\(\int \frac {\sin ^3(a+b x)}{(c+d x)^2} \, dx\) [21]

Optimal result

Integrand size = 16, antiderivative size = 145 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^2} \, dx=\frac {3 b \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{4 d^2}-\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {\sin ^3(a+b x)}{d (c+d x)}-\frac {3 b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^2} \, dx=\frac {3 b (c+d x) \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right )-3 b (c+d x) \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b (c+d x)}{d}\right )-3 d \cos (b x) \sin (a)+d \cos (3 b x) \sin (3 a)-3 d \cos (a) \sin (b x)+d \cos (3 a) \sin (3 b x)-3 b (c+d x) \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )+3 b (c+d x) \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )}{4 d^2 (c+d x)} \] Input:

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

Output:

(3*b*(c + d*x)*Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)] - 3*b*(c + d*x)*C 
os[3*a - (3*b*c)/d]*CosIntegral[(3*b*(c + d*x))/d] - 3*d*Cos[b*x]*Sin[a] + 
 d*Cos[3*b*x]*Sin[3*a] - 3*d*Cos[a]*Sin[b*x] + d*Cos[3*a]*Sin[3*b*x] - 3*b 
*(c + d*x)*Sin[a - (b*c)/d]*SinIntegral[b*(c/d + x)] + 3*b*(c + d*x)*Sin[3 
*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d])/(4*d^2*(c + d*x))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 3794, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3794

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-(Sin[a + b*x]^3/(d*(c + d*x))) + (3*b*((Cos[a - (b*c)/d]*CosIntegral[(b*c 
)/d + b*x])/(4*d) - (Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*b*x])/ 
(4*d) - (Sin[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(4*d) + (Sin[3*a - ( 
3*b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(4*d)))/d
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.69

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

Input:

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

Output:

1/b*(-1/12*b^2*(-3*sin(3*b*x+3*a)/(-a*d+b*c+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)*co 
s(3*(-a*d+b*c)/d)/d)/d)+3/4*b^2*(-sin(b*x+a)/(-a*d+b*c+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))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^2} \, dx=-\frac {3 \, {\left (b d x + b c\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - 3 \, {\left (b d x + b c\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) - 3 \, {\left (b d x + b c\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + 3 \, {\left (b d x + b c\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - 4 \, {\left (d \cos \left (b x + a\right )^{2} - d\right )} \sin \left (b x + a\right )}{4 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sin(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 = 306, normalized size of antiderivative = 2.11 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^2} \, dx=-\frac {3 \, b^{2} {\left (i \, E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - b^{2} {\left (-i \, E_{2}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{2}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 3 \, b^{2} {\left (E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{2} {\left (E_{2}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{2}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \] Input:

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

Output:

-1/8*(3*b^2*(I*exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*ex 
p_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_i 
ntegral_e(2, -3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-3*(b*c - a*d)/d) 
 + 3*b^2*(exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integ 
ral_e(2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b^2*(e 
xp_integral_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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1000 vs. \(2 (137) = 274\).

Time = 0.51 (sec) , antiderivative size = 1000, normalized size of antiderivative = 6.90 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos(-3*(b*c - a* 
d)/d)*cos_integral(3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c 
- a*d)/d) + 3*b^3*c*cos(-3*(b*c - a*d)/d)*cos_integral(3*((d*x + c)*(b - b 
*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) - 3*a*b^2*d*cos(-3*(b*c - a* 
d)/d)*cos_integral(3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c 
- a*d)/d) - 3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos(-(b*c 
- a*d)/d)*cos_integral(((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b* 
c - a*d)/d) - 3*b^3*c*cos(-(b*c - a*d)/d)*cos_integral(((d*x + c)*(b - b*c 
/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + 3*a*b^2*d*cos(-(b*c - a*d)/d 
)*cos_integral(((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d) 
/d) - 3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*sin(-(b*c - a*d) 
/d)*sin_integral(-((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a 
*d)/d) - 3*b^3*c*sin(-(b*c - a*d)/d)*sin_integral(-((d*x + c)*(b - b*c/(d* 
x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + 3*a*b^2*d*sin(-(b*c - a*d)/d)*si 
n_integral(-((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) 
 + 3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*sin(-3*(b*c - a*d)/ 
d)*sin_integral(-3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - 
a*d)/d) + 3*b^3*c*sin(-3*(b*c - a*d)/d)*sin_integral(-3*((d*x + c)*(b - b* 
c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) - 3*a*b^2*d*sin(-3*(b*c - a*d 
)/d)*sin_integral(-3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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