Integrand size = 20, antiderivative size = 144 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=-\frac {b \cos (2 a+2 b x)}{6 d^2 (c+d x)^2}-\frac {2 b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {\sin (2 a+2 b x)}{6 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{3 d^3 (c+d x)}+\frac {2 b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4} \]
-2/3*b^3*Ci(2*b*c/d+2*b*x)*cos(2*a-2*b*c/d)/d^4-1/6*b*cos(2*b*x+2*a)/d^2/( d*x+c)^2+2/3*b^3*Si(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^4-1/6*sin(2*b*x+2*a) /d/(d*x+c)^3+1/3*b^2*sin(2*b*x+2*a)/d^3/(d*x+c)
Time = 0.62 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.14 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=\frac {-d \cos (2 b x) \left (b d (c+d x) \cos (2 a)+\left (d^2-2 b^2 (c+d x)^2\right ) \sin (2 a)\right )+d \left (\left (-d^2+2 b^2 (c+d x)^2\right ) \cos (2 a)+b d (c+d x) \sin (2 a)\right ) \sin (2 b x)-4 b^3 (c+d x)^3 \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )-\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )\right )}{6 d^4 (c+d x)^3} \]
(-(d*Cos[2*b*x]*(b*d*(c + d*x)*Cos[2*a] + (d^2 - 2*b^2*(c + d*x)^2)*Sin[2* a])) + d*((-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*a] + b*d*(c + d*x)*Sin[2*a])*Si n[2*b*x] - 4*b^3*(c + d*x)^3*(Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d *x))/d] - Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d]))/(6*d^4*(c + d*x)^3)
Time = 0.79 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4906, 27, 3042, 3778, 3042, 3778, 25, 3042, 3778, 3042, 3784, 3042, 3780, 3783}
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.
| \(\displaystyle \int \frac {\sin (a+b x) \cos (a+b x)}{(c+d x)^4} \, dx\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \int \frac {\sin (2 a+2 b x)}{2 (c+d x)^4}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^4}dx\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \int \frac {\cos (2 a+2 b x)}{(c+d x)^3}dx}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \int \frac {\sin \left (2 a+2 b x+\frac {\pi }{2}\right )}{(c+d x)^3}dx}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (\frac {b \int -\frac {\sin (2 a+2 b x)}{(c+d x)^2}dx}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \int \frac {\sin (2 a+2 b x)}{(c+d x)^2}dx}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \int \frac {\sin (2 a+2 b x)}{(c+d x)^2}dx}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \left (\frac {2 b \int \frac {\cos (2 a+2 b x)}{c+d x}dx}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \left (\frac {2 b \int \frac {\sin \left (2 a+2 b x+\frac {\pi }{2}\right )}{c+d x}dx}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3784 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \left (\frac {2 b \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \left (\frac {2 b \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3780 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \left (\frac {2 b \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
| \(\Big \downarrow \) 3783 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (-\frac {b \left (\frac {2 b \left (\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d}-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\cos (2 a+2 b x)}{2 d (c+d x)^2}\right )}{3 d}-\frac {\sin (2 a+2 b x)}{3 d (c+d x)^3}\right )\) |
(-1/3*Sin[2*a + 2*b*x]/(d*(c + d*x)^3) + (2*b*(-1/2*Cos[2*a + 2*b*x]/(d*(c + d*x)^2) - (b*(-(Sin[2*a + 2*b*x]/(d*(c + d*x))) + (2*b*((Cos[2*a - (2*b *c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/d - (Sin[2*a - (2*b*c)/d]*SinIntegr al[(2*b*c)/d + 2*b*x])/d))/d))/d))/(3*d))/2
3.1.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* f)/d] Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] && NeQ[d*e - c*f, 0]
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 = 0.77 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {b^{3} \left (-\frac {2 \sin \left (2 x b +2 a \right )}{3 \left (-a d +c b +d \left (x b +a \right )\right )^{3} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{3 \left (-a d +c b +d \left (x b +a \right )\right )^{2} d}-\frac {2 \left (-\frac {2 \sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {-\frac {4 \,\operatorname {Si}\left (-2 x b -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}}{d}\right )}{3 d}}{d}\right )}{4}\) | \(200\) |
| default | \(\frac {b^{3} \left (-\frac {2 \sin \left (2 x b +2 a \right )}{3 \left (-a d +c b +d \left (x b +a \right )\right )^{3} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{3 \left (-a d +c b +d \left (x b +a \right )\right )^{2} d}-\frac {2 \left (-\frac {2 \sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {-\frac {4 \,\operatorname {Si}\left (-2 x b -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}}{d}\right )}{3 d}}{d}\right )}{4}\) | \(200\) |
| risch | \(\frac {b^{3} {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{3 d^{4}}+\frac {b^{3} {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{3 d^{4}}+\frac {i \left (2 i b^{4} d^{5} x^{4}+8 i b^{4} c \,d^{4} x^{3}+12 i b^{4} c^{2} d^{3} x^{2}+8 i b^{4} c^{3} d^{2} x +2 i b^{4} c^{4} d \right ) \cos \left (2 x b +2 a \right )}{12 d^{3} \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 b^{3} c^{2} d x +b^{3} c^{3}\right ) \left (d x +c \right )^{3}}-\frac {\left (-4 b^{5} d^{5} x^{5}-20 b^{5} c \,d^{4} x^{4}-40 b^{5} c^{2} d^{3} x^{3}-40 b^{5} c^{3} d^{2} x^{2}-20 b^{5} c^{4} d x +2 b^{3} d^{5} x^{3}-4 b^{5} c^{5}+6 b^{3} c \,d^{4} x^{2}+6 b^{3} c^{2} d^{3} x +2 b^{3} c^{3} d^{2}\right ) \sin \left (2 x b +2 a \right )}{12 d^{3} \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 b^{3} c^{2} d x +b^{3} c^{3}\right ) \left (d x +c \right )^{3}}\) | \(409\) |
1/4*b^3*(-2/3*sin(2*b*x+2*a)/(-a*d+c*b+d*(b*x+a))^3/d+2/3*(-cos(2*b*x+2*a) /(-a*d+c*b+d*(b*x+a))^2/d-(-2*sin(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)*sin(2*(-a*d+b*c)/d)/d+2*Ci(2*x*b+2*a+2*(-a*d +b*c)/d)*cos(2*(-a*d+b*c)/d)/d)/d)/d)/d)
Time = 0.24 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.83 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=\frac {b d^{3} x + b c d^{2} - 2 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 4 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{6 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]
1/6*(b*d^3*x + b*c*d^2 - 2*(b*d^3*x + b*c*d^2)*cos(b*x + a)^2 - 4*(b^3*d^3 *x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos(-2*(b*c - a*d)/d)*co s_integral(2*(b*d*x + b*c)/d) + 2*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c ^2*d - d^3)*cos(b*x + a)*sin(b*x + a) + 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*sin(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b* c)/d))/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^4)
\[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=\int \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \]
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.74 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=-\frac {b^{4} {\left (-i \, E_{4}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{4}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (E_{4}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{4}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} b} \]
-1/4*(b^4*(-I*exp_integral_e(4, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + I* exp_integral_e(4, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-2*(b*c - a* d)/d) + b^4*(exp_integral_e(4, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp _integral_e(4, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-2*(b*c - a*d)/ d))/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + (b*x + a)^3*d^4 - a^3* d^4 + 3*(b*c*d^3 - a*d^4)*(b*x + a)^2 + 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2 *d^4)*(b*x + a))*b)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.58 (sec) , antiderivative size = 7592, normalized size of antiderivative = 52.72 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=\text {Too large to display} \]
-1/6*(2*b^3*d^3*x^3*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*ta n(a)^2*tan(b*c/d)^2 + 2*b^3*d^3*x^3*real_part(cos_integral(-2*b*x - 2*b*c/ d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 4*b^3*d^3*x^3*imag_part(cos_integra l(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 4*b^3*d^3*x^3*imag_pa rt(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 8*b^3* d^3*x^3*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 4 *b^3*d^3*x^3*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*ta n(b*c/d)^2 - 4*b^3*d^3*x^3*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b *x)^2*tan(a)*tan(b*c/d)^2 + 8*b^3*d^3*x^3*sin_integral(2*(b*d*x + b*c)/d)* tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 6*b^3*c*d^2*x^2*real_part(cos_integral(2* b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + 6*b^3*c*d^2*x^2*real_pa rt(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 2*b^ 3*d^3*x^3*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - 2 *b^3*d^3*x^3*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 + 8*b^3*d^3*x^3*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a )*tan(b*c/d) + 8*b^3*d^3*x^3*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan (b*x)^2*tan(a)*tan(b*c/d) - 12*b^3*c*d^2*x^2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 12*b^3*c*d^2*x^2*imag_part(co s_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 24*b^3*c*d^ 2*x^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d) - ...
Timed out. \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^4} \, dx=\int \frac {\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \]