\(\int \sin ^3(\frac {a+b x}{c+d x}) \, dx\) [40]

Optimal result

Integrand size = 16, antiderivative size = 194 \[ \int \sin ^3\left (\frac {a+b x}{c+d x}\right ) \, dx=\frac {3 (b c-a d) \cos \left (\frac {b}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 (b c-a d) \cos \left (\frac {3 b}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d}+\frac {3 (b c-a d) \sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 (b c-a d) \sin \left (\frac {3 b}{d}\right ) \text {Si}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.75 (sec) , antiderivative size = 845, normalized size of antiderivative = 4.36 \[ \int \sin ^3\left (\frac {a+b x}{c+d x}\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(((3*I)/8)*c)/(d*E^((I*(a + b*x))/(c + d*x))) - (((3*I)/8)*c*E^((I*(a + b* 
x))/(c + d*x)))/d - ((I/8)*c)/(d*E^(((3*I)*(a + b*x))/(c + d*x))) + ((I/8) 
*c*E^(((3*I)*(a + b*x))/(c + d*x)))/d + (3*x*Cos[(-(b*c) + a*d)/(d*(c + d* 
x))]*Sin[b/d])/4 - (x*Cos[(3*(-(b*c) + a*d))/(d*(c + d*x))]*Sin[(3*b)/d])/ 
4 + (3*x*Cos[b/d]*Sin[(-(b*c) + a*d)/(d*(c + d*x))])/4 - (x*Cos[(3*b)/d]*S 
in[(3*(-(b*c) + a*d))/(d*(c + d*x))])/4 + (3*(-(b*c) + a*d)*(-(Cos[b/d]*Co 
sIntegral[(I*(I*b*c - I*a*d))/(c*d + d^2*x)]) - Cos[b/d]*CosIntegral[(I*(( 
-I)*b*c + I*a*d))/(c*d + d^2*x)] + Cos[(3*b)/d]*CosIntegral[(I*((3*I)*b*c 
- (3*I)*a*d))/(c*d + d^2*x)] + Cos[(3*b)/d]*CosIntegral[(I*((-3*I)*b*c + ( 
3*I)*a*d))/(c*d + d^2*x)] + I*CosIntegral[(I*(I*b*c - I*a*d))/(c*d + d^2*x 
)]*Sin[b/d] - I*CosIntegral[(I*((-I)*b*c + I*a*d))/(c*d + d^2*x)]*Sin[b/d] 
 - I*CosIntegral[(I*((3*I)*b*c - (3*I)*a*d))/(c*d + d^2*x)]*Sin[(3*b)/d] + 
 I*CosIntegral[(I*((-3*I)*b*c + (3*I)*a*d))/(c*d + d^2*x)]*Sin[(3*b)/d] - 
Cos[b/d]*SinhIntegral[(I*b*c - I*a*d)/(c*d + d^2*x)] + I*Sin[b/d]*SinhInte 
gral[(I*b*c - I*a*d)/(c*d + d^2*x)] - Cos[b/d]*SinhIntegral[((-I)*b*c + I* 
a*d)/(c*d + d^2*x)] - I*Sin[b/d]*SinhIntegral[((-I)*b*c + I*a*d)/(c*d + d^ 
2*x)] + Cos[(3*b)/d]*SinhIntegral[((3*I)*b*c - (3*I)*a*d)/(c*d + d^2*x)] - 
 I*Sin[(3*b)/d]*SinhIntegral[((3*I)*b*c - (3*I)*a*d)/(c*d + d^2*x)] + Cos[ 
(3*b)/d]*SinhIntegral[((-3*I)*b*c + (3*I)*a*d)/(c*d + d^2*x)] + I*Sin[(3*b 
)/d]*SinhIntegral[((-3*I)*b*c + (3*I)*a*d)/(c*d + d^2*x)]))/(8*d^2)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5074, 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.

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

\(\Big \downarrow \) 5074

\(\displaystyle -\frac {\int (c+d x)^2 \sin ^3\left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int (c+d x)^2 \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )^3d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3794

\(\displaystyle -\frac {-\frac {3 (b c-a d) \int \left (\frac {1}{4} (c+d x) \cos \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )-\frac {1}{4} (c+d x) \cos \left (\frac {3 b}{d}-\frac {3 (b c-a d)}{d (c+d x)}\right )\right )d\frac {1}{c+d x}}{d}-\left ((c+d x) \sin ^3\left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {3 (b c-a d) \left (\frac {1}{4} \cos \left (\frac {b}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c-a d}{d (c+d x)}\right )-\frac {1}{4} \cos \left (\frac {3 b}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )+\frac {1}{4} \sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )-\frac {1}{4} \sin \left (\frac {3 b}{d}\right ) \text {Si}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )\right )}{d}-\left ((c+d x) \sin ^3\left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

Input:

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

Output:

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

rule 5074

Int[Sin[((e_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_))]^(n_.), x_Symbol] 
 :> Simp[-d^(-1)   Subst[Int[Sin[b*(e/d) - e*(b*c - a*d)*(x/d)]^n/x^2, x], 
x, 1/(c + d*x)], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && NeQ[b*c - a* 
d, 0]
 

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.52

method	result	size
derivativedivides	\(-\frac {\left (a d -c b \right ) \left (-\frac {d^{2} \left (-\frac {3 \sin \left (\frac {3 a d -3 c b}{d \left (d x +c \right )}+\frac {3 b}{d}\right )}{\left (\left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {9 \,\operatorname {Si}\left (\frac {3 a d -3 c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {3 b}{d}\right )}{d}+\frac {9 \,\operatorname {Ci}\left (\frac {3 a d -3 c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {3 b}{d}\right )}{d}}{d}\right )}{12}+\frac {3 d^{2} \left (-\frac {\sin \left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right )}{\left (\left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {\operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {b}{d}\right )}{d}}{d}\right )}{4}\right )}{d^{2}}\)	\(295\)
default	\(-\frac {\left (a d -c b \right ) \left (-\frac {d^{2} \left (-\frac {3 \sin \left (\frac {3 a d -3 c b}{d \left (d x +c \right )}+\frac {3 b}{d}\right )}{\left (\left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {9 \,\operatorname {Si}\left (\frac {3 a d -3 c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {3 b}{d}\right )}{d}+\frac {9 \,\operatorname {Ci}\left (\frac {3 a d -3 c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {3 b}{d}\right )}{d}}{d}\right )}{12}+\frac {3 d^{2} \left (-\frac {\sin \left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right )}{\left (\left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {\operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {b}{d}\right )}{d}}{d}\right )}{4}\right )}{d^{2}}\)	\(295\)
risch	\(\frac {3 i {\mathrm e}^{-\frac {3 i b}{d}} \operatorname {Si}\left (\frac {3 a d -3 c b}{d \left (d x +c \right )}\right ) b c}{4 d^{2}}+\frac {3 i {\mathrm e}^{-\frac {i b}{d}} \pi \,\operatorname {csgn}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) b c}{8 d^{2}}-\frac {3 i {\mathrm e}^{-\frac {3 i b}{d}} \operatorname {Si}\left (\frac {3 a d -3 c b}{d \left (d x +c \right )}\right ) a}{4 d}+\frac {3 i {\mathrm e}^{-\frac {3 i b}{d}} \pi \,\operatorname {csgn}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) a}{8 d}-\frac {3 \,{\mathrm e}^{-\frac {3 i b}{d}} \operatorname {expIntegral}_{1}\left (-\frac {3 i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) a}{8 d}+\frac {3 \,{\mathrm e}^{-\frac {3 i b}{d}} \operatorname {expIntegral}_{1}\left (-\frac {3 i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) b c}{8 d^{2}}-\frac {3 \,\operatorname {expIntegral}_{1}\left (-\frac {3 i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) {\mathrm e}^{\frac {3 i b}{d}} a}{8 d}+\frac {3 \,\operatorname {expIntegral}_{1}\left (-\frac {3 i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) {\mathrm e}^{\frac {3 i b}{d}} c b}{8 d^{2}}+\frac {3 \,\operatorname {expIntegral}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) {\mathrm e}^{\frac {i b}{d}} a}{8 d}-\frac {3 \,\operatorname {expIntegral}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) {\mathrm e}^{\frac {i b}{d}} c b}{8 d^{2}}-\frac {3 i {\mathrm e}^{-\frac {i b}{d}} \operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) b c}{4 d^{2}}-\frac {3 i {\mathrm e}^{-\frac {i b}{d}} \pi \,\operatorname {csgn}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) a}{8 d}-\frac {3 i {\mathrm e}^{-\frac {3 i b}{d}} \pi \,\operatorname {csgn}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) b c}{8 d^{2}}+\frac {3 i {\mathrm e}^{-\frac {i b}{d}} \operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) a}{4 d}+\frac {3 \,{\mathrm e}^{-\frac {i b}{d}} \operatorname {expIntegral}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) a}{8 d}-\frac {3 \,{\mathrm e}^{-\frac {i b}{d}} \operatorname {expIntegral}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) b c}{8 d^{2}}+\frac {3 \sin \left (\frac {b x +a}{d x +c}\right ) x}{4}+\frac {3 \sin \left (\frac {b x +a}{d x +c}\right ) c}{4 d}-\frac {\sin \left (\frac {3 b x +3 a}{d x +c}\right ) x}{4}-\frac {\sin \left (\frac {3 b x +3 a}{d x +c}\right ) c}{4 d}\)	\(668\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.09 \[ \int \sin ^3\left (\frac {a+b x}{c+d x}\right ) \, dx=\frac {3 \, {\left (b c - a d\right )} \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b c - a d}{d^{2} x + c d}\right ) - 3 \, {\left (b c - a d\right )} \cos \left (\frac {3 \, b}{d}\right ) \operatorname {Ci}\left (-\frac {3 \, {\left (b c - a d\right )}}{d^{2} x + c d}\right ) - 3 \, {\left (b c - a d\right )} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (-\frac {b c - a d}{d^{2} x + c d}\right ) + 3 \, {\left (b c - a d\right )} \sin \left (\frac {3 \, b}{d}\right ) \operatorname {Si}\left (-\frac {3 \, {\left (b c - a d\right )}}{d^{2} x + c d}\right ) + 4 \, {\left (d^{2} x - {\left (d^{2} x + c d\right )} \cos \left (\frac {b x + a}{d x + c}\right )^{2} + c d\right )} \sin \left (\frac {b x + a}{d x + c}\right )}{4 \, d^{2}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1239 vs. \(2 (186) = 372\).

Time = 68.95 (sec) , antiderivative size = 1239, normalized size of antiderivative = 6.39 \[ \int \sin ^3\left (\frac {a+b x}{c+d x}\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(3*b^3*c^2*cos(b/d)*cos_integral(-(b - (b*x + a)*d/(d*x + c))/d) - 6*a 
*b^2*c*d*cos(b/d)*cos_integral(-(b - (b*x + a)*d/(d*x + c))/d) - 3*(b*x + 
a)*b^2*c^2*d*cos(b/d)*cos_integral(-(b - (b*x + a)*d/(d*x + c))/d)/(d*x + 
c) + 3*a^2*b*d^2*cos(b/d)*cos_integral(-(b - (b*x + a)*d/(d*x + c))/d) + 6 
*(b*x + a)*a*b*c*d^2*cos(b/d)*cos_integral(-(b - (b*x + a)*d/(d*x + c))/d) 
/(d*x + c) - 3*(b*x + a)*a^2*d^3*cos(b/d)*cos_integral(-(b - (b*x + a)*d/( 
d*x + c))/d)/(d*x + c) - 3*b^3*c^2*cos(3*b/d)*cos_integral(-3*(b - (b*x + 
a)*d/(d*x + c))/d) + 6*a*b^2*c*d*cos(3*b/d)*cos_integral(-3*(b - (b*x + a) 
*d/(d*x + c))/d) + 3*(b*x + a)*b^2*c^2*d*cos(3*b/d)*cos_integral(-3*(b - ( 
b*x + a)*d/(d*x + c))/d)/(d*x + c) - 3*a^2*b*d^2*cos(3*b/d)*cos_integral(- 
3*(b - (b*x + a)*d/(d*x + c))/d) - 6*(b*x + a)*a*b*c*d^2*cos(3*b/d)*cos_in 
tegral(-3*(b - (b*x + a)*d/(d*x + c))/d)/(d*x + c) + 3*(b*x + a)*a^2*d^3*c 
os(3*b/d)*cos_integral(-3*(b - (b*x + a)*d/(d*x + c))/d)/(d*x + c) - 3*b^3 
*c^2*sin(3*b/d)*sin_integral(3*(b - (b*x + a)*d/(d*x + c))/d) + 6*a*b^2*c* 
d*sin(3*b/d)*sin_integral(3*(b - (b*x + a)*d/(d*x + c))/d) + 3*(b*x + a)*b 
^2*c^2*d*sin(3*b/d)*sin_integral(3*(b - (b*x + a)*d/(d*x + c))/d)/(d*x + c 
) - 3*a^2*b*d^2*sin(3*b/d)*sin_integral(3*(b - (b*x + a)*d/(d*x + c))/d) - 
 6*(b*x + a)*a*b*c*d^2*sin(3*b/d)*sin_integral(3*(b - (b*x + a)*d/(d*x + c 
))/d)/(d*x + c) + 3*(b*x + a)*a^2*d^3*sin(3*b/d)*sin_integral(3*(b - (b*x 
+ a)*d/(d*x + c))/d)/(d*x + c) + 3*b^3*c^2*sin(b/d)*sin_integral((b - (...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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