d x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )-3 i b^2 d^2 x^4}{6 d^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.75, size = 795, normalized size = 3.29 \[ \frac {a^{2} d^{3} x^{6} \cos \left (d x^{2} + c\right ) + 3 \, b^{2} d^{2} x^{4} \sin \left (d x^{2} + c\right ) - 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + {\left (-6 i \, a b d x^{2} + 3 i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + {\left (-6 i \, a b d x^{2} - 3 i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + {\left (6 i \, a b d x^{2} - 3 i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + {\left (6 i \, a b d x^{2} + 3 i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 3 \, {\left (a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, {\left (a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 3 \, {\left (a b d^{2} x^{4} + b^{2} d x^{2} - a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (a b d^{2} x^{4} - b^{2} d x^{2} - a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \, {\left (a b d^{2} x^{4} + b^{2} d x^{2} - a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (a b d^{2} x^{4} - b^{2} d x^{2} - a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \, {\left (a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, {\left (a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right )}{6 \, d^{3} \cos \left (d x^{2} + c\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 \[ \int {\left (b \sec \left (d x^{2} + c\right ) + a\right )}^{2} x^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.10, size = 0, normalized size = 0.00 \[ \int x^{5} \left (a +b \sec \left (d \,x^{2}+c \right )\right )^{2}\, dx \]
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 \[ \frac {1}{6} \, a^{2} x^{6} + \frac {b^{2} x^{4} \sin \left (2 \, d x^{2} + 2 \, c\right ) + 4 \, {\left (d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d\right )} \int \frac {a b d x^{5} \cos \left (2 \, d x^{2} + 2 \, c\right ) \cos \left (d x^{2} + c\right ) + a b d x^{5} \cos \left (d x^{2} + c\right ) + {\left (a b d x^{5} \sin \left (d x^{2} + c\right ) - b^{2} x^{3}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d}\,{d x}}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d} \]
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 \[ \int x^5\,{\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right )}^2 \,d x \]
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 \[ \int x^{5} \left (a + b \sec {\left (c + d x^{2} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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