Integrand size = 23, antiderivative size = 173 \[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {2 c d x}{b}-\frac {d^2 x^2}{b}-\frac {i (c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac {d^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {2 d (c+d x) \cos (a+b x) \sin (a+b x)}{b^2}-\frac {d^2 \sin ^2(a+b x)}{b^3}+\frac {2 (c+d x)^2 \sin ^2(a+b x)}{b} \]
-2*c*d*x/b-d^2*x^2/b-1/3*I*(d*x+c)^3/d+(d*x+c)^2*ln(1+exp(2*I*(b*x+a)))/b- I*d*(d*x+c)*polylog(2,-exp(2*I*(b*x+a)))/b^2+1/2*d^2*polylog(3,-exp(2*I*(b *x+a)))/b^3+2*d*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b^2-d^2*sin(b*x+a)^2/b^3+2*( d*x+c)^2*sin(b*x+a)^2/b
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(516\) vs. \(2(173)=346\).
Time = 6.37 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.98 \[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {i d^2 e^{-i a} \left (2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 b \left (1+e^{2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )\right ) \sec (a)}{12 b^3}+\frac {c^2 \sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a))}{b \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac {c d \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {\cos (2 b x) \left (2 b^2 c^2 \cos (2 a)-d^2 \cos (2 a)+4 b^2 c d x \cos (2 a)+2 b^2 d^2 x^2 \cos (2 a)-2 b c d \sin (2 a)-2 b d^2 x \sin (2 a)\right )}{2 b^3}+\frac {\left (2 b c d \cos (2 a)+2 b d^2 x \cos (2 a)+2 b^2 c^2 \sin (2 a)-d^2 \sin (2 a)+4 b^2 c d x \sin (2 a)+2 b^2 d^2 x^2 \sin (2 a)\right ) \sin (2 b x)}{2 b^3}-\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right ) \tan (a) \]
((I/12)*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)* (a + b*x))]) + 6*b*(1 + E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^3*E ^(I*a)) + (c^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x *Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) + (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[C ot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b *x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*P olyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/ (b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (Cos[2*b*x]*(2*b^2*c^2*Cos[2* a] - d^2*Cos[2*a] + 4*b^2*c*d*x*Cos[2*a] + 2*b^2*d^2*x^2*Cos[2*a] - 2*b*c* d*Sin[2*a] - 2*b*d^2*x*Sin[2*a]))/(2*b^3) + ((2*b*c*d*Cos[2*a] + 2*b*d^2*x *Cos[2*a] + 2*b^2*c^2*Sin[2*a] - d^2*Sin[2*a] + 4*b^2*c*d*x*Sin[2*a] + 2*b ^2*d^2*x^2*Sin[2*a])*Sin[2*b*x])/(2*b^3) - (x*(3*c^2 + 3*c*d*x + d^2*x^2)* Tan[a])/3
Time = 0.50 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4931, 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.
| \(\displaystyle \int (c+d x)^2 \sin (3 a+3 b x) \sec (a+b x) \, dx\) |
| \(\Big \downarrow \) 4931 |
| \(\displaystyle \int \left (3 (c+d x)^2 \sin (a+b x) \cos (a+b x)-(c+d x)^2 \sin ^2(a+b x) \tan (a+b x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d^2 \sin ^2(a+b x)}{b^3}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac {2 d (c+d x) \sin (a+b x) \cos (a+b x)}{b^2}+\frac {(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {2 (c+d x)^2 \sin ^2(a+b x)}{b}-\frac {(c+d x)^2}{b}-\frac {i (c+d x)^3}{3 d}\) |
-((c + d*x)^2/b) - ((I/3)*(c + d*x)^3)/d + ((c + d*x)^2*Log[1 + E^((2*I)*( a + b*x))])/b - (I*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 + (d^ 2*PolyLog[3, -E^((2*I)*(a + b*x))])/(2*b^3) + (2*d*(c + d*x)*Cos[a + b*x]* Sin[a + b*x])/b^2 - (d^2*Sin[a + b*x]^2)/b^3 + (2*(c + d*x)^2*Sin[a + b*x] ^2)/b
3.4.84.3.1 Defintions of rubi rules used
Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int[ExpandTrigExpand[(e + f*x)^m*G[c + d*x] ^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Member Q[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && E qQ[b*c - a*d, 0] && IGtQ[b/d, 1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (161 ) = 322\).
Time = 1.39 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.23
| method | result | size |
| risch | \(i c^{2} x -\frac {i d^{2} x^{3}}{3}-\frac {2 i d c \,a^{2}}{b^{2}}-\frac {4 i d c x a}{b}-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 i b \,d^{2} x +2 b^{2} c^{2}+2 i b c d -d^{2}\right ) {\mathrm e}^{2 i \left (x b +a \right )}}{4 b^{3}}-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x -2 i b \,d^{2} x +2 b^{2} c^{2}-2 i b c d -d^{2}\right ) {\mathrm e}^{-2 i \left (x b +a \right )}}{4 b^{3}}-\frac {2 d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {i c^{3}}{3 d}+\frac {2 c d \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b}-\frac {i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right ) x}{b^{2}}+\frac {4 c d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {i c d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{b^{2}}+\frac {2 i d^{2} a^{2} x}{b^{2}}-i d c \,x^{2}+\frac {4 i d^{2} a^{3}}{3 b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{3}}+\frac {c^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b}-\frac {2 c^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}\) | \(386\) |
I*c^2*x-1/3*I*d^2*x^3-2*I/b^2*d*c*a^2-4*I/b*d*c*x*a-1/4*(2*x^2*d^2*b^2+2*I *b*d^2*x+4*b^2*c*d*x+2*I*b*c*d+2*b^2*c^2-d^2)/b^3*exp(2*I*(b*x+a))-1/4*(2* x^2*d^2*b^2-2*I*b*d^2*x+4*b^2*c*d*x-2*I*b*c*d+2*b^2*c^2-d^2)/b^3*exp(-2*I* (b*x+a))-2/b^3*d^2*a^2*ln(exp(I*(b*x+a)))+1/3*I/d*c^3+2/b*c*d*ln(exp(2*I*( b*x+a))+1)*x-I/b^2*d^2*polylog(2,-exp(2*I*(b*x+a)))*x+4/b^2*c*d*a*ln(exp(I *(b*x+a)))-I/b^2*c*d*polylog(2,-exp(2*I*(b*x+a)))+2*I/b^2*d^2*a^2*x-I*d*c* x^2+4/3*I/b^3*d^2*a^3+1/b*d^2*ln(exp(2*I*(b*x+a))+1)*x^2+1/2*d^2*polylog(3 ,-exp(2*I*(b*x+a)))/b^3+1/b*c^2*ln(exp(2*I*(b*x+a))+1)-2/b*c^2*ln(exp(I*(b *x+a)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 681 vs. \(2 (158) = 316\).
Time = 0.30 (sec) , antiderivative size = 681, normalized size of antiderivative = 3.94 \[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x - 2 \, {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{3}} \]
1/2*(2*b^2*d^2*x^2 + 4*b^2*c*d*x - 2*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2* c^2 - d^2)*cos(b*x + a)^2 + 2*d^2*polylog(3, I*cos(b*x + a) + sin(b*x + a) ) + 2*d^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 2*d^2*polylog(3, -I* cos(b*x + a) + sin(b*x + a)) + 2*d^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a)) + 4*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a) - 2*(-I*b*d^2*x - I* b*c*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*dilo g(I*cos(b*x + a) - sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*dilog(-I*cos(b* x + a) + sin(b*x + a)) - 2*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x + a) + I*sin(b *x + a) + I) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x + a) - I*sin(b* x + a) + I) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I*cos( b*x + a) + sin(b*x + a) + 1) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^ 2*d^2)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b^2*d^2 *x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) - I*sin(b*x + a) + I))/b^3
\[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \left (c + d x\right )^{2} \sin {\left (3 a + 3 b x \right )} \sec {\left (a + b x \right )}\, dx \]
Time = 0.32 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.75 \[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {c^{2} {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right )\right )}}{2 \, b} + \frac {-2 i \, b^{3} d^{2} x^{3} - 6 i \, b^{3} c d x^{2} + 3 \, d^{2} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 6 \, {\left (-i \, b^{2} d^{2} x^{2} - 2 i \, b^{2} c d x\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{6 \, b^{3}} \]
-1/2*c^2*(2*cos(2*b*x + 2*a) - log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2))/b + 1/6*( -2*I*b^3*d^2*x^3 - 6*I*b^3*c*d*x^2 + 3*d^2*polylog(3, -e^(2*I*b*x + 2*I*a) ) - 6*(-I*b^2*d^2*x^2 - 2*I*b^2*c*d*x)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 3*(2*b^2*d^2*x^2 + 4*b^2*c*d*x - d^2)*cos(2*b*x + 2*a) - 6* (I*b*d^2*x + I*b*c*d)*dilog(-e^(2*I*b*x + 2*I*a)) + 3*(b^2*d^2*x^2 + 2*b^2 *c*d*x)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 6*(b*d^2*x + b*c*d)*sin(2*b*x + 2*a))/b^3
\[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )}^{2} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \]
Timed out. \[ \int (c+d x)^2 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )\,{\left (c+d\,x\right )}^2}{\cos \left (a+b\,x\right )} \,d x \]