3.2.4 \(\int x \sin (a+\frac {b}{x}) \, dx\) [104]

Optimal. Leaf size=60 \[ \frac {1}{2} b x \cos \left (a+\frac {b}{x}\right )+\frac {1}{2} b^2 \text {Ci}\left (\frac {b}{x}\right ) \sin (a)+\frac {1}{2} x^2 \sin \left (a+\frac {b}{x}\right )+\frac {1}{2} b^2 \cos (a) \text {Si}\left (\frac {b}{x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3460, 3378, 3384, 3380, 3383} \begin {gather*} \frac {1}{2} b^2 \sin (a) \text {CosIntegral}\left (\frac {b}{x}\right )+\frac {1}{2} b^2 \cos (a) \text {Si}\left (\frac {b}{x}\right )+\frac {1}{2} x^2 \sin \left (a+\frac {b}{x}\right )+\frac {1}{2} b x \cos \left (a+\frac {b}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b/x],x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 52, normalized size = 0.87 \begin {gather*} \frac {1}{2} \left (b^2 \text {Ci}\left (\frac {b}{x}\right ) \sin (a)+x \left (b \cos \left (a+\frac {b}{x}\right )+x \sin \left (a+\frac {b}{x}\right )\right )+b^2 \cos (a) \text {Si}\left (\frac {b}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b/x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.06, size = 57, normalized size = 0.95

method	result	size
derivativedivides	\(-b^{2} \left (-\frac {\sin \left (a +\frac {b}{x}\right ) x^{2}}{2 b^{2}}-\frac {\cos \left (a +\frac {b}{x}\right ) x}{2 b}-\frac {\cos \left (a \right ) \sinIntegral \left (\frac {b}{x}\right )}{2}-\frac {\cosineIntegral \left (\frac {b}{x}\right ) \sin \left (a \right )}{2}\right )\)	\(57\)
default	\(-b^{2} \left (-\frac {\sin \left (a +\frac {b}{x}\right ) x^{2}}{2 b^{2}}-\frac {\cos \left (a +\frac {b}{x}\right ) x}{2 b}-\frac {\cos \left (a \right ) \sinIntegral \left (\frac {b}{x}\right )}{2}-\frac {\cosineIntegral \left (\frac {b}{x}\right ) \sin \left (a \right )}{2}\right )\)	\(57\)
risch	\(-\frac {\pi \,\mathrm {csgn}\left (\frac {b}{x}\right ) {\mathrm e}^{-i a} b^{2}}{4}+\frac {\sinIntegral \left (\frac {b}{x}\right ) {\mathrm e}^{-i a} b^{2}}{2}-\frac {i \expIntegral \left (1, -\frac {i b}{x}\right ) {\mathrm e}^{-i a} b^{2}}{4}+\frac {i b^{2} \expIntegral \left (1, -\frac {i b}{x}\right ) {\mathrm e}^{i a}}{4}+\frac {b x \cos \left (\frac {a x +b}{x}\right )}{2}+\frac {x^{2} \sin \left (\frac {a x +b}{x}\right )}{2}\)	\(104\)
meijerg	\(-\frac {b^{2} \sqrt {\pi }\, \cos \left (a \right ) \left (-\frac {4 x \cos \left (\frac {b}{x}\right )}{b \sqrt {\pi }}-\frac {4 x^{2} \sin \left (\frac {b}{x}\right )}{b^{2} \sqrt {\pi }}-\frac {4 \sinIntegral \left (\frac {b}{x}\right )}{\sqrt {\pi }}\right )}{8}-\frac {b^{2} \sqrt {\pi }\, \sin \left (a \right ) \left (-\frac {4 x^{2}}{\sqrt {\pi }\, b^{2}}-\frac {2 \left (2 \gamma -3-2 \ln \left (x \right )+\ln \left (b^{2}\right )\right )}{\sqrt {\pi }}+\frac {4 x^{2} \left (-\frac {9 b^{2}}{2 x^{2}}+3\right )}{3 \sqrt {\pi }\, b^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {b}{2 x}\right )}{\sqrt {\pi }}-\frac {4 x^{2} \cos \left (\frac {b}{x}\right )}{\sqrt {\pi }\, b^{2}}+\frac {4 x \sin \left (\frac {b}{x}\right )}{\sqrt {\pi }\, b}-\frac {4 \cosineIntegral \left (\frac {b}{x}\right )}{\sqrt {\pi }}\right )}{8}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.35, size = 76, normalized size = 1.27 \begin {gather*} \frac {1}{4} \, {\left ({\left (-i \, {\rm Ei}\left (\frac {i \, b}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, b}{x}\right )\right )} \cos \left (a\right ) + {\left ({\rm Ei}\left (\frac {i \, b}{x}\right ) + {\rm Ei}\left (-\frac {i \, b}{x}\right )\right )} \sin \left (a\right )\right )} b^{2} + \frac {1}{2} \, b x \cos \left (\frac {a x + b}{x}\right ) + \frac {1}{2} \, x^{2} \sin \left (\frac {a x + b}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((-I*Ei(I*b/x) + I*Ei(-I*b/x))*cos(a) + (Ei(I*b/x) + Ei(-I*b/x))*sin(a))*b^2 + 1/2*b*x*cos((a*x + b)/x) +
1/2*x^2*sin((a*x + b)/x)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 69, normalized size = 1.15 \begin {gather*} \frac {1}{2} \, b^{2} \cos \left (a\right ) \operatorname {Si}\left (\frac {b}{x}\right ) + \frac {1}{2} \, b x \cos \left (\frac {a x + b}{x}\right ) + \frac {1}{2} \, x^{2} \sin \left (\frac {a x + b}{x}\right ) + \frac {1}{4} \, {\left (b^{2} \operatorname {Ci}\left (\frac {b}{x}\right ) + b^{2} \operatorname {Ci}\left (-\frac {b}{x}\right )\right )} \sin \left (a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*cos(a)*sin_integral(b/x) + 1/2*b*x*cos((a*x + b)/x) + 1/2*x^2*sin((a*x + b)/x) + 1/4*(b^2*cos_integral
(b/x) + b^2*cos_integral(-b/x))*sin(a)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(a + b/x), x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (52) = 104\).
time = 5.25, size = 251, normalized size = 4.18 \begin {gather*} \frac {a^{2} b^{3} \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right ) \sin \left (a\right ) - a^{2} b^{3} \cos \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right ) - \frac {2 \, {\left (a x + b\right )} a b^{3} \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right ) \sin \left (a\right )}{x} + \frac {2 \, {\left (a x + b\right )} a b^{3} \cos \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right )}{x} - a b^{3} \cos \left (\frac {a x + b}{x}\right ) + \frac {{\left (a x + b\right )}^{2} b^{3} \operatorname {Ci}\left (-a + \frac {a x + b}{x}\right ) \sin \left (a\right )}{x^{2}} - \frac {{\left (a x + b\right )}^{2} b^{3} \cos \left (a\right ) \operatorname {Si}\left (a - \frac {a x + b}{x}\right )}{x^{2}} + \frac {{\left (a x + b\right )} b^{3} \cos \left (\frac {a x + b}{x}\right )}{x} + b^{3} \sin \left (\frac {a x + b}{x}\right )}{2 \, {\left (a^{2} - \frac {2 \, {\left (a x + b\right )} a}{x} + \frac {{\left (a x + b\right )}^{2}}{x^{2}}\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,\sin \left (a+\frac {b}{x}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b/x),x)

[Out]

int(x*sin(a + b/x), x)

________________________________________________________________________________________