3.112 \(\int x^5 \text{FresnelC}(b x) \, dx\)

Optimal. Leaf size=99 \[ -\frac{5 S(b x)}{2 \pi ^3 b^6}-\frac{x^5 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{6 \pi b}+\frac{5 x \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi ^3 b^5}-\frac{5 x^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{6 \pi ^2 b^3}+\frac{1}{6} x^6 \text{FresnelC}(b x) \]

[Out]

(-5*x^3*Cos[(b^2*Pi*x^2)/2])/(6*b^3*Pi^2) + (x^6*FresnelC[b*x])/6 - (5*FresnelS[b*x])/(2*b^6*Pi^3) + (5*x*Sin[
(b^2*Pi*x^2)/2])/(2*b^5*Pi^3) - (x^5*Sin[(b^2*Pi*x^2)/2])/(6*b*Pi)

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Rubi [A]  time = 0.0624975, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6427, 3386, 3385, 3351} \[ -\frac{5 S(b x)}{2 \pi ^3 b^6}-\frac{x^5 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{6 \pi b}+\frac{5 x \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi ^3 b^5}-\frac{5 x^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{6 \pi ^2 b^3}+\frac{1}{6} x^6 \text{FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^5*FresnelC[b*x],x]

[Out]

(-5*x^3*Cos[(b^2*Pi*x^2)/2])/(6*b^3*Pi^2) + (x^6*FresnelC[b*x])/6 - (5*FresnelS[b*x])/(2*b^6*Pi^3) + (5*x*Sin[
(b^2*Pi*x^2)/2])/(2*b^5*Pi^3) - (x^5*Sin[(b^2*Pi*x^2)/2])/(6*b*Pi)

Rule 6427

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

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 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 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^5 C(b x) \, dx &=\frac{1}{6} x^6 C(b x)-\frac{1}{6} b \int x^6 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{6} x^6 C(b x)-\frac{x^5 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b \pi }+\frac{5 \int x^4 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{6 b \pi }\\ &=-\frac{5 x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b^3 \pi ^2}+\frac{1}{6} x^6 C(b x)-\frac{x^5 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b \pi }+\frac{5 \int x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{2 b^3 \pi ^2}\\ &=-\frac{5 x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b^3 \pi ^2}+\frac{1}{6} x^6 C(b x)+\frac{5 x \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{2 b^5 \pi ^3}-\frac{x^5 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b \pi }-\frac{5 \int \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{2 b^5 \pi ^3}\\ &=-\frac{5 x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b^3 \pi ^2}+\frac{1}{6} x^6 C(b x)-\frac{5 S(b x)}{2 b^6 \pi ^3}+\frac{5 x \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{2 b^5 \pi ^3}-\frac{x^5 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{6 b \pi }\\ \end{align*}

Mathematica [A]  time = 0.0634354, size = 80, normalized size = 0.81 \[ \frac{\pi ^3 b^6 x^6 \text{FresnelC}(b x)+b x \left (15-\pi ^2 b^4 x^4\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )-5 \pi b^3 x^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )-15 S(b x)}{6 \pi ^3 b^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*FresnelC[b*x],x]

[Out]

(-5*b^3*Pi*x^3*Cos[(b^2*Pi*x^2)/2] + b^6*Pi^3*x^6*FresnelC[b*x] - 15*FresnelS[b*x] + b*x*(15 - b^4*Pi^2*x^4)*S
in[(b^2*Pi*x^2)/2])/(6*b^6*Pi^3)

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Maple [A]  time = 0.048, size = 97, normalized size = 1. \begin{align*}{\frac{1}{{b}^{6}} \left ({\frac{{b}^{6}{x}^{6}{\it FresnelC} \left ( bx \right ) }{6}}-{\frac{{b}^{5}{x}^{5}}{6\,\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{5}{6\,\pi } \left ( -{\frac{{x}^{3}{b}^{3}}{\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+3\,{\frac{1}{\pi } \left ({\frac{bx\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}-{\frac{{\it FresnelS} \left ( bx \right ) }{\pi }} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*FresnelC(b*x),x)

[Out]

1/b^6*(1/6*b^6*x^6*FresnelC(b*x)-1/6/Pi*b^5*x^5*sin(1/2*b^2*Pi*x^2)+5/6/Pi*(-1/Pi*b^3*x^3*cos(1/2*b^2*Pi*x^2)+
3/Pi*(1/Pi*b*x*sin(1/2*b^2*Pi*x^2)-1/Pi*FresnelS(b*x))))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*fresnelc(b*x),x, algorithm="maxima")

[Out]

integrate(x^5*fresnelc(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{5}{\rm fresnelc}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*fresnelc(b*x),x, algorithm="fricas")

[Out]

integral(x^5*fresnelc(b*x), x)

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Sympy [A]  time = 1.12886, size = 49, normalized size = 0.49 \begin{align*} \frac{b x^{7} \Gamma \left (\frac{1}{4}\right ) \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{1}{4}, \frac{7}{4} \\ \frac{1}{2}, \frac{5}{4}, \frac{11}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma \left (\frac{5}{4}\right ) \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*fresnelc(b*x),x)

[Out]

b*x**7*gamma(1/4)*gamma(7/4)*hyper((1/4, 7/4), (1/2, 5/4, 11/4), -pi**2*b**4*x**4/16)/(16*gamma(5/4)*gamma(11/
4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*fresnelc(b*x),x, algorithm="giac")

[Out]

integrate(x^5*fresnelc(b*x), x)