Integrand size = 20, antiderivative size = 114 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=-\frac {b \cos (2 a+2 b x)}{2 d^2 (c+d x)}-\frac {b^2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^3}-\frac {\sin (2 a+2 b x)}{4 d (c+d x)^2}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3} \]
-1/2*b*cos(2*b*x+2*a)/d^2/(d*x+c)-b^2*cos(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/d ^3-b^2*Ci(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^3-1/4*sin(2*b*x+2*a)/d/(d*x+c) ^2
Time = 1.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=-\frac {4 b^2 \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )+\frac {d (2 b (c+d x) \cos (2 (a+b x))+d \sin (2 (a+b x)))}{(c+d x)^2}+4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{4 d^3} \]
-1/4*(4*b^2*CosIntegral[(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])/d^3
Time = 0.66 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4906, 27, 3042, 3778, 3042, 3778, 25, 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.
| \(\displaystyle \int \frac {\sin (a+b x) \cos (a+b x)}{(c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \int \frac {\sin (2 a+2 b x)}{2 (c+d x)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3}dx\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \int \frac {\cos (2 a+2 b x)}{(c+d x)^2}dx}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \int \frac {\sin \left (2 a+2 b x+\frac {\pi }{2}\right )}{(c+d x)^2}dx}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (\frac {2 b \int -\frac {\sin (2 a+2 b x)}{c+d x}dx}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (-\frac {2 b \int \frac {\sin (2 a+2 b x)}{c+d x}dx}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (-\frac {2 b \int \frac {\sin (2 a+2 b x)}{c+d x}dx}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3784 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (-\frac {2 b \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (-\frac {2 b \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3780 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (-\frac {2 b \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
| \(\Big \downarrow \) 3783 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (-\frac {2 b \left (\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d}+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )}{d}-\frac {\cos (2 a+2 b x)}{d (c+d x)}\right )}{d}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)^2}\right )\) |
(-1/2*Sin[2*a + 2*b*x]/(d*(c + d*x)^2) + (b*(-(Cos[2*a + 2*b*x]/(d*(c + d* x))) - (2*b*((CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/d + (Co s[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/d))/d))/d)/2
3.1.8.3.1 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 = 0.58 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.42
| method | result | size |
| derivativedivides | \(\frac {b^{2} \left (-\frac {\sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {2 \left (-\frac {2 \,\operatorname {Si}\left (-2 x b -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{4}\) | \(162\) |
| default | \(\frac {b^{2} \left (-\frac {\sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {2 \left (-\frac {2 \,\operatorname {Si}\left (-2 x b -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{4}\) | \(162\) |
| risch | \(\frac {i b^{2} {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{2 d^{3}}-\frac {i b^{2} {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{2 d^{3}}+\frac {i \left (4 i b^{3} d^{3} x^{3}+12 i b^{3} c \,d^{2} x^{2}+12 i b^{3} c^{2} d x +4 i b^{3} c^{3}\right ) \cos \left (2 x b +2 a \right )}{8 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (2 x b +2 a \right )}{8 d \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}\) | \(277\) |
1/4*b^2*(-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)
Time = 0.24 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.64 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=\frac {b d^{2} x - d^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + b c d - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]
1/2*(b*d^2*x - d^2*cos(b*x + a)*sin(b*x + a) + b*c*d - 2*(b*d^2*x + b*c*d) *cos(b*x + a)^2 - 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(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d))/(d^5*x^2 + 2* c*d^4*x + c^2*d^3)
\[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=\int \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \]
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.76 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=-\frac {b^{3} {\left (-i \, E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]
-1/4*(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*(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^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 = 0.46 (sec) , antiderivative size = 5398, normalized size of antiderivative = 47.35 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=\text {Too large to display} \]
-1/2*(b^2*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)^2 - b^2*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)^2 + 2*b^2*d^2*x^2*sin_integral(2*(b*d*x + b *c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + 2*b^2*d^2*x^2*real_part(cos_inte gral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b^2*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*b ^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan( b*c/d)^2 - 2*b^2*d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x )^2*tan(a)*tan(b*c/d)^2 + 2*b^2*c*d*x*imag_part(cos_integral(2*b*x + 2*b*c /d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 2*b^2*c*d*x*imag_part(cos_integral (-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + 4*b^2*c*d*x*sin_int egral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b^2*d^2*x^2*im ag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 + b^2*d^2*x^2*i mag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 - 2*b^2*d^2*x ^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2 + 4*b^2*d^2*x^2*ima g_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) - 4*b^2 *d^2*x^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b *c/d) + 8*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*ta n(b*c/d) + 4*b^2*c*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2 *tan(a)^2*tan(b*c/d) + 4*b^2*c*d*x*real_part(cos_integral(-2*b*x - 2*b*...
Timed out. \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx=\int \frac {\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^3} \,d x \]