3.1.15 \(\int \frac {S(b x)}{x^7} \, dx\) [15]

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

[Out]

-1/90*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^3-1/90*b^6*Pi^3*FresnelC(b*x)-1/6*FresnelS(b*x)/x^6-1/30*b*sin(1/2*b^2*Pi*x
^2)/x^5+1/90*b^5*Pi^2*sin(1/2*b^2*Pi*x^2)/x

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Rubi [A]
time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3468, 3469, 3433} \begin {gather*} -\frac {1}{90} \pi ^3 b^6 \text {FresnelC}(b x)-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{30 x^5}+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{90 x}-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/90*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/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 3433

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

Rule 3468

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 3469

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 6561

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/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

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*}

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Mathematica [A]
time = 0.04, size = 76, normalized size = 0.81 \begin {gather*} \frac {1}{90} \left (-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^3}-b^6 \pi ^3 \text {FresnelC}(b x)-\frac {15 S(b x)}{x^6}+\frac {b \left (-3+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5}\right ) \end {gather*}

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 = 0.34, size = 86, normalized size = 0.91

method	result	size
meijerg	\(-\frac {\pi \,b^{3} \hypergeom \left (\left [-\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {1}{4}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{18 x^{3}}\)	\(29\)
derivativedivides	\(b^{6} \left (-\frac {\mathrm {S}\left (b x \right )}{6 b^{6} x^{6}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{30 b^{5} x^{5}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}+\pi \FresnelC \left (b x \right )\right )}{3}\right )}{30}\right )\)	\(86\)
default	\(b^{6} \left (-\frac {\mathrm {S}\left (b x \right )}{6 b^{6} x^{6}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{30 b^{5} x^{5}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}+\pi \FresnelC \left (b x \right )\right )}{3}\right )}{30}\right )\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^7,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.53, size = 61, normalized size = 0.65 \begin {gather*} -\frac {\sqrt {\frac {1}{2}} \left (\pi x^{2}\right )^{\frac {5}{2}} {\left (-\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{6}}{192 \, x^{5}} - \frac {\operatorname {S}\left (b x\right )}{6 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*sqrt(1/2)*(pi*x^2)^(5/2)*(-(I - 1)*sqrt(2)*gamma(-5/2, 1/2*I*pi*b^2*x^2) + (I + 1)*sqrt(2)*gamma(-5/2,
-1/2*I*pi*b^2*x^2))*b^6/x^5 - 1/6*fresnel_sin(b*x)/x^6

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Fricas [A]
time = 0.36, size = 80, normalized size = 0.85 \begin {gather*} -\frac {\pi ^{3} \sqrt {b^{2}} b^{5} x^{6} \operatorname {C}\left (\sqrt {b^{2}} x\right ) + \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{5} x^{5} - 3 \, b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 15 \, \operatorname {S}\left (b x\right )}{90 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/90*(pi^3*sqrt(b^2)*b^5*x^6*fresnel_cos(sqrt(b^2)*x) + pi*b^3*x^3*cos(1/2*pi*b^2*x^2) - (pi^2*b^5*x^5 - 3*b*
x)*sin(1/2*pi*b^2*x^2) + 15*fresnel_sin(b*x))/x^6

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Sympy [A]
time = 0.81, size = 56, normalized size = 0.60 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnel_sin(b*x)/x^7, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelS(b*x)/x^7, x)

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