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

Optimal. Leaf size=78 \[ \frac{1}{6} b^3 \cos (a) \text{CosIntegral}\left (\frac{b}{x}\right )-\frac{1}{6} b^3 \sin (a) \text{Si}\left (\frac{b}{x}\right )-\frac{1}{6} b^2 x \sin \left (a+\frac{b}{x}\right )+\frac{1}{3} x^3 \sin \left (a+\frac{b}{x}\right )+\frac{1}{6} b x^2 \cos \left (a+\frac{b}{x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.131339, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3379, 3297, 3303, 3299, 3302} \[ \frac{1}{6} b^3 \cos (a) \text{CosIntegral}\left (\frac{b}{x}\right )-\frac{1}{6} b^3 \sin (a) \text{Si}\left (\frac{b}{x}\right )-\frac{1}{6} b^2 x \sin \left (a+\frac{b}{x}\right )+\frac{1}{3} x^3 \sin \left (a+\frac{b}{x}\right )+\frac{1}{6} b x^2 \cos \left (a+\frac{b}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[a + b/x],x]

[Out]

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

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

Mathematica [A]  time = 0.073134, size = 70, normalized size = 0.9 \[ \frac{1}{6} \left (b^3 \cos (a) \text{CosIntegral}\left (\frac{b}{x}\right )-b^3 \sin (a) \text{Si}\left (\frac{b}{x}\right )+x \left (b^2 \left (-\sin \left (a+\frac{b}{x}\right )\right )+2 x^2 \sin \left (a+\frac{b}{x}\right )+b x \cos \left (a+\frac{b}{x}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[a + b/x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 73, normalized size = 0.9 \begin{align*} -{b}^{3} \left ( -{\frac{{x}^{3}}{3\,{b}^{3}}\sin \left ( a+{\frac{b}{x}} \right ) }-{\frac{{x}^{2}}{6\,{b}^{2}}\cos \left ( a+{\frac{b}{x}} \right ) }+{\frac{x}{6\,b}\sin \left ( a+{\frac{b}{x}} \right ) }+{\frac{\sin \left ( a \right ) }{6}{\it Si} \left ({\frac{b}{x}} \right ) }-{\frac{\cos \left ( a \right ) }{6}{\it Ci} \left ({\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),x)

[Out]

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

________________________________________________________________________________________

Maxima [C]  time = 1.14623, size = 116, normalized size = 1.49 \begin{align*} \frac{1}{12} \,{\left ({\left ({\rm Ei}\left (\frac{i \, b}{x}\right ) +{\rm Ei}\left (-\frac{i \, b}{x}\right )\right )} \cos \left (a\right ) +{\left (i \,{\rm Ei}\left (\frac{i \, b}{x}\right ) - i \,{\rm Ei}\left (-\frac{i \, b}{x}\right )\right )} \sin \left (a\right )\right )} b^{3} + \frac{1}{6} \, b x^{2} \cos \left (\frac{a x + b}{x}\right ) - \frac{1}{6} \,{\left (b^{2} x - 2 \, x^{3}\right )} \sin \left (\frac{a x + b}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((Ei(I*b/x) + Ei(-I*b/x))*cos(a) + (I*Ei(I*b/x) - I*Ei(-I*b/x))*sin(a))*b^3 + 1/6*b*x^2*cos((a*x + b)/x)
- 1/6*(b^2*x - 2*x^3)*sin((a*x + b)/x)

________________________________________________________________________________________

Fricas [A]  time = 2.04289, size = 224, normalized size = 2.87 \begin{align*} -\frac{1}{6} \, b^{3} \sin \left (a\right ) \operatorname{Si}\left (\frac{b}{x}\right ) + \frac{1}{6} \, b x^{2} \cos \left (\frac{a x + b}{x}\right ) + \frac{1}{12} \,{\left (b^{3} \operatorname{Ci}\left (\frac{b}{x}\right ) + b^{3} \operatorname{Ci}\left (-\frac{b}{x}\right )\right )} \cos \left (a\right ) - \frac{1}{6} \,{\left (b^{2} x - 2 \, x^{3}\right )} \sin \left (\frac{a x + b}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b^3*sin(a)*sin_integral(b/x) + 1/6*b*x^2*cos((a*x + b)/x) + 1/12*(b^3*cos_integral(b/x) + b^3*cos_integra
l(-b/x))*cos(a) - 1/6*(b^2*x - 2*x^3)*sin((a*x + b)/x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sin{\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),x)

[Out]

Integral(x**2*sin(a + b/x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*sin(a + b/x), x)