\(\int x^{-1-2 n} \sin ^3(a+b x^n) \, dx\) [152]

Optimal result

Integrand size = 18, antiderivative size = 167 \[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=-\frac {3 b x^{-n} \cos \left (a+b x^n\right )}{8 n}+\frac {3 b x^{-n} \cos \left (3 a+3 b x^n\right )}{8 n}-\frac {3 b^2 \operatorname {CosIntegral}\left (b x^n\right ) \sin (a)}{8 n}+\frac {9 b^2 \operatorname {CosIntegral}\left (3 b x^n\right ) \sin (3 a)}{8 n}-\frac {3 x^{-2 n} \sin \left (a+b x^n\right )}{8 n}+\frac {x^{-2 n} \sin \left (3 a+3 b x^n\right )}{8 n}-\frac {3 b^2 \cos (a) \text {Si}\left (b x^n\right )}{8 n}+\frac {9 b^2 \cos (3 a) \text {Si}\left (3 b x^n\right )}{8 n} \] Output:

-3/8*b*cos(a+b*x^n)/n/(x^n)+3/8*b*cos(3*a+3*b*x^n)/n/(x^n)-3/8*b^2*Ci(b*x^ 
n)*sin(a)/n+9/8*b^2*Ci(3*b*x^n)*sin(3*a)/n-3/8*sin(a+b*x^n)/n/(x^(2*n))+1/ 
8*sin(3*a+3*b*x^n)/n/(x^(2*n))-3/8*b^2*cos(a)*Si(b*x^n)/n+9/8*b^2*cos(3*a) 
*Si(3*b*x^n)/n
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.84 \[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=\frac {x^{-2 n} \left (-3 b x^n \cos \left (a+b x^n\right )+3 b x^n \cos \left (3 \left (a+b x^n\right )\right )-3 b^2 x^{2 n} \operatorname {CosIntegral}\left (b x^n\right ) \sin (a)+9 b^2 x^{2 n} \operatorname {CosIntegral}\left (3 b x^n\right ) \sin (3 a)-3 \sin \left (a+b x^n\right )+\sin \left (3 \left (a+b x^n\right )\right )-3 b^2 x^{2 n} \cos (a) \text {Si}\left (b x^n\right )+9 b^2 x^{2 n} \cos (3 a) \text {Si}\left (3 b x^n\right )\right )}{8 n} \] Input:

Integrate[x^(-1 - 2*n)*Sin[a + b*x^n]^3,x]
 

Output:

(-3*b*x^n*Cos[a + b*x^n] + 3*b*x^n*Cos[3*(a + b*x^n)] - 3*b^2*x^(2*n)*CosI 
ntegral[b*x^n]*Sin[a] + 9*b^2*x^(2*n)*CosIntegral[3*b*x^n]*Sin[3*a] - 3*Si 
n[a + b*x^n] + Sin[3*(a + b*x^n)] - 3*b^2*x^(2*n)*Cos[a]*SinIntegral[b*x^n 
] + 9*b^2*x^(2*n)*Cos[3*a]*SinIntegral[3*b*x^n])/(8*n*x^(2*n))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3906, 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.

\(\displaystyle \int x^{-2 n-1} \sin ^3\left (a+b x^n\right ) \, dx\)

\(\Big \downarrow \) 3906

\(\displaystyle \int \left (\frac {3}{4} x^{-2 n-1} \sin \left (a+b x^n\right )-\frac {1}{4} x^{-2 n-1} \sin \left (3 a+3 b x^n\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 b^2 \sin (a) \operatorname {CosIntegral}\left (b x^n\right )}{8 n}+\frac {9 b^2 \sin (3 a) \operatorname {CosIntegral}\left (3 b x^n\right )}{8 n}-\frac {3 b^2 \cos (a) \text {Si}\left (b x^n\right )}{8 n}+\frac {9 b^2 \cos (3 a) \text {Si}\left (3 b x^n\right )}{8 n}-\frac {3 x^{-2 n} \sin \left (a+b x^n\right )}{8 n}+\frac {x^{-2 n} \sin \left (3 \left (a+b x^n\right )\right )}{8 n}-\frac {3 b x^{-n} \cos \left (a+b x^n\right )}{8 n}+\frac {3 b x^{-n} \cos \left (3 \left (a+b x^n\right )\right )}{8 n}\)

Input:

Int[x^(-1 - 2*n)*Sin[a + b*x^n]^3,x]
 

Output:

(-3*b*Cos[a + b*x^n])/(8*n*x^n) + (3*b*Cos[3*(a + b*x^n)])/(8*n*x^n) - (3* 
b^2*CosIntegral[b*x^n]*Sin[a])/(8*n) + (9*b^2*CosIntegral[3*b*x^n]*Sin[3*a 
])/(8*n) - (3*Sin[a + b*x^n])/(8*n*x^(2*n)) + Sin[3*(a + b*x^n)]/(8*n*x^(2 
*n)) - (3*b^2*Cos[a]*SinIntegral[b*x^n])/(8*n) + (9*b^2*Cos[3*a]*SinIntegr 
al[3*b*x^n])/(8*n)
 

Defintions of rubi rules used

rule 2009

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

rule 3906

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x 
_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

Maple [A] (verified)

Time = 5.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {3 b^{2} \left (-\frac {\sin \left (a +b \,x^{n}\right ) x^{-2 n}}{2 b^{2}}-\frac {\cos \left (a +b \,x^{n}\right ) x^{-n}}{2 b}-\frac {\operatorname {Si}\left (b \,x^{n}\right ) \cos \left (a \right )}{2}-\frac {\operatorname {Ci}\left (b \,x^{n}\right ) \sin \left (a \right )}{2}\right )}{4 n}-\frac {9 b^{2} \left (-\frac {\sin \left (3 a +3 b \,x^{n}\right ) x^{-2 n}}{18 b^{2}}-\frac {\cos \left (3 a +3 b \,x^{n}\right ) x^{-n}}{6 b}-\frac {\operatorname {Si}\left (3 b \,x^{n}\right ) \cos \left (3 a \right )}{2}-\frac {\operatorname {Ci}\left (3 b \,x^{n}\right ) \sin \left (3 a \right )}{2}\right )}{4 n}\)	\(144\)
risch	\(-\frac {\left (-3 i b^{2} {\mathrm e}^{-i a} \operatorname {expIntegral}_{1}\left (-i b \,x^{n}\right ) x^{2 n}-3 b^{2} {\mathrm e}^{-i a} \pi \,\operatorname {csgn}\left (b \,x^{n}\right ) x^{2 n}-9 i b^{2} {\mathrm e}^{3 i a} \operatorname {expIntegral}_{1}\left (-3 i b \,x^{n}\right ) x^{2 n}+9 i b^{2} {\mathrm e}^{-3 i a} \operatorname {expIntegral}_{1}\left (-3 i b \,x^{n}\right ) x^{2 n}+3 i b^{2} {\mathrm e}^{i a} \operatorname {expIntegral}_{1}\left (-i b \,x^{n}\right ) x^{2 n}+9 b^{2} {\mathrm e}^{-3 i a} \pi \,\operatorname {csgn}\left (b \,x^{n}\right ) x^{2 n}+6 b^{2} {\mathrm e}^{-i a} \operatorname {Si}\left (b \,x^{n}\right ) x^{2 n}-18 b^{2} {\mathrm e}^{-3 i a} \operatorname {Si}\left (3 b \,x^{n}\right ) x^{2 n}+6 x^{n} \cos \left (a +b \,x^{n}\right ) b -6 \cos \left (3 a +3 b \,x^{n}\right ) b \,x^{n}+6 \sin \left (a +b \,x^{n}\right )-2 \sin \left (3 a +3 b \,x^{n}\right )\right ) x^{-2 n}}{16 n}\)	\(253\)

Input:

int(x^(-1-2*n)*sin(a+b*x^n)^3,x,method=_RETURNVERBOSE)
 

Output:

3/4/n*b^2*(-1/2*sin(a+b*x^n)/b^2/(x^n)^2-1/2*cos(a+b*x^n)/b/(x^n)-1/2*Si(b 
*x^n)*cos(a)-1/2*Ci(b*x^n)*sin(a))-9/4/n*b^2*(-1/18*sin(3*a+3*b*x^n)/b^2/( 
x^n)^2-1/6*cos(3*a+3*b*x^n)/b/(x^n)-1/2*Si(3*b*x^n)*cos(3*a)-1/2*Ci(3*b*x^ 
n)*sin(3*a))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86 \[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=\frac {12 \, b x^{n} \cos \left (b x^{n} + a\right )^{3} + 9 \, b^{2} x^{2 \, n} \operatorname {Ci}\left (3 \, b x^{n}\right ) \sin \left (3 \, a\right ) - 3 \, b^{2} x^{2 \, n} \operatorname {Ci}\left (b x^{n}\right ) \sin \left (a\right ) + 9 \, b^{2} x^{2 \, n} \cos \left (3 \, a\right ) \operatorname {Si}\left (3 \, b x^{n}\right ) - 3 \, b^{2} x^{2 \, n} \cos \left (a\right ) \operatorname {Si}\left (b x^{n}\right ) - 12 \, b x^{n} \cos \left (b x^{n} + a\right ) + 4 \, {\left (\cos \left (b x^{n} + a\right )^{2} - 1\right )} \sin \left (b x^{n} + a\right )}{8 \, n x^{2 \, n}} \] Input:

integrate(x^(-1-2*n)*sin(a+b*x^n)^3,x, algorithm="fricas")
 

Output:

1/8*(12*b*x^n*cos(b*x^n + a)^3 + 9*b^2*x^(2*n)*cos_integral(3*b*x^n)*sin(3 
*a) - 3*b^2*x^(2*n)*cos_integral(b*x^n)*sin(a) + 9*b^2*x^(2*n)*cos(3*a)*si 
n_integral(3*b*x^n) - 3*b^2*x^(2*n)*cos(a)*sin_integral(b*x^n) - 12*b*x^n* 
cos(b*x^n + a) + 4*(cos(b*x^n + a)^2 - 1)*sin(b*x^n + a))/(n*x^(2*n))
 

Sympy [F]

\[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=\int x^{- 2 n - 1} \sin ^{3}{\left (a + b x^{n} \right )}\, dx \] Input:

integrate(x**(-1-2*n)*sin(a+b*x**n)**3,x)
 

Output:

Integral(x**(-2*n - 1)*sin(a + b*x**n)**3, x)
 

Maxima [F]

\[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=\int { x^{-2 \, n - 1} \sin \left (b x^{n} + a\right )^{3} \,d x } \] Input:

integrate(x^(-1-2*n)*sin(a+b*x^n)^3,x, algorithm="maxima")
 

Output:

integrate(x^(-2*n - 1)*sin(b*x^n + a)^3, x)
 

Giac [F]

\[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=\int { x^{-2 \, n - 1} \sin \left (b x^{n} + a\right )^{3} \,d x } \] Input:

integrate(x^(-1-2*n)*sin(a+b*x^n)^3,x, algorithm="giac")
 

Output:

integrate(x^(-2*n - 1)*sin(b*x^n + a)^3, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx=\int \frac {{\sin \left (a+b\,x^n\right )}^3}{x^{2\,n+1}} \,d x \] Input:

int(sin(a + b*x^n)^3/x^(2*n + 1),x)
 

Output:

int(sin(a + b*x^n)^3/x^(2*n + 1), x)
 

Reduce [F]

\[ \int x^{-1-2 n} \sin ^3\left (a+b x^n\right ) \, dx =\text {Too large to display} \] Input:

int(x^(-1-2*n)*sin(a+b*x^n)^3,x)
 

Output:

( - 9*cos(x**n*b + a)*sin(x**n*b + a)*tan((x**n*b + a)/2)**6 - 27*cos(x**n 
*b + a)*sin(x**n*b + a)*tan((x**n*b + a)/2)**4 - 27*cos(x**n*b + a)*sin(x* 
*n*b + a)*tan((x**n*b + a)/2)**2 - 9*cos(x**n*b + a)*sin(x**n*b + a) + 160 
*x**(2*n)*int(tan((x**n*b + a)/2)**3/(x**(2*n)*tan((x**n*b + a)/2)**6*x + 
3*x**(2*n)*tan((x**n*b + a)/2)**4*x + 3*x**(2*n)*tan((x**n*b + a)/2)**2*x 
+ x**(2*n)*x),x)*tan((x**n*b + a)/2)**6*n + 480*x**(2*n)*int(tan((x**n*b + 
 a)/2)**3/(x**(2*n)*tan((x**n*b + a)/2)**6*x + 3*x**(2*n)*tan((x**n*b + a) 
/2)**4*x + 3*x**(2*n)*tan((x**n*b + a)/2)**2*x + x**(2*n)*x),x)*tan((x**n* 
b + a)/2)**4*n + 480*x**(2*n)*int(tan((x**n*b + a)/2)**3/(x**(2*n)*tan((x* 
*n*b + a)/2)**6*x + 3*x**(2*n)*tan((x**n*b + a)/2)**4*x + 3*x**(2*n)*tan(( 
x**n*b + a)/2)**2*x + x**(2*n)*x),x)*tan((x**n*b + a)/2)**2*n + 160*x**(2* 
n)*int(tan((x**n*b + a)/2)**3/(x**(2*n)*tan((x**n*b + a)/2)**6*x + 3*x**(2 
*n)*tan((x**n*b + a)/2)**4*x + 3*x**(2*n)*tan((x**n*b + a)/2)**2*x + x**(2 
*n)*x),x)*n + 2*sin(x**n*b + a)**3*tan((x**n*b + a)/2)**6 + 6*sin(x**n*b + 
 a)**3*tan((x**n*b + a)/2)**4 + 6*sin(x**n*b + a)**3*tan((x**n*b + a)/2)** 
2 + 2*sin(x**n*b + a)**3 - 24*sin(x**n*b + a)*tan((x**n*b + a)/2)**6 - 72* 
sin(x**n*b + a)*tan((x**n*b + a)/2)**4 - 72*sin(x**n*b + a)*tan((x**n*b + 
a)/2)**2 - 24*sin(x**n*b + a) + 30*tan((x**n*b + a)/2)**5 + 80*tan((x**n*b 
 + a)/2)**3 + 66*tan((x**n*b + a)/2))/(20*x**(2*n)*n*(tan((x**n*b + a)/2)* 
*6 + 3*tan((x**n*b + a)/2)**4 + 3*tan((x**n*b + a)/2)**2 + 1))