3.116 \(\int \frac{\sin ^2(a+\frac{b}{x})}{x^3} \, dx\)

Optimal. Leaf size=51 \[ -\frac{\sin ^2\left (a+\frac{b}{x}\right )}{4 b^2}+\frac{\sin \left (a+\frac{b}{x}\right ) \cos \left (a+\frac{b}{x}\right )}{2 b x}-\frac{1}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.039533, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3379, 3310, 30} \[ -\frac{\sin ^2\left (a+\frac{b}{x}\right )}{4 b^2}+\frac{\sin \left (a+\frac{b}{x}\right ) \cos \left (a+\frac{b}{x}\right )}{2 b x}-\frac{1}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x]^2/x^3,x]

[Out]

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

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 0.0793681, size = 43, normalized size = 0.84 \[ \frac{x^2 \cos \left (2 \left (a+\frac{b}{x}\right )\right )-2 b \left (b-x \sin \left (2 \left (a+\frac{b}{x}\right )\right )\right )}{8 b^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]^2/x^3,x]

[Out]

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

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Maple [B]  time = 0.011, size = 97, normalized size = 1.9 \begin{align*} -{\frac{1}{{b}^{2}} \left ( \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{a}{2}}+{\frac{b}{2\,x}} \right ) -{\frac{1}{4} \left ( a+{\frac{b}{x}} \right ) ^{2}}+{\frac{1}{4} \left ( \sin \left ( a+{\frac{b}{x}} \right ) \right ) ^{2}}-a \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^3,x)

[Out]

-1/b^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*(-1/2*cos(a+b/x)*s
in(a+b/x)+1/2*a+1/2*b/x))

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Maxima [C]  time = 1.12844, size = 92, normalized size = 1.8 \begin{align*} \frac{{\left ({\left (\Gamma \left (2, \frac{2 i \, b}{x}\right ) + \Gamma \left (2, -\frac{2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) -{\left (i \, \Gamma \left (2, \frac{2 i \, b}{x}\right ) - i \, \Gamma \left (2, -\frac{2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{2} - 4 \, b^{2}}{16 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)^2/x^3,x, algorithm="maxima")

[Out]

1/16*(((gamma(2, 2*I*b/x) + gamma(2, -2*I*b/x))*cos(2*a) - (I*gamma(2, 2*I*b/x) - I*gamma(2, -2*I*b/x))*sin(2*
a))*x^2 - 4*b^2)/(b^2*x^2)

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Fricas [A]  time = 1.32934, size = 132, normalized size = 2.59 \begin{align*} \frac{2 \, x^{2} \cos \left (\frac{a x + b}{x}\right )^{2} + 4 \, b x \cos \left (\frac{a x + b}{x}\right ) \sin \left (\frac{a x + b}{x}\right ) - 2 \, b^{2} - x^{2}}{8 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)^2/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.38814, size = 445, normalized size = 8.73 \begin{align*} \begin{cases} - \frac{b^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac{2 b^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac{b^{2}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac{4 b x \tan ^{3}{\left (\frac{a}{2} + \frac{b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} + \frac{4 b x \tan{\left (\frac{a}{2} + \frac{b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} + \frac{2 x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} + \frac{2 x^{2}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac{a}{2} + \frac{b}{2 x} \right )} + 4 b^{2} x^{2}} & \text{for}\: b \neq 0 \\- \frac{\sin ^{2}{\left (a \right )}}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)**2/x**3,x)

[Out]

Piecewise((-b**2*tan(a/2 + b/(2*x))**4/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*x))**2
+ 4*b**2*x**2) - 2*b**2*tan(a/2 + b/(2*x))**2/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*
x))**2 + 4*b**2*x**2) - b**2/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*x))**2 + 4*b**2*x
**2) - 4*b*x*tan(a/2 + b/(2*x))**3/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*x))**2 + 4*
b**2*x**2) + 4*b*x*tan(a/2 + b/(2*x))/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*x))**2 +
4*b**2*x**2) + 2*x**2*tan(a/2 + b/(2*x))**4/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*x
))**2 + 4*b**2*x**2) + 2*x**2/(4*b**2*x**2*tan(a/2 + b/(2*x))**4 + 8*b**2*x**2*tan(a/2 + b/(2*x))**2 + 4*b**2*
x**2), Ne(b, 0)), (-sin(a)**2/(2*x**2), 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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)^2/x^3,x, algorithm="giac")

[Out]

integrate(sin(a + b/x)^2/x^3, x)