\(\int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx\) [142]

Optimal result

Integrand size = 22, antiderivative size = 179 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^2}+\frac {b \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{2 d^2}-\frac {\sin (2 a+2 b x)}{4 d (c+d x)}-\frac {\sin (4 a+4 b x)}{8 d (c+d x)}-\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^2}-\frac {b \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{2 d^2} \] Output:

1/2*b*cos(2*a-2*b*c/d)*Ci(2*b*c/d+2*b*x)/d^2+1/2*b*cos(4*a-4*b*c/d)*Ci(4*b 
*c/d+4*b*x)/d^2-1/4*sin(2*b*x+2*a)/d/(d*x+c)-1/8*sin(4*b*x+4*a)/d/(d*x+c)- 
1/2*b*sin(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/d^2-1/2*b*sin(4*a-4*b*c/d)*Si(4*b 
*c/d+4*b*x)/d^2
 

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=-\frac {-4 b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )-4 b \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b (c+d x)}{d}\right )+\frac {2 d \sin (2 (a+b x))}{c+d x}+\frac {d \sin (4 (a+b x))}{c+d x}+4 b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+4 b \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )}{8 d^2} \] Input:

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

Output:

-1/8*(-4*b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d*x))/d] - 4*b*Cos[4 
*a - (4*b*c)/d]*CosIntegral[(4*b*(c + d*x))/d] + (2*d*Sin[2*(a + b*x)])/(c 
 + d*x) + (d*Sin[4*(a + b*x)])/(c + d*x) + 4*b*Sin[2*a - (2*b*c)/d]*SinInt 
egral[(2*b*(c + d*x))/d] + 4*b*Sin[4*a - (4*b*c)/d]*SinIntegral[(4*b*(c + 
d*x))/d])/d^2
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 179, 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.

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

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {\sin (2 a+2 b x)}{4 (c+d x)^2}+\frac {\sin (4 a+4 b x)}{8 (c+d x)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^2}+\frac {b \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{2 d^2}-\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^2}-\frac {b \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{2 d^2}-\frac {\sin (2 a+2 b x)}{4 d (c+d x)}-\frac {\sin (4 a+4 b x)}{8 d (c+d x)}\)

Input:

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

Output:

(b*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/(2*d^2) + (b*Cos[4 
*a - (4*b*c)/d]*CosIntegral[(4*b*c)/d + 4*b*x])/(2*d^2) - Sin[2*a + 2*b*x] 
/(4*d*(c + d*x)) - Sin[4*a + 4*b*x]/(8*d*(c + d*x)) - (b*Sin[2*a - (2*b*c) 
/d]*SinIntegral[(2*b*c)/d + 2*b*x])/(2*d^2) - (b*Sin[4*a - (4*b*c)/d]*SinI 
ntegral[(4*b*c)/d + 4*b*x])/(2*d^2)
 

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 = 1.27 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \left (-\frac {4 \sin \left (4 b x +4 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {16 \,\operatorname {Si}\left (-4 b x -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}+\frac {16 \,\operatorname {Ci}\left (4 b x +4 a +\frac {-4 a d +4 c b}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}}{d}\right )}{32}+\frac {b^{2} \left (-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {4 \,\operatorname {Si}\left (-2 b x -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}}{d}\right )}{8}}{b}\)	\(256\)
default	\(\frac {\frac {b^{2} \left (-\frac {4 \sin \left (4 b x +4 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {16 \,\operatorname {Si}\left (-4 b x -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}+\frac {16 \,\operatorname {Ci}\left (4 b x +4 a +\frac {-4 a d +4 c b}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}}{d}\right )}{32}+\frac {b^{2} \left (-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {-\frac {4 \,\operatorname {Si}\left (-2 b x -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}}{d}\right )}{8}}{b}\)	\(256\)
risch	\(-\frac {b \,{\mathrm e}^{-\frac {4 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (4 i b x +4 i a -\frac {4 i \left (a d -c b \right )}{d}\right )}{4 d^{2}}-\frac {b \,{\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{4 d^{2}}-\frac {b \,{\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{4 d^{2}}-\frac {b \,{\mathrm e}^{\frac {4 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-4 i b x -4 i a -\frac {4 \left (-i a d +i c b \right )}{d}\right )}{4 d^{2}}-\frac {\left (-2 b d x -2 c b \right ) \sin \left (4 b x +4 a \right )}{16 d \left (-b d x -c b \right ) \left (d x +c \right )}-\frac {\left (-2 b d x -2 c b \right ) \sin \left (2 b x +2 a \right )}{8 d \left (-b d x -c b \right ) \left (d x +c \right )}\)	\(280\)

Input:

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

Output:

1/b*(1/32*b^2*(-4*sin(4*b*x+4*a)/(-a*d+c*b+d*(b*x+a))/d+4*(-4*Si(-4*b*x-4* 
a-4*(-a*d+b*c)/d)*sin(4*(-a*d+b*c)/d)/d+4*Ci(4*b*x+4*a+4*(-a*d+b*c)/d)*cos 
(4*(-a*d+b*c)/d)/d)/d)+1/8*b^2*(-2*sin(2*b*x+2*a)/(-a*d+c*b+d*(b*x+a))/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))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=-\frac {2 \, d \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) - {\left (b d x + b c\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (b d x + b c\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

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

Output:

-1/2*(2*d*cos(b*x + a)^3*sin(b*x + a) - (b*d*x + b*c)*cos(-4*(b*c - a*d)/d 
)*cos_integral(4*(b*d*x + b*c)/d) - (b*d*x + b*c)*cos(-2*(b*c - a*d)/d)*co 
s_integral(2*(b*d*x + b*c)/d) + (b*d*x + b*c)*sin(-4*(b*c - a*d)/d)*sin_in 
tegral(4*(b*d*x + b*c)/d) + (b*d*x + b*c)*sin(-2*(b*c - a*d)/d)*sin_integr 
al(2*(b*d*x + b*c)/d))/(d^3*x + c*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.72 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\frac {2 \, b^{2} {\left (i \, E_{2}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{2}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{2} {\left (-i \, E_{2}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{2}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b^{2} {\left (E_{2}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{2}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{2} {\left (E_{2}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{2}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \] Input:

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

Output:

1/16*(2*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*(-I*exp_integral_e(2, 4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + 
 I*exp_integral_e(2, -4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-4*(b*c - 
 a*d)/d) - 2*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^2*(exp_integral_e(2, 4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + e 
xp_integral_e(2, -4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-4*(b*c - a*d 
)/d))/((b*c*d + (b*x + a)*d^2 - a*d^2)*b)
 

Giac [C] (verification not implemented)

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

Time = 2.56 (sec) , antiderivative size = 63510, normalized size of antiderivative = 354.80 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(b*d*x*real_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*b*x)^2*tan(b*x)^ 
2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 + b*d*x*real_part(cos_in 
tegral(2*b*x + 2*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2 
*b*c/d)^2*tan(b*c/d)^2 + b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*t 
an(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 + b 
*d*x*real_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan 
(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 - 2*b*d*x*imag_part(cos_integ 
ral(2*b*x + 2*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b* 
c/d)^2*tan(b*c/d) + 2*b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan( 
2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d) - 4*b*d* 
x*sin_integral(2*(b*d*x + b*c)/d)*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a 
)^2*tan(2*b*c/d)^2*tan(b*c/d) - 2*b*d*x*imag_part(cos_integral(4*b*x + 4*b 
*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)*tan(b*c/d) 
^2 + 2*b*d*x*imag_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*b*x)^2*tan(b* 
x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)*tan(b*c/d)^2 - 4*b*d*x*sin_integral( 
4*(b*d*x + b*c)/d)*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2*tan(2*b*c/d 
)*tan(b*c/d)^2 + 2*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(2*b* 
x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)*tan(2*b*c/d)^2*tan(b*c/d)^2 - 2*b*d*x*im 
ag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2 
*tan(a)*tan(2*b*c/d)^2*tan(b*c/d)^2 + 4*b*d*x*sin_integral(2*(b*d*x + b...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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