\(\int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx\) [20]

Optimal result

Integrand size = 22, antiderivative size = 221 \[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}+\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3} \] Output:

-1/8*cos(b*x+a)/d/(d*x+c)^2+1/8*cos(3*b*x+3*a)/d/(d*x+c)^2-1/8*b^2*cos(a-b 
*c/d)*Ci(b*c/d+b*x)/d^3+9/8*b^2*cos(3*a-3*b*c/d)*Ci(3*b*c/d+3*b*x)/d^3+1/8 
*b*sin(b*x+a)/d^2/(d*x+c)-3/8*b*sin(3*b*x+3*a)/d^2/(d*x+c)+1/8*b^2*sin(a-b 
*c/d)*Si(b*c/d+b*x)/d^3-9/8*b^2*sin(3*a-3*b*c/d)*Si(3*b*c/d+3*b*x)/d^3
 

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.83 \[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=\frac {-b^2 \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right )+9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b (c+d x)}{d}\right )+\frac {d (-d \cos (a+b x)+b (c+d x) \sin (a+b x))}{(c+d x)^2}+\frac {d (d \cos (3 (a+b x))-3 b (c+d x) \sin (3 (a+b x)))}{(c+d x)^2}+b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )-9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )}{8 d^3} \] Input:

Integrate[(Cos[a + b*x]*Sin[a + b*x]^2)/(c + d*x)^3,x]
 

Output:

(-(b^2*Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)]) + 9*b^2*Cos[3*a - (3*b*c 
)/d]*CosIntegral[(3*b*(c + d*x))/d] + (d*(-(d*Cos[a + b*x]) + b*(c + d*x)* 
Sin[a + b*x]))/(c + d*x)^2 + (d*(d*Cos[3*(a + b*x)] - 3*b*(c + d*x)*Sin[3* 
(a + b*x)]))/(c + d*x)^2 + b^2*Sin[a - (b*c)/d]*SinIntegral[b*(c/d + x)] - 
 9*b^2*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d])/(8*d^3)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4906, 2009}

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.

\(\displaystyle \int \frac {\sin ^2(a+b x) \cos (a+b x)}{(c+d x)^3} \, dx\)

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {\cos (a+b x)}{4 (c+d x)^3}-\frac {\cos (3 a+3 b x)}{4 (c+d x)^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2}\)

Input:

Int[(Cos[a + b*x]*Sin[a + b*x]^2)/(c + d*x)^3,x]
 

Output:

-1/8*Cos[a + b*x]/(d*(c + d*x)^2) + Cos[3*a + 3*b*x]/(8*d*(c + d*x)^2) - ( 
b^2*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(8*d^3) + (9*b^2*Cos[3*a 
- (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*b*x])/(8*d^3) + (b*Sin[a + b*x])/(8 
*d^2*(c + d*x)) - (3*b*Sin[3*a + 3*b*x])/(8*d^2*(c + d*x)) + (b^2*Sin[a - 
(b*c)/d]*SinIntegral[(b*c)/d + b*x])/(8*d^3) - (9*b^2*Sin[3*a - (3*b*c)/d] 
*SinIntegral[(3*b*c)/d + 3*b*x])/(8*d^3)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

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]
 

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \left (-\frac {\cos \left (b x +a \right )}{2 \left (-a d +c b +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}}{d}}{2 d}\right )}{4}-\frac {b^{3} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (-a d +c b +d \left (b x +a \right )\right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {9 \,\operatorname {Si}\left (-3 b x -3 a -\frac {3 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-3 a d +3 c b}{d}\right )}{d}+\frac {9 \,\operatorname {Ci}\left (3 b x +3 a +\frac {-3 a d +3 c b}{d}\right ) \cos \left (\frac {-3 a d +3 c b}{d}\right )}{d}}{d}\right )}{2 d}\right )}{12}}{b}\)	\(316\)
default	\(\frac {\frac {b^{3} \left (-\frac {\cos \left (b x +a \right )}{2 \left (-a d +c b +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}}{d}}{2 d}\right )}{4}-\frac {b^{3} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (-a d +c b +d \left (b x +a \right )\right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {9 \,\operatorname {Si}\left (-3 b x -3 a -\frac {3 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-3 a d +3 c b}{d}\right )}{d}+\frac {9 \,\operatorname {Ci}\left (3 b x +3 a +\frac {-3 a d +3 c b}{d}\right ) \cos \left (\frac {-3 a d +3 c b}{d}\right )}{d}}{d}\right )}{2 d}\right )}{12}}{b}\)	\(316\)
risch	\(-\frac {9 b^{2} {\mathrm e}^{-\frac {3 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (3 i b x +3 i a -\frac {3 i \left (a d -c b \right )}{d}\right )}{16 d^{3}}+\frac {b^{2} {\mathrm e}^{-\frac {i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (i b x +i a -\frac {i \left (a d -c b \right )}{d}\right )}{16 d^{3}}+\frac {b^{2} {\mathrm e}^{\frac {i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-i b x -i a -\frac {-i a d +i c b}{d}\right )}{16 d^{3}}-\frac {9 b^{2} {\mathrm e}^{\frac {3 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-3 i b x -3 i a -\frac {3 \left (-i a d +i c b \right )}{d}\right )}{16 d^{3}}+\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (b x +a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {i \left (2 i b^{3} d^{3} x^{3}+6 i b^{3} c \,d^{2} x^{2}+6 i b^{3} c^{2} d x +2 i b^{3} c^{3}\right ) \sin \left (b x +a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (3 b x +3 a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}+\frac {i \left (6 i b^{3} d^{3} x^{3}+18 i b^{3} c \,d^{2} x^{2}+18 i b^{3} c^{2} d x +6 i b^{3} c^{3}\right ) \sin \left (3 b x +3 a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}\)	\(548\)

Input:

int(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

1/b*(1/4*b^3*(-1/2*cos(b*x+a)/(-a*d+c*b+d*(b*x+a))^2/d-1/2*(-sin(b*x+a)/(- 
a*d+c*b+d*(b*x+a))/d+(-Si(-b*x-a-(-a*d+b*c)/d)*sin((-a*d+b*c)/d)/d+Ci(b*x+ 
a+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d)/d)/d)-1/12*b^3*(-3/2*cos(3*b*x+3*a)/( 
-a*d+c*b+d*(b*x+a))^2/d-3/2*(-3*sin(3*b*x+3*a)/(-a*d+c*b+d*(b*x+a))/d+3*(- 
3*Si(-3*b*x-3*a-3*(-a*d+b*c)/d)*sin(3*(-a*d+b*c)/d)/d+3*Ci(3*b*x+3*a+3*(-a 
*d+b*c)/d)*cos(3*(-a*d+b*c)/d)/d)/d)/d))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.41 \[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=\frac {4 \, d^{2} \cos \left (b x + a\right )^{3} - 4 \, d^{2} \cos \left (b x + a\right ) + 9 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) - 9 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 4 \, {\left (b d^{2} x + b c d - 3 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{8 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

integrate(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x, algorithm="fricas")
 

Output:

1/8*(4*d^2*cos(b*x + a)^3 - 4*d^2*cos(b*x + a) + 9*(b^2*d^2*x^2 + 2*b^2*c* 
d*x + b^2*c^2)*cos(-3*(b*c - a*d)/d)*cos_integral(3*(b*d*x + b*c)/d) - (b^ 
2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-(b*c - a*d)/d)*cos_integral((b*d*x 
 + b*c)/d) - 9*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*sin(-3*(b*c - a*d)/d) 
*sin_integral(3*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*s 
in(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) + 4*(b*d^2*x + b*c*d - 3* 
(b*d^2*x + b*c*d)*cos(b*x + a)^2)*sin(b*x + a))/(d^5*x^2 + 2*c*d^4*x + c^2 
*d^3)
 

Sympy [F]

\[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=\int \frac {\sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \] Input:

integrate(cos(b*x+a)*sin(b*x+a)**2/(d*x+c)**3,x)
 

Output:

Integral(sin(a + b*x)**2*cos(a + b*x)/(c + d*x)**3, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.54 \[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=-\frac {b^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - b^{3} {\left (E_{3}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - b^{3} {\left (i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{3} {\left (i \, E_{3}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{3}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \] Input:

integrate(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x, algorithm="maxima")
 

Output:

-1/8*(b^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_inte 
gral_e(3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) - b^3*( 
exp_integral_e(3, 3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_integral_e(3 
, -3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-3*(b*c - a*d)/d) - b^3*(I*e 
xp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_integral_e(3, 
-(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b^3*(I*exp_inte 
gral_e(3, 3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) - I*exp_integral_e(3, -3*( 
-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-3*(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)
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 3.15 (sec) , antiderivative size = 114422, normalized size of antiderivative = 517.75 \[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x, algorithm="giac")
 

Output:

1/16*(9*b^2*d^2*x^2*real_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^ 
2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) 
^2 - b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1 
/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 - b^ 
2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x 
)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 9*b^2*d^ 
2*x^2*real_part(cos_integral(-3*b*x - 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x 
)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^ 
2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*t 
an(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) - 2*b^2*d^2*x^2*i 
mag_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2 
*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) + 4*b^2*d^2*x^2*sin_int 
egral((b*d*x + b*c)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2* 
a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) - 18*b^2*d^2*x^2*imag_part(cos_integr 
al(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a) 
^2*tan(3/2*b*c/d)*tan(1/2*b*c/d)^2 + 18*b^2*d^2*x^2*imag_part(cos_integral 
(-3*b*x - 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^ 
2*tan(3/2*b*c/d)*tan(1/2*b*c/d)^2 - 36*b^2*d^2*x^2*sin_integral(3*(b*d*x + 
 b*c)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b 
*c/d)*tan(1/2*b*c/d)^2 - 2*b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=\int \frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^3} \,d x \] Input:

int((cos(a + b*x)*sin(a + b*x)^2)/(c + d*x)^3,x)
 

Output:

int((cos(a + b*x)*sin(a + b*x)^2)/(c + d*x)^3, x)
 

Reduce [F]

\[ \int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx=\int \frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \] Input:

int(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x)
 

Output:

int((cos(a + b*x)*sin(a + b*x)**2)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d* 
*3*x**3),x)