3.7 \(\int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=85 \[ \frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)} \]

[Out]

b*Ci(2*b*c/d+2*b*x)*cos(2*a-2*b*c/d)/d^2-b*Si(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^2-1/2*sin(2*b*x+2*a)/d/(d*x+c)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4406, 12, 3297, 3303, 3299, 3302} \[ \frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {\sin (2 a+2 b x)}{2 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]*Sin[a + b*x])/(c + d*x)^2,x]

[Out]

(b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/d^2 - Sin[2*a + 2*b*x]/(2*d*(c + d*x)) - (b*Sin[2*a -
(2*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/d^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3297

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] - Dist[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]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
 e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[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]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(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]
&& IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx &=\int \frac {\sin (2 a+2 b x)}{2 (c+d x)^2} \, dx\\ &=\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^2} \, dx\\ &=-\frac {\sin (2 a+2 b x)}{2 d (c+d x)}+\frac {b \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{d}\\ &=-\frac {\sin (2 a+2 b x)}{2 d (c+d x)}+\frac {\left (b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}-\frac {\left (b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}\\ &=\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.31, size = 80, normalized size = 0.94 \[ \frac {2 b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )-2 b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )-\frac {d \sin (2 (a+b x))}{c+d x}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]*Sin[a + b*x])/(c + d*x)^2,x]

[Out]

(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)

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 132, normalized size = 1.55 \[ -\frac {2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/2*(2*d*cos(b*x + a)*sin(b*x + a) + 2*(b*d*x + b*c)*sin(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) -
((b*d*x + b*c)*cos_integral(2*(b*d*x + b*c)/d) + (b*d*x + b*c)*cos_integral(-2*(b*d*x + b*c)/d))*cos(-2*(b*c -
 a*d)/d))/(d^3*x + c*d^2)

________________________________________________________________________________________

giac [C]  time = 0.56, size = 2870, normalized size = 33.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x, algorithm="giac")

[Out]

1/2*(b*d*x*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*real_part(cos_int
egral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 2*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*t
an(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_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2
*tan(a)*tan(b*c/d)^2 + 4*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*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*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*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - b*d*
x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 + 4*b*d*x*real_part(cos_integral(2*b*x + 2*b*c
/d))*tan(b*x)^2*tan(a)*tan(b*c/d) + 4*b*d*x*real_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_int
egral(-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(co
s_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 + 2*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x
)^2*tan(a)*tan(b*c/d)^2 - 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 4*b
*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + b*d*x*real_part(cos_integral(2*b*x + 2*b*c
/d))*tan(a)^2*tan(b*c/d)^2 + b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 - 2*b*d*x*i
mag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a) + 2*b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*
tan(b*x)^2*tan(a) - 4*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a) - b*c*real_part(cos_integral(2*b
*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 + 2*b*d
*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^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(b*c/d) + 4*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(b*c/d) + 4*b*c*real_par
t(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) + 4*b*c*real_part(cos_integral(-2*b*x - 2*b*c/d)
)*tan(b*x)^2*tan(a)*tan(b*c/d) - 2*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d) + 2*b*d*
x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d) - 4*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(
a)^2*tan(b*c/d) - b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 - b*c*real_part(cos_int
egral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 + 2*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan
(b*c/d)^2 - 2*b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d)^2 + 4*b*d*x*sin_integral(2*(b*
d*x + b*c)/d)*tan(a)*tan(b*c/d)^2 + b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 + b*c*r
eal_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 + b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d)
)*tan(b*x)^2 + b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2 - 2*b*c*imag_part(cos_integral(2*b*x
 + 2*b*c/d))*tan(b*x)^2*tan(a) + 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a) - 4*b*c*sin
_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a) - b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2 - b*d
*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2 + 2*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x
)^2*tan(b*c/d) - 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d) + 4*b*c*sin_integral(2*
(b*d*x + b*c)/d)*tan(b*x)^2*tan(b*c/d) + 4*b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d) +
4*b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d) - 2*b*c*imag_part(cos_integral(2*b*x + 2*b
*c/d))*tan(a)^2*tan(b*c/d) + 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d) - 4*b*c*sin_i
ntegral(2*(b*d*x + b*c)/d)*tan(a)^2*tan(b*c/d) - b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d)^2 -
 b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d)^2 + 2*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))
*tan(a)*tan(b*c/d)^2 - 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d)^2 + 4*b*c*sin_integra
l(2*(b*d*x + b*c)/d)*tan(a)*tan(b*c/d)^2 + 2*d*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 2*d*tan(b*x)*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 + b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))
*tan(b*x)^2 - 2*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a) + 2*b*d*x*imag_part(cos_integral(-2*b*x
- 2*b*c/d))*tan(a) - 4*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(a) - b*c*real_part(cos_integral(2*b*x + 2*b*c
/d))*tan(a)^2 - b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2 + 2*b*d*x*imag_part(cos_integral(2*b*x
+ 2*b*c/d))*tan(b*c/d) - 2*b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d) + 4*b*d*x*sin_integral(2
*(b*d*x + b*c)/d)*tan(b*c/d) + 4*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d) + 4*b*c*real_p
art(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d) - b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d
)^2 - b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d)^2 + b*d*x*real_part(cos_integral(2*b*x + 2*b*c/
d)) + b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d)) - 2*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)
+ 2*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a) - 4*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(a) + 2*d*
tan(b*x)^2*tan(a) + 2*d*tan(b*x)*tan(a)^2 + 2*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d) - 2*b*c*
imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d) + 4*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*c/d) - 2*d*
tan(b*x)*tan(b*c/d)^2 - 2*d*tan(a)*tan(b*c/d)^2 + b*c*real_part(cos_integral(2*b*x + 2*b*c/d)) + b*c*real_part
(cos_integral(-2*b*x - 2*b*c/d)) - 2*d*tan(b*x) - 2*d*tan(a))/(d^3*x*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + c*d^2*
tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + d^3*x*tan(b*x)^2*tan(a)^2 + d^3*x*tan(b*x)^2*tan(b*c/d)^2 + d^3*x*tan(a)^2*
tan(b*c/d)^2 + c*d^2*tan(b*x)^2*tan(a)^2 + c*d^2*tan(b*x)^2*tan(b*c/d)^2 + c*d^2*tan(a)^2*tan(b*c/d)^2 + d^3*x
*tan(b*x)^2 + d^3*x*tan(a)^2 + d^3*x*tan(b*c/d)^2 + c*d^2*tan(b*x)^2 + c*d^2*tan(a)^2 + c*d^2*tan(b*c/d)^2 + d
^3*x + c*d^2)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 124, normalized size = 1.46 \[ \frac {b \left (-\frac {2 \sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {4 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}+\frac {4 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}}{d}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x)

[Out]

1/4*b*(-2*sin(2*b*x+2*a)/((b*x+a)*d-d*a+c*b)/d+2*(2*Si(2*b*x+2*a+2*(-a*d+b*c)/d)*sin(2*(-a*d+b*c)/d)/d+2*Ci(2*
b*x+2*a+2*(-a*d+b*c)/d)*cos(2*(-a*d+b*c)/d)/d)/d)

________________________________________________________________________________________

maxima [C]  time = 0.44, size = 164, normalized size = 1.93 \[ -\frac {b^{2} {\left (i \, E_{2}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - i \, E_{2}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{2}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)^2,x, algorithm="maxima")

[Out]

-1/4*(b^2*(I*exp_integral_e(2, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) - I*exp_integral_e(2, -(2*I*b*c + 2*I*
(b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) + b^2*(exp_integral_e(2, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d
)/d) + exp_integral_e(2, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d))*sin(-2*(b*c - a*d)/d))/((b*c*d + (b*x + a)
*d^2 - a*d^2)*b)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(a + b*x)*sin(a + b*x))/(c + d*x)^2,x)

[Out]

int((cos(a + b*x)*sin(a + b*x))/(c + d*x)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)/(d*x+c)**2,x)

[Out]

Integral(sin(a + b*x)*cos(a + b*x)/(c + d*x)**2, x)

________________________________________________________________________________________