Optimal. Leaf size=157 \[ \frac {a (c+d x)^3}{3 d}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac {2 b d^2 \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3} \]
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Rubi [A] time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4190, 4181, 2531, 2282, 6589} \[ \frac {2 i b d (c+d x) \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {2 b d^2 \text {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4181
Rule 4190
Rule 6589
Rubi steps
\begin {align*} \int (c+d x)^2 (a+b \sec (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \sec (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^3}{3 d}+b \int (c+d x)^2 \sec (e+f x) \, dx\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac {(2 b d) \int (c+d x) \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac {(2 b d) \int (c+d x) \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {\left (2 i b d^2\right ) \int \text {Li}_2\left (-i e^{i (e+f x)}\right ) \, dx}{f^2}+\frac {\left (2 i b d^2\right ) \int \text {Li}_2\left (i e^{i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {\left (2 b d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3}+\frac {\left (2 b d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3}\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {2 b d^2 \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3}\\ \end {align*}
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Mathematica [A] time = 0.27, size = 203, normalized size = 1.29 \[ a c^2 x+a c d x^2+\frac {1}{3} a d^2 x^3+\frac {b c^2 \tanh ^{-1}(\sin (e+f x))}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {4 i b c d x \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac {2 b d^2 \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3}-\frac {2 i b d^2 x^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.80, size = 671, normalized size = 4.27 \[ \frac {2 \, a d^{2} f^{3} x^{3} + 6 \, a c d f^{3} x^{2} + 6 \, a c^{2} f^{3} x - 6 \, b d^{2} {\rm polylog}\left (3, i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 6 \, b d^{2} {\rm polylog}\left (3, i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) - 6 \, b d^{2} {\rm polylog}\left (3, -i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 6 \, b d^{2} {\rm polylog}\left (3, -i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + {\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + {\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + {\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + {\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right )}{6 \, f^{3}} \]
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 \[ \int {\left (d x + c\right )}^{2} {\left (b \sec \left (f x + e\right ) + a\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.94, size = 431, normalized size = 2.75 \[ \frac {a \,d^{2} x^{3}}{3}+a c d \,x^{2}+a \,c^{2} x -\frac {b \,d^{2} e^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {2 b c d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 i b \,d^{2} \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {2 b c d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {2 i b \,c^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {2 i b \,d^{2} e^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 b c d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {4 i b c d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b c d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}+\frac {2 b \,d^{2} \polylog \left (3, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {b \,d^{2} e^{2} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{3}}-\frac {2 b \,d^{2} \polylog \left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {b \,d^{2} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x^{2}}{f}+\frac {2 i b \,d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {b \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}+\frac {2 i b c d \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 i b c d \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.23, size = 510, normalized size = 3.25 \[ \frac {6 \, {\left (f x + e\right )} a c^{2} + \frac {2 \, {\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac {6 \, {\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac {6 \, {\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac {6 \, {\left (f x + e\right )}^{2} a c d}{f} - \frac {12 \, {\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + \frac {6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f^{2}} - \frac {12 \, b c d e \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f} + \frac {3 \, {\left (4 \, b d^{2} {\rm Li}_{3}(i \, e^{\left (i \, f x + i \, e\right )}) - 4 \, b d^{2} {\rm Li}_{3}(-i \, e^{\left (i \, f x + i \, e\right )}) + {\left (-2 i \, {\left (f x + e\right )}^{2} b d^{2} + {\left (4 i \, b d^{2} e - 4 i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + {\left (-2 i \, {\left (f x + e\right )}^{2} b d^{2} + {\left (4 i \, b d^{2} e - 4 i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + {\left (-4 i \, {\left (f x + e\right )} b d^{2} + 4 i \, b d^{2} e - 4 i \, b c d f\right )} {\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) + {\left (4 i \, {\left (f x + e\right )} b d^{2} - 4 i \, b d^{2} e + 4 i \, b c d f\right )} {\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) + {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{f^{2}}}{6 \, f} \]
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 \[ \int \left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )\,{\left (c+d\,x\right )}^2 \,d x \]
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 \[ \int \left (a + b \sec {\left (e + f x \right )}\right ) \left (c + d x\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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