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

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

[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)

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Rubi [A]  time = 0.0653922, antiderivative size = 87, normalized size of antiderivative = 1., 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 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]))

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 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 8

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

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 = 0.126866, 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)

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Maple [B]  time = 0.013, size = 197, normalized size = 2.3 \begin{align*} -{\frac{1}{{b}^{3}} \left ( \left ( a+{\frac{b}{x}} \right ) ^{2} \left ( -{\frac{1}{2}\cos \left ( a+{\frac{b}{x}} \right ) \sin \left ( a+{\frac{b}{x}} \right ) }+{\frac{a}{2}}+{\frac{b}{2\,x}} \right ) -{\frac{1}{2} \left ( a+{\frac{b}{x}} \right ) \left ( \cos \left ( a+{\frac{b}{x}} \right ) \right ) ^{2}}+{\frac{1}{4}\cos \left ( a+{\frac{b}{x}} \right ) \sin \left ( a+{\frac{b}{x}} \right ) }+{\frac{b}{4\,x}}+{\frac{a}{4}}-{\frac{1}{3} \left ( a+{\frac{b}{x}} \right ) ^{3}}-2\,a \left ( \left ( a+{\frac{b}{x}} \right ) \left ( -1/2\,\cos \left ( a+{\frac{b}{x}} \right ) \sin \left ( a+{\frac{b}{x}} \right ) +a/2+1/2\,{\frac{b}{x}} \right ) -1/4\, \left ( a+{\frac{b}{x}} \right ) ^{2}+1/4\, \left ( \sin \left ( a+{\frac{b}{x}} \right ) \right ) ^{2} \right ) +{a}^{2} \left ( -{\frac{1}{2}\cos \left ( a+{\frac{b}{x}} \right ) \sin \left ( a+{\frac{b}{x}} \right ) }+{\frac{a}{2}}+{\frac{b}{2\,x}} \right ) \right ) } \end{align*}

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))

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Maxima [C]  time = 1.13232, size = 92, normalized size = 1.06 \begin{align*} -\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}} \end{align*}

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)

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Fricas [A]  time = 1.28512, size = 158, normalized size = 1.82 \begin{align*} \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}} \end{align*}

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)

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Sympy [A]  time = 9.59835, size = 654, normalized size = 7.52 \begin{align*} \text{result too large to display} \end{align*}

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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (a + \frac{b}{x}\right )^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/x)^2/x^4, x)