\(\int \frac {\sin ^3(a+b x)}{(c+d x)^3} \, dx\) [22]

Optimal result

Integrand size = 16, antiderivative size = 184 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^3} \, dx=\frac {9 b^2 \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{8 d^3}-\frac {3 b^2 \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{8 d^3}-\frac {3 b \cos (a+b x) \sin ^2(a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}-\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3} \] Output:

9/8*b^2*Ci(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^3-3/8*b^2*Ci(b*c/d+b*x)*sin(a 
-b*c/d)/d^3-3/2*b*cos(b*x+a)*sin(b*x+a)^2/d^2/(d*x+c)-1/2*sin(b*x+a)^3/d/( 
d*x+c)^2-3/8*b^2*cos(a-b*c/d)*Si(b*c/d+b*x)/d^3+9/8*b^2*cos(3*a-3*b*c/d)*S 
i(3*b*c/d+3*b*x)/d^3
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.20 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^3} \, dx=\frac {-6 d \cos (b x) (b (c+d x) \cos (a)+d \sin (a))+2 d \cos (3 b x) (3 b (c+d x) \cos (3 a)+d \sin (3 a))+6 d (-d \cos (a)+b (c+d x) \sin (a)) \sin (b x)+2 d (d \cos (3 a)-3 b (c+d x) \sin (3 a)) \sin (3 b x)+6 b^2 (c+d x)^2 \left (3 \operatorname {CosIntegral}\left (\frac {3 b (c+d x)}{d}\right ) \sin \left (3 a-\frac {3 b c}{d}\right )-\operatorname {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right ) \sin \left (a-\frac {b c}{d}\right )-\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )+3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )\right )}{16 d^3 (c+d x)^2} \] Input:

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

Output:

(-6*d*Cos[b*x]*(b*(c + d*x)*Cos[a] + d*Sin[a]) + 2*d*Cos[3*b*x]*(3*b*(c + 
d*x)*Cos[3*a] + d*Sin[3*a]) + 6*d*(-(d*Cos[a]) + b*(c + d*x)*Sin[a])*Sin[b 
*x] + 2*d*(d*Cos[3*a] - 3*b*(c + d*x)*Sin[3*a])*Sin[3*b*x] + 6*b^2*(c + d* 
x)^2*(3*CosIntegral[(3*b*(c + d*x))/d]*Sin[3*a - (3*b*c)/d] - CosIntegral[ 
b*(c/d + x)]*Sin[a - (b*c)/d] - Cos[a - (b*c)/d]*SinIntegral[b*(c/d + x)] 
+ 3*Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d]))/(16*d^3*(c + d*x 
)^2)
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.32, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 3795, 3042, 3784, 3042, 3780, 3783, 3793, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3795

\(\displaystyle -\frac {9 b^2 \int \frac {\sin ^3(a+b x)}{c+d x}dx}{2 d^2}+\frac {3 b^2 \int \frac {\sin (a+b x)}{c+d x}dx}{d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 b^2 \int \frac {\sin (a+b x)}{c+d x}dx}{d^2}-\frac {9 b^2 \int \frac {\sin (a+b x)^3}{c+d x}dx}{2 d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 3784

\(\displaystyle -\frac {9 b^2 \int \frac {\sin (a+b x)^3}{c+d x}dx}{2 d^2}+\frac {3 b^2 \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x}dx+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x}dx\right )}{d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9 b^2 \int \frac {\sin (a+b x)^3}{c+d x}dx}{2 d^2}+\frac {3 b^2 \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x+\frac {\pi }{2}\right )}{c+d x}dx+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x}dx\right )}{d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 3780

\(\displaystyle \frac {3 b^2 \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x+\frac {\pi }{2}\right )}{c+d x}dx+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d}\right )}{d^2}-\frac {9 b^2 \int \frac {\sin (a+b x)^3}{c+d x}dx}{2 d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 3783

\(\displaystyle -\frac {9 b^2 \int \frac {\sin (a+b x)^3}{c+d x}dx}{2 d^2}+\frac {3 b^2 \left (\frac {\sin \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{d}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d}\right )}{d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 3793

\(\displaystyle -\frac {9 b^2 \int \left (\frac {3 \sin (a+b x)}{4 (c+d x)}-\frac {\sin (3 a+3 b x)}{4 (c+d x)}\right )dx}{2 d^2}+\frac {3 b^2 \left (\frac {\sin \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{d}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d}\right )}{d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 b^2 \left (\frac {\sin \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{d}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d}\right )}{d^2}-\frac {9 b^2 \left (-\frac {\sin \left (3 a-\frac {3 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{4 d}+\frac {3 \sin \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{4 d}+\frac {3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{4 d}-\frac {\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{4 d}\right )}{2 d^2}-\frac {3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin ^3(a+b x)}{2 d (c+d x)^2}\)

Input:

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

Output:

(-3*b*Cos[a + b*x]*Sin[a + b*x]^2)/(2*d^2*(c + d*x)) - Sin[a + b*x]^3/(2*d 
*(c + d*x)^2) + (3*b^2*((CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/d + 
(Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/d))/d^2 - (9*b^2*(-1/4*(CosI 
ntegral[(3*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/d + (3*CosIntegral[(b*c)/ 
d + b*x]*Sin[a - (b*c)/d])/(4*d) + (3*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d 
 + b*x])/(4*d) - (Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(4* 
d)))/(2*d^2)
 

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 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* 
f)/d]   Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] 
&& NeQ[d*e - c*f, 0]
 

rule 3793

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

rule 3795

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), x] + (-Simp[ 
b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1) 
*(m + 2))), x] + Simp[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2)))   Int[(c + 
d*x)^(m + 2)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[f^2*(n^2/(d^2*(m + 1)* 
(m + 2)))   Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, 
c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
 

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {-\frac {b^{3} \left (-\frac {3 \sin \left (3 b x +3 a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {9 \cos \left (3 b x +3 a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right ) d}-\frac {9 \left (-\frac {3 \,\operatorname {Si}\left (-3 b x -3 a -\frac {3 \left (-a d +b c \right )}{d}\right ) \cos \left (\frac {-3 a d +3 b c}{d}\right )}{d}-\frac {3 \,\operatorname {Ci}\left (3 b x +3 a +\frac {-3 a d +3 b c}{d}\right ) \sin \left (\frac {-3 a d +3 b c}{d}\right )}{d}\right )}{2 d}}{d}\right )}{12}+\frac {3 b^{3} \left (-\frac {\sin \left (b x +a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}-\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +b c}{d}\right ) \cos \left (\frac {-a d +b c}{d}\right )}{d}-\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +b c}{d}\right ) \sin \left (\frac {-a d +b c}{d}\right )}{d}}{d}}{2 d}\right )}{4}}{b}\)	\(318\)
default	\(\frac {-\frac {b^{3} \left (-\frac {3 \sin \left (3 b x +3 a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {9 \cos \left (3 b x +3 a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right ) d}-\frac {9 \left (-\frac {3 \,\operatorname {Si}\left (-3 b x -3 a -\frac {3 \left (-a d +b c \right )}{d}\right ) \cos \left (\frac {-3 a d +3 b c}{d}\right )}{d}-\frac {3 \,\operatorname {Ci}\left (3 b x +3 a +\frac {-3 a d +3 b c}{d}\right ) \sin \left (\frac {-3 a d +3 b c}{d}\right )}{d}\right )}{2 d}}{d}\right )}{12}+\frac {3 b^{3} \left (-\frac {\sin \left (b x +a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}-\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +b c}{d}\right ) \cos \left (\frac {-a d +b c}{d}\right )}{d}-\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +b c}{d}\right ) \sin \left (\frac {-a d +b c}{d}\right )}{d}}{d}}{2 d}\right )}{4}}{b}\)	\(318\)
risch	\(\frac {9 i b^{2} {\mathrm e}^{\frac {3 i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (-3 i b x -3 i a -\frac {3 \left (-i a d +i b c \right )}{d}\right )}{16 d^{3}}-\frac {9 i b^{2} {\mathrm e}^{-\frac {3 i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (3 i b x +3 i a -\frac {3 i \left (a d -b c \right )}{d}\right )}{16 d^{3}}+\frac {3 i b^{2} {\mathrm e}^{-\frac {i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (i b x +i a -\frac {i \left (a d -b c \right )}{d}\right )}{16 d^{3}}-\frac {3 i b^{2} {\mathrm e}^{\frac {i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (-i b x -i a -\frac {-i a d +i b c}{d}\right )}{16 d^{3}}+\frac {3 i \left (2 i b^{3} d^{3} x^{3}+6 i b^{3} c \,d^{2} x^{2}+6 i b^{3} c^{2} d x +2 i b^{3} c^{3}\right ) \cos \left (b x +a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {3 \left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (b x +a \right )}{16 d \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {i \left (6 i b^{3} d^{3} x^{3}+18 i b^{3} c \,d^{2} x^{2}+18 i b^{3} c^{2} d x +6 i b^{3} c^{3}\right ) \cos \left (3 b x +3 a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}+\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (3 b x +3 a \right )}{16 d \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}\)	\(546\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.71 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^3} \, dx=\frac {12 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + 9 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 9 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - 12 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) + 4 \, {\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sin \left (b x + a\right )}{8 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

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

Output:

1/8*(12*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 3*(b^2*d^2*x^2 + 2*b^2*c*d*x + 
b^2*c^2)*cos_integral((b*d*x + b*c)/d)*sin(-(b*c - a*d)/d) + 9*(b^2*d^2*x^ 
2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(3*(b*d*x + b*c)/d)*sin(-3*(b*c - a 
*d)/d) + 9*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-3*(b*c - a*d)/d)*sin 
_integral(3*(b*d*x + b*c)/d) - 3*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos 
(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) - 12*(b*d^2*x + b*c*d)*cos( 
b*x + a) + 4*(d^2*cos(b*x + a)^2 - d^2)*sin(b*x + a))/(d^5*x^2 + 2*c*d^4*x 
 + c^2*d^3)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.85 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^3} \, dx=-\frac {3 \, b^{3} {\left (i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - b^{3} {\left (-i \, E_{3}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + i \, E_{3}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 3 \, b^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{3} {\left (E_{3}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \] Input:

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

Output:

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

Giac [C] (verification not implemented)

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

Time = 3.35 (sec) , antiderivative size = 116534, normalized size of antiderivative = 633.34 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(9*b^2*d^2*x^2*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)^2*tan(1/2*b*c/d) 
^2 - 3*b^2*d^2*x^2*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 + 
3*b^2*d^2*x^2*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 - 9*b^ 
2*d^2*x^2*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)^2*tan(1/2*b*c/d)^2 + 18*b 
^2*d^2*x^2*sin_integral(3*(b*d*x + b*c)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*t 
an(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 - 6*b^2*d^2*x^2 
*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)^2 - 6*b^2*d^2*x^2*real_part(c 
os_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) - 6*b^2*d^2*x^2*real_part(cos_inte 
gral(-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) + 18*b^2*d^2*x^2*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)*tan(1/2*b*c/d)^2 + 18*b^2*d^2*x^2*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*t 
an(3/2*b*c/d)*tan(1/2*b*c/d)^2 + 6*b^2*d^2*x^2*real_part(cos_integral(b...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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