Integrand size = 22, antiderivative size = 229 \[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}+\frac {b^2 \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{d^3}-\frac {b^2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{2 d^3}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}+\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3} \]
-1/4*b*cos(2*b*x+2*a)/d^2/(d*x+c)+1/4*b*cos(4*b*x+4*a)/d^2/(d*x+c)-1/2*b^2 *cos(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/d^3+b^2*cos(4*a-4*b*c/d)*Si(4*b*c/d+4* b*x)/d^3+b^2*Ci(4*b*c/d+4*b*x)*sin(4*a-4*b*c/d)/d^3-1/2*b^2*Ci(2*b*c/d+2*b *x)*sin(2*a-2*b*c/d)/d^3-1/8*sin(2*b*x+2*a)/d/(d*x+c)^2+1/16*sin(4*b*x+4*a )/d/(d*x+c)^2
Time = 3.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=\frac {16 b^2 \operatorname {CosIntegral}\left (\frac {4 b (c+d x)}{d}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )+\frac {d (4 b (c+d x) \cos (4 (a+b x))+d \sin (4 (a+b x)))}{(c+d x)^2}-2 \left (4 b^2 \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )+\frac {d (2 b (c+d x) \cos (2 (a+b x))+d \sin (2 (a+b x)))}{(c+d x)^2}+4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )\right )+16 b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )}{16 d^3} \]
(16*b^2*CosIntegral[(4*b*(c + d*x))/d]*Sin[4*a - (4*b*c)/d] + (d*(4*b*(c + d*x)*Cos[4*(a + b*x)] + d*Sin[4*(a + b*x)]))/(c + d*x)^2 - 2*(4*b^2*CosIn tegral[(2*b*(c + d*x))/d]*Sin[2*a - (2*b*c)/d] + (d*(2*b*(c + d*x)*Cos[2*( a + b*x)] + d*Sin[2*(a + b*x)]))/(c + d*x)^2 + 4*b^2*Cos[2*a - (2*b*c)/d]* SinIntegral[(2*b*(c + d*x))/d]) + 16*b^2*Cos[4*a - (4*b*c)/d]*SinIntegral[ (4*b*(c + d*x))/d])/(16*d^3)
Time = 0.57 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4906, 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.
| \(\displaystyle \int \frac {\sin ^3(a+b x) \cos (a+b x)}{(c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \int \left (\frac {\sin (2 a+2 b x)}{4 (c+d x)^3}-\frac {\sin (4 a+4 b x)}{8 (c+d x)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}+\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}\) |
-1/4*(b*Cos[2*a + 2*b*x])/(d^2*(c + d*x)) + (b*Cos[4*a + 4*b*x])/(4*d^2*(c + d*x)) + (b^2*CosIntegral[(4*b*c)/d + 4*b*x]*Sin[4*a - (4*b*c)/d])/d^3 - (b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/(2*d^3) - Sin[2 *a + 2*b*x]/(8*d*(c + d*x)^2) + Sin[4*a + 4*b*x]/(16*d*(c + d*x)^2) - (b^2 *Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/(2*d^3) + (b^2*Cos[4 *a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/d^3
3.1.29.3.1 Defintions of rubi rules used
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Time = 1.64 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.44
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \left (-\frac {\sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {2 \left (-\frac {2 \,\operatorname {Si}\left (-2 x b -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{8}-\frac {b^{3} \left (-\frac {2 \sin \left (4 x b +4 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {8 \cos \left (4 x b +4 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {8 \left (-\frac {4 \,\operatorname {Si}\left (-4 x b -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}-\frac {4 \,\operatorname {Ci}\left (4 x b +4 a +\frac {-4 a d +4 c b}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{32}}{b}\) | \(329\) |
| default | \(\frac {\frac {b^{3} \left (-\frac {\sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {2 \left (-\frac {2 \,\operatorname {Si}\left (-2 x b -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{8}-\frac {b^{3} \left (-\frac {2 \sin \left (4 x b +4 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {8 \cos \left (4 x b +4 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {8 \left (-\frac {4 \,\operatorname {Si}\left (-4 x b -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}-\frac {4 \,\operatorname {Ci}\left (4 x b +4 a +\frac {-4 a d +4 c b}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{32}}{b}\) | \(329\) |
| risch | \(\frac {i b^{2} {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{4 d^{3}}-\frac {i b^{2} {\mathrm e}^{-\frac {4 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (4 i b x +4 i a -\frac {4 i \left (a d -c b \right )}{d}\right )}{2 d^{3}}-\frac {i b^{2} {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{4 d^{3}}+\frac {i b^{2} {\mathrm e}^{\frac {4 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-4 i b x -4 i a -\frac {4 \left (-i a d +i c b \right )}{d}\right )}{2 d^{3}}-\frac {i \left (8 i b^{3} d^{3} x^{3}+24 i b^{3} c \,d^{2} x^{2}+24 i b^{3} c^{2} d x +8 i b^{3} c^{3}\right ) \cos \left (4 x b +4 a \right )}{32 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}+\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (4 x b +4 a \right )}{32 d \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}+\frac {i \left (4 i b^{3} d^{3} x^{3}+12 i b^{3} c \,d^{2} x^{2}+12 i b^{3} c^{2} d x +4 i b^{3} c^{3}\right ) \cos \left (2 x b +2 a \right )}{16 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (2 x b +2 a \right )}{16 d \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}\) | \(552\) |
1/b*(1/8*b^3*(-sin(2*b*x+2*a)/(-a*d+c*b+d*(b*x+a))^2/d+(-2*cos(2*b*x+2*a)/ (-a*d+c*b+d*(b*x+a))/d-2*(-2*Si(-2*x*b-2*a-2*(-a*d+b*c)/d)*cos(2*(-a*d+b*c )/d)/d-2*Ci(2*x*b+2*a+2*(-a*d+b*c)/d)*sin(2*(-a*d+b*c)/d)/d)/d)/d)-1/32*b^ 3*(-2*sin(4*b*x+4*a)/(-a*d+c*b+d*(b*x+a))^2/d+2*(-4*cos(4*b*x+4*a)/(-a*d+c *b+d*(b*x+a))/d-4*(-4*Si(-4*x*b-4*a-4*(-a*d+b*c)/d)*cos(4*(-a*d+b*c)/d)/d- 4*Ci(4*x*b+4*a+4*(-a*d+b*c)/d)*sin(4*(-a*d+b*c)/d)/d)/d)/d))
Time = 0.26 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.46 \[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=\frac {4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} + b d^{2} x + b c d - 5 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (d^{2} \cos \left (b x + a\right )^{3} - d^{2} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]
1/2*(4*(b*d^2*x + b*c*d)*cos(b*x + a)^4 + b*d^2*x + b*c*d - 5*(b*d^2*x + b *c*d)*cos(b*x + a)^2 - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral( 2*(b*d*x + b*c)/d)*sin(-2*(b*c - a*d)/d) + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(4*(b*d*x + b*c)/d)*sin(-4*(b*c - a*d)/d) + 2*(b^2*d^ 2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-4*(b*c - a*d)/d)*sin_integral(4*(b*d*x + b*c)/d) - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-2*(b*c - a*d)/d)*s in_integral(2*(b*d*x + b*c)/d) + (d^2*cos(b*x + a)^3 - d^2*cos(b*x + a))*s in(b*x + a))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)
\[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=\int \frac {\sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \]
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.50 \[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=-\frac {2 \, b^{3} {\left (-i \, E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{3} {\left (-i \, E_{3}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{3}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{3} {\left (E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{3} {\left (E_{3}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]
-1/16*(2*b^3*(-I*exp_integral_e(3, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + I*exp_integral_e(3, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-2*(b*c - a*d)/d) - b^3*(-I*exp_integral_e(3, 4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + I*exp_integral_e(3, -4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-4*(b*c - a*d)/d) + 2*b^3*(exp_integral_e(3, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d ) + exp_integral_e(3, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-2*(b*c - a*d)/d) - b^3*(exp_integral_e(3, 4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_integral_e(3, -4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-4*(b*c - a *d)/d))/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*b)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.34 (sec) , antiderivative size = 111694, normalized size of antiderivative = 487.75 \[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=\text {Too large to display} \]
1/8*(4*b^2*d^2*x^2*imag_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*b*x)^2*t an(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 - 2*b^2*d^2*x^2* imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^ 2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 + 2*b^2*d^2*x^2*imag_part(cos_integ ral(-2*b*x - 2*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b *c/d)^2*tan(b*c/d)^2 - 4*b^2*d^2*x^2*imag_part(cos_integral(-4*b*x - 4*b*c /d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d) ^2 + 8*b^2*d^2*x^2*sin_integral(4*(b*d*x + b*c)/d)*tan(2*b*x)^2*tan(b*x)^2 *tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 - 4*b^2*d^2*x^2*sin_integ ral(2*(b*d*x + b*c)/d)*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b *c/d)^2*tan(b*c/d)^2 - 4*b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/ d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d) - 4*b^2*d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(2*b*x)^2*tan (b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d) + 8*b^2*d^2*x^2*real _part(cos_integral(4*b*x + 4*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*ta n(a)^2*tan(2*b*c/d)*tan(b*c/d)^2 + 8*b^2*d^2*x^2*real_part(cos_integral(-4 *b*x - 4*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)* tan(b*c/d)^2 + 4*b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan( 2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)*tan(2*b*c/d)^2*tan(b*c/d)^2 + 4*b^2* d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(2*b*x)^2*tan(b*x)...
Timed out. \[ \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx=\int \frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^3} \,d x \]