3.152 \(\int \frac {C(b x)^2}{x^5} \, dx\)

Optimal. Leaf size=127 \[ -\frac {1}{12} \pi ^2 b^4 C(b x)^2-\frac {b C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^3}-\frac {b^2}{24 x^2}-\frac {b^2 \cos \left (\pi b^2 x^2\right )}{24 x^2}-\frac {1}{12} \pi b^4 \text {Si}\left (b^2 \pi x^2\right )+\frac {\pi b^3 C(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x}-\frac {C(b x)^2}{4 x^4} \]

[Out]

-1/24*b^2/x^2-1/24*b^2*cos(b^2*Pi*x^2)/x^2-1/6*b*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/x^3-1/12*b^4*Pi^2*FresnelC(
b*x)^2-1/4*FresnelC(b*x)^2/x^4-1/12*b^4*Pi*Si(b^2*Pi*x^2)+1/6*b^3*Pi*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/x

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Rubi [A]  time = 0.14, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6431, 6457, 6465, 6441, 30, 3375, 3380, 3297, 3299} \[ \frac {\pi b^3 \text {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x}-\frac {b \text {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^3}-\frac {1}{12} \pi ^2 b^4 \text {FresnelC}(b x)^2-\frac {1}{12} \pi b^4 \text {Si}\left (b^2 \pi x^2\right )-\frac {b^2}{24 x^2}-\frac {b^2 \cos \left (\pi b^2 x^2\right )}{24 x^2}-\frac {\text {FresnelC}(b x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[FresnelC[b*x]^2/x^5,x]

[Out]

-b^2/(24*x^2) - (b^2*Cos[b^2*Pi*x^2])/(24*x^2) - (b*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(6*x^3) - (b^4*Pi^2*Fre
snelC[b*x]^2)/12 - FresnelC[b*x]^2/(4*x^4) + (b^3*Pi*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(6*x) - (b^4*Pi*SinInt
egral[b^2*Pi*x^2])/12

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3297

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3375

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 6431

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

Rule 6441

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]^(n_.), x_Symbol] :> Dist[(Pi*b)/(2*d), Subst[Int[x^n, x], x, Fresne
lC[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6457

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m + 1)*Cos[d*x^2]*FresnelC[b*x])/(
m + 1), x] + (Dist[(2*d)/(m + 1), Int[x^(m + 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(2*(m + 1)), Int[x^(
m + 1)*Cos[2*d*x^2], x], x] - Simp[(b*x^(m + 2))/(2*(m + 1)*(m + 2)), x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^
2*b^4)/4] && ILtQ[m, -2]

Rule 6465

Int[FresnelC[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(x^(m + 1)*Sin[d*x^2]*FresnelC[b*x])/(
m + 1), x] + (-Dist[(2*d)/(m + 1), Int[x^(m + 2)*Cos[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(2*(m + 1)), Int[x^
(m + 1)*Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {C(b x)^2}{x^5} \, dx &=-\frac {C(b x)^2}{4 x^4}+\frac {1}{2} b \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{x^4} \, dx\\ &=-\frac {b^2}{24 x^2}-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{6 x^3}-\frac {C(b x)^2}{4 x^4}+\frac {1}{12} b^2 \int \frac {\cos \left (b^2 \pi x^2\right )}{x^3} \, dx-\frac {1}{6} \left (b^3 \pi \right ) \int \frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac {b^2}{24 x^2}-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{6 x^3}-\frac {C(b x)^2}{4 x^4}+\frac {b^3 \pi C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x}+\frac {1}{24} b^2 \operatorname {Subst}\left (\int \frac {\cos \left (b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )-\frac {1}{12} \left (b^4 \pi \right ) \int \frac {\sin \left (b^2 \pi x^2\right )}{x} \, dx-\frac {1}{6} \left (b^5 \pi ^2\right ) \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=-\frac {b^2}{24 x^2}-\frac {b^2 \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{6 x^3}-\frac {C(b x)^2}{4 x^4}+\frac {b^3 \pi C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x}-\frac {1}{24} b^4 \pi \text {Si}\left (b^2 \pi x^2\right )-\frac {1}{24} \left (b^4 \pi \right ) \operatorname {Subst}\left (\int \frac {\sin \left (b^2 \pi x\right )}{x} \, dx,x,x^2\right )-\frac {1}{6} \left (b^4 \pi ^2\right ) \operatorname {Subst}(\int x \, dx,x,C(b x))\\ &=-\frac {b^2}{24 x^2}-\frac {b^2 \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{6 x^3}-\frac {1}{12} b^4 \pi ^2 C(b x)^2-\frac {C(b x)^2}{4 x^4}+\frac {b^3 \pi C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x}-\frac {1}{12} b^4 \pi \text {Si}\left (b^2 \pi x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 127, normalized size = 1.00 \[ -\frac {1}{12} \pi ^2 b^4 C(b x)^2-\frac {b C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^3}-\frac {b^2}{24 x^2}-\frac {b^2 \cos \left (\pi b^2 x^2\right )}{24 x^2}-\frac {1}{12} \pi b^4 \text {Si}\left (b^2 \pi x^2\right )+\frac {\pi b^3 C(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x}-\frac {C(b x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[b*x]^2/x^5,x]

[Out]

-1/24*b^2/x^2 - (b^2*Cos[b^2*Pi*x^2])/(24*x^2) - (b*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(6*x^3) - (b^4*Pi^2*Fre
snelC[b*x]^2)/12 - FresnelC[b*x]^2/(4*x^4) + (b^3*Pi*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(6*x) - (b^4*Pi*SinInt
egral[b^2*Pi*x^2])/12

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnelc}\left (b x\right )^{2}}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)^2/x^5,x, algorithm="fricas")

[Out]

integral(fresnelc(b*x)^2/x^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnelc}\left (b x\right )^{2}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)^2/x^5,x, algorithm="giac")

[Out]

integrate(fresnelc(b*x)^2/x^5, x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {\FresnelC \left (b x \right )^{2}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)^2/x^5,x)

[Out]

int(FresnelC(b*x)^2/x^5,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnelc}\left (b x\right )^{2}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)^2/x^5,x, algorithm="maxima")

[Out]

integrate(fresnelc(b*x)^2/x^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {FresnelC}\left (b\,x\right )}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)^2/x^5,x)

[Out]

int(FresnelC(b*x)^2/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C^{2}\left (b x\right )}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)**2/x**5,x)

[Out]

Integral(fresnelc(b*x)**2/x**5, x)

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