Optimal. Leaf size=277 \[ \frac {a^3 (c+d x)^2}{2 d}-\frac {3 a^2 b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i a^2 b (c+d x)^2}{2 d}+\frac {3 i a^2 b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}-3 a b^2 c x+\frac {3 a b^2 d \log (\cos (e+f x))}{f^2}-\frac {3}{2} a b^2 d x^2+\frac {b^3 (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}-\frac {i b^3 (c+d x)^2}{2 d}-\frac {i b^3 d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b^3 d \tan (e+f x)}{2 f^2}+\frac {b^3 d x}{2 f} \]
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Rubi [A] time = 0.34, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3722, 3719, 2190, 2279, 2391, 3720, 3475, 3473, 8} \[ -\frac {3 a^2 b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i a^2 b (c+d x)^2}{2 d}+\frac {3 i a^2 b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}+\frac {a^3 (c+d x)^2}{2 d}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}-3 a b^2 c x+\frac {3 a b^2 d \log (\cos (e+f x))}{f^2}-\frac {3}{2} a b^2 d x^2+\frac {b^3 (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}-\frac {i b^3 (c+d x)^2}{2 d}-\frac {i b^3 d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b^3 d \tan (e+f x)}{2 f^2}+\frac {b^3 d x}{2 f} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2190
Rule 2279
Rule 2391
Rule 3473
Rule 3475
Rule 3719
Rule 3720
Rule 3722
Rubi steps
\begin {align*} \int (c+d x) (a+b \tan (e+f x))^3 \, dx &=\int \left (a^3 (c+d x)+3 a^2 b (c+d x) \tan (e+f x)+3 a b^2 (c+d x) \tan ^2(e+f x)+b^3 (c+d x) \tan ^3(e+f x)\right ) \, dx\\ &=\frac {a^3 (c+d x)^2}{2 d}+\left (3 a^2 b\right ) \int (c+d x) \tan (e+f x) \, dx+\left (3 a b^2\right ) \int (c+d x) \tan ^2(e+f x) \, dx+b^3 \int (c+d x) \tan ^3(e+f x) \, dx\\ &=\frac {a^3 (c+d x)^2}{2 d}+\frac {3 i a^2 b (c+d x)^2}{2 d}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}-\left (6 i a^2 b\right ) \int \frac {e^{2 i (e+f x)} (c+d x)}{1+e^{2 i (e+f x)}} \, dx-\left (3 a b^2\right ) \int (c+d x) \, dx-b^3 \int (c+d x) \tan (e+f x) \, dx-\frac {\left (3 a b^2 d\right ) \int \tan (e+f x) \, dx}{f}-\frac {\left (b^3 d\right ) \int \tan ^2(e+f x) \, dx}{2 f}\\ &=-3 a b^2 c x-\frac {3}{2} a b^2 d x^2+\frac {a^3 (c+d x)^2}{2 d}+\frac {3 i a^2 b (c+d x)^2}{2 d}-\frac {i b^3 (c+d x)^2}{2 d}-\frac {3 a^2 b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 a b^2 d \log (\cos (e+f x))}{f^2}-\frac {b^3 d \tan (e+f x)}{2 f^2}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}+\left (2 i b^3\right ) \int \frac {e^{2 i (e+f x)} (c+d x)}{1+e^{2 i (e+f x)}} \, dx+\frac {\left (3 a^2 b d\right ) \int \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}+\frac {\left (b^3 d\right ) \int 1 \, dx}{2 f}\\ &=-3 a b^2 c x+\frac {b^3 d x}{2 f}-\frac {3}{2} a b^2 d x^2+\frac {a^3 (c+d x)^2}{2 d}+\frac {3 i a^2 b (c+d x)^2}{2 d}-\frac {i b^3 (c+d x)^2}{2 d}-\frac {3 a^2 b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 a b^2 d \log (\cos (e+f x))}{f^2}-\frac {b^3 d \tan (e+f x)}{2 f^2}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}-\frac {\left (3 i a^2 b d\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^2}-\frac {\left (b^3 d\right ) \int \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=-3 a b^2 c x+\frac {b^3 d x}{2 f}-\frac {3}{2} a b^2 d x^2+\frac {a^3 (c+d x)^2}{2 d}+\frac {3 i a^2 b (c+d x)^2}{2 d}-\frac {i b^3 (c+d x)^2}{2 d}-\frac {3 a^2 b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 a b^2 d \log (\cos (e+f x))}{f^2}+\frac {3 i a^2 b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b^3 d \tan (e+f x)}{2 f^2}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}+\frac {\left (i b^3 d\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^2}\\ &=-3 a b^2 c x+\frac {b^3 d x}{2 f}-\frac {3}{2} a b^2 d x^2+\frac {a^3 (c+d x)^2}{2 d}+\frac {3 i a^2 b (c+d x)^2}{2 d}-\frac {i b^3 (c+d x)^2}{2 d}-\frac {3 a^2 b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 a b^2 d \log (\cos (e+f x))}{f^2}+\frac {3 i a^2 b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {i b^3 d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b^3 d \tan (e+f x)}{2 f^2}+\frac {3 a b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^3 (c+d x) \tan ^2(e+f x)}{2 f}\\ \end {align*}
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Mathematica [A] time = 3.69, size = 277, normalized size = 1.00 \[ \frac {\cos (e+f x) (a+b \tan (e+f x))^3 \left (-i b d \left (b^2-3 a^2\right ) \text {Li}_2\left (-e^{2 i (e+f x)}\right ) \cos ^2(e+f x)+\cos ^2(e+f x) \left (2 b \log (\cos (e+f x)) \left (3 a^2 (d e-c f)+3 a b d+b^2 (c f-d e)\right )+2 b d \left (b^2-3 a^2\right ) (e+f x) \log \left (1+e^{2 i (e+f x)}\right )-\left ((e+f x) \left (a^3 (d (e-f x)-2 c f)-3 i a^2 b d (e+f x)+3 a b^2 (2 c f-d e+d f x)+i b^3 d (e+f x)\right )\right )\right )+\frac {1}{2} b^2 (\sin (2 (e+f x)) (6 a f (c+d x)-b d)+2 b f (c+d x))\right )}{2 f^2 (a \cos (e+f x)+b \sin (e+f x))^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 327, normalized size = 1.18 \[ \frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} d f^{2} x^{2} - i \, {\left (3 \, a^{2} b - b^{3}\right )} d {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + i \, {\left (3 \, a^{2} b - b^{3}\right )} d {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + 2 \, {\left (b^{3} d f x + b^{3} c f\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left (b^{3} d f + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} c f^{2}\right )} x + 2 \, {\left (3 \, a b^{2} d - {\left (3 \, a^{2} b - b^{3}\right )} d f x - {\left (3 \, a^{2} b - b^{3}\right )} c f\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (3 \, a b^{2} d - {\left (3 \, a^{2} b - b^{3}\right )} d f x - {\left (3 \, a^{2} b - b^{3}\right )} c f\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (6 \, a b^{2} d f x + 6 \, a b^{2} c f - b^{3} d\right )} \tan \left (f x + e\right )}{4 \, f^{2}} \]
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 \[ \int {\left (d x + c\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.77, size = 493, normalized size = 1.78 \[ \frac {3 i a^{2} b d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{2}}+\frac {6 i b \,a^{2} d e x}{f}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) d x}{f}-\frac {3 b \,a^{2} c \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {6 b \,a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {3 b^{2} a d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f^{2}}-\frac {i b^{3} d \,e^{2}}{f^{2}}+\frac {3 i a^{2} b d \,x^{2}}{2}-\frac {i b^{3} d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{2}}-3 i a^{2} b c x -\frac {6 b^{2} a d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {2 b^{3} d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {b^{3} c \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}-\frac {2 b^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {b^{2} \left (6 i a d f x \,{\mathrm e}^{2 i \left (f x +e \right )}+6 i a c f \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b d f x \,{\mathrm e}^{2 i \left (f x +e \right )}+6 i a d f x -i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b c f \,{\mathrm e}^{2 i \left (f x +e \right )}+6 i a c f -i d b \right )}{f^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+i b^{3} c x -\frac {i b^{3} d \,x^{2}}{2}-\frac {6 b \,a^{2} d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a^{2} d x}{f}-\frac {2 i b^{3} d e x}{f}+\frac {3 i b \,a^{2} d \,e^{2}}{f^{2}}+\frac {a^{3} d \,x^{2}}{2}+a^{3} c x -3 a \,b^{2} c x -\frac {3 a \,b^{2} d \,x^{2}}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.68, size = 1327, normalized size = 4.79 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3\,\left (c+d\,x\right ) \,d x \]
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 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{3} \left (c + d x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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