\(\int \frac {\sin ^2(a+\frac {b}{x})}{x^4} \, dx\) [117]

Optimal result

Integrand size = 14, antiderivative size = 87 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=-\frac {1}{6 x^3}+\frac {1}{4 b^2 x}-\frac {\cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{4 b^3}+\frac {\cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{2 b x^2}-\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x} \] Output:

-1/6/x^3+1/4/b^2/x-1/4*cos(a+b/x)*sin(a+b/x)/b^3+1/2*cos(a+b/x)*sin(a+b/x) 
/b/x^2-1/2*sin(a+b/x)^2/b^2/x
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.62 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {-4 b^3+6 b x^2 \cos \left (2 \left (a+\frac {b}{x}\right )\right )-3 \left (-2 b^2 x+x^3\right ) \sin \left (2 \left (a+\frac {b}{x}\right )\right )}{24 b^3 x^3} \] Input:

Integrate[Sin[a + b/x]^2/x^4,x]
 

Output:

(-4*b^3 + 6*b*x^2*Cos[2*(a + b/x)] - 3*(-2*b^2*x + x^3)*Sin[2*(a + b/x)])/ 
(24*b^3*x^3)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3860, 3042, 3792, 15, 3042, 3115, 24}

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.

\(\displaystyle \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx\)

\(\Big \downarrow \) 3860

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3792

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {\int \sin ^2\left (a+\frac {b}{x}\right )d\frac {1}{x}}{2 b^2}-\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^2}-\frac {1}{6 x^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (a+\frac {b}{x}\right )^2d\frac {1}{x}}{2 b^2}-\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^2}-\frac {1}{6 x^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {\int 1d\frac {1}{x}}{2}-\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b}}{2 b^2}-\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^2}-\frac {1}{6 x^3}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x}+\frac {\frac {1}{2 x}-\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b}}{2 b^2}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^2}-\frac {1}{6 x^3}\)

Input:

Int[Sin[a + b/x]^2/x^4,x]
 

Output:

-1/6*1/x^3 + (Cos[a + b/x]*Sin[a + b/x])/(2*b*x^2) - Sin[a + b/x]^2/(2*b^2 
*x) + (1/(2*x) - (Cos[a + b/x]*Sin[a + b/x])/(2*b))/(2*b^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3792

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

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64

method	result	size
risch	\(-\frac {1}{6 x^{3}}+\frac {\cos \left (\frac {2 a x +2 b}{x}\right )}{4 b^{2} x}+\frac {\left (2 b^{2}-x^{2}\right ) \sin \left (\frac {2 a x +2 b}{x}\right )}{8 b^{3} x^{2}}\)	\(56\)
parallelrisch	\(\frac {6 x^{2} b \cos \left (\frac {2 a x +2 b}{x}\right )+6 b^{2} x \sin \left (\frac {2 a x +2 b}{x}\right )-3 x^{3} \sin \left (\frac {2 a x +2 b}{x}\right )-4 b^{3}}{24 x^{3} b^{3}}\)	\(71\)
norman	\(\frac {-\frac {1}{6}+\frac {x \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b}-\frac {\tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{2}}{3}-\frac {\tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{4}}{6}+\frac {x^{2}}{4 b^{2}}-\frac {x^{3} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{2 b^{3}}+\frac {x^{3} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{3}}{2 b^{3}}-\frac {3 x^{2} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{2}}{2 b^{2}}+\frac {x^{2} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{4}}{4 b^{2}}-\frac {x \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{3}}{b}}{\left (1+\tan \left (\frac {a}{2}+\frac {b}{2 x}\right )^{2}\right )^{2} x^{3}}\)	\(179\)
orering	\(-\frac {\left (b^{4}+8 x^{2} b^{2}-9 x^{4}\right ) \sin \left (a +\frac {b}{x}\right )^{2}}{3 x^{3} b^{4}}-\frac {x^{4} \left (26 b^{2}-33 x^{2}\right ) \left (-\frac {2 \sin \left (a +\frac {b}{x}\right ) b \cos \left (a +\frac {b}{x}\right )}{x^{6}}-\frac {4 \sin \left (a +\frac {b}{x}\right )^{2}}{x^{5}}\right )}{24 b^{4}}-\frac {\left (2 b^{2}-3 x^{2}\right ) x^{5} \left (\frac {2 b^{2} \cos \left (a +\frac {b}{x}\right )^{2}}{x^{8}}+\frac {20 \sin \left (a +\frac {b}{x}\right ) b \cos \left (a +\frac {b}{x}\right )}{x^{7}}-\frac {2 \sin \left (a +\frac {b}{x}\right )^{2} b^{2}}{x^{8}}+\frac {20 \sin \left (a +\frac {b}{x}\right )^{2}}{x^{6}}\right )}{24 b^{4}}\)	\(187\)
derivativedivides	\(-\frac {a^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-2 a \left (\left (a +\frac {b}{x}\right ) \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right )^{2}}{4}+\frac {\sin \left (a +\frac {b}{x}\right )^{2}}{4}\right )+\left (a +\frac {b}{x}\right )^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )^{2}}{2}+\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{4}+\frac {b}{4 x}+\frac {a}{4}-\frac {\left (a +\frac {b}{x}\right )^{3}}{3}}{b^{3}}\)	\(197\)
default	\(-\frac {a^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-2 a \left (\left (a +\frac {b}{x}\right ) \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right )^{2}}{4}+\frac {\sin \left (a +\frac {b}{x}\right )^{2}}{4}\right )+\left (a +\frac {b}{x}\right )^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )^{2}}{2}+\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{4}+\frac {b}{4 x}+\frac {a}{4}-\frac {\left (a +\frac {b}{x}\right )^{3}}{3}}{b^{3}}\)	\(197\)

Input:

int(sin(a+b/x)^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/6/x^3+1/4/b^2/x*cos(2*(a*x+b)/x)+1/8*(2*b^2-x^2)/b^3/x^2*sin(2*(a*x+b)/ 
x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {6 \, b x^{2} \cos \left (\frac {a x + b}{x}\right )^{2} - 2 \, b^{3} - 3 \, b x^{2} + 3 \, {\left (2 \, b^{2} x - x^{3}\right )} \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right )}{12 \, b^{3} x^{3}} \] Input:

integrate(sin(a+b/x)^2/x^4,x, algorithm="fricas")
 

Output:

1/12*(6*b*x^2*cos((a*x + b)/x)^2 - 2*b^3 - 3*b*x^2 + 3*(2*b^2*x - x^3)*cos 
((a*x + b)/x)*sin((a*x + b)/x))/(b^3*x^3)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 654 vs. \(2 (68) = 136\).

Time = 1.73 (sec) , antiderivative size = 654, normalized size of antiderivative = 7.52 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx =\text {Too large to display} \] Input:

integrate(sin(a+b/x)**2/x**4,x)
 

Output:

Piecewise((-2*b**3*tan(a/2 + b/(2*x))**4/(12*b**3*x**3*tan(a/2 + b/(2*x))* 
*4 + 24*b**3*x**3*tan(a/2 + b/(2*x))**2 + 12*b**3*x**3) - 4*b**3*tan(a/2 + 
 b/(2*x))**2/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 24*b**3*x**3*tan(a/2 + 
b/(2*x))**2 + 12*b**3*x**3) - 2*b**3/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 
 24*b**3*x**3*tan(a/2 + b/(2*x))**2 + 12*b**3*x**3) - 12*b**2*x*tan(a/2 + 
b/(2*x))**3/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 24*b**3*x**3*tan(a/2 + b 
/(2*x))**2 + 12*b**3*x**3) + 12*b**2*x*tan(a/2 + b/(2*x))/(12*b**3*x**3*ta 
n(a/2 + b/(2*x))**4 + 24*b**3*x**3*tan(a/2 + b/(2*x))**2 + 12*b**3*x**3) + 
 3*b*x**2*tan(a/2 + b/(2*x))**4/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 24*b 
**3*x**3*tan(a/2 + b/(2*x))**2 + 12*b**3*x**3) - 18*b*x**2*tan(a/2 + b/(2* 
x))**2/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 24*b**3*x**3*tan(a/2 + b/(2*x 
))**2 + 12*b**3*x**3) + 3*b*x**2/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 24* 
b**3*x**3*tan(a/2 + b/(2*x))**2 + 12*b**3*x**3) + 6*x**3*tan(a/2 + b/(2*x) 
)**3/(12*b**3*x**3*tan(a/2 + b/(2*x))**4 + 24*b**3*x**3*tan(a/2 + b/(2*x)) 
**2 + 12*b**3*x**3) - 6*x**3*tan(a/2 + b/(2*x))/(12*b**3*x**3*tan(a/2 + b/ 
(2*x))**4 + 24*b**3*x**3*tan(a/2 + b/(2*x))**2 + 12*b**3*x**3), Ne(b, 0)), 
 (-sin(a)**2/(3*x**3), True))
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {3 \, {\left ({\left (-i \, \Gamma \left (3, \frac {2 i \, b}{x}\right ) + i \, \Gamma \left (3, -\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) - {\left (\Gamma \left (3, \frac {2 i \, b}{x}\right ) + \Gamma \left (3, -\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{3} - 16 \, b^{3}}{96 \, b^{3} x^{3}} \] Input:

integrate(sin(a+b/x)^2/x^4,x, algorithm="maxima")
 

Output:

1/96*(3*((-I*gamma(3, 2*I*b/x) + I*gamma(3, -2*I*b/x))*cos(2*a) - (gamma(3 
, 2*I*b/x) + gamma(3, -2*I*b/x))*sin(2*a))*x^3 - 16*b^3)/(b^3*x^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.76 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {6 \, a^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a^{2}}{x} - 6 \, a \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {12 \, {\left (a x + b\right )}^{2} a}{x^{2}} + \frac {6 \, {\left (a x + b\right )} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {6 \, {\left (a x + b\right )}^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3}}{x^{3}} - 3 \, \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{24 \, b^{3}} \] Input:

integrate(sin(a+b/x)^2/x^4,x, algorithm="giac")
 

Output:

1/24*(6*a^2*sin(2*(a*x + b)/x) - 12*(a*x + b)*a^2/x - 6*a*cos(2*(a*x + b)/ 
x) - 12*(a*x + b)*a*sin(2*(a*x + b)/x)/x + 12*(a*x + b)^2*a/x^2 + 6*(a*x + 
 b)*cos(2*(a*x + b)/x)/x + 6*(a*x + b)^2*sin(2*(a*x + b)/x)/x^2 - 4*(a*x + 
 b)^3/x^3 - 3*sin(2*(a*x + b)/x))/b^3
 

Mupad [B] (verification not implemented)

Time = 42.89 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {\frac {b\,x^2\,\cos \left (2\,a+\frac {2\,b}{x}\right )}{4}-\frac {b^3}{6}+\frac {b^2\,x\,\sin \left (2\,a+\frac {2\,b}{x}\right )}{4}}{b^3\,x^3}-\frac {\sin \left (2\,a+\frac {2\,b}{x}\right )}{8\,b^3} \] Input:

int(sin(a + b/x)^2/x^4,x)
                                                                                    
                                                                                    
 

Output:

((b*x^2*cos(2*a + (2*b)/x))/4 - b^3/6 + (b^2*x*sin(2*a + (2*b)/x))/4)/(b^3 
*x^3) - sin(2*a + (2*b)/x)/(8*b^3)
 

Reduce [F]

\[ \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx=\int \frac {\sin \left (\frac {a x +b}{x}\right )^{2}}{x^{4}}d x \] Input:

int(sin(a+b/x)^2/x^4,x)
 

Output:

int(sin((a*x + b)/x)**2/x**4,x)