3.2.15 \(\int x^2 \text {FresnelC}(b x) \, dx\) [115]

Optimal. Leaf size=59 \[ -\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac {1}{3} x^3 \text {FresnelC}(b x)-\frac {x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi } \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 59, 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 = {6562, 3461, 3377, 2718} \begin {gather*} -\frac {x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi b}-\frac {2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 \text {FresnelC}(b x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3461

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 6562

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 59, normalized size = 1.00 \begin {gather*} -\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac {1}{3} x^3 \text {FresnelC}(b x)-\frac {x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.30, size = 54, normalized size = 0.92

method	result	size
meijerg	\(\frac {b \,x^{4} \hypergeom \left (\left [\frac {1}{4}, 1\right ], \left [\frac {1}{2}, \frac {5}{4}, 2\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{4}\)	\(26\)
derivativedivides	\(\frac {\frac {\FresnelC \left (b x \right ) b^{3} x^{3}}{3}-\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(54\)
default	\(\frac {\frac {\FresnelC \left (b x \right ) b^{3} x^{3}}{3}-\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelC(b*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 49, normalized size = 0.83 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {C}\left (b x\right ) - \frac {\pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*fresnel_cos(b*x),x, algorithm="maxima")

[Out]

1/3*x^3*fresnel_cos(b*x) - 1/3*(pi*b^2*x^2*sin(1/2*pi*b^2*x^2) + 2*cos(1/2*pi*b^2*x^2))/(pi^2*b^3)

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Fricas [A]
time = 0.35, size = 54, normalized size = 0.92 \begin {gather*} \frac {\pi ^{2} b^{3} x^{3} \operatorname {C}\left (b x\right ) - \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*fresnel_cos(b*x),x, algorithm="fricas")

[Out]

1/3*(pi^2*b^3*x^3*fresnel_cos(b*x) - pi*b^2*x^2*sin(1/2*pi*b^2*x^2) - 2*cos(1/2*pi*b^2*x^2))/(pi^2*b^3)

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Sympy [A]
time = 0.54, size = 80, normalized size = 1.36 \begin {gather*} \frac {x^{3} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{12 \Gamma \left (\frac {5}{4}\right )} - \frac {x^{2} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{12 \pi b \Gamma \left (\frac {5}{4}\right )} - \frac {\cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{6 \pi ^{2} b^{3} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*fresnelc(b*x)*gamma(1/4)/(12*gamma(5/4)) - x**2*sin(pi*b**2*x**2/2)*gamma(1/4)/(12*pi*b*gamma(5/4)) - cos
(pi*b**2*x**2/2)*gamma(1/4)/(6*pi**2*b**3*gamma(5/4))

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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(x^2*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x^2*fresnel_cos(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*FresnelC(b*x), x)

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