Optimal. Leaf size=114 \[ \frac {3 d^3 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {i (c+d x)^3}{b} \]
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Rubi [A] time = 0.22, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4184, 3719, 2190, 2531, 2282, 6589} \[ -\frac {3 i d^2 (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac {3 d^3 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {i (c+d x)^3}{b} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3719
Rule 4184
Rule 6589
Rubi steps
\begin {align*} \int (c+d x)^3 \sec ^2(a+b x) \, dx &=\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {(3 d) \int (c+d x)^2 \tan (a+b x) \, dx}{b}\\ &=-\frac {i (c+d x)^3}{b}+\frac {(c+d x)^3 \tan (a+b x)}{b}+\frac {(6 i d) \int \frac {e^{2 i (a+b x)} (c+d x)^2}{1+e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac {i (c+d x)^3}{b}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {i (c+d x)^3}{b}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \tan (a+b x)}{b}+\frac {\left (3 i d^3\right ) \int \text {Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {i (c+d x)^3}{b}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \tan (a+b x)}{b}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}\\ &=-\frac {i (c+d x)^3}{b}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \tan (a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 0.51, size = 109, normalized size = 0.96 \[ \frac {2 b^2 (c+d x)^2 \left (b (c+d x) \tan (a+b x)+3 d \log \left (1+e^{2 i (a+b x)}\right )-i b (c+d x)\right )-6 i b d^2 (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )+3 d^3 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 1.22, size = 786, normalized size = 6.89 \[ \frac {6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (b x + a\right )}{2 \, b^{4} \cos \left (b x + a\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 \[ \int {\left (d x + c\right )}^{3} \sec \left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.20, size = 316, normalized size = 2.77 \[ -\frac {2 i d^{3} x^{3}}{b}+\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 i d^{2} c \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}-\frac {12 i d^{2} c a x}{b^{2}}+\frac {6 d^{2} c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}-\frac {3 i d^{3} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {3 d^{3} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{4}}+\frac {12 d^{2} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 i d^{2} c \,a^{2}}{b^{3}}-\frac {6 i d^{2} c \,x^{2}}{b}+\frac {6 i d^{3} a^{2} x}{b^{3}}+\frac {4 i d^{3} a^{3}}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.42, size = 1056, normalized size = 9.26 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (c+d\,x\right )}^3}{{\cos \left (a+b\,x\right )}^2} \,d x \]
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 \[ \int \left (c + d x\right )^{3} \sec ^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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