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

Optimal. Leaf size=87 \[ -\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{4 b^3}-\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^2}+\frac {1}{4 b^2 x}-\frac {1}{6 x^3} \]

[Out]

-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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3379, 3311, 30, 2635, 8} \[ -\frac {\sin ^2\left (a+\frac {b}{x}\right )}{2 b^2 x}-\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{4 b^3}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^2}+\frac {1}{4 b^2 x}-\frac {1}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x]^2/x^4,x]

[Out]

-1/(6*x^3) + 1/(4*b^2*x) - (Cos[a + b/x]*Sin[a + b/x])/(4*b^3) + (Cos[a + b/x]*Sin[a + b/x])/(2*b*x^2) - Sin[a
+ b/x]^2/(2*b^2*x)

Rule 8

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

Rule 30

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

Rule 2635

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] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3311

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

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

\begin {align*} \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^4} \, dx &=-\operatorname {Subst}\left (\int x^2 \sin ^2(a+b x) \, dx,x,\frac {1}{x}\right )\\ &=\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}-\frac {1}{2} \operatorname {Subst}\left (\int x^2 \, dx,x,\frac {1}{x}\right )+\frac {\operatorname {Subst}\left (\int \sin ^2(a+b x) \, dx,x,\frac {1}{x}\right )}{2 b^2}\\ &=-\frac {1}{6 x^3}-\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}+\frac {\operatorname {Subst}\left (\int 1 \, dx,x,\frac {1}{x}\right )}{4 b^2}\\ &=-\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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 54, normalized size = 0.62 \[ \frac {-3 \left (x^3-2 b^2 x\right ) \sin \left (2 \left (a+\frac {b}{x}\right )\right )+6 b x^2 \cos \left (2 \left (a+\frac {b}{x}\right )\right )-4 b^3}{24 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]^2/x^4,x]

[Out]

(-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)

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 72, normalized size = 0.83 \[ \frac {6 \, b x^{2} \cos \left (\frac {a x + b}{x}\right )^{2} - 2 \, b^{3} - 3 \, b x^{2} + 3 \, {\left (2 \, b^{2} x - x^{3}\right )} \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right )}{12 \, b^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

giac [A]  time = 0.38, size = 153, normalized size = 1.76 \[ \frac {6 \, a^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a^{2}}{x} - 6 \, a \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {12 \, {\left (a x + b\right )}^{2} a}{x^{2}} + \frac {6 \, {\left (a x + b\right )} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {6 \, {\left (a x + b\right )}^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3}}{x^{3}} - 3 \, \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{24 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.07, size = 197, normalized size = 2.26 \[ -\frac {\left (a +\frac {b}{x}\right )^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right ) \left (\cos ^{2}\left (a +\frac {b}{x}\right )\right )}{2}+\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{4}+\frac {b}{4 x}+\frac {a}{4}-\frac {\left (a +\frac {b}{x}\right )^{3}}{3}-2 a \left (\left (a +\frac {b}{x}\right ) \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right )^{2}}{4}+\frac {\left (\sin ^{2}\left (a +\frac {b}{x}\right )\right )}{4}\right )+a^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [C]  time = 0.39, size = 68, normalized size = 0.78 \[ -\frac {{\left ({\left (3 i \, \Gamma \left (3, \frac {2 i \, b}{x}\right ) - 3 i \, \Gamma \left (3, -\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) + 3 \, {\left (\Gamma \left (3, \frac {2 i \, b}{x}\right ) + \Gamma \left (3, -\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{3} + 16 \, b^{3}}{96 \, b^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.70, size = 64, normalized size = 0.74 \[ \frac {\frac {b\,x^2\,\cos \left (2\,a+\frac {2\,b}{x}\right )}{4}-\frac {b^3}{6}+\frac {b^2\,x\,\sin \left (2\,a+\frac {2\,b}{x}\right )}{4}}{b^3\,x^3}-\frac {\sin \left (2\,a+\frac {2\,b}{x}\right )}{8\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x)^2/x^4,x)

[Out]

((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)

________________________________________________________________________________________

sympy [A]  time = 5.97, size = 654, normalized size = 7.52 \[ \begin {cases} - \frac {2 b^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} - \frac {4 b^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} - \frac {2 b^{3}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} - \frac {12 b^{2} x \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} + \frac {12 b^{2} x \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} + \frac {3 b x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} - \frac {18 b x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} + \frac {3 b x^{2}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} + \frac {6 x^{3} \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} - \frac {6 x^{3} \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{12 b^{3} x^{3} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 24 b^{3} x^{3} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 12 b^{3} x^{3}} & \text {for}\: b \neq 0 \\- \frac {\sin ^{2}{\relax (a )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*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*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*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), Ne(b, 0)), (-sin(a)**2/(3*x**3), Tru
e))

________________________________________________________________________________________