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

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 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 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^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*}

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Mathematica [A]  time = 0.20, size = 65, normalized size = 0.61 \[ -\frac {3 \left (x^4-2 b^2 x^2\right ) \cos \left (2 \left (a+\frac {b}{x}\right )\right )+2 b \left (\left (3 x^3-2 b^2 x\right ) \sin \left (2 \left (a+\frac {b}{x}\right )\right )+b^3\right )}{16 b^4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]^2/x^5,x]

[Out]

-1/16*(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)]))/(b^4*x^4)

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fricas [A]  time = 0.50, size = 90, normalized size = 0.84 \[ -\frac {2 \, b^{4} + 6 \, b^{2} x^{2} - 3 \, x^{4} - 6 \, {\left (2 \, b^{2} x^{2} - x^{4}\right )} \cos \left (\frac {a x + b}{x}\right )^{2} - 4 \, {\left (2 \, b^{3} x - 3 \, b x^{3}\right )} \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right )}{16 \, b^{4} x^{4}} \]

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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giac [B]  time = 0.69, size = 255, normalized size = 2.38 \[ -\frac {4 \, a^{3} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {8 \, {\left (a x + b\right )} a^{3}}{x} - 6 \, a^{2} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {12 \, {\left (a x + b\right )}^{2} a^{2}}{x^{2}} + \frac {12 \, {\left (a x + b\right )} a \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - 6 \, a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {12 \, {\left (a x + b\right )}^{2} a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {8 \, {\left (a x + b\right )}^{3} a}{x^{3}} - \frac {6 \, {\left (a x + b\right )}^{2} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{3}} + \frac {6 \, {\left (a x + b\right )} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {2 \, {\left (a x + b\right )}^{4}}{x^{4}} + 3 \, \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{16 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a^3*sin(2*(a*x + b)/x) - 8*(a*x + b)*a^3/x - 6*a^2*cos(2*(a*x + b)/x) - 12*(a*x + b)*a^2*sin(2*(a*x +
b)/x)/x + 12*(a*x + b)^2*a^2/x^2 + 12*(a*x + b)*a*cos(2*(a*x + b)/x)/x - 6*a*sin(2*(a*x + b)/x) + 12*(a*x + b
)^2*a*sin(2*(a*x + b)/x)/x^2 - 8*(a*x + b)^3*a/x^3 - 6*(a*x + b)^2*cos(2*(a*x + b)/x)/x^2 - 4*(a*x + b)^3*sin(
2*(a*x + b)/x)/x^3 + 6*(a*x + b)*sin(2*(a*x + b)/x)/x + 2*(a*x + b)^4/x^4 + 3*cos(2*(a*x + b)/x))/b^4

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maple [B]  time = 0.08, size = 334, normalized size = 3.12 \[ -\frac {\left (a +\frac {b}{x}\right )^{3} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {3 \left (a +\frac {b}{x}\right )^{2} \left (\cos ^{2}\left (a +\frac {b}{x}\right )\right )}{4}+\frac {3 \left (a +\frac {b}{x}\right ) \left (\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {b}{2 x}+\frac {a}{2}\right )}{2}-\frac {3 \left (a +\frac {b}{x}\right )^{2}}{8}-\frac {3 \left (\sin ^{2}\left (a +\frac {b}{x}\right )\right )}{8}-\frac {3 \left (a +\frac {b}{x}\right )^{4}}{8}-3 a \left (\left (a +\frac {b}{x}\right )^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right ) \left (\cos ^{2}\left (a +\frac {b}{x}\right )\right )}{2}+\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{4}+\frac {b}{4 x}+\frac {a}{4}-\frac {\left (a +\frac {b}{x}\right )^{3}}{3}\right )+3 a^{2} \left (\left (a +\frac {b}{x}\right ) \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right )^{2}}{4}+\frac {\left (\sin ^{2}\left (a +\frac {b}{x}\right )\right )}{4}\right )-a^{3} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )}{b^{4}} \]

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 = 0.37, size = 68, normalized size = 0.64 \[ -\frac {{\left ({\left (\Gamma \left (4, \frac {2 i \, b}{x}\right ) + \Gamma \left (4, -\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) - {\left (i \, \Gamma \left (4, \frac {2 i \, b}{x}\right ) - i \, \Gamma \left (4, -\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{4} + 8 \, b^{4}}{64 \, b^{4} x^{4}} \]

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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mupad [B]  time = 4.72, size = 84, normalized size = 0.79 \[ -\frac {3\,\cos \left (2\,a+\frac {2\,b}{x}\right )}{16\,b^4}-\frac {\frac {b^4}{8}-\frac {3\,b^2\,x^2\,\cos \left (2\,a+\frac {2\,b}{x}\right )}{8}+\frac {3\,b\,x^3\,\sin \left (2\,a+\frac {2\,b}{x}\right )}{8}-\frac {b^3\,x\,\sin \left (2\,a+\frac {2\,b}{x}\right )}{4}}{b^4\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x)^2/x^5,x)

[Out]

- (3*cos(2*a + (2*b)/x))/(16*b^4) - (b^4/8 - (3*b^2*x^2*cos(2*a + (2*b)/x))/8 + (3*b*x^3*sin(2*a + (2*b)/x))/8
- (b^3*x*sin(2*a + (2*b)/x))/4)/(b^4*x^4)

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sympy [A]  time = 8.21, size = 726, normalized size = 6.79 \[ \begin {cases} - \frac {b^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {2 b^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {b^{4}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {8 b^{3} x \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {8 b^{3} x \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {3 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {18 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {3 b^{2} x^{2}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {12 b x^{3} \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {12 b x^{3} \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {12 x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} & \text {for}\: b \neq 0 \\- \frac {\sin ^{2}{\relax (a )}}{4 x^{4}} & \text {otherwise} \end {cases} \]

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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