3.136 \(\int x \text{FresnelC}(a+b x) \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^2 \text{FresnelC}(a+b x)}{2 b^2}+\frac{S(a+b x)}{2 \pi b^2}+\frac{a \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac{(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac{1}{2} x^2 \text{FresnelC}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.0707863, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351} \[ -\frac{a^2 \text{FresnelC}(a+b x)}{2 b^2}+\frac{S(a+b x)}{2 \pi b^2}+\frac{a \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac{(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac{1}{2} x^2 \text{FresnelC}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x*FresnelC[a + b*x],x]

[Out]

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

Rule 6429

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

Rule 3434

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Cos[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 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 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 2637

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

Rule 3386

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

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]

Rubi steps

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

Mathematica [A]  time = 0.148444, size = 59, normalized size = 0.62 \[ \frac{\left (\pi b^2 x^2-\pi a^2\right ) \text{FresnelC}(a+b x)+S(a+b x)+(a-b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*FresnelC[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.052, size = 79, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{2}} \left ({\it FresnelC} \left ( bx+a \right ) \left ({\frac{ \left ( bx+a \right ) ^{2}}{2}}-a \left ( bx+a \right ) \right ) -{\frac{bx+a}{2\,\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{\frac{{\it FresnelS} \left ( bx+a \right ) }{2\,\pi }}+{\frac{a}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*fresnelc(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*fresnelc(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*fresnelc(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*fresnelc(b*x + a), x)