3.19 \(\int x^2 \text{Si}(a+b x) \, dx\)

Optimal. Leaf size=118 \[ \frac{a^3 \text{Si}(a+b x)}{3 b^3}+\frac{a^2 \cos (a+b x)}{3 b^3}+\frac{a \sin (a+b x)}{3 b^3}-\frac{2 x \sin (a+b x)}{3 b^2}-\frac{a x \cos (a+b x)}{3 b^2}-\frac{2 \cos (a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{Si}(a+b x)+\frac{x^2 \cos (a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.276629, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6503, 6742, 2638, 3296, 2637, 3299} \[ \frac{a^3 \text{Si}(a+b x)}{3 b^3}+\frac{a^2 \cos (a+b x)}{3 b^3}+\frac{a \sin (a+b x)}{3 b^3}-\frac{2 x \sin (a+b x)}{3 b^2}-\frac{a x \cos (a+b x)}{3 b^2}-\frac{2 \cos (a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{Si}(a+b x)+\frac{x^2 \cos (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*SinIntegral[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.141367, size = 63, normalized size = 0.53 \[ \frac{\left (a^3+b^3 x^3\right ) \text{Si}(a+b x)+\left (a^2-a b x+b^2 x^2-2\right ) \cos (a+b x)+(a-2 b x) \sin (a+b x)}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*SinIntegral[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.053, size = 99, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3}{\it Si} \left ( bx+a \right ) }{3}}+{\frac{ \left ( bx+a \right ) ^{2}\cos \left ( bx+a \right ) }{3}}-{\frac{2\,\cos \left ( bx+a \right ) }{3}}-{\frac{2\,\sin \left ( bx+a \right ) \left ( bx+a \right ) }{3}}+a \left ( - \left ( bx+a \right ) \cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) \right ) +{a}^{2}\cos \left ( bx+a \right ) +{\frac{{a}^{3}{\it Si} \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Si(b*x+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Si(b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{Si}\left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*sin_integral(b*x + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{Si}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*Si(b*x+a),x)

[Out]

Integral(x**2*Si(a + b*x), x)

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Giac [C]  time = 1.21799, size = 74, normalized size = 0.63 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{Si}\left (b x + a\right ) + \frac{a^{3} \Im \left ( \operatorname{Ci}\left (b x + a\right ) \right ) - a^{3} \Im \left ( \operatorname{Ci}\left (-b x - a\right ) \right ) + 2 \, a^{3} \operatorname{Si}\left (b x + a\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*sin_integral(b*x + a) + 1/6*(a^3*imag_part(cos_integral(b*x + a)) - a^3*imag_part(cos_integral(-b*x -
a)) + 2*a^3*sin_integral(b*x + a))/b^3