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

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

[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

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Rubi [A]  time = 0.169148, antiderivative size = 97, normalized size of antiderivative = 1., 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 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]

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 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 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 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]

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*}

Mathematica [A]  time = 0.169807, size = 86, normalized size = 0.89 \[ \frac{1}{6} \left (4 b^3 \sin (2 a) \text{CosIntegral}\left (\frac{2 b}{x}\right )+4 b^3 \cos (2 a) \text{Si}\left (\frac{2 b}{x}\right )+x \left (2 b^2 \cos \left (2 \left (a+\frac{b}{x}\right )\right )-x^2 \cos \left (2 \left (a+\frac{b}{x}\right )\right )+b x \sin \left (2 \left (a+\frac{b}{x}\right )\right )+x^2\right )\right ) \]

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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Maple [A]  time = 0.014, size = 96, normalized size = 1. \begin{align*} -{b}^{3} \left ( -{\frac{{x}^{3}}{6\,{b}^{3}}}+{\frac{{x}^{3}}{6\,{b}^{3}}\cos \left ( 2\,a+2\,{\frac{b}{x}} \right ) }-{\frac{{x}^{2}}{6\,{b}^{2}}\sin \left ( 2\,a+2\,{\frac{b}{x}} \right ) }-{\frac{x}{3\,b}\cos \left ( 2\,a+2\,{\frac{b}{x}} \right ) }-{\frac{2\,\cos \left ( 2\,a \right ) }{3}{\it Si} \left ( 2\,{\frac{b}{x}} \right ) }-{\frac{2\,\sin \left ( 2\,a \right ) }{3}{\it Ci} \left ( 2\,{\frac{b}{x}} \right ) } \right ) \end{align*}

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 = 1.15476, size = 134, normalized size = 1.38 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 1.44585, size = 290, normalized size = 2.99 \begin{align*} \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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sin ^{2}{\left (a + \frac{b}{x} \right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sin \left (a + \frac{b}{x}\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*sin(a + b/x)^2, x)