Integrand size = 20, antiderivative size = 436 \[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx=\frac {b^3 c d x}{f}+\frac {b^3 d^2 x^2}{2 f}-\frac {3 i a b^2 (c+d x)^2}{f}+\frac {a^3 (c+d x)^3}{3 d}+\frac {i a^2 b (c+d x)^3}{d}-\frac {a b^2 (c+d x)^3}{d}-\frac {i b^3 (c+d x)^3}{3 d}+\frac {6 a b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac {3 a^2 b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {b^3 d^2 \log (\cos (e+f x))}{f^3}-\frac {3 i a b^2 d^2 \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}+\frac {3 i a^2 b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {i b^3 d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {3 a^2 b d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {b^3 d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {b^3 d (c+d x) \tan (e+f x)}{f^2}+\frac {3 a b^2 (c+d x)^2 \tan (e+f x)}{f}+\frac {b^3 (c+d x)^2 \tan ^2(e+f x)}{2 f} \]
b^3*c*d*x/f+1/2*b^3*d^2*x^2/f+3*I*a^2*b*d*(d*x+c)*polylog(2,-exp(2*I*(f*x+ e)))/f^2+1/3*a^3*(d*x+c)^3/d-3*I*a*b^2*d^2*polylog(2,-exp(2*I*(f*x+e)))/f^ 3-a*b^2*(d*x+c)^3/d+I*a^2*b*(d*x+c)^3/d+6*a*b^2*d*(d*x+c)*ln(1+exp(2*I*(f* x+e)))/f^2-3*a^2*b*(d*x+c)^2*ln(1+exp(2*I*(f*x+e)))/f+b^3*(d*x+c)^2*ln(1+e xp(2*I*(f*x+e)))/f-b^3*d^2*ln(cos(f*x+e))/f^3-3*I*a*b^2*(d*x+c)^2/f-I*b^3* d*(d*x+c)*polylog(2,-exp(2*I*(f*x+e)))/f^2-1/3*I*b^3*(d*x+c)^3/d-3/2*a^2*b *d^2*polylog(3,-exp(2*I*(f*x+e)))/f^3+1/2*b^3*d^2*polylog(3,-exp(2*I*(f*x+ e)))/f^3-b^3*d*(d*x+c)*tan(f*x+e)/f^2+3*a*b^2*(d*x+c)^2*tan(f*x+e)/f+1/2*b ^3*(d*x+c)^2*tan(f*x+e)^2/f
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1846\) vs. \(2(436)=872\).
Time = 7.16 (sec) , antiderivative size = 1846, normalized size of antiderivative = 4.23 \[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx =\text {Too large to display} \]
((-1/4*I)*a^2*b*d^2*(2*f^2*x^2*(2*f*x - (3*I)*(1 + E^((2*I)*e))*Log[1 + E^ ((-2*I)*(e + f*x))]) + 6*(1 + E^((2*I)*e))*f*x*PolyLog[2, -E^((-2*I)*(e + f*x))] - (3*I)*(1 + E^((2*I)*e))*PolyLog[3, -E^((-2*I)*(e + f*x))])*Sec[e] )/(E^(I*e)*f^3) + ((I/12)*b^3*d^2*(2*f^2*x^2*(2*f*x - (3*I)*(1 + E^((2*I)* e))*Log[1 + E^((-2*I)*(e + f*x))]) + 6*(1 + E^((2*I)*e))*f*x*PolyLog[2, -E ^((-2*I)*(e + f*x))] - (3*I)*(1 + E^((2*I)*e))*PolyLog[3, -E^((-2*I)*(e + f*x))])*Sec[e])/(E^(I*e)*f^3) - (b^3*d^2*Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f*x ] - Sin[e]*Sin[f*x]] + f*x*Sin[e]))/(f^3*(Cos[e]^2 + Sin[e]^2)) + (6*a*b^2 *c*d*Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f*x] - Sin[e]*Sin[f*x]] + f*x*Sin[e]))/ (f^2*(Cos[e]^2 + Sin[e]^2)) - (3*a^2*b*c^2*Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f *x] - Sin[e]*Sin[f*x]] + f*x*Sin[e]))/(f*(Cos[e]^2 + Sin[e]^2)) + (b^3*c^2 *Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f*x] - Sin[e]*Sin[f*x]] + f*x*Sin[e]))/(f*( Cos[e]^2 + Sin[e]^2)) + (3*a*b^2*d^2*Csc[e]*((f^2*x^2)/E^(I*ArcTan[Cot[e]] ) - (Cot[e]*(I*f*x*(-Pi - 2*ArcTan[Cot[e]]) - Pi*Log[1 + E^((-2*I)*f*x)] - 2*(f*x - ArcTan[Cot[e]])*Log[1 - E^((2*I)*(f*x - ArcTan[Cot[e]]))] + Pi*L og[Cos[f*x]] - 2*ArcTan[Cot[e]]*Log[Sin[f*x - ArcTan[Cot[e]]]] + I*PolyLog [2, E^((2*I)*(f*x - ArcTan[Cot[e]]))]))/Sqrt[1 + Cot[e]^2])*Sec[e])/(f^3*S qrt[Csc[e]^2*(Cos[e]^2 + Sin[e]^2)]) - (3*a^2*b*c*d*Csc[e]*((f^2*x^2)/E^(I *ArcTan[Cot[e]]) - (Cot[e]*(I*f*x*(-Pi - 2*ArcTan[Cot[e]]) - Pi*Log[1 + E^ ((-2*I)*f*x)] - 2*(f*x - ArcTan[Cot[e]])*Log[1 - E^((2*I)*(f*x - ArcTan...
Time = 0.91 (sec) , antiderivative size = 427, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4205, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x)^2 (a+b \tan (e+f x))^3dx\) |
\(\Big \downarrow \) 4205 |
\(\displaystyle \int \left (a^3 (c+d x)^2+3 a^2 b (c+d x)^2 \tan (e+f x)+3 a b^2 (c+d x)^2 \tan ^2(e+f x)+b^3 (c+d x)^2 \tan ^3(e+f x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 (c+d x)^3}{3 d}+\frac {3 i a^2 b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {3 a^2 b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i a^2 b (c+d x)^3}{d}-\frac {3 a^2 b d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {6 a b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac {3 a b^2 (c+d x)^2 \tan (e+f x)}{f}-\frac {3 i a b^2 (c+d x)^2}{f}-\frac {a b^2 (c+d x)^3}{d}-\frac {3 i a b^2 d^2 \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}-\frac {i b^3 d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {b^3 d (c+d x) \tan (e+f x)}{f^2}+\frac {b^3 (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^3 (c+d x)^2 \tan ^2(e+f x)}{2 f}+\frac {b^3 (c+d x)^2}{2 f}-\frac {i b^3 (c+d x)^3}{3 d}+\frac {b^3 d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {b^3 d^2 \log (\cos (e+f x))}{f^3}\) |
((-3*I)*a*b^2*(c + d*x)^2)/f + (b^3*(c + d*x)^2)/(2*f) + (a^3*(c + d*x)^3) /(3*d) + (I*a^2*b*(c + d*x)^3)/d - (a*b^2*(c + d*x)^3)/d - ((I/3)*b^3*(c + d*x)^3)/d + (6*a*b^2*d*(c + d*x)*Log[1 + E^((2*I)*(e + f*x))])/f^2 - (3*a ^2*b*(c + d*x)^2*Log[1 + E^((2*I)*(e + f*x))])/f + (b^3*(c + d*x)^2*Log[1 + E^((2*I)*(e + f*x))])/f - (b^3*d^2*Log[Cos[e + f*x]])/f^3 - ((3*I)*a*b^2 *d^2*PolyLog[2, -E^((2*I)*(e + f*x))])/f^3 + ((3*I)*a^2*b*d*(c + d*x)*Poly Log[2, -E^((2*I)*(e + f*x))])/f^2 - (I*b^3*d*(c + d*x)*PolyLog[2, -E^((2*I )*(e + f*x))])/f^2 - (3*a^2*b*d^2*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^3 ) + (b^3*d^2*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^3) - (b^3*d*(c + d*x)* Tan[e + f*x])/f^2 + (3*a*b^2*(c + d*x)^2*Tan[e + f*x])/f + (b^3*(c + d*x)^ 2*Tan[e + f*x]^2)/(2*f)
3.1.50.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1137 vs. \(2 (402 ) = 804\).
Time = 1.06 (sec) , antiderivative size = 1138, normalized size of antiderivative = 2.61
3*I*d*a^2*b*c*x^2+12*I/f*b*a^2*c*d*e*x+6/f^2*b^2*a*c*d*ln(exp(2*I*(f*x+e)) +1)-12/f^2*b^2*a*c*d*ln(exp(I*(f*x+e)))+4/f^2*b^3*e*d*c*ln(exp(I*(f*x+e))) +6/f^2*b^2*a*d^2*ln(exp(2*I*(f*x+e))+1)*x+2/f*b^3*c*d*ln(exp(2*I*(f*x+e))+ 1)*x+6/f^3*b*e^2*d^2*a^2*ln(exp(I*(f*x+e)))-3/f*b*d^2*a^2*ln(exp(2*I*(f*x+ e))+1)*x^2+12/f^3*b^2*e*a*d^2*ln(exp(I*(f*x+e)))-6*I/f^3*b^2*a*d^2*e^2-I/f ^2*b^3*d^2*polylog(2,-exp(2*I*(f*x+e)))*x-2*I/f^2*b^3*c*d*e^2-I/f^2*b^3*c* d*polylog(2,-exp(2*I*(f*x+e)))+2*I/f^2*b^3*d^2*e^2*x-6*I/f*b^2*a*d^2*x^2-4 *I/f^3*b*d^2*a^2*e^3+1/f*b^3*d^2*ln(exp(2*I*(f*x+e))+1)*x^2-3/f*b*a^2*c^2* ln(exp(2*I*(f*x+e))+1)+6/f*b*a^2*c^2*ln(exp(I*(f*x+e)))-2/f^3*b^3*e^2*d^2* ln(exp(I*(f*x+e)))+4/3*I/f^3*b^3*d^2*e^3-I*d*b^3*c*x^2-3*d*a*b^2*c*x^2+I*d ^2*a^2*b*x^3-3*I*a^2*b*c^2*x-I/d*a^2*b*c^3+d*a^3*c*x^2+a^3*c^2*x-d^2*a*b^2 *x^3-3*a*b^2*c^2*x-1/d*a*b^2*c^3+I*b^3*c^2*x+1/3*I/d*b^3*c^3+1/3*d^2*a^3*x ^3+1/3/d*c^3*a^3-1/3*I*b^3*d^2*x^3+1/f*b^3*c^2*ln(exp(2*I*(f*x+e))+1)-2/f* b^3*c^2*ln(exp(I*(f*x+e)))-1/f^3*b^3*d^2*ln(exp(2*I*(f*x+e))+1)+2/f^3*b^3* d^2*ln(exp(I*(f*x+e)))+2*b^2*(3*I*a*d^2*f*x^2*exp(2*I*(f*x+e))+6*I*a*c*d*f *x*exp(2*I*(f*x+e))+b*d^2*f*x^2*exp(2*I*(f*x+e))+3*I*a*c^2*f*exp(2*I*(f*x+ e))+3*I*a*d^2*f*x^2-I*b*d^2*x*exp(2*I*(f*x+e))+2*b*c*d*f*x*exp(2*I*(f*x+e) )+6*I*a*c*d*f*x-I*b*c*d*exp(2*I*(f*x+e))+b*c^2*f*exp(2*I*(f*x+e))+3*I*a*c^ 2*f-I*d^2*x*b-I*c*d*b)/f^2/(exp(2*I*(f*x+e))+1)^2-3*I*a*b^2*d^2*polylog(2, -exp(2*I*(f*x+e)))/f^3-3/2*a^2*b*d^2*polylog(3,-exp(2*I*(f*x+e)))/f^3+1...
Time = 0.29 (sec) , antiderivative size = 688, normalized size of antiderivative = 1.58 \[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx=\frac {4 \, {\left (a^{3} - 3 \, a b^{2}\right )} d^{2} f^{3} x^{3} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} 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 \, {\left (3 \, a^{2} b - b^{3}\right )} 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 (b^{3} d^{2} f^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} c d f^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} f^{2} x^{2} + 2 \, b^{3} c d f^{2} x + b^{3} c^{2} f^{2}\right )} \tan \left (f x + e\right )^{2} + 12 \, {\left (b^{3} c d f^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} c^{2} f^{3}\right )} x - 6 \, {\left (-3 i \, a b^{2} d^{2} + i \, {\left (3 \, a^{2} b - b^{3}\right )} d^{2} f x + i \, {\left (3 \, a^{2} b - b^{3}\right )} 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 (3 i \, a b^{2} d^{2} - i \, {\left (3 \, a^{2} b - b^{3}\right )} d^{2} f x - i \, {\left (3 \, a^{2} b - b^{3}\right )} 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 ({\left (3 \, a^{2} b - b^{3}\right )} d^{2} f^{2} x^{2} - 6 \, a b^{2} c d f + b^{3} d^{2} + {\left (3 \, a^{2} b - b^{3}\right )} c^{2} f^{2} - 2 \, {\left (3 \, a b^{2} d^{2} f - {\left (3 \, a^{2} b - b^{3}\right )} c d f^{2}\right )} x\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 ({\left (3 \, a^{2} b - b^{3}\right )} d^{2} f^{2} x^{2} - 6 \, a b^{2} c d f + b^{3} d^{2} + {\left (3 \, a^{2} b - b^{3}\right )} c^{2} f^{2} - 2 \, {\left (3 \, a b^{2} d^{2} f - {\left (3 \, a^{2} b - b^{3}\right )} c d f^{2}\right )} x\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 \, {\left (3 \, a b^{2} d^{2} f^{2} x^{2} + 3 \, a b^{2} c^{2} f^{2} - b^{3} c d f + {\left (6 \, a b^{2} c d f^{2} - b^{3} d^{2} f\right )} x\right )} \tan \left (f x + e\right )}{12 \, f^{3}} \]
1/12*(4*(a^3 - 3*a*b^2)*d^2*f^3*x^3 - 3*(3*a^2*b - b^3)*d^2*polylog(3, (ta n(f*x + e)^2 + 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 3*(3*a^2*b - b^3)*d^2*polylog(3, (tan(f*x + e)^2 - 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^ 2 + 1)) + 6*(b^3*d^2*f^2 + 2*(a^3 - 3*a*b^2)*c*d*f^3)*x^2 + 6*(b^3*d^2*f^2 *x^2 + 2*b^3*c*d*f^2*x + b^3*c^2*f^2)*tan(f*x + e)^2 + 12*(b^3*c*d*f^2 + ( a^3 - 3*a*b^2)*c^2*f^3)*x - 6*(-3*I*a*b^2*d^2 + I*(3*a^2*b - b^3)*d^2*f*x + I*(3*a^2*b - b^3)*c*d*f)*dilog(2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) - 6*(3*I*a*b^2*d^2 - I*(3*a^2*b - b^3)*d^2*f*x - I*(3*a^2*b - b^3) *c*d*f)*dilog(2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) - 6*((3*a^ 2*b - b^3)*d^2*f^2*x^2 - 6*a*b^2*c*d*f + b^3*d^2 + (3*a^2*b - b^3)*c^2*f^2 - 2*(3*a*b^2*d^2*f - (3*a^2*b - b^3)*c*d*f^2)*x)*log(-2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*((3*a^2*b - b^3)*d^2*f^2*x^2 - 6*a*b^2*c*d*f + b^3*d^2 + (3*a^2*b - b^3)*c^2*f^2 - 2*(3*a*b^2*d^2*f - (3*a^2*b - b^3)* c*d*f^2)*x)*log(-2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) + 12*(3*a*b ^2*d^2*f^2*x^2 + 3*a*b^2*c^2*f^2 - b^3*c*d*f + (6*a*b^2*c*d*f^2 - b^3*d^2* f)*x)*tan(f*x + e))/f^3
\[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{3} \left (c + d x\right )^{2}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3407 vs. \(2 (393) = 786\).
Time = 1.97 (sec) , antiderivative size = 3407, normalized size of antiderivative = 7.81 \[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx=\text {Too large to display} \]
1/3*(3*(f*x + e)*a^3*c^2 + (f*x + e)^3*a^3*d^2/f^2 - 3*(f*x + e)^2*a^3*d^2 *e/f^2 + 3*(f*x + e)*a^3*d^2*e^2/f^2 + 3*(f*x + e)^2*a^3*c*d/f - 6*(f*x + e)*a^3*c*d*e/f + 9*a^2*b*c^2*log(sec(f*x + e)) + 9*a^2*b*d^2*e^2*log(sec(f *x + e))/f^2 - 18*a^2*b*c*d*e*log(sec(f*x + e))/f + 3*(36*a*b^2*d^2*e^2 + 36*a*b^2*c^2*f^2 + 2*(3*a^2*b + 3*I*a*b^2 - b^3)*(f*x + e)^3*d^2 + 12*b^3* d^2*e - 6*((3*a^2*b + 3*I*a*b^2 - b^3)*d^2*e - (3*a^2*b + 3*I*a*b^2 - b^3) *c*d*f)*(f*x + e)^2 - 6*((-3*I*a*b^2 + b^3)*d^2*e^2 + 2*(3*I*a*b^2 - b^3)* c*d*e*f + (-3*I*a*b^2 + b^3)*c^2*f^2)*(f*x + e) - 12*(6*a*b^2*c*d*e + b^3* c*d)*f + 6*(b^3*d^2*e^2 + b^3*c^2*f^2 - 6*a*b^2*d^2*e - (3*a^2*b - b^3)*(f *x + e)^2*d^2 - b^3*d^2 + 2*(3*a*b^2*d^2 + (3*a^2*b - b^3)*d^2*e - (3*a^2* b - b^3)*c*d*f)*(f*x + e) - 2*(b^3*c*d*e - 3*a*b^2*c*d)*f + (b^3*d^2*e^2 + b^3*c^2*f^2 - 6*a*b^2*d^2*e - (3*a^2*b - b^3)*(f*x + e)^2*d^2 - b^3*d^2 + 2*(3*a*b^2*d^2 + (3*a^2*b - b^3)*d^2*e - (3*a^2*b - b^3)*c*d*f)*(f*x + e) - 2*(b^3*c*d*e - 3*a*b^2*c*d)*f)*cos(4*f*x + 4*e) + 2*(b^3*d^2*e^2 + b^3* c^2*f^2 - 6*a*b^2*d^2*e - (3*a^2*b - b^3)*(f*x + e)^2*d^2 - b^3*d^2 + 2*(3 *a*b^2*d^2 + (3*a^2*b - b^3)*d^2*e - (3*a^2*b - b^3)*c*d*f)*(f*x + e) - 2* (b^3*c*d*e - 3*a*b^2*c*d)*f)*cos(2*f*x + 2*e) - (-I*b^3*d^2*e^2 - I*b^3*c^ 2*f^2 + 6*I*a*b^2*d^2*e + (3*I*a^2*b - I*b^3)*(f*x + e)^2*d^2 + I*b^3*d^2 + 2*(-3*I*a*b^2*d^2 + (-3*I*a^2*b + I*b^3)*d^2*e + (3*I*a^2*b - I*b^3)*c*d *f)*(f*x + e) + 2*(I*b^3*c*d*e - 3*I*a*b^2*c*d)*f)*sin(4*f*x + 4*e) - 2...
\[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx=\int { {\left (d x + c\right )}^{2} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (c+d x)^2 (a+b \tan (e+f x))^3 \, dx=\int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3\,{\left (c+d\,x\right )}^2 \,d x \]