Optimal. Leaf size=382 \[ \frac {i b \text {Li}_3\left (-\frac {a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {i b \text {Li}_3\left (-\frac {a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d \sqrt {b^2-a^2}}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^6}{6 a} \]
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Rubi [A] time = 0.88, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4204, 4191, 3321, 2264, 2190, 2531, 2282, 6589} \[ \frac {b x^2 \text {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^2 \text {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b \text {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {i b \text {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d \sqrt {b^2-a^2}}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^6}{6 a} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3321
Rule 4191
Rule 4204
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{a+b \sec (c+d x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {x^2}{a}-\frac {b x^2}{a (b+a \cos (c+d x))}\right ) \, dx,x,x^2\right )\\ &=\frac {x^6}{6 a}-\frac {b \operatorname {Subst}\left (\int \frac {x^2}{b+a \cos (c+d x)} \, dx,x,x^2\right )}{2 a}\\ &=\frac {x^6}{6 a}-\frac {b \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{a+2 b e^{i (c+d x)}+a e^{2 i (c+d x)}} \, dx,x,x^2\right )}{a}\\ &=\frac {x^6}{6 a}-\frac {b \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{2 b-2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {-a^2+b^2}}+\frac {b \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{2 b+2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {-a^2+b^2}}\\ &=\frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {(i b) \operatorname {Subst}\left (\int x \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d}+\frac {(i b) \operatorname {Subst}\left (\int x \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d}\\ &=\frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {2 a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {b \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {2 a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d^2}\\ &=\frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a x}{-b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {a x}{b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{a \sqrt {-a^2+b^2} d^3}\\ &=\frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \text {Li}_2\left (-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {i b \text {Li}_3\left (-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {i b \text {Li}_3\left (-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}\\ \end {align*}
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Mathematica [A] time = 1.12, size = 472, normalized size = 1.24 \[ \frac {d^3 x^6 \sqrt {e^{2 i c} \left (b^2-a^2\right )}+3 i b e^{i c} d^2 x^4 \log \left (1+\frac {a e^{i \left (2 c+d x^2\right )}}{b e^{i c}-\sqrt {e^{2 i c} \left (b^2-a^2\right )}}\right )-3 i b e^{i c} d^2 x^4 \log \left (1+\frac {a e^{i \left (2 c+d x^2\right )}}{\sqrt {e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )+6 b e^{i c} d x^2 \text {Li}_2\left (-\frac {a e^{i \left (d x^2+2 c\right )}}{b e^{i c}-\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right )-6 b e^{i c} d x^2 \text {Li}_2\left (-\frac {a e^{i \left (d x^2+2 c\right )}}{e^{i c} b+\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right )+6 i b e^{i c} \text {Li}_3\left (-\frac {a e^{i \left (d x^2+2 c\right )}}{b e^{i c}-\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right )-6 i b e^{i c} \text {Li}_3\left (-\frac {a e^{i \left (d x^2+2 c\right )}}{e^{i c} b+\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right )}{6 a d^3 \sqrt {e^{2 i c} \left (b^2-a^2\right )}} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.95, size = 1457, normalized size = 3.81 \[ \text {result too large to display} \]
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 \[ \int \frac {x^{5}}{b \sec \left (d x^{2} + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.85, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{a +b \sec \left (d \,x^{2}+c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {x^5}{a+\frac {b}{\cos \left (d\,x^2+c\right )}} \,d x \]
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 \[ \int \frac {x^{5}}{a + b \sec {\left (c + d x^{2} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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