3.2.17 \(\int \text {FresnelC}(b x) \, dx\) [117]

Optimal. Leaf size=27 \[ x \text {FresnelC}(b x)-\frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi } \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6554} \begin {gather*} x \text {FresnelC}(b x)-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelC[b*x],x]

[Out]

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

Rule 6554

Int[FresnelC[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(a + b*x)*(FresnelC[a + b*x]/b), x] - Simp[Sin[(Pi/2)*(a +
 b*x)^2]/(b*Pi), x] /; FreeQ[{a, b}, x]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 27, normalized size = 1.00 \begin {gather*} x \text {FresnelC}(b x)-\frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[b*x],x]

[Out]

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

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Maple [A]
time = 0.39, size = 28, normalized size = 1.04

method result size
derivativedivides \(\frac {\FresnelC \left (b x \right ) b x -\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }}{b}\) \(28\)
default \(\frac {\FresnelC \left (b x \right ) b x -\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }}{b}\) \(28\)
meijerg \(\frac {-\frac {4 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\sqrt {\pi }}+4 \sqrt {\pi }\, b x \FresnelC \left (b x \right )}{4 \sqrt {\pi }\, b}\) \(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 27, normalized size = 1.00 \begin {gather*} \frac {b x \operatorname {C}\left (b x\right ) - \frac {\sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{\pi }}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 28, normalized size = 1.04 \begin {gather*} \frac {\pi b x \operatorname {C}\left (b x\right ) - \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{\pi b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
time = 0.36, size = 44, normalized size = 1.63 \begin {gather*} \frac {x C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{4 \Gamma \left (\frac {5}{4}\right )} - \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{4 \pi b \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x),x)

[Out]

x*fresnelc(b*x)*gamma(1/4)/(4*gamma(5/4)) - sin(pi*b**2*x**2/2)*gamma(1/4)/(4*pi*b*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(fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \mathrm {FresnelC}\left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x),x)

[Out]

int(FresnelC(b*x), x)

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