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

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, 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 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]

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^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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.61, size = 60, normalized size = 1.18 \[ \frac {2 \, x^{2} \cos \left (\frac {a x + b}{x}\right )^{2} + 4 \, b x \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right ) - 2 \, b^{2} - x^{2}}{8 \, b^{2} x^{2}} \]

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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giac [A]  time = 0.71, size = 77, normalized size = 1.51 \[ -\frac {2 \, a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {4 \, {\left (a x + b\right )} a}{x} - \frac {2 \, {\left (a x + b\right )} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {2 \, {\left (a x + b\right )}^{2}}{x^{2}} - \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 97, normalized size = 1.90 \[ -\frac {\left (a +\frac {b}{x}\right ) \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right )^{2}}{4}+\frac {\left (\sin ^{2}\left (a +\frac {b}{x}\right )\right )}{4}-a \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )}{b^{2}} \]

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 = 0.37, size = 68, normalized size = 1.33 \[ \frac {{\left ({\left (\Gamma \left (2, \frac {2 i \, b}{x}\right ) + \Gamma \left (2, -\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) - {\left (i \, \Gamma \left (2, \frac {2 i \, b}{x}\right ) - i \, \Gamma \left (2, -\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{2} - 4 \, b^{2}}{16 \, b^{2} x^{2}} \]

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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mupad [B]  time = 4.62, size = 41, normalized size = 0.80 \[ \frac {\cos \left (2\,a+\frac {2\,b}{x}\right )}{8\,b^2}-\frac {1}{4\,x^2}+\frac {\sin \left (2\,a+\frac {2\,b}{x}\right )}{4\,b\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x)^2/x^3,x)

[Out]

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

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sympy [A]  time = 4.21, size = 391, normalized size = 7.67 \[ \begin {cases} - \frac {b^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac {2 b^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac {b^{2}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac {4 b x \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 4 b^{2} x^{2}} + \frac {4 b x \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 4 b^{2} x^{2}} - \frac {4 x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{4 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 4 b^{2} x^{2}} & \text {for}\: b \neq 0 \\- \frac {\sin ^{2}{\relax (a )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

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) - 4*x**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), Ne(b, 0)), (-sin(a)**2/(2*x**2), True))

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