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

Optimal. Leaf size=65 \[ b^2 (-\cos (2 a)) \text {Ci}\left (\frac {2 b}{x}\right )+b^2 \sin (2 a) \text {Si}\left (\frac {2 b}{x}\right )+\frac {1}{2} x^2 \sin ^2\left (a+\frac {b}{x}\right )+\frac {1}{2} b x \sin \left (2 \left (a+\frac {b}{x}\right )\right ) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3393, 4573, 3373, 3361, 3297, 3303, 3299, 3302} \[ b^2 (-\cos (2 a)) \text {CosIntegral}\left (\frac {2 b}{x}\right )+b^2 \sin (2 a) \text {Si}\left (\frac {2 b}{x}\right )+\frac {1}{2} x^2 \sin ^2\left (a+\frac {b}{x}\right )+\frac {1}{2} b x \sin \left (2 \left (a+\frac {b}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b/x]^2,x]

[Out]

-(b^2*Cos[2*a]*CosIntegral[(2*b)/x]) + (x^2*Sin[a + b/x]^2)/2 + (b*x*Sin[2*(a + b/x)])/2 + b^2*Sin[2*a]*SinInt
egral[(2*b)/x]

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 3361

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Sin[c + d*x])^p, x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && In
tegerQ[1/n]

Rule 3373

Int[((a_.) + (b_.)*Sin[u_])^(p_.), x_Symbol] :> Int[(a + b*Sin[ExpandToSum[u, x]])^p, x] /; FreeQ[{a, b, p}, x
] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rule 3393

Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> Simp[(x^(m + 1)*Sin[a + b*x^n]^p)/(m + 1), x] -
Dist[(b*n*p)/(m + 1), Int[Sin[a + b*x^n]^(p - 1)*Cos[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] &&
EqQ[m + n, 0] && NeQ[n, 1] && IntegerQ[n]

Rule 4573

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 65, normalized size = 1.00 \[ b^2 (-\cos (2 a)) \text {Ci}\left (\frac {2 b}{x}\right )+b^2 \sin (2 a) \text {Si}\left (\frac {2 b}{x}\right )+\frac {1}{4} x \left (2 b \sin \left (2 \left (a+\frac {b}{x}\right )\right )+x \left (-\cos \left (2 \left (a+\frac {b}{x}\right )\right )\right )+x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b/x]^2,x]

[Out]

-(b^2*Cos[2*a]*CosIntegral[(2*b)/x]) + (x*(x - x*Cos[2*(a + b/x)] + 2*b*Sin[2*(a + b/x)]))/4 + b^2*Sin[2*a]*Si
nIntegral[(2*b)/x]

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fricas [A]  time = 0.53, size = 90, normalized size = 1.38 \[ -\frac {1}{2} \, x^{2} \cos \left (\frac {a x + b}{x}\right )^{2} + b x \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right ) + b^{2} \sin \left (2 \, a\right ) \operatorname {Si}\left (\frac {2 \, b}{x}\right ) + \frac {1}{2} \, x^{2} - \frac {1}{2} \, {\left (b^{2} \operatorname {Ci}\left (\frac {2 \, b}{x}\right ) + b^{2} \operatorname {Ci}\left (-\frac {2 \, b}{x}\right )\right )} \cos \left (2 \, a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.37, size = 283, normalized size = 4.35 \[ -\frac {4 \, a^{2} b^{3} \cos \left (2 \, a\right ) \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) + 4 \, a^{2} b^{3} \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {8 \, {\left (a x + b\right )} a b^{3} \cos \left (2 \, a\right ) \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - \frac {8 \, {\left (a x + b\right )} a b^{3} \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {4 \, {\left (a x + b\right )}^{2} b^{3} \cos \left (2 \, a\right ) \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} + 2 \, a b^{3} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {4 \, {\left (a x + b\right )}^{2} b^{3} \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} + b^{3} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {2 \, {\left (a x + b\right )} b^{3} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - b^{3}}{4 \, {\left (a^{2} - \frac {2 \, {\left (a x + b\right )} a}{x} + \frac {{\left (a x + b\right )}^{2}}{x^{2}}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*a^2*b^3*cos(2*a)*cos_integral(-2*a + 2*(a*x + b)/x) + 4*a^2*b^3*sin(2*a)*sin_integral(2*a - 2*(a*x + b
)/x) - 8*(a*x + b)*a*b^3*cos(2*a)*cos_integral(-2*a + 2*(a*x + b)/x)/x - 8*(a*x + b)*a*b^3*sin(2*a)*sin_integr
al(2*a - 2*(a*x + b)/x)/x + 4*(a*x + b)^2*b^3*cos(2*a)*cos_integral(-2*a + 2*(a*x + b)/x)/x^2 + 2*a*b^3*sin(2*
(a*x + b)/x) + 4*(a*x + b)^2*b^3*sin(2*a)*sin_integral(2*a - 2*(a*x + b)/x)/x^2 + b^3*cos(2*(a*x + b)/x) - 2*(
a*x + b)*b^3*sin(2*(a*x + b)/x)/x - b^3)/((a^2 - 2*(a*x + b)*a/x + (a*x + b)^2/x^2)*b)

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maple [A]  time = 0.04, size = 76, normalized size = 1.17 \[ -b^{2} \left (-\frac {x^{2}}{4 b^{2}}+\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x^{2}}{4 b^{2}}-\frac {\sin \left (2 a +\frac {2 b}{x}\right ) x}{2 b}-\Si \left (\frac {2 b}{x}\right ) \sin \left (2 a \right )+\Ci \left (\frac {2 b}{x}\right ) \cos \left (2 a \right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*(-1/4*x^2/b^2+1/4*cos(2*a+2*b/x)*x^2/b^2-1/2*sin(2*a+2*b/x)*x/b-Si(2*b/x)*sin(2*a)+Ci(2*b/x)*cos(2*a))

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maxima [C]  time = 0.39, size = 89, normalized size = 1.37 \[ -\frac {1}{4} \, {\left (2 \, {\left ({\rm Ei}\left (\frac {2 i \, b}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) - {\left (-2 i \, {\rm Ei}\left (\frac {2 i \, b}{x}\right ) + 2 i \, {\rm Ei}\left (-\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} b^{2} - \frac {1}{4} \, x^{2} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {1}{2} \, b x \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {1}{4} \, x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,{\sin \left (a+\frac {b}{x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b/x)^2,x)

[Out]

int(x*sin(a + b/x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(a + b/x)**2, x)

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