\(\int x^{-1-2 n} \sin ^2(a+b x^n) \, dx\) [151]

Optimal result

Integrand size = 18, antiderivative size = 97 \[ \int x^{-1-2 n} \sin ^2\left (a+b x^n\right ) \, dx=-\frac {x^{-2 n}}{4 n}+\frac {x^{-2 n} \cos \left (2 a+2 b x^n\right )}{4 n}+\frac {b^2 \cos (2 a) \operatorname {CosIntegral}\left (2 b x^n\right )}{n}-\frac {b x^{-n} \sin \left (2 a+2 b x^n\right )}{2 n}-\frac {b^2 \sin (2 a) \text {Si}\left (2 b x^n\right )}{n} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int x^{-1-2 n} \sin ^2\left (a+b x^n\right ) \, dx=\frac {x^{-2 n} \left (-1+\cos \left (2 \left (a+b x^n\right )\right )+4 b^2 x^{2 n} \cos (2 a) \operatorname {CosIntegral}\left (2 b x^n\right )-2 b x^n \sin \left (2 \left (a+b x^n\right )\right )-4 b^2 x^{2 n} \sin (2 a) \text {Si}\left (2 b x^n\right )\right )}{4 n} \] Input:

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

Output:

(-1 + Cos[2*(a + b*x^n)] + 4*b^2*x^(2*n)*Cos[2*a]*CosIntegral[2*b*x^n] - 2 
*b*x^n*Sin[2*(a + b*x^n)] - 4*b^2*x^(2*n)*Sin[2*a]*SinIntegral[2*b*x^n])/( 
4*n*x^(2*n))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, 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.

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

\(\Big \downarrow \) 3906

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/4*1/(n*x^(2*n)) + Cos[2*(a + b*x^n)]/(4*n*x^(2*n)) + (b^2*Cos[2*a]*CosI 
ntegral[2*b*x^n])/n - (b*Sin[2*(a + b*x^n)])/(2*n*x^n) - (b^2*Sin[2*a]*Sin 
Integral[2*b*x^n])/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 = 1.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {x^{-2 n}}{4 n}-\frac {2 b^{2} \left (-\frac {x^{-2 n} \cos \left (2 a +2 b \,x^{n}\right )}{8 b^{2}}+\frac {\sin \left (2 a +2 b \,x^{n}\right ) x^{-n}}{4 b}+\frac {\operatorname {Si}\left (2 b \,x^{n}\right ) \sin \left (2 a \right )}{2}-\frac {\operatorname {Ci}\left (2 b \,x^{n}\right ) \cos \left (2 a \right )}{2}\right )}{n}\)	\(89\)
risch	\(\frac {\left (2 i b^{2} {\mathrm e}^{-2 i a} \pi \,\operatorname {csgn}\left (b \,x^{n}\right ) x^{2 n}-4 i b^{2} {\mathrm e}^{-2 i a} \operatorname {Si}\left (2 b \,x^{n}\right ) x^{2 n}-2 b^{2} {\mathrm e}^{2 i a} \operatorname {expIntegral}_{1}\left (-2 i b \,x^{n}\right ) x^{2 n}-2 b^{2} {\mathrm e}^{-2 i a} \operatorname {expIntegral}_{1}\left (-2 i b \,x^{n}\right ) x^{2 n}-2 \sin \left (2 a +2 b \,x^{n}\right ) b \,x^{n}+\cos \left (2 a +2 b \,x^{n}\right )-1\right ) x^{-2 n}}{4 n}\)	\(135\)

Input:

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

Output:

-1/4/n/(x^n)^2-2/n*b^2*(-1/8/b^2/(x^n)^2*cos(2*a+2*b*x^n)+1/4*sin(2*a+2*b* 
x^n)/b/(x^n)+1/2*Si(2*b*x^n)*sin(2*a)-1/2*Ci(2*b*x^n)*cos(2*a))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int x^{-1-2 n} \sin ^2\left (a+b x^n\right ) \, dx=\frac {2 \, b^{2} x^{2 \, n} \cos \left (2 \, a\right ) \operatorname {Ci}\left (2 \, b x^{n}\right ) - 2 \, b^{2} x^{2 \, n} \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x^{n}\right ) - 2 \, b x^{n} \cos \left (b x^{n} + a\right ) \sin \left (b x^{n} + a\right ) + \cos \left (b x^{n} + a\right )^{2} - 1}{2 \, n x^{2 \, n}} \] Input:

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

Output:

1/2*(2*b^2*x^(2*n)*cos(2*a)*cos_integral(2*b*x^n) - 2*b^2*x^(2*n)*sin(2*a) 
*sin_integral(2*b*x^n) - 2*b*x^n*cos(b*x^n + a)*sin(b*x^n + a) + cos(b*x^n 
 + a)^2 - 1)/(n*x^(2*n))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-1/4*(2*n*x^(2*n)*integrate(cos(2*b*x^n + 2*a)/(x*x^(2*n)), x) + 1)/(n*x^( 
2*n))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*x**n*cos(x**n*b + a)*sin(x**n*b + a)*tan((x**n*b + a)/2)**4*b + 4*x**n* 
cos(x**n*b + a)*sin(x**n*b + a)*tan((x**n*b + a)/2)**2*b + 2*x**n*cos(x**n 
*b + a)*sin(x**n*b + a)*b - 8*cos(x**n*b + a)*tan((x**n*b + a)/2)**4 - 16* 
cos(x**n*b + a)*tan((x**n*b + a)/2)**2 - 8*cos(x**n*b + a) + 32*x**(2*n)*i 
nt(tan((x**n*b + a)/2)/(x**n*tan((x**n*b + a)/2)**4*x + 2*x**n*tan((x**n*b 
 + a)/2)**2*x + x**n*x),x)*tan((x**n*b + a)/2)**4*b*n + 64*x**(2*n)*int(ta 
n((x**n*b + a)/2)/(x**n*tan((x**n*b + a)/2)**4*x + 2*x**n*tan((x**n*b + a) 
/2)**2*x + x**n*x),x)*tan((x**n*b + a)/2)**2*b*n + 32*x**(2*n)*int(tan((x* 
*n*b + a)/2)/(x**n*tan((x**n*b + a)/2)**4*x + 2*x**n*tan((x**n*b + a)/2)** 
2*x + x**n*x),x)*b*n - 16*x**(2*n)*int(1/(tan((x**n*b + a)/2)**4*x + 2*tan 
((x**n*b + a)/2)**2*x + x),x)*tan((x**n*b + a)/2)**4*b**2*n - 32*x**(2*n)* 
int(1/(tan((x**n*b + a)/2)**4*x + 2*tan((x**n*b + a)/2)**2*x + x),x)*tan(( 
x**n*b + a)/2)**2*b**2*n - 16*x**(2*n)*int(1/(tan((x**n*b + a)/2)**4*x + 2 
*tan((x**n*b + a)/2)**2*x + x),x)*b**2*n + 6*x**(2*n)*log(x)*tan((x**n*b + 
 a)/2)**4*b**2*n + 12*x**(2*n)*log(x)*tan((x**n*b + a)/2)**2*b**2*n + 6*x* 
*(2*n)*log(x)*b**2*n + 8*x**n*sin(x**n*b + a)*tan((x**n*b + a)/2)**4*b + 1 
6*x**n*sin(x**n*b + a)*tan((x**n*b + a)/2)**2*b + 8*x**n*sin(x**n*b + a)*b 
 + sin(x**n*b + a)**2*tan((x**n*b + a)/2)**4 + 2*sin(x**n*b + a)**2*tan((x 
**n*b + a)/2)**2 + sin(x**n*b + a)**2 - 8*tan((x**n*b + a)/2)**4 - 16*tan( 
(x**n*b + a)/2)**2 + 8)/(6*x**(2*n)*n*(tan((x**n*b + a)/2)**4 + 2*tan((...