3.118 \(\int \frac{\sin ^2(a+\frac{b}{x})}{x^5} \, dx\)

Optimal. Leaf size=107 \[ -\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{4 b^2 x^2}+\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{8 b^4}-\frac{3 \sin \left (a+\frac{b}{x}\right ) \cos \left (a+\frac{b}{x}\right )}{4 b^3 x}+\frac{\sin \left (a+\frac{b}{x}\right ) \cos \left (a+\frac{b}{x}\right )}{2 b x^3}+\frac{3}{8 b^2 x^2}-\frac{1}{8 x^4} \]

[Out]

-1/(8*x^4) + 3/(8*b^2*x^2) + (Cos[a + b/x]*Sin[a + b/x])/(2*b*x^3) - (3*Cos[a + b/x]*Sin[a + b/x])/(4*b^3*x) +
(3*Sin[a + b/x]^2)/(8*b^4) - (3*Sin[a + b/x]^2)/(4*b^2*x^2)

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Rubi [A]  time = 0.0807238, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3379, 3311, 30, 3310} \[ -\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{4 b^2 x^2}+\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{8 b^4}-\frac{3 \sin \left (a+\frac{b}{x}\right ) \cos \left (a+\frac{b}{x}\right )}{4 b^3 x}+\frac{\sin \left (a+\frac{b}{x}\right ) \cos \left (a+\frac{b}{x}\right )}{2 b x^3}+\frac{3}{8 b^2 x^2}-\frac{1}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x]^2/x^5,x]

[Out]

-1/(8*x^4) + 3/(8*b^2*x^2) + (Cos[a + b/x]*Sin[a + b/x])/(2*b*x^3) - (3*Cos[a + b/x]*Sin[a + b/x])/(4*b^3*x) +
(3*Sin[a + b/x]^2)/(8*b^4) - (3*Sin[a + b/x]^2)/(4*b^2*x^2)

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 3310

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

Rubi steps

\begin{align*} \int \frac{\sin ^2\left (a+\frac{b}{x}\right )}{x^5} \, dx &=-\operatorname{Subst}\left (\int x^3 \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^3}-\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{4 b^2 x^2}-\frac{1}{2} \operatorname{Subst}\left (\int x^3 \, dx,x,\frac{1}{x}\right )+\frac{3 \operatorname{Subst}\left (\int x \sin ^2(a+b x) \, dx,x,\frac{1}{x}\right )}{2 b^2}\\ &=-\frac{1}{8 x^4}+\frac{\cos \left (a+\frac{b}{x}\right ) \sin \left (a+\frac{b}{x}\right )}{2 b x^3}-\frac{3 \cos \left (a+\frac{b}{x}\right ) \sin \left (a+\frac{b}{x}\right )}{4 b^3 x}+\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{8 b^4}-\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{4 b^2 x^2}+\frac{3 \operatorname{Subst}\left (\int x \, dx,x,\frac{1}{x}\right )}{4 b^2}\\ &=-\frac{1}{8 x^4}+\frac{3}{8 b^2 x^2}+\frac{\cos \left (a+\frac{b}{x}\right ) \sin \left (a+\frac{b}{x}\right )}{2 b x^3}-\frac{3 \cos \left (a+\frac{b}{x}\right ) \sin \left (a+\frac{b}{x}\right )}{4 b^3 x}+\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{8 b^4}-\frac{3 \sin ^2\left (a+\frac{b}{x}\right )}{4 b^2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.179913, size = 65, normalized size = 0.61 \[ -\frac{2 b \left (\left (3 x^3-2 b^2 x\right ) \sin \left (2 \left (a+\frac{b}{x}\right )\right )+b^3\right )+3 \left (x^4-2 b^2 x^2\right ) \cos \left (2 \left (a+\frac{b}{x}\right )\right )}{16 b^4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]^2/x^5,x]

[Out]

-(3*(-2*b^2*x^2 + x^4)*Cos[2*(a + b/x)] + 2*b*(b^3 + (-2*b^2*x + 3*x^3)*Sin[2*(a + b/x)]))/(16*b^4*x^4)

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Maple [B]  time = 0.012, size = 334, normalized size = 3.1 \begin{align*} -{\frac{1}{{b}^{4}} \left ( \left ( a+{\frac{b}{x}} \right ) ^{3} \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{3}{4} \left ( a+{\frac{b}{x}} \right ) ^{2} \left ( \cos \left ( a+{\frac{b}{x}} \right ) \right ) ^{2}}+{\frac{3}{2} \left ( a+{\frac{b}{x}} \right ) \left ({\frac{1}{2}\cos \left ( a+{\frac{b}{x}} \right ) \sin \left ( a+{\frac{b}{x}} \right ) }+{\frac{b}{2\,x}}+{\frac{a}{2}} \right ) }-{\frac{3}{8} \left ( a+{\frac{b}{x}} \right ) ^{2}}-{\frac{3}{8} \left ( \sin \left ( a+{\frac{b}{x}} \right ) \right ) ^{2}}-{\frac{3}{8} \left ( a+{\frac{b}{x}} \right ) ^{4}}-3\,a \left ( \left ( a+{\frac{b}{x}} \right ) ^{2} \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 ) -1/2\, \left ( a+{\frac{b}{x}} \right ) \left ( \cos \left ( a+{\frac{b}{x}} \right ) \right ) ^{2}+1/4\,\cos \left ( a+{\frac{b}{x}} \right ) \sin \left ( a+{\frac{b}{x}} \right ) +1/4\,{\frac{b}{x}}+a/4-1/3\, \left ( a+{\frac{b}{x}} \right ) ^{3} \right ) +3\,{a}^{2} \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}^{3} \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^5,x)

[Out]

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

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Maxima [C]  time = 1.13318, size = 92, normalized size = 0.86 \begin{align*} -\frac{{\left ({\left (\Gamma \left (4, \frac{2 i \, b}{x}\right ) + \Gamma \left (4, -\frac{2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) -{\left (i \, \Gamma \left (4, \frac{2 i \, b}{x}\right ) - i \, \Gamma \left (4, -\frac{2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{4} + 8 \, b^{4}}{64 \, b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)^2/x^5,x, algorithm="maxima")

[Out]

-1/64*(((gamma(4, 2*I*b/x) + gamma(4, -2*I*b/x))*cos(2*a) - (I*gamma(4, 2*I*b/x) - I*gamma(4, -2*I*b/x))*sin(2
*a))*x^4 + 8*b^4)/(b^4*x^4)

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Fricas [A]  time = 1.56288, size = 194, normalized size = 1.81 \begin{align*} -\frac{2 \, b^{4} + 6 \, b^{2} x^{2} - 3 \, x^{4} - 6 \,{\left (2 \, b^{2} x^{2} - x^{4}\right )} \cos \left (\frac{a x + b}{x}\right )^{2} - 4 \,{\left (2 \, b^{3} x - 3 \, b x^{3}\right )} \cos \left (\frac{a x + b}{x}\right ) \sin \left (\frac{a x + b}{x}\right )}{16 \, b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)^2/x^5,x, algorithm="fricas")

[Out]

-1/16*(2*b^4 + 6*b^2*x^2 - 3*x^4 - 6*(2*b^2*x^2 - x^4)*cos((a*x + b)/x)^2 - 4*(2*b^3*x - 3*b*x^3)*cos((a*x + b
)/x)*sin((a*x + b)/x))/(b^4*x^4)

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Sympy [A]  time = 16.8054, size = 726, normalized size = 6.79 \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**5,x)

[Out]

Piecewise((-b**4*tan(a/2 + b/(2*x))**4/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/(2*x))**2
+ 8*b**4*x**4) - 2*b**4*tan(a/2 + b/(2*x))**2/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/(
2*x))**2 + 8*b**4*x**4) - b**4/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/(2*x))**2 + 8*b**
4*x**4) - 8*b**3*x*tan(a/2 + b/(2*x))**3/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/(2*x))*
*2 + 8*b**4*x**4) + 8*b**3*x*tan(a/2 + b/(2*x))/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/
(2*x))**2 + 8*b**4*x**4) + 3*b**2*x**2*tan(a/2 + b/(2*x))**4/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4*x**4
*tan(a/2 + b/(2*x))**2 + 8*b**4*x**4) - 18*b**2*x**2*tan(a/2 + b/(2*x))**2/(8*b**4*x**4*tan(a/2 + b/(2*x))**4
+ 16*b**4*x**4*tan(a/2 + b/(2*x))**2 + 8*b**4*x**4) + 3*b**2*x**2/(8*b**4*x**4*tan(a/2 + b/(2*x))**4 + 16*b**4
*x**4*tan(a/2 + b/(2*x))**2 + 8*b**4*x**4) + 12*b*x**3*tan(a/2 + b/(2*x))**3/(8*b**4*x**4*tan(a/2 + b/(2*x))**
4 + 16*b**4*x**4*tan(a/2 + b/(2*x))**2 + 8*b**4*x**4) - 12*b*x**3*tan(a/2 + b/(2*x))/(8*b**4*x**4*tan(a/2 + b/
(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/(2*x))**2 + 8*b**4*x**4) + 12*x**4*tan(a/2 + b/(2*x))**2/(8*b**4*x**4*tan
(a/2 + b/(2*x))**4 + 16*b**4*x**4*tan(a/2 + b/(2*x))**2 + 8*b**4*x**4), Ne(b, 0)), (-sin(a)**2/(4*x**4), True)
)

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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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)^2/x^5,x, algorithm="giac")

[Out]

integrate(sin(a + b/x)^2/x^5, x)