3.124 \(\int \frac {C(b x)}{x^7} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, 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 = {6427, 3388, 3387, 3351} \[ \frac {1}{90} \pi ^3 b^6 S(b x)+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{90 x^3}+\frac {\pi ^2 b^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{90 x}-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{30 x^5}-\frac {\text {FresnelC}(b x)}{6 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 74, normalized size = 0.79 \[ \frac {1}{90} \left (\pi ^3 b^6 S(b x)+\frac {b \left (\pi ^2 b^4 x^4-3\right ) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^5}+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^3}-\frac {15 C(b x)}{x^6}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnelc}\left (b x\right )}{x^{7}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(fresnelc(b*x)/x^7, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnelc}\left (b x\right )}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x)/x^7, x)

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maple [A]  time = 0.00, size = 87, normalized size = 0.93 \[ b^{6} \left (-\frac {\FresnelC \left (b x \right )}{6 b^{6} x^{6}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{30 b^{5} x^{5}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\mathrm {S}\left (b x \right )\right )}{3}\right )}{30}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnelc}\left (b x\right )}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x)/x^7, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelC(b*x)/x^7, x)

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sympy [A]  time = 1.42, size = 56, normalized size = 0.60 \[ \frac {b \Gamma \left (- \frac {5}{4}\right ) \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{4} \\ - \frac {1}{4}, \frac {1}{2}, \frac {5}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 x^{5} \Gamma \left (- \frac {1}{4}\right ) \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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