3.122 \(\int \frac {C(b x)}{x^5} \, dx\)

Optimal. Leaf size=69 \[ -\frac {1}{12} \pi ^2 b^4 C(b x)-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x^3}+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x}-\frac {C(b x)}{4 x^4} \]

[Out]

-1/12*b*cos(1/2*b^2*Pi*x^2)/x^3-1/12*b^4*Pi^2*FresnelC(b*x)-1/4*FresnelC(b*x)/x^4+1/12*b^3*Pi*sin(1/2*b^2*Pi*x
^2)/x

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6427, 3388, 3387, 3352} \[ -\frac {1}{12} \pi ^2 b^4 \text {FresnelC}(b x)+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x}-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x^3}-\frac {\text {FresnelC}(b x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[(b^2*Pi*x^2)/2])/(12*x^3) - (b^4*Pi^2*FresnelC[b*x])/12 - FresnelC[b*x]/(4*x^4) + (b^3*Pi*Sin[(b^2*Pi*
x^2)/2])/(12*x)

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 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^5} \, dx &=-\frac {C(b x)}{4 x^4}+\frac {1}{4} b \int \frac {\cos \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 )}{12 x^3}-\frac {C(b x)}{4 x^4}-\frac {1}{12} \left (b^3 \pi \right ) \int \frac {\sin \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 )}{12 x^3}-\frac {C(b x)}{4 x^4}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} \left (b^5 \pi ^2\right ) \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3}-\frac {1}{12} b^4 \pi ^2 C(b x)-\frac {C(b x)}{4 x^4}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 69, normalized size = 1.00 \[ -\frac {1}{12} \pi ^2 b^4 C(b x)-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x^3}+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x}-\frac {C(b x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*Cos[(b^2*Pi*x^2)/2])/x^3 - (b^4*Pi^2*FresnelC[b*x])/12 - FresnelC[b*x]/(4*x^4) + (b^3*Pi*Sin[(b^2*Pi*
x^2)/2])/(12*x)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnelc}\left (b x\right )}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(fresnelc(b*x)/x^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnelc}\left (b x\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x)/x^5, x)

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maple [A]  time = 0.00, size = 64, normalized size = 0.93 \[ b^{4} \left (-\frac {\FresnelC \left (b x \right )}{4 b^{4} x^{4}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}+\pi \FresnelC \left (b x \right )\right )}{12}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^4*(-1/4*FresnelC(b*x)/b^4/x^4-1/12/b^3/x^3*cos(1/2*b^2*Pi*x^2)-1/12*Pi*(-sin(1/2*b^2*Pi*x^2)/b/x+Pi*FresnelC
(b*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnelc}\left (b x\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x)/x^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelC(b*x)/x^5, x)

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sympy [A]  time = 1.35, size = 110, normalized size = 1.59 \[ \frac {\pi ^{2} b^{4} C\left (b x\right ) \Gamma \left (- \frac {3}{4}\right )}{64 \Gamma \left (\frac {5}{4}\right )} - \frac {\pi b^{3} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {3}{4}\right )}{64 x \Gamma \left (\frac {5}{4}\right )} + \frac {b \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {3}{4}\right )}{64 x^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {3 C\left (b x\right ) \Gamma \left (- \frac {3}{4}\right )}{64 x^{4} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

pi**2*b**4*fresnelc(b*x)*gamma(-3/4)/(64*gamma(5/4)) - pi*b**3*sin(pi*b**2*x**2/2)*gamma(-3/4)/(64*x*gamma(5/4
)) + b*cos(pi*b**2*x**2/2)*gamma(-3/4)/(64*x**3*gamma(5/4)) + 3*fresnelc(b*x)*gamma(-3/4)/(64*x**4*gamma(5/4))

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