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

Optimal. Leaf size=97 \[ \frac {2}{3} b^3 \sin (2 a) \text {Ci}\left (\frac {2 b}{x}\right )+\frac {2}{3} b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {x^3}{6} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3425, 3380, 3297, 3303, 3299, 3302} \[ \frac {2}{3} b^3 \sin (2 a) \text {CosIntegral}\left (\frac {2 b}{x}\right )+\frac {2}{3} b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[a + b/x]^2,x]

[Out]

x^3/6 + (b^2*x*Cos[2*(a + b/x)])/3 - (x^3*Cos[2*(a + b/x)])/6 + (2*b^3*CosIntegral[(2*b)/x]*Sin[2*a])/3 + (b*x
^2*Sin[2*(a + b/x)])/6 + (2*b^3*Cos[2*a]*SinIntegral[(2*b)/x])/3

Rule 3297

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 3425

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^2 \sin ^2\left (a+\frac {b}{x}\right ) \, dx &=\int \left (\frac {x^2}{2}-\frac {1}{2} x^2 \cos \left (2 a+\frac {2 b}{x}\right )\right ) \, dx\\ &=\frac {x^3}{6}-\frac {1}{2} \int x^2 \cos \left (2 a+\frac {2 b}{x}\right ) \, dx\\ &=\frac {x^3}{6}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\cos (2 a+2 b x)}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{3} b \operatorname {Subst}\left (\int \frac {\sin (2 a+2 b x)}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{3} b^2 \operatorname {Subst}\left (\int \frac {\cos (2 a+2 b x)}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{3} \left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {\sin (2 a+2 b x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{3} \left (2 b^3 \cos (2 a)\right ) \operatorname {Subst}\left (\int \frac {\sin (2 b x)}{x} \, dx,x,\frac {1}{x}\right )+\frac {1}{3} \left (2 b^3 \sin (2 a)\right ) \operatorname {Subst}\left (\int \frac {\cos (2 b x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {2}{3} b^3 \text {Ci}\left (\frac {2 b}{x}\right ) \sin (2 a)+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {2}{3} b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 86, normalized size = 0.89 \[ \frac {1}{6} \left (4 b^3 \sin (2 a) \text {Ci}\left (\frac {2 b}{x}\right )+4 b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )+x \left (2 b^2 \cos \left (2 \left (a+\frac {b}{x}\right )\right )-x^2 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+b x \sin \left (2 \left (a+\frac {b}{x}\right )\right )+x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[a + b/x]^2,x]

[Out]

(4*b^3*CosIntegral[(2*b)/x]*Sin[2*a] + x*(x^2 + 2*b^2*Cos[2*(a + b/x)] - x^2*Cos[2*(a + b/x)] + b*x*Sin[2*(a +
b/x)]) + 4*b^3*Cos[2*a]*SinIntegral[(2*b)/x])/6

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fricas [A]  time = 0.80, size = 109, normalized size = 1.12 \[ \frac {1}{3} \, b x^{2} \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right ) + \frac {2}{3} \, b^{3} \cos \left (2 \, a\right ) \operatorname {Si}\left (\frac {2 \, b}{x}\right ) - \frac {1}{3} \, b^{2} x + \frac {1}{3} \, x^{3} + \frac {1}{3} \, {\left (2 \, b^{2} x - x^{3}\right )} \cos \left (\frac {a x + b}{x}\right )^{2} + \frac {1}{3} \, {\left (b^{3} \operatorname {Ci}\left (\frac {2 \, b}{x}\right ) + b^{3} \operatorname {Ci}\left (-\frac {2 \, b}{x}\right )\right )} \sin \left (2 \, a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^2*cos((a*x + b)/x)*sin((a*x + b)/x) + 2/3*b^3*cos(2*a)*sin_integral(2*b/x) - 1/3*b^2*x + 1/3*x^3 + 1/3
*(2*b^2*x - x^3)*cos((a*x + b)/x)^2 + 1/3*(b^3*cos_integral(2*b/x) + b^3*cos_integral(-2*b/x))*sin(2*a)

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giac [B]  time = 1.12, size = 442, normalized size = 4.56 \[ \frac {4 \, a^{3} b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right ) - 4 \, a^{3} b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a^{2} b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right )}{x} + \frac {12 \, {\left (a x + b\right )} a^{2} b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - 2 \, a^{2} b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {12 \, {\left (a x + b\right )}^{2} a b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right )}{x^{2}} - \frac {12 \, {\left (a x + b\right )}^{2} a b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} + \frac {4 \, {\left (a x + b\right )} a b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - \frac {4 \, {\left (a x + b\right )}^{3} b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right )}{x^{3}} + a b^{4} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {4 \, {\left (a x + b\right )}^{3} b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{3}} + b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {2 \, {\left (a x + b\right )}^{2} b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {{\left (a x + b\right )} b^{4} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - b^{4}}{6 \, {\left (a^{3} - \frac {3 \, {\left (a x + b\right )} a^{2}}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a}{x^{2}} - \frac {{\left (a x + b\right )}^{3}}{x^{3}}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*a^3*b^4*cos_integral(-2*a + 2*(a*x + b)/x)*sin(2*a) - 4*a^3*b^4*cos(2*a)*sin_integral(2*a - 2*(a*x + b)
/x) - 12*(a*x + b)*a^2*b^4*cos_integral(-2*a + 2*(a*x + b)/x)*sin(2*a)/x + 12*(a*x + b)*a^2*b^4*cos(2*a)*sin_i
ntegral(2*a - 2*(a*x + b)/x)/x - 2*a^2*b^4*cos(2*(a*x + b)/x) + 12*(a*x + b)^2*a*b^4*cos_integral(-2*a + 2*(a*
x + b)/x)*sin(2*a)/x^2 - 12*(a*x + b)^2*a*b^4*cos(2*a)*sin_integral(2*a - 2*(a*x + b)/x)/x^2 + 4*(a*x + b)*a*b
^4*cos(2*(a*x + b)/x)/x - 4*(a*x + b)^3*b^4*cos_integral(-2*a + 2*(a*x + b)/x)*sin(2*a)/x^3 + a*b^4*sin(2*(a*x
+ b)/x) + 4*(a*x + b)^3*b^4*cos(2*a)*sin_integral(2*a - 2*(a*x + b)/x)/x^3 + b^4*cos(2*(a*x + b)/x) - 2*(a*x
+ b)^2*b^4*cos(2*(a*x + b)/x)/x^2 - (a*x + b)*b^4*sin(2*(a*x + b)/x)/x - b^4)/((a^3 - 3*(a*x + b)*a^2/x + 3*(a
*x + b)^2*a/x^2 - (a*x + b)^3/x^3)*b)

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maple [A]  time = 0.04, size = 96, normalized size = 0.99 \[ -b^{3} \left (-\frac {x^{3}}{6 b^{3}}+\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x^{3}}{6 b^{3}}-\frac {\sin \left (2 a +\frac {2 b}{x}\right ) x^{2}}{6 b^{2}}-\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x}{3 b}-\frac {2 \Si \left (\frac {2 b}{x}\right ) \cos \left (2 a \right )}{3}-\frac {2 \Ci \left (\frac {2 b}{x}\right ) \sin \left (2 a \right )}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(a+b/x)^2,x)

[Out]

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

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maxima [C]  time = 0.45, size = 99, normalized size = 1.02 \[ \frac {1}{6} \, {\left ({\left (-2 i \, {\rm Ei}\left (\frac {2 i \, b}{x}\right ) + 2 i \, {\rm Ei}\left (-\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) + 2 \, {\left ({\rm Ei}\left (\frac {2 i \, b}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} b^{3} + \frac {1}{6} \, b x^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {1}{6} \, x^{3} + \frac {1}{6} \, {\left (2 \, b^{2} x - x^{3}\right )} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\sin \left (a+\frac {b}{x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(a + b/x)^2,x)

[Out]

int(x^2*sin(a + b/x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sin ^{2}{\left (a + \frac {b}{x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sin(a + b/x)**2, x)

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