Integrand size = 18, antiderivative size = 115 \[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3} \]
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Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3803, 3800, 2221, 2611, 2320, 6724} \[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}+\frac {i b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^3}{3 d}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3} \]
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Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 3803
Rule 6724
Rubi steps \begin{align*} \text {integral}& = \int \left (a (c+d x)^2+b (c+d x)^2 \tan (e+f x)\right ) \, dx \\ & = \frac {a (c+d x)^3}{3 d}+b \int (c+d x)^2 \tan (e+f x) \, dx \\ & = \frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)^2}{1+e^{2 i (e+f x)}} \, dx \\ & = \frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {(2 b d) \int (c+d x) \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f} \\ & = \frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {\left (i b d^2\right ) \int \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right ) \, dx}{f^2} \\ & = \frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^3} \\ & = \frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.66 \[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=a c^2 x+a c d x^2+i b c d x^2+\frac {1}{3} a d^2 x^3+\frac {1}{3} i b d^2 x^3-\frac {2 b c d x \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {b d^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {b c^2 \log (\cos (e+f x))}{f}+\frac {i b c d \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}+\frac {i b d^2 x \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (101 ) = 202\).
Time = 0.58 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.73
method | result | size |
risch | \(\frac {i d^{2} b \,x^{3}}{3}-i b \,c^{2} x +\frac {d^{2} a \,x^{3}}{3}+\frac {4 i b d c e x}{f}+d a c \,x^{2}+\frac {2 i b d c \,e^{2}}{f^{2}}+a \,c^{2} x +\frac {a \,c^{3}}{3 d}-\frac {b \,c^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}-\frac {2 b d c \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f}+i d b c \,x^{2}+\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 i b \,d^{2} e^{2} x}{f^{2}}-\frac {b \,d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{2}}{f}-\frac {b \,d^{2} \operatorname {Li}_{3}\left (-{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{3}}+\frac {i b d c \,\operatorname {Li}_{2}\left (-{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}+\frac {i b \,d^{2} \operatorname {Li}_{2}\left (-{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {i b \,c^{3}}{3 d}-\frac {4 i b \,d^{2} e^{3}}{3 f^{3}}-\frac {4 b c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}\) | \(314\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (98) = 196\).
Time = 0.26 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.76 \[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\frac {4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x - 3 \, b d^{2} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b d^{2} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (i \, b d^{2} f x + i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \, {\left (-i \, b d^{2} f x - i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, f^{3}} \]
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\[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right ) \left (c + d x\right )^{2}\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (98) = 196\).
Time = 0.52 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.24 \[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\frac {6 \, {\left (f x + e\right )} a c^{2} + \frac {2 \, {\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac {6 \, {\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac {6 \, {\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac {6 \, {\left (f x + e\right )}^{2} a c d}{f} - \frac {12 \, {\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right )\right ) + \frac {6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac {12 \, b c d e \log \left (\sec \left (f x + e\right )\right )}{f} - \frac {-2 i \, {\left (f x + e\right )}^{3} b d^{2} + 3 \, b d^{2} {\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )}) - 6 \, {\left (-i \, b d^{2} e + i \, b c d f\right )} {\left (f x + e\right )}^{2} - 6 \, {\left (-i \, {\left (f x + e\right )}^{2} b d^{2} + 2 \, {\left (i \, b d^{2} e - i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 6 \, {\left (i \, {\left (f x + e\right )} b d^{2} - i \, b d^{2} e + i \, b c d f\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 3 \, {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}{f^{2}}}{6 \, f} \]
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\[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\int { {\left (d x + c\right )}^{2} {\left (b \tan \left (f x + e\right ) + a\right )} \,d x } \]
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Timed out. \[ \int (c+d x)^2 (a+b \tan (e+f x)) \, dx=\int \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^2 \,d x \]
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