\(\int x \operatorname {FresnelC}(b x) \, dx\) [116]

Optimal result

Integrand size = 6, antiderivative size = 49 \[ \int x \operatorname {FresnelC}(b x) \, dx=\frac {1}{2} x^2 \operatorname {FresnelC}(b x)+\frac {\operatorname {FresnelS}(b x)}{2 b^2 \pi }-\frac {x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi } \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6562, 3467, 3432} \[ \int x \operatorname {FresnelC}(b x) \, dx=\frac {\operatorname {FresnelS}(b x)}{2 \pi b^2}-\frac {x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac {1}{2} x^2 \operatorname {FresnelC}(b x) \]

[In]

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

[Out]

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

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

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*} \text {integral}& = \frac {1}{2} x^2 \operatorname {FresnelC}(b x)-\frac {1}{2} b \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {1}{2} x^2 \operatorname {FresnelC}(b x)-\frac {x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi }+\frac {\int \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{2 b \pi } \\ & = \frac {1}{2} x^2 \operatorname {FresnelC}(b x)+\frac {\operatorname {FresnelS}(b x)}{2 b^2 \pi }-\frac {x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi } \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int x \operatorname {FresnelC}(b x) \, dx=\frac {1}{2} x^2 \operatorname {FresnelC}(b x)+\frac {\operatorname {FresnelS}(b x)}{2 b^2 \pi }-\frac {x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi } \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.53

method	result	size
meijerg	\(\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {1}{2}, \frac {5}{4}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{3}\)	\(26\)
derivativedivides	\(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{2} x^{2}}{2}-\frac {b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 \pi }+\frac {\operatorname {FresnelS}\left (b x \right )}{2 \pi }}{b^{2}}\)	\(44\)
default	\(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{2} x^{2}}{2}-\frac {b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 \pi }+\frac {\operatorname {FresnelS}\left (b x \right )}{2 \pi }}{b^{2}}\)	\(44\)
parts	\(\frac {x^{2} \operatorname {FresnelC}\left (b x \right )}{2}-\frac {b \left (\frac {x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {\operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{2}\)	\(64\)

[In]

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

[Out]

1/3*b*x^3*hypergeom([1/4,3/4],[1/2,5/4,7/4],-1/16*x^4*Pi^2*b^4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int x \operatorname {FresnelC}(b x) \, dx=\frac {\pi b^{3} x^{2} \operatorname {C}\left (b x\right ) - b^{2} x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \sqrt {b^{2}} \operatorname {S}\left (\sqrt {b^{2}} x\right )}{2 \, \pi b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int x \operatorname {FresnelC}(b x) \, dx=\frac {b x^{3} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {1}{2}, \frac {5}{4}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

b*x**3*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2, 5/4, 7/4), -pi**2*b**4*x**4/16)/(16*gamma(5/4)*gamma(7/4)
)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.43 \[ \int x \operatorname {FresnelC}(b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {C}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \left (i + 1\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) + \left (i - 1\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{4 \, \pi ^{2} b^{2}} \]

[In]

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

[Out]

1/2*x^2*fresnel_cos(b*x) - 1/4*sqrt(1/2)*(4*sqrt(1/2)*pi*b*x*sin(1/2*pi*b^2*x^2) - (I + 1)*(1/4)^(1/4)*pi*erf(
sqrt(1/2*I*pi)*b*x) + (I - 1)*(1/4)^(1/4)*pi*erf(sqrt(-1/2*I*pi)*b*x))/(pi^2*b^2)

Giac [F]

\[ \int x \operatorname {FresnelC}(b x) \, dx=\int { x \operatorname {C}\left (b x\right ) \,d x } \]

[In]

integrate(x*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x*fresnel_cos(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x \operatorname {FresnelC}(b x) \, dx=\int x\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \]

[In]

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

[Out]

int(x*FresnelC(b*x), x)