3.2.3 \(\int x^2 \sin (a+\frac {b}{x}) \, dx\) [103]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 = {3460, 3378, 3384, 3380, 3383} \begin {gather*} \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 ) \end {gather*}

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 3378

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 3380

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 3383

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 3384

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 3460

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

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Mathematica [A]
time = 0.05, size = 70, normalized size = 0.90 \begin {gather*} \frac {1}{6} \left (b^3 \cos (a) \text {Ci}\left (\frac {b}{x}\right )+x \left (b x \cos \left (a+\frac {b}{x}\right )-b^2 \sin \left (a+\frac {b}{x}\right )+2 x^2 \sin \left (a+\frac {b}{x}\right )\right )-b^3 \sin (a) \text {Si}\left (\frac {b}{x}\right )\right ) \end {gather*}

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

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Maple [A]
time = 0.09, size = 73, normalized size = 0.94

method	result	size
derivativedivides	\(-b^{3} \left (-\frac {\sin \left (a +\frac {b}{x}\right ) x^{3}}{3 b^{3}}-\frac {\cos \left (a +\frac {b}{x}\right ) x^{2}}{6 b^{2}}+\frac {\sin \left (a +\frac {b}{x}\right ) x}{6 b}+\frac {\sinIntegral \left (\frac {b}{x}\right ) \sin \left (a \right )}{6}-\frac {\cosineIntegral \left (\frac {b}{x}\right ) \cos \left (a \right )}{6}\right )\)	\(73\)
default	\(-b^{3} \left (-\frac {\sin \left (a +\frac {b}{x}\right ) x^{3}}{3 b^{3}}-\frac {\cos \left (a +\frac {b}{x}\right ) x^{2}}{6 b^{2}}+\frac {\sin \left (a +\frac {b}{x}\right ) x}{6 b}+\frac {\sinIntegral \left (\frac {b}{x}\right ) \sin \left (a \right )}{6}-\frac {\cosineIntegral \left (\frac {b}{x}\right ) \cos \left (a \right )}{6}\right )\)	\(73\)
risch	\(\frac {i \pi \,\mathrm {csgn}\left (\frac {b}{x}\right ) {\mathrm e}^{-i a} b^{3}}{12}-\frac {i \sinIntegral \left (\frac {b}{x}\right ) {\mathrm e}^{-i a} b^{3}}{6}-\frac {\expIntegral \left (1, -\frac {i b}{x}\right ) {\mathrm e}^{-i a} b^{3}}{12}-\frac {b^{3} \expIntegral \left (1, -\frac {i b}{x}\right ) {\mathrm e}^{i a}}{12}+\frac {x^{2} b \cos \left (\frac {a x +b}{x}\right )}{6}-\frac {\sin \left (\frac {a x +b}{x}\right ) b^{2} x}{6}+\frac {\sin \left (\frac {a x +b}{x}\right ) x^{3}}{3}\)	\(122\)
meijerg	\(-\frac {b^{3} \sqrt {\pi }\, \cos \left (a \right ) \left (-\frac {8 x^{2}}{\sqrt {\pi }\, b^{2}}-\frac {4 \left (2 \gamma -\frac {11}{3}-2 \ln \left (x \right )+2 \ln \left (b \right )\right )}{3 \sqrt {\pi }}+\frac {8 x^{2} \left (-\frac {55 b^{2}}{2 x^{2}}+45\right )}{45 \sqrt {\pi }\, b^{2}}+\frac {8 \gamma }{3 \sqrt {\pi }}+\frac {8 \ln \left (2\right )}{3 \sqrt {\pi }}+\frac {8 \ln \left (\frac {b}{2 x}\right )}{3 \sqrt {\pi }}-\frac {8 x^{2} \cos \left (\frac {b}{x}\right )}{3 \sqrt {\pi }\, b^{2}}-\frac {16 x^{3} \left (-\frac {5 b^{2}}{2 x^{2}}+5\right ) \sin \left (\frac {b}{x}\right )}{15 \sqrt {\pi }\, b^{3}}-\frac {8 \cosineIntegral \left (\frac {b}{x}\right )}{3 \sqrt {\pi }}\right )}{16}-\frac {b^{2} \sqrt {\pi }\, \sin \left (a \right ) \sqrt {b^{2}}\, \left (-\frac {8 \left (-\frac {b^{2}}{x^{2}}+2\right ) x^{3} b^{2} \cos \left (\frac {\sqrt {b^{2}}}{x}\right )}{3 \left (b^{2}\right )^{\frac {5}{2}} \sqrt {\pi }}+\frac {8 x^{2} \sin \left (\frac {\sqrt {b^{2}}}{x}\right )}{3 b^{2} \sqrt {\pi }}+\frac {8 \sinIntegral \left (\frac {\sqrt {b^{2}}}{x}\right )}{3 \sqrt {\pi }}\right )}{16}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.36, size = 86, normalized size = 1.10 \begin {gather*} \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 {gather*}

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)

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Fricas [A]
time = 0.38, size = 79, normalized size = 1.01 \begin {gather*} -\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 {gather*}

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)

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

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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (68) = 136\).
time = 6.68, size = 400, normalized size = 5.13 \begin {gather*} \frac {a^{3} b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right ) + a^{3} b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right ) - \frac {3 \, {\left (a x + b\right )} a^{2} b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right )}{x} - \frac {3 \, {\left (a x + b\right )} a^{2} b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right )}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right )}{x^{2}} + a^{2} b^{4} \sin \left (\frac {a x + b}{x}\right ) + \frac {3 \, {\left (a x + b\right )}^{2} a b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right )}{x^{2}} + a b^{4} \cos \left (\frac {a x + b}{x}\right ) - \frac {{\left (a x + b\right )}^{3} b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right )}{x^{3}} - \frac {2 \, {\left (a x + b\right )} a b^{4} \sin \left (\frac {a x + b}{x}\right )}{x} - \frac {{\left (a x + b\right )}^{3} b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right )}{x^{3}} - \frac {{\left (a x + b\right )} b^{4} \cos \left (\frac {a x + b}{x}\right )}{x} - 2 \, b^{4} \sin \left (\frac {a x + b}{x}\right ) + \frac {{\left (a x + b\right )}^{2} b^{4} \sin \left (\frac {a x + b}{x}\right )}{x^{2}}}{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^3*b^4*cos(a)*cos_integral(-a + (a*x + b)/x) + a^3*b^4*sin(a)*sin_integral(a - (a*x + b)/x) - 3*(a*x + b
)*a^2*b^4*cos(a)*cos_integral(-a + (a*x + b)/x)/x - 3*(a*x + b)*a^2*b^4*sin(a)*sin_integral(a - (a*x + b)/x)/x
+ 3*(a*x + b)^2*a*b^4*cos(a)*cos_integral(-a + (a*x + b)/x)/x^2 + a^2*b^4*sin((a*x + b)/x) + 3*(a*x + b)^2*a*
b^4*sin(a)*sin_integral(a - (a*x + b)/x)/x^2 + a*b^4*cos((a*x + b)/x) - (a*x + b)^3*b^4*cos(a)*cos_integral(-a
+ (a*x + b)/x)/x^3 - 2*(a*x + b)*a*b^4*sin((a*x + b)/x)/x - (a*x + b)^3*b^4*sin(a)*sin_integral(a - (a*x + b)
/x)/x^3 - (a*x + b)*b^4*cos((a*x + b)/x)/x - 2*b^4*sin((a*x + b)/x) + (a*x + b)^2*b^4*sin((a*x + b)/x)/x^2)/((
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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\sin \left (a+\frac {b}{x}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(a + b/x),x)

[Out]

int(x^2*sin(a + b/x), x)

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