Integrand size = 20, antiderivative size = 85 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)}-\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2} \] Output:
b*cos(2*a-2*b*c/d)*Ci(2*b*c/d+2*b*x)/d^2-1/2*sin(2*b*x+2*a)/d/(d*x+c)-b*si n(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/d^2
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\frac {2 b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )-\frac {d \sin (2 (a+b x))}{c+d x}-2 b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{2 d^2} \] Input:
Integrate[(Cos[a + b*x]*Sin[a + b*x])/(c + d*x)^2,x]
Output:
(2*b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d*x))/d] - (d*Sin[2*(a + b *x)])/(c + d*x) - 2*b*Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d]) /(2*d^2)
Time = 0.57 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {4906, 27, 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.
| \(\displaystyle \int \frac {\sin (a+b x) \cos (a+b x)}{(c+d x)^2} \, dx\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \int \frac {\sin (2 a+2 b x)}{2 (c+d x)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^2}dx\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \int \frac {\cos (2 a+2 b x)}{c+d x}dx}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \int \frac {\sin \left (2 a+2 b x+\frac {\pi }{2}\right )}{c+d x}dx}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )\) |
| \(\Big \downarrow \) 3784 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx-\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )\) |
| \(\Big \downarrow \) 3780 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )\) |
| \(\Big \downarrow \) 3783 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 b \left (\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d}-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )}{d}-\frac {\sin (2 a+2 b x)}{d (c+d x)}\right )\) |
Input:
Int[(Cos[a + b*x]*Sin[a + b*x])/(c + d*x)^2,x]
Output:
(-(Sin[2*a + 2*b*x]/(d*(c + d*x))) + (2*b*((Cos[2*a - (2*b*c)/d]*CosIntegr al[(2*b*c)/d + 2*b*x])/d - (Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2 *b*x])/d))/d)/2
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 = 0.54 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.45
| method | result | size |
| derivativedivides | \(\frac {\frac {\sin \left (2 b x +2 a \right ) b^{2}}{2 d \left (a d -c b -d \left (b x +a \right )\right )}+\frac {b^{2} \left (\operatorname {Si}\left (-2 b x -2 a +\frac {2 a d -2 c b}{d}\right ) \sin \left (\frac {2 a d -2 c b}{d}\right )+\operatorname {Ci}\left (2 b x +2 a -\frac {2 \left (a d -c b \right )}{d}\right ) \cos \left (\frac {2 a d -2 c b}{d}\right )\right )}{d^{2}}}{b}\) | \(123\) |
| default | \(\frac {\frac {\sin \left (2 b x +2 a \right ) b^{2}}{2 d \left (a d -c b -d \left (b x +a \right )\right )}+\frac {b^{2} \left (\operatorname {Si}\left (-2 b x -2 a +\frac {2 a d -2 c b}{d}\right ) \sin \left (\frac {2 a d -2 c b}{d}\right )+\operatorname {Ci}\left (2 b x +2 a -\frac {2 \left (a d -c b \right )}{d}\right ) \cos \left (\frac {2 a d -2 c b}{d}\right )\right )}{d^{2}}}{b}\) | \(123\) |
| risch | \(-\frac {b \,{\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{2 d^{2}}-\frac {b \,{\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{2 d^{2}}-\frac {\left (-2 b d x -2 c b \right ) \sin \left (2 b x +2 a \right )}{4 d \left (-b d x -c b \right ) \left (d x +c \right )}\) | \(141\) |
Input:
int(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(1/2*sin(2*b*x+2*a)*b^2/d/(a*d-c*b-d*(b*x+a))+b^2/d^2*(Si(-2*b*x-2*a+2 *(a*d-b*c)/d)*sin(2*(a*d-b*c)/d)+Ci(2*b*x+2*a-2*(a*d-b*c)/d)*cos(2*(a*d-b* c)/d)))
Time = 0.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\frac {{\left (b d x + b c\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - {\left (b d x + b c\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{d^{3} x + c d^{2}} \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x, algorithm="fricas")
Output:
((b*d*x + b*c)*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*d*x + b*c)/d) - d*c os(b*x + a)*sin(b*x + a) - (b*d*x + b*c)*sin(-2*(b*c - a*d)/d)*sin_integra l(2*(b*d*x + b*c)/d))/(d^3*x + c*d^2)
\[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\int \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)**2,x)
Output:
Integral(sin(a + b*x)*cos(a + b*x)/(c + d*x)**2, x)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.95 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=-\frac {b^{2} {\left (-i \, E_{2}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{2}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{2}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x, algorithm="maxima")
Output:
-1/4*(b^2*(-I*exp_integral_e(2, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + I* exp_integral_e(2, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-2*(b*c - a* d)/d) + b^2*(exp_integral_e(2, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp _integral_e(2, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-2*(b*c - a*d)/ d))/((b*c*d + (b*x + a)*d^2 - a*d^2)*b)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.38 (sec) , antiderivative size = 2870, normalized size of antiderivative = 33.76 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x, algorithm="giac")
Output:
1/2*(b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*ta n(b*c/d)^2 + b*d*x*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*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*b*d*x*imag_part(cos_integral(-2*b*x - 2*b* c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 4*b*d*x*sin_integral(2*(b*d*x + b*c )/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 - 2*b*d*x*imag_part(cos_integr al(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 4*b*d*x*sin_integra l(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + b*c*real_part(cos_in tegral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + b*c*real_part( cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b*d*x*r eal_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - b*d*x*real_p art(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 + 4*b*d*x*real_par t(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) + 4*b*d*x*re al_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) - 2*b *c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan (b*c/d) - 4*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b* c/d) - b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d )^2 - b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*...
Timed out. \[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\int \frac {\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^2} \,d x \] Input:
int((cos(a + b*x)*sin(a + b*x))/(c + d*x)^2,x)
Output:
int((cos(a + b*x)*sin(a + b*x))/(c + d*x)^2, x)
\[ \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\int \frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{d^{2} x^{2}+2 c d x +c^{2}}d x \] Input:
int(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x)
Output:
int((cos(a + b*x)*sin(a + b*x))/(c**2 + 2*c*d*x + d**2*x**2),x)