3.15 \(\int \frac{S(b x)}{x^7} \, dx\)

Optimal. Leaf size=94 \[ -\frac{1}{90} \pi ^3 b^6 \text{FresnelC}(b x)+\frac{\pi ^2 b^5 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{90 x}-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{30 x^5}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{90 x^3}-\frac{S(b x)}{6 x^6} \]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(90*x^3) - (b^6*Pi^3*FresnelC[b*x])/90 - FresnelS[b*x]/(6*x^6) - (b*Sin[(b^2*Pi*
x^2)/2])/(30*x^5) + (b^5*Pi^2*Sin[(b^2*Pi*x^2)/2])/(90*x)

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Rubi [A]  time = 0.0593807, antiderivative size = 94, 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 = {6426, 3387, 3388, 3352} \[ -\frac{1}{90} \pi ^3 b^6 \text{FresnelC}(b x)+\frac{\pi ^2 b^5 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{90 x}-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{30 x^5}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{90 x^3}-\frac{S(b x)}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^7,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(90*x^3) - (b^6*Pi^3*FresnelC[b*x])/90 - FresnelS[b*x]/(6*x^6) - (b*Sin[(b^2*Pi*
x^2)/2])/(30*x^5) + (b^5*Pi^2*Sin[(b^2*Pi*x^2)/2])/(90*x)

Rule 6426

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

Rule 3387

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

Rule 3388

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

Rubi steps

\begin{align*} \int \frac{S(b x)}{x^7} \, dx &=-\frac{S(b x)}{6 x^6}+\frac{1}{6} b \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^6} \, dx\\ &=-\frac{S(b x)}{6 x^6}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{30 x^5}+\frac{1}{30} \left (b^3 \pi \right ) \int \frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{90 x^3}-\frac{S(b x)}{6 x^6}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{30 x^5}-\frac{1}{90} \left (b^5 \pi ^2\right ) \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{90 x^3}-\frac{S(b x)}{6 x^6}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{30 x^5}+\frac{b^5 \pi ^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{90 x}-\frac{1}{90} \left (b^7 \pi ^3\right ) \int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{90 x^3}-\frac{1}{90} b^6 \pi ^3 C(b x)-\frac{S(b x)}{6 x^6}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{30 x^5}+\frac{b^5 \pi ^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{90 x}\\ \end{align*}

Mathematica [A]  time = 0.0574427, size = 76, normalized size = 0.81 \[ \frac{1}{90} \left (-\pi ^3 b^6 \text{FresnelC}(b x)+\frac{b \left (\pi ^2 b^4 x^4-3\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{x^5}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{x^3}-\frac{15 S(b x)}{x^6}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^7,x]

[Out]

(-((b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x^3) - b^6*Pi^3*FresnelC[b*x] - (15*FresnelS[b*x])/x^6 + (b*(-3 + b^4*Pi^2*x^4
)*Sin[(b^2*Pi*x^2)/2])/x^5)/90

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Maple [A]  time = 0.05, size = 86, normalized size = 0.9 \begin{align*}{b}^{6} \left ( -{\frac{{\it FresnelS} \left ( bx \right ) }{6\,{b}^{6}{x}^{6}}}-{\frac{1}{30\,{b}^{5}{x}^{5}}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\pi }{30} \left ( -{\frac{1}{3\,{x}^{3}{b}^{3}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{\pi }{3} \left ( -{\frac{1}{bx}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+\pi \,{\it FresnelC} \left ( bx \right ) \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^7,x)

[Out]

b^6*(-1/6*FresnelS(b*x)/b^6/x^6-1/30*sin(1/2*b^2*Pi*x^2)/b^5/x^5+1/30*Pi*(-1/3/b^3/x^3*cos(1/2*b^2*Pi*x^2)-1/3
*Pi*(-sin(1/2*b^2*Pi*x^2)/b/x+Pi*FresnelC(b*x))))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )}{x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^7,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^7, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\rm fresnels}\left (b x\right )}{x^{7}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^7,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^7, x)

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Sympy [A]  time = 1.52855, size = 56, normalized size = 0.6 \begin{align*} \frac{\pi b^{3} \Gamma \left (- \frac{3}{4}\right ) \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{4} \\ \frac{1}{4}, \frac{3}{2}, \frac{7}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 x^{3} \Gamma \left (\frac{1}{4}\right ) \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**7,x)

[Out]

pi*b**3*gamma(-3/4)*gamma(3/4)*hyper((-3/4, 3/4), (1/4, 3/2, 7/4), -pi**2*b**4*x**4/16)/(32*x**3*gamma(1/4)*ga
mma(7/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )}{x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^7,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^7, x)