Integrand size = 20, antiderivative size = 65 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {\operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{2 d}+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d} \] Output:
1/2*Ci(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d+1/2*cos(2*a-2*b*c/d)*Si(2*b*c/d+2 *b*x)/d
Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {\operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )+\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d} \] Input:
Integrate[(Cos[a + b*x]*Sin[a + b*x])/(c + d*x),x]
Output:
(CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d] + Cos[2*a - (2*b*c)/d ]*SinIntegral[(2*b*c)/d + 2*b*x])/(2*d)
Time = 0.47 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4906, 27, 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.
\(\displaystyle \int \frac {\sin (a+b x) \cos (a+b x)}{c+d x} \, dx\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \int \frac {\sin (2 a+2 b x)}{2 (c+d x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{c+d x}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{c+d x}dx\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \frac {1}{2} \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x}dx\right )\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {1}{2} \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{c+d x}dx+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {1}{2} \left (\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d}+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\right )\) |
Input:
Int[(Cos[a + b*x]*Sin[a + b*x])/(c + d*x),x]
Output:
((CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/d + (Cos[2*a - (2*b *c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/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[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 = 0.41 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {-\operatorname {Si}\left (-2 b x -2 a +\frac {2 a d -2 c b}{d}\right ) \cos \left (\frac {2 a d -2 c b}{d}\right )+\operatorname {Ci}\left (2 b x +2 a -\frac {2 \left (a d -c b \right )}{d}\right ) \sin \left (\frac {2 a d -2 c b}{d}\right )}{2 d}\) | \(82\) |
default | \(\frac {-\operatorname {Si}\left (-2 b x -2 a +\frac {2 a d -2 c b}{d}\right ) \cos \left (\frac {2 a d -2 c b}{d}\right )+\operatorname {Ci}\left (2 b x +2 a -\frac {2 \left (a d -c b \right )}{d}\right ) \sin \left (\frac {2 a d -2 c b}{d}\right )}{2 d}\) | \(82\) |
risch | \(-\frac {i {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{4 d}+\frac {i {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{4 d}\) | \(98\) |
Input:
int(cos(b*x+a)*sin(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/2/d*(-Si(-2*b*x-2*a+2*(a*d-b*c)/d)*cos(2*(a*d-b*c)/d)+Ci(2*b*x+2*a-2*(a* d-b*c)/d)*sin(2*(a*d-b*c)/d))
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {\operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{2 \, d} \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c),x, algorithm="fricas")
Output:
1/2*(cos_integral(2*(b*d*x + b*c)/d)*sin(-2*(b*c - a*d)/d) + cos(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d))/d
\[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=\int \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{c + d x}\, dx \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c),x)
Output:
Integral(sin(a + b*x)*cos(a + b*x)/(c + d*x), x)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.20 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=-\frac {b {\left (-i \, E_{1}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{1}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b {\left (E_{1}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{1}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, b d} \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c),x, algorithm="maxima")
Output:
-1/4*(b*(-I*exp_integral_e(1, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + I*ex p_integral_e(1, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-2*(b*c - a*d) /d) + b*(exp_integral_e(1, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_int egral_e(1, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-2*(b*c - a*d)/d))/ (b*d)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 569, normalized size of antiderivative = 8.75 \[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx =\text {Too large to display} \] Input:
integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c),x, algorithm="giac")
Output:
1/4*(imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 - imag _part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 + 2*sin_integr al(2*(b*d*x + b*c)/d)*tan(a)^2*tan(b*c/d)^2 + 2*real_part(cos_integral(2*b *x + 2*b*c/d))*tan(a)^2*tan(b*c/d) + 2*real_part(cos_integral(-2*b*x - 2*b *c/d))*tan(a)^2*tan(b*c/d) - 2*real_part(cos_integral(2*b*x + 2*b*c/d))*ta n(a)*tan(b*c/d)^2 - 2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan (b*c/d)^2 - imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2 + imag_part( cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2 - 2*sin_integral(2*(b*d*x + b*c)/ d)*tan(a)^2 + 4*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d) - 4*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d) + 8*sin_i ntegral(2*(b*d*x + b*c)/d)*tan(a)*tan(b*c/d) - imag_part(cos_integral(2*b* x + 2*b*c/d))*tan(b*c/d)^2 + imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan (b*c/d)^2 - 2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*c/d)^2 + 2*real_part(c os_integral(2*b*x + 2*b*c/d))*tan(a) + 2*real_part(cos_integral(-2*b*x - 2 *b*c/d))*tan(a) - 2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d) - 2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d) + imag_part(cos_int egral(2*b*x + 2*b*c/d)) - imag_part(cos_integral(-2*b*x - 2*b*c/d)) + 2*si n_integral(2*(b*d*x + b*c)/d))/(d*tan(a)^2*tan(b*c/d)^2 + d*tan(a)^2 + d*t an(b*c/d)^2 + d)
Timed out. \[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=\int \frac {\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )}{c+d\,x} \,d x \] Input:
int((cos(a + b*x)*sin(a + b*x))/(c + d*x),x)
Output:
int((cos(a + b*x)*sin(a + b*x))/(c + d*x), x)
\[ \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx=\int \frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{d x +c}d x \] Input:
int(cos(b*x+a)*sin(b*x+a)/(d*x+c),x)
Output:
int((cos(a + b*x)*sin(a + b*x))/(c + d*x),x)