3.1.76 \(\int \frac {\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx\) [76]

3.1.76.1 Optimal result
3.1.76.2 Mathematica [A] (verified)
3.1.76.3 Rubi [A] (verified)
3.1.76.4 Maple [A] (verified)
3.1.76.5 Fricas [A] (verification not implemented)
3.1.76.6 Sympy [F]
3.1.76.7 Maxima [C] (verification not implemented)
3.1.76.8 Giac [C] (verification not implemented)
3.1.76.9 Mupad [F(-1)]

3.1.76.1 Optimal result

Integrand size = 22, antiderivative size = 168 \[ \int \frac {\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\frac {b \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {\sin (a+b x)}{4 d (c+d x)}-\frac {\sin (3 a+3 b x)}{4 d (c+d x)}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{4 d^2}-\frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2} \]

output
3/4*b*Ci(3*b*c/d+3*b*x)*cos(3*a-3*b*c/d)/d^2+1/4*b*Ci(b*c/d+b*x)*cos(a-b*c 
/d)/d^2-3/4*b*Si(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^2-1/4*b*Si(b*c/d+b*x)*s 
in(a-b*c/d)/d^2-1/4*sin(b*x+a)/d/(d*x+c)-1/4*sin(3*b*x+3*a)/d/(d*x+c)
 
3.1.76.2 Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=-\frac {-b \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right )-3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b (c+d x)}{d}\right )+\frac {d \sin (a+b x)}{c+d x}+\frac {d \sin (3 (a+b x))}{c+d x}+b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )+3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )}{4 d^2} \]

input
Integrate[(Cos[a + b*x]^2*Sin[a + b*x])/(c + d*x)^2,x]
 
output
-1/4*(-(b*Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)]) - 3*b*Cos[3*a - (3*b* 
c)/d]*CosIntegral[(3*b*(c + d*x))/d] + (d*Sin[a + b*x])/(c + d*x) + (d*Sin 
[3*(a + b*x)])/(c + d*x) + b*Sin[a - (b*c)/d]*SinIntegral[b*(c/d + x)] + 3 
*b*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d])/d^2
 
3.1.76.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 168, 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.

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

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {\sin (a+b x)}{4 (c+d x)^2}+\frac {\sin (3 a+3 b x)}{4 (c+d x)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{4 d^2}-\frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {\sin (a+b x)}{4 d (c+d x)}-\frac {\sin (3 a+3 b x)}{4 d (c+d x)}\)

input
Int[(Cos[a + b*x]^2*Sin[a + b*x])/(c + d*x)^2,x]
 
output
(b*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(4*d^2) + (3*b*Cos[3*a - ( 
3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*b*x])/(4*d^2) - Sin[a + b*x]/(4*d*(c + 
 d*x)) - Sin[3*a + 3*b*x]/(4*d*(c + d*x)) - (b*Sin[a - (b*c)/d]*SinIntegra 
l[(b*c)/d + b*x])/(4*d^2) - (3*b*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/ 
d + 3*b*x])/(4*d^2)
 

3.1.76.3.1 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]
 
3.1.76.4 Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.46

method result size
derivativedivides \(\frac {\frac {b^{2} \left (-\frac {3 \sin \left (3 x b +3 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {-\frac {9 \,\operatorname {Si}\left (-3 x b -3 a -\frac {3 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-3 a d +3 c b}{d}\right )}{d}+\frac {9 \,\operatorname {Ci}\left (3 x b +3 a +\frac {-3 a d +3 c b}{d}\right ) \cos \left (\frac {-3 a d +3 c b}{d}\right )}{d}}{d}\right )}{12}+\frac {b^{2} \left (-\frac {\sin \left (x b +a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {-\frac {\operatorname {Si}\left (-x b -a -\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (x b +a +\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}}{d}\right )}{4}}{b}\) \(245\)
default \(\frac {\frac {b^{2} \left (-\frac {3 \sin \left (3 x b +3 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {-\frac {9 \,\operatorname {Si}\left (-3 x b -3 a -\frac {3 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-3 a d +3 c b}{d}\right )}{d}+\frac {9 \,\operatorname {Ci}\left (3 x b +3 a +\frac {-3 a d +3 c b}{d}\right ) \cos \left (\frac {-3 a d +3 c b}{d}\right )}{d}}{d}\right )}{12}+\frac {b^{2} \left (-\frac {\sin \left (x b +a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {-\frac {\operatorname {Si}\left (-x b -a -\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (x b +a +\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}}{d}\right )}{4}}{b}\) \(245\)
risch \(-\frac {3 b \,{\mathrm e}^{-\frac {3 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (3 i b x +3 i a -\frac {3 i \left (a d -c b \right )}{d}\right )}{8 d^{2}}-\frac {b \,{\mathrm e}^{-\frac {i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (i b x +i a -\frac {i \left (a d -c b \right )}{d}\right )}{8 d^{2}}-\frac {b \,{\mathrm e}^{\frac {i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-i b x -i a -\frac {-i a d +i c b}{d}\right )}{8 d^{2}}-\frac {3 b \,{\mathrm e}^{\frac {3 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-3 i b x -3 i a -\frac {3 \left (-i a d +i c b \right )}{d}\right )}{8 d^{2}}-\frac {\left (-2 d x b -2 c b \right ) \sin \left (x b +a \right )}{8 d \left (d x +c \right ) \left (-d x b -c b \right )}-\frac {\left (-2 d x b -2 c b \right ) \sin \left (3 x b +3 a \right )}{8 d \left (d x +c \right ) \left (-d x b -c b \right )}\) \(277\)

input
int(cos(b*x+a)^2*sin(b*x+a)/(d*x+c)^2,x,method=_RETURNVERBOSE)
 
output
1/b*(1/12*b^2*(-3*sin(3*b*x+3*a)/(-a*d+c*b+d*(b*x+a))/d+3*(-3*Si(-3*x*b-3* 
a-3*(-a*d+b*c)/d)*sin(3*(-a*d+b*c)/d)/d+3*Ci(3*x*b+3*a+3*(-a*d+b*c)/d)*cos 
(3*(-a*d+b*c)/d)/d)/d)+1/4*b^2*(-sin(b*x+a)/(-a*d+c*b+d*(b*x+a))/d+(-Si(-x 
*b-a-(-a*d+b*c)/d)*sin((-a*d+b*c)/d)/d+Ci(x*b+a+(-a*d+b*c)/d)*cos((-a*d+b* 
c)/d)/d)/d))
 
3.1.76.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=-\frac {4 \, d \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - 3 \, {\left (b d x + b c\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (b d x + b c\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + 3 \, {\left (b d x + b c\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right )}{4 \, {\left (d^{3} x + c d^{2}\right )}} \]

input
integrate(cos(b*x+a)^2*sin(b*x+a)/(d*x+c)^2,x, algorithm="fricas")
 
output
-1/4*(4*d*cos(b*x + a)^2*sin(b*x + a) - 3*(b*d*x + b*c)*cos(-3*(b*c - a*d) 
/d)*cos_integral(3*(b*d*x + b*c)/d) - (b*d*x + b*c)*cos(-(b*c - a*d)/d)*co 
s_integral((b*d*x + b*c)/d) + 3*(b*d*x + b*c)*sin(-3*(b*c - a*d)/d)*sin_in 
tegral(3*(b*d*x + b*c)/d) + (b*d*x + b*c)*sin(-(b*c - a*d)/d)*sin_integral 
((b*d*x + b*c)/d))/(d^3*x + c*d^2)
 
3.1.76.6 Sympy [F]

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

input
integrate(cos(b*x+a)**2*sin(b*x+a)/(d*x+c)**2,x)
 
output
Integral(sin(a + b*x)*cos(a + b*x)**2/(c + d*x)**2, x)
 
3.1.76.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=-\frac {b^{2} {\left (i \, E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{2} {\left (-i \, E_{2}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{2}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{2}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

input
integrate(cos(b*x+a)^2*sin(b*x+a)/(d*x+c)^2,x, algorithm="maxima")
 
output
-1/8*(b^2*(I*exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_ 
integral_e(2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b 
^2*(-I*exp_integral_e(2, 3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + I*exp_int 
egral_e(2, -3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-3*(b*c - a*d)/d) + 
 b^2*(exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_ 
e(2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) + b^2*(exp_i 
ntegral_e(2, 3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_integral_e(2, -3* 
(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-3*(b*c - a*d)/d))/((b*c*d + (b*x 
 + a)*d^2 - a*d^2)*b)
 
3.1.76.8 Giac [C] (verification not implemented)

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

Time = 2.45 (sec) , antiderivative size = 66726, normalized size of antiderivative = 397.18 \[ \int \frac {\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx=\text {Too large to display} \]

input
integrate(cos(b*x+a)^2*sin(b*x+a)/(d*x+c)^2,x, algorithm="giac")
 
output
1/8*(3*b*d*x*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* 
d*x*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*d*x*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*ta 
n(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 3*b*d*x*real_part(cos_integ 
ral(-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*d*x*imag_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*d*x*imag_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*t 
an(1/2*b*c/d) - 4*b*d*x*sin_integral((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) - 6*b* 
d*x*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 + 6*b*d*x*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 - 12*b*d*x*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*d*x*imag_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)*tan(3/2*b*...
 
3.1.76.9 Mupad [F(-1)]

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

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