Optimal. Leaf size=217 \[ \frac {a^2 x^6}{6}+\frac {2 i a b \text {Li}_3\left (-i e^{d x^2+c}\right )}{d^3}-\frac {2 i a b \text {Li}_3\left (i e^{d x^2+c}\right )}{d^3}-\frac {2 i a b x^2 \text {Li}_2\left (-i e^{d x^2+c}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (i e^{d x^2+c}\right )}{d^2}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 \text {Li}_2\left (-e^{2 \left (d x^2+c\right )}\right )}{2 d^3}-\frac {b^2 x^2 \log \left (e^{2 \left (c+d x^2\right )}+1\right )}{d^2}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac {b^2 x^4}{2 d} \]
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Rubi [A] time = 0.36, antiderivative size = 217, 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 = {5436, 4190, 4180, 2531, 2282, 6589, 4184, 3718, 2190, 2279, 2391} \[ -\frac {2 i a b x^2 \text {PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b x^2 \text {PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b \text {PolyLog}\left (3,-i e^{c+d x^2}\right )}{d^3}-\frac {2 i a b \text {PolyLog}\left (3,i e^{c+d x^2}\right )}{d^3}-\frac {b^2 \text {PolyLog}\left (2,-e^{2 \left (c+d x^2\right )}\right )}{2 d^3}+\frac {a^2 x^6}{6}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 x^2 \log \left (e^{2 \left (c+d x^2\right )}+1\right )}{d^2}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac {b^2 x^4}{2 d} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3718
Rule 4180
Rule 4184
Rule 4190
Rule 5436
Rule 6589
Rubi steps
\begin {align*} \int x^5 \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 (a+b \text {sech}(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 \text {sech}(c+d x)+b^2 x^2 \text {sech}^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac {a^2 x^6}{6}+(a b) \operatorname {Subst}\left (\int x^2 \text {sech}(c+d x) \, dx,x,x^2\right )+\frac {1}{2} b^2 \operatorname {Subst}\left (\int x^2 \text {sech}^2(c+d x) \, dx,x,x^2\right )\\ &=\frac {a^2 x^6}{6}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}-\frac {(2 i a b) \operatorname {Subst}\left (\int x \log \left (1-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}+\frac {(2 i a b) \operatorname {Subst}\left (\int x \log \left (1+i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}-\frac {b^2 \operatorname {Subst}\left (\int x \tanh (c+d x) \, dx,x,x^2\right )}{d}\\ &=\frac {b^2 x^4}{2 d}+\frac {a^2 x^6}{6}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {2 i a b x^2 \text {Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac {(2 i a b) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{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^{c+d x}\right ) \, dx,x,x^2\right )}{d^2}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 (c+d x)} x}{1+e^{2 (c+d x)}} \, dx,x,x^2\right )}{d}\\ &=\frac {b^2 x^4}{2 d}+\frac {a^2 x^6}{6}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 x^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac {(2 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}-\frac {(2 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}+\frac {b^2 \operatorname {Subst}\left (\int \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}\\ &=\frac {b^2 x^4}{2 d}+\frac {a^2 x^6}{6}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 x^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b \text {Li}_3\left (-i e^{c+d x^2}\right )}{d^3}-\frac {2 i a b \text {Li}_3\left (i e^{c+d x^2}\right )}{d^3}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \left (c+d x^2\right )}\right )}{2 d^3}\\ &=\frac {b^2 x^4}{2 d}+\frac {a^2 x^6}{6}+\frac {2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 x^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac {2 i a b x^2 \text {Li}_2\left (i e^{c+d x^2}\right )}{d^2}-\frac {b^2 \text {Li}_2\left (-e^{2 \left (c+d x^2\right )}\right )}{2 d^3}+\frac {2 i a b \text {Li}_3\left (-i e^{c+d x^2}\right )}{d^3}-\frac {2 i a b \text {Li}_3\left (i e^{c+d x^2}\right )}{d^3}+\frac {b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}\\ \end {align*}
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Mathematica [A] time = 4.80, size = 294, normalized size = 1.35 \[ \frac {\cosh \left (c+d x^2\right ) \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \left (a^2 x^6 \cosh \left (c+d x^2\right )+\frac {3 b \cosh \left (c+d x^2\right ) \left (2 i a d^2 x^4 \log \left (1-i e^{c+d x^2}\right )-2 i a d^2 x^4 \log \left (1+i e^{c+d x^2}\right )-4 i a d x^2 \text {Li}_2\left (-i e^{d x^2+c}\right )+4 i a d x^2 \text {Li}_2\left (i e^{d x^2+c}\right )+4 i a \text {Li}_3\left (-i e^{d x^2+c}\right )-4 i a \text {Li}_3\left (i e^{d x^2+c}\right )+\frac {2 b e^{2 c} d^2 x^4}{e^{2 c}+1}-b \text {Li}_2\left (-e^{2 \left (d x^2+c\right )}\right )-2 b d x^2 \log \left (e^{2 \left (c+d x^2\right )}+1\right )\right )}{d^3}+\frac {3 b^2 x^4 \text {sech}(c) \sinh \left (d x^2\right )}{d}\right )}{6 \left (a \cosh \left (c+d x^2\right )+b\right )^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.50, size = 1210, normalized size = 5.58 \[ \text {result too large to display} \]
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 \[ \int {\left (b \operatorname {sech}\left (d x^{2} + c\right ) + a\right )}^{2} x^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.38, size = 0, normalized size = 0.00 \[ \int x^{5} \left (a +b \,\mathrm {sech}\left (d \,x^{2}+c \right )\right )^{2}\, dx \]
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 \[ \frac {1}{6} \, a^{2} x^{6} - \frac {b^{2} x^{4}}{d e^{\left (2 \, d x^{2} + 2 \, c\right )} + d} + \int \frac {4 \, {\left (a b d x^{5} e^{\left (d x^{2} + c\right )} + b^{2} x^{3}\right )}}{d e^{\left (2 \, d x^{2} + 2 \, c\right )} + d}\,{d x} \]
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 \[ \int x^5\,{\left (a+\frac {b}{\mathrm {cosh}\left (d\,x^2+c\right )}\right )}^2 \,d x \]
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 \[ \int x^{5} \left (a + b \operatorname {sech}{\left (c + d x^{2} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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