Optimal. Leaf size=92 \[ \frac {a x^4}{4}+\frac {i b \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]
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Rubi [A] time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {14, 4204, 4181, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {i b \text {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{2 d^2}+\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2279
Rule 2391
Rule 4181
Rule 4204
Rubi steps
\begin {align*} \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx &=\int \left (a x^3+b x^3 \sec \left (c+d x^2\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \sec \left (c+d x^2\right ) \, dx\\ &=\frac {a x^4}{4}+\frac {1}{2} b \operatorname {Subst}\left (\int x \sec (c+d x) \, dx,x,x^2\right )\\ &=\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {b \operatorname {Subst}\left (\int \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{2 d}+\frac {b \operatorname {Subst}\left (\int \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{2 d}\\ &=\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 d^2}\\ &=\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{2 d^2}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 95, normalized size = 1.03 \[ \frac {a x^4}{4}+\frac {i b \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b x^2 \tan ^{-1}\left (e^{i c+i d x^2}\right )}{d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.79, size = 346, normalized size = 3.76 \[ \frac {a d^{2} x^{4} - b c \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + b c \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) - b c \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + b c \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) - i \, b {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - i \, b {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + i \, b {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + i \, b {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + {\left (b d x^{2} + b c\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (b d x^{2} + b c\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + {\left (b d x^{2} + b c\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (b d x^{2} + b c\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right )}{4 \, d^{2}} \]
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 \[ \int {\left (b \sec \left (d x^{2} + c\right ) + a\right )} x^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.74, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \sec \left (d \,x^{2}+c \right )\right )\, dx \]
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 \[ \frac {1}{4} \, a x^{4} + 2 \, b \int \frac {x^{3} \cos \left (2 \, d x^{2} + 2 \, c\right ) \cos \left (d x^{2} + c\right ) + x^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) \sin \left (d x^{2} + c\right ) + x^{3} \cos \left (d x^{2} + c\right )}{\cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x^{2} + 2 \, c\right ) + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right ) \,d x \]
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 \[ \int x^{3} \left (a + b \sec {\left (c + d x^{2} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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