3.26 \(\int x^2 S(a+b x) \, dx\)

Optimal. Leaf size=147 \[ \frac{a^3 S(a+b x)}{3 b^3}+\frac{a^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{a \text{FresnelC}(a+b x)}{\pi b^3}-\frac{2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac{a (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{(a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac{1}{3} x^3 S(a+b x) \]

[Out]

(a^2*Cos[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (a*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(b^3*Pi) + ((a + b*x)^2*Cos[(Pi
*(a + b*x)^2)/2])/(3*b^3*Pi) + (a*FresnelC[a + b*x])/(b^3*Pi) + (a^3*FresnelS[a + b*x])/(3*b^3) + (x^3*Fresnel
S[a + b*x])/3 - (2*Sin[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi^2)

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Rubi [A]  time = 0.128683, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6428, 3433, 3351, 3379, 2638, 3385, 3352, 3296, 2637} \[ \frac{a^3 S(a+b x)}{3 b^3}+\frac{a^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{a \text{FresnelC}(a+b x)}{\pi b^3}-\frac{2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac{a (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{(a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac{1}{3} x^3 S(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*FresnelS[a + b*x],x]

[Out]

(a^2*Cos[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (a*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(b^3*Pi) + ((a + b*x)^2*Cos[(Pi
*(a + b*x)^2)/2])/(3*b^3*Pi) + (a*FresnelC[a + b*x])/(b^3*Pi) + (a^3*FresnelS[a + b*x])/(3*b^3) + (x^3*Fresnel
S[a + b*x])/3 - (2*Sin[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi^2)

Rule 6428

Int[FresnelS[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*FresnelS[a +
b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Sin[(Pi*(a + b*x)^2)/2], x], x] /; FreeQ[{a
, b, c, d}, x] && IGtQ[m, 0]

Rule 3433

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3379

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]))

Rule 2638

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

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, 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]

Rubi steps

\begin{align*} \int x^2 S(a+b x) \, dx &=\frac{1}{3} x^3 S(a+b x)-\frac{1}{3} b \int x^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac{1}{3} x^3 S(a+b x)-\frac{\operatorname{Subst}\left (\int \left (-a^3 \sin \left (\frac{\pi x^2}{2}\right )+3 a^2 x \sin \left (\frac{\pi x^2}{2}\right )-3 a x^2 \sin \left (\frac{\pi x^2}{2}\right )+x^3 \sin \left (\frac{\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{3 b^3}\\ &=\frac{1}{3} x^3 S(a+b x)-\frac{\operatorname{Subst}\left (\int x^3 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}+\frac{a \operatorname{Subst}\left (\int x^2 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac{a^2 \operatorname{Subst}\left (\int x \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}+\frac{a^3 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}\\ &=-\frac{a (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{a^3 S(a+b x)}{3 b^3}+\frac{1}{3} x^3 S(a+b x)-\frac{\operatorname{Subst}\left (\int x \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{6 b^3}-\frac{a^2 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^3}+\frac{a \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3 \pi }\\ &=\frac{a^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac{a (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{(a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac{a C(a+b x)}{b^3 \pi }+\frac{a^3 S(a+b x)}{3 b^3}+\frac{1}{3} x^3 S(a+b x)-\frac{\operatorname{Subst}\left (\int \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{3 b^3 \pi }\\ &=\frac{a^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac{a (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{(a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac{a C(a+b x)}{b^3 \pi }+\frac{a^3 S(a+b x)}{3 b^3}+\frac{1}{3} x^3 S(a+b x)-\frac{2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}\\ \end{align*}

Mathematica [A]  time = 0.23802, size = 115, normalized size = 0.78 \[ \frac{\pi ^2 \left (a^3+b^3 x^3\right ) S(a+b x)+\pi a^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi b^2 x^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+3 \pi a \text{FresnelC}(a+b x)-2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-\pi a b x \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*FresnelS[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.05, size = 121, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3}{\it FresnelS} \left ( bx+a \right ) }{3}}+{\frac{ \left ( bx+a \right ) ^{2}}{3\,\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{2}{3\,{\pi }^{2}}\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{a \left ( bx+a \right ) }{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{\frac{a{\it FresnelC} \left ( bx+a \right ) }{\pi }}+{\frac{{a}^{2}}{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{\frac{{a}^{3}{\it FresnelS} \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*b^3*x^3*FresnelS(b*x+a)+1/3/Pi*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)-2/3/Pi^2*sin(1/2*Pi*(b*x+a)^2)-a/Pi*
(b*x+a)*cos(1/2*Pi*(b*x+a)^2)+a/Pi*FresnelC(b*x+a)+a^2/Pi*cos(1/2*Pi*(b*x+a)^2)+1/3*a^3*FresnelS(b*x+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnels(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*fresnels(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*fresnels(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnels(b*x + a), x)