\(\int \frac {\cos ^3(a+b x) \sin (a+b x)}{c+d x} \, dx\) [141]

Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {\operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{8 d}+\frac {\operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{4 d}+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{4 d}+\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{8 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {\operatorname {CosIntegral}\left (\frac {4 b (c+d x)}{d}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )+2 \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )+2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+\cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )}{8 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 129, 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.

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 1.03 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {\frac {b \left (-\frac {4 \,\operatorname {Si}\left (-4 b x -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}-\frac {4 \,\operatorname {Ci}\left (4 b x +4 a +\frac {-4 a d +4 c b}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}\right )}{32}+\frac {b \left (-\frac {2 \,\operatorname {Si}\left (-2 b x -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{8}}{b}\)	\(178\)
default	\(\frac {\frac {b \left (-\frac {4 \,\operatorname {Si}\left (-4 b x -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}-\frac {4 \,\operatorname {Ci}\left (4 b x +4 a +\frac {-4 a d +4 c b}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}\right )}{32}+\frac {b \left (-\frac {2 \,\operatorname {Si}\left (-2 b x -2 a -\frac {2 \left (-a d +c b \right )}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{8}}{b}\)	\(178\)
risch	\(-\frac {i {\mathrm e}^{-\frac {4 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (4 i b x +4 i a -\frac {4 i \left (a d -c b \right )}{d}\right )}{16 d}-\frac {i {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{8 d}+\frac {i {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{8 d}+\frac {i {\mathrm e}^{\frac {4 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-4 i b x -4 i a -\frac {4 \left (-i a d +i c b \right )}{d}\right )}{16 d}\)	\(194\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {2 \, \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{8 \, d} \] Input:

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

Output:

1/8*(2*cos_integral(2*(b*d*x + b*c)/d)*sin(-2*(b*c - a*d)/d) + cos_integra 
l(4*(b*d*x + b*c)/d)*sin(-4*(b*c - a*d)/d) + cos(-4*(b*c - a*d)/d)*sin_int 
egral(4*(b*d*x + b*c)/d) + 2*cos(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + 
 b*c)/d))/d
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.18 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{c+d x} \, dx=\frac {2 \, b {\left (i \, E_{1}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{1}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b {\left (-i \, E_{1}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{1}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b {\left (E_{1}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{1}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b {\left (E_{1}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{1}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, b d} \] Input:

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

Output:

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

Giac [C] (verification not implemented)

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

Time = 0.35 (sec) , antiderivative size = 6046, normalized size of antiderivative = 46.87 \[ \int \frac {\cos ^3(a+b x) \sin (a+b x)}{c+d x} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(imag_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*a)^2*tan(a)^2*tan(2*b 
*c/d)^2*tan(b*c/d)^2 + 2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(2*a) 
^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 - 2*imag_part(cos_integral(-2*b*x 
- 2*b*c/d))*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 - imag_part(co 
s_integral(-4*b*x - 4*b*c/d))*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d 
)^2 + 2*sin_integral(4*(b*d*x + b*c)/d)*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2 
*tan(b*c/d)^2 + 4*sin_integral(2*(b*d*x + b*c)/d)*tan(2*a)^2*tan(a)^2*tan( 
2*b*c/d)^2*tan(b*c/d)^2 + 4*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(2 
*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d) + 4*real_part(cos_integral(-2*b*x 
 - 2*b*c/d))*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d) + 2*real_part(c 
os_integral(4*b*x + 4*b*c/d))*tan(2*a)^2*tan(a)^2*tan(2*b*c/d)*tan(b*c/d)^ 
2 + 2*real_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)^2*tan(a)^2*tan(2* 
b*c/d)*tan(b*c/d)^2 - 4*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(2*a)^ 
2*tan(a)*tan(2*b*c/d)^2*tan(b*c/d)^2 - 4*real_part(cos_integral(-2*b*x - 2 
*b*c/d))*tan(2*a)^2*tan(a)*tan(2*b*c/d)^2*tan(b*c/d)^2 - 2*real_part(cos_i 
ntegral(4*b*x + 4*b*c/d))*tan(2*a)*tan(a)^2*tan(2*b*c/d)^2*tan(b*c/d)^2 - 
2*real_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)*tan(a)^2*tan(2*b*c/d) 
^2*tan(b*c/d)^2 + imag_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*a)^2*tan( 
a)^2*tan(2*b*c/d)^2 - 2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(2*a)^ 
2*tan(a)^2*tan(2*b*c/d)^2 + 2*imag_part(cos_integral(-2*b*x - 2*b*c/d))...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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