Integrand size = 21, antiderivative size = 107 \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {d x}{b}-\frac {i (c+d x)^2}{2 d}+\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {d \cos (a+b x) \sin (a+b x)}{b^2}+\frac {2 (c+d x) \sin ^2(a+b x)}{b} \]
-d*x/b-1/2*I*(d*x+c)^2/d+(d*x+c)*ln(1+exp(2*I*(b*x+a)))/b-1/2*I*d*polylog( 2,-exp(2*I*(b*x+a)))/b^2+d*cos(b*x+a)*sin(b*x+a)/b^2+2*(d*x+c)*sin(b*x+a)^ 2/b
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(107)=214\).
Time = 3.79 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.40 \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {c \log (\cos (a+b x))}{b}+\frac {d \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {d \cos (2 b x) (2 b x \cos (2 a)-\sin (2 a))}{2 b^2}+\frac {d (\cos (2 a)+2 b x \sin (2 a)) \sin (2 b x)}{2 b^2}+\frac {2 c \sin ^2(a+b x)}{b}-\frac {1}{2} d x^2 \tan (a) \]
(c*Log[Cos[a + b*x]])/b + (d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[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*PolyLog[2, E^(( 2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^2*Sqrt[Cs c[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (d*Cos[2*b*x]*(2*b*x*Cos[2*a] - Sin[2*a]) )/(2*b^2) + (d*(Cos[2*a] + 2*b*x*Sin[2*a])*Sin[2*b*x])/(2*b^2) + (2*c*Sin[ a + b*x]^2)/b - (d*x^2*Tan[a])/2
Time = 0.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int (c+d x) \sin (3 a+3 b x) \sec (a+b x) \, dx\) |
\(\Big \downarrow \) 4931 |
\(\displaystyle \int \left (3 (c+d x) \sin (a+b x) \cos (a+b x)-(c+d x) \sin ^2(a+b x) \tan (a+b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {d \sin (a+b x) \cos (a+b x)}{b^2}+\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {2 (c+d x) \sin ^2(a+b x)}{b}-\frac {d x}{b}-\frac {i (c+d x)^2}{2 d}\) |
-((d*x)/b) - ((I/2)*(c + d*x)^2)/d + ((c + d*x)*Log[1 + E^((2*I)*(a + b*x) )])/b - ((I/2)*d*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 + (d*Cos[a + b*x]*S in[a + b*x])/b^2 + (2*(c + d*x)*Sin[a + b*x]^2)/b
3.4.85.3.1 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]
Time = 2.03 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.65
method | result | size |
risch | \(-\frac {i d \,x^{2}}{2}+i c x -\frac {\left (2 d x b +2 c b +i d \right ) {\mathrm e}^{2 i \left (x b +a \right )}}{4 b^{2}}-\frac {\left (2 d x b +2 c b -i d \right ) {\mathrm e}^{-2 i \left (x b +a \right )}}{4 b^{2}}+\frac {c \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b}-\frac {2 c \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}-\frac {2 i d x a}{b}-\frac {i d \,a^{2}}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b}-\frac {i d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{2}}+\frac {2 d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}\) | \(177\) |
-1/2*I*d*x^2+I*c*x-1/4*(2*d*x*b+I*d+2*c*b)/b^2*exp(2*I*(b*x+a))-1/4*(2*d*x *b-I*d+2*c*b)/b^2*exp(-2*I*(b*x+a))+1/b*c*ln(exp(2*I*(b*x+a))+1)-2/b*c*ln( exp(I*(b*x+a)))-2*I/b*d*x*a-I/b^2*d*a^2+1/b*d*ln(exp(2*I*(b*x+a))+1)*x-1/2 *I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2+2/b^2*d*a*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. 340 vs. \(2 (94) = 188\).
Time = 0.28 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.18 \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {2 \, b d x - 4 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{2}} \]
1/2*(2*b*d*x - 4*(b*d*x + b*c)*cos(b*x + a)^2 + 2*d*cos(b*x + a)*sin(b*x + a) + I*d*dilog(I*cos(b*x + a) + sin(b*x + a)) - I*d*dilog(I*cos(b*x + a) - sin(b*x + a)) - I*d*dilog(-I*cos(b*x + a) + sin(b*x + a)) + I*d*dilog(-I *cos(b*x + a) - sin(b*x + a)) + (b*c - a*d)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b*c - a*d)*log(cos(b*x + a) - I*sin(b*x + a) + I) + (b*d*x + a *d)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d*x + a*d)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b*d*x + a*d)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d*x + a*d)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b*c - a*d )*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b*c - a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + I))/b^2
\[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \left (c + d x\right ) \sin {\left (3 a + 3 b x \right )} \sec {\left (a + b x \right )}\, dx \]
\[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \]
-1/2*c*(2*cos(2*b*x + 2*a) - log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + co s(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2))/b - 1/2*(2* b*x*cos(2*b*x + 2*a) + 4*b^2*integrate(x*sin(2*b*x + 2*a)/(cos(2*b*x + 2*a )^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1), x) - sin(2*b*x + 2*a)) *d/b^2
\[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \]
Timed out. \[ \int (c+d x) \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )\,\left (c+d\,x\right )}{\cos \left (a+b\,x\right )} \,d x \]