3.20 \(\int x \text{Si}(a+b x) \, dx\)

Optimal. Leaf size=71 \[ -\frac{a^2 \text{Si}(a+b x)}{2 b^2}-\frac{\sin (a+b x)}{2 b^2}-\frac{a \cos (a+b x)}{2 b^2}+\frac{1}{2} x^2 \text{Si}(a+b x)+\frac{x \cos (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.206409, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6503, 6742, 2638, 3296, 2637, 3299} \[ -\frac{a^2 \text{Si}(a+b x)}{2 b^2}-\frac{\sin (a+b x)}{2 b^2}-\frac{a \cos (a+b x)}{2 b^2}+\frac{1}{2} x^2 \text{Si}(a+b x)+\frac{x \cos (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*SinIntegral[a + b*x],x]

[Out]

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

Rule 6503

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*SinIntegr
al[a + b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[((c + d*x)^(m + 1)*Sin[a + b*x])/(a + b*x), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.0998636, size = 50, normalized size = 0.7 \[ \frac{\left (b^2 x^2-a^2\right ) \text{Si}(a+b x)-\sin (a+b x)+(b x-a) \cos (a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*SinIntegral[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.051, size = 61, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\it Si} \left ( bx+a \right ) \left ({\frac{ \left ( bx+a \right ) ^{2}}{2}}-a \left ( bx+a \right ) \right ) +{\frac{ \left ( bx+a \right ) \cos \left ( bx+a \right ) }{2}}-{\frac{\sin \left ( bx+a \right ) }{2}}-a\cos \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*Si(b*x+a),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Si}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Si(b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x \operatorname{Si}\left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*sin_integral(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Si(b*x+a),x)

[Out]

Integral(x*Si(a + b*x), x)

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Giac [C]  time = 1.26429, size = 74, normalized size = 1.04 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{Si}\left (b x + a\right ) - \frac{a^{2} \Im \left ( \operatorname{Ci}\left (b x + a\right ) \right ) - a^{2} \Im \left ( \operatorname{Ci}\left (-b x - a\right ) \right ) + 2 \, a^{2} \operatorname{Si}\left (b x + a\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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