3.2.14 \(\int x^3 \text {FresnelC}(b x) \, dx\) [114]

Optimal. Leaf size=74 \[ -\frac {3 x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {3 \text {FresnelC}(b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \text {FresnelC}(b x)-\frac {x^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi } \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 74, 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, 3467, 3466, 3433} \begin {gather*} \frac {3 \text {FresnelC}(b x)}{4 \pi ^2 b^4}-\frac {x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 \pi b}-\frac {3 x \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {1}{4} x^4 \text {FresnelC}(b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*FresnelC[b*x],x]

[Out]

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

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 3466

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^(n - 1))*(e*x)^(m - n + 1)*(Cos[c +
d*x^n]/(d*n)), x] + Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3467

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sin[c + d*
x^n]/(d*n)), x] - Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

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

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Mathematica [A]
time = 0.01, size = 74, normalized size = 1.00 \begin {gather*} -\frac {3 x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {3 \text {FresnelC}(b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \text {FresnelC}(b x)-\frac {x^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*FresnelC[b*x],x]

[Out]

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

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Maple [A]
time = 0.30, size = 70, normalized size = 0.95

method	result	size
meijerg	\(\frac {-\frac {3 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) b x}{4}-\frac {\pi \,b^{3} x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}+\frac {\left (5 x^{4} \pi ^{2} b^{4}+15\right ) \FresnelC \left (b x \right )}{20}}{b^{4} \pi ^{2}}\)	\(62\)
derivativedivides	\(\frac {\frac {\FresnelC \left (b x \right ) b^{4} x^{4}}{4}-\frac {b^{3} x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 \pi }+\frac {-\frac {3 b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 \pi }+\frac {3 \FresnelC \left (b x \right )}{4 \pi }}{\pi }}{b^{4}}\)	\(70\)
default	\(\frac {\frac {\FresnelC \left (b x \right ) b^{4} x^{4}}{4}-\frac {b^{3} x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 \pi }+\frac {-\frac {3 b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 \pi }+\frac {3 \FresnelC \left (b x \right )}{4 \pi }}{\pi }}{b^{4}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelC(b*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.48, size = 94, normalized size = 1.27 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {C}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 12 \, \sqrt {\frac {1}{2}} \pi b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (3 i - 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (3 i + 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{8 \, \pi ^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_cos(b*x),x, algorithm="maxima")

[Out]

1/4*x^4*fresnel_cos(b*x) - 1/8*sqrt(1/2)*(4*sqrt(1/2)*pi^2*b^3*x^3*sin(1/2*pi*b^2*x^2) + 12*sqrt(1/2)*pi*b*x*c
os(1/2*pi*b^2*x^2) + (3*I - 3)*(1/4)^(1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) - (3*I + 3)*(1/4)^(1/4)*pi*erf(sqrt(-1/2
*I*pi)*b*x))/(pi^3*b^4)

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Fricas [A]
time = 0.33, size = 59, normalized size = 0.80 \begin {gather*} -\frac {\pi b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 3 \, b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {C}\left (b x\right )}{4 \, \pi ^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_cos(b*x),x, algorithm="fricas")

[Out]

-1/4*(pi*b^3*x^3*sin(1/2*pi*b^2*x^2) + 3*b*x*cos(1/2*pi*b^2*x^2) - (pi^2*b^4*x^4 + 3)*fresnel_cos(b*x))/(pi^2*
b^4)

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Sympy [A]
time = 0.51, size = 112, normalized size = 1.51 \begin {gather*} \frac {5 x^{4} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{64 \Gamma \left (\frac {9}{4}\right )} - \frac {5 x^{3} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{64 \pi b \Gamma \left (\frac {9}{4}\right )} - \frac {15 x \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{64 \pi ^{2} b^{3} \Gamma \left (\frac {9}{4}\right )} + \frac {15 C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{64 \pi ^{2} b^{4} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*fresnelc(b*x),x)

[Out]

5*x**4*fresnelc(b*x)*gamma(1/4)/(64*gamma(9/4)) - 5*x**3*sin(pi*b**2*x**2/2)*gamma(1/4)/(64*pi*b*gamma(9/4)) -
15*x*cos(pi*b**2*x**2/2)*gamma(1/4)/(64*pi**2*b**3*gamma(9/4)) + 15*fresnelc(b*x)*gamma(1/4)/(64*pi**2*b**4*g
amma(9/4))

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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^3*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x^3*fresnel_cos(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelC(b*x),x)

[Out]

int(x^3*FresnelC(b*x), x)

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