\(\int x^3 \cos (\frac {1}{2} b^2 \pi x^2) \operatorname {FresnelC}(b x) \, dx\) [185]

Optimal result

Integrand size = 20, antiderivative size = 104 \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=-\frac {x}{b^3 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^4 \pi ^2}-\frac {5 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi } \]

[Out]

-x/b^3/Pi^2+1/4*x*cos(b^2*Pi*x^2)/b^3/Pi^2+2*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/b^4/Pi^2+x^2*FresnelC(b*x)*sin(
1/2*b^2*Pi*x^2)/b^2/Pi-5/8*FresnelC(b*x*2^(1/2))/b^4/Pi^2*2^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6590, 6596, 3439, 3433, 3466} \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=-\frac {5 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} \pi ^2 b^4}-\frac {x}{\pi ^2 b^3}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {2 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]

[In]

Int[x^3*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x],x]

[Out]

-(x/(b^3*Pi^2)) + (x*Cos[b^2*Pi*x^2])/(4*b^3*Pi^2) + (2*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^4*Pi^2) - (5*Fre
snelC[Sqrt[2]*b*x])/(4*Sqrt[2]*b^4*Pi^2) + (x^2*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*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 3439

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Cos[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

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 6590

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

Rule 6596

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

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {2 \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac {\int x^2 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi } \\ & = \frac {x \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int \cos \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}-\frac {2 \int \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2} \\ & = \frac {x \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^4 \pi ^2}-\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {2 \int \left (\frac {1}{2}+\frac {1}{2} \cos \left (b^2 \pi x^2\right )\right ) \, dx}{b^3 \pi ^2} \\ & = -\frac {x}{b^3 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^4 \pi ^2}-\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int \cos \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2} \\ & = -\frac {x}{b^3 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^4 \pi ^2}-\frac {5 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi } \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.80 \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {2 b x \left (-4+\cos \left (b^2 \pi x^2\right )\right )-5 \sqrt {2} \operatorname {FresnelC}\left (\sqrt {2} b x\right )+8 \operatorname {FresnelC}(b x) \left (2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )+b^2 \pi x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )}{8 b^4 \pi ^2} \]

[In]

Integrate[x^3*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x],x]

[Out]

(2*b*x*(-4 + Cos[b^2*Pi*x^2]) - 5*Sqrt[2]*FresnelC[Sqrt[2]*b*x] + 8*FresnelC[b*x]*(2*Cos[(b^2*Pi*x^2)/2] + b^2
*Pi*x^2*Sin[(b^2*Pi*x^2)/2]))/(8*b^4*Pi^2)

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{b^{3}}-\frac {\frac {b x}{\pi ^{2}}+\frac {\sqrt {2}\, \operatorname {FresnelC}\left (b x \sqrt {2}\right )}{2 \pi ^{2}}+\frac {-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \operatorname {FresnelC}\left (b x \sqrt {2}\right )}{4 \pi }}{2 \pi }}{b^{3}}}{b}\)	\(114\)

[In]

int(x^3*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {8 \, \pi b^{3} x^{2} \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} - 10 \, b^{2} x + 16 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) - 5 \, \sqrt {2} \sqrt {b^{2}} \operatorname {C}\left (\sqrt {2} \sqrt {b^{2}} x\right )}{8 \, \pi ^{2} b^{5}} \]

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnel_cos(b*x),x, algorithm="fricas")

[Out]

1/8*(8*pi*b^3*x^2*fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2) + 4*b^2*x*cos(1/2*pi*b^2*x^2)^2 - 10*b^2*x + 16*b*cos(1
/2*pi*b^2*x^2)*fresnel_cos(b*x) - 5*sqrt(2)*sqrt(b^2)*fresnel_cos(sqrt(2)*sqrt(b^2)*x))/(pi^2*b^5)

Sympy [F]

\[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )\, dx \]

[In]

integrate(x**3*cos(1/2*b**2*pi*x**2)*fresnelc(b*x),x)

[Out]

Integral(x**3*cos(pi*b**2*x**2/2)*fresnelc(b*x), x)

Maxima [F]

\[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int { x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) \,d x } \]

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnel_cos(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x), x)

Giac [F]

\[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int { x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) \,d x } \]

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x^3*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int x^3\,\mathrm {FresnelC}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

[In]

int(x^3*FresnelC(b*x)*cos((Pi*b^2*x^2)/2),x)

[Out]

int(x^3*FresnelC(b*x)*cos((Pi*b^2*x^2)/2), x)