3.131 \(\int C(a+b x) \, dx\)

Optimal. Leaf size=37 \[ \frac {(a+b x) C(a+b x)}{b}-\frac {\sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6419} \[ \frac {(a+b x) \text {FresnelC}(a+b x)}{b}-\frac {\sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b} \]

Antiderivative was successfully verified.

[In]

Int[FresnelC[a + b*x],x]

[Out]

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

Rule 6419

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

Rubi steps

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

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Mathematica [B]  time = 0.04, size = 90, normalized size = 2.43 \[ -\frac {\sin \left (\frac {\pi a^2}{2}\right ) \cos \left (\pi a b x+\frac {1}{2} \pi b^2 x^2\right )}{\pi b}-\frac {\cos \left (\frac {\pi a^2}{2}\right ) \sin \left (\pi a b x+\frac {1}{2} \pi b^2 x^2\right )}{\pi b}+x C(a+b x)+\frac {a C(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[a + b*x],x]

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\rm fresnelc}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x+a),x, algorithm="fricas")

[Out]

integral(fresnelc(b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm fresnelc}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x+a),x, algorithm="giac")

[Out]

integrate(fresnelc(b*x + a), x)

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maple [A]  time = 0.00, size = 34, normalized size = 0.92 \[ \frac {\left (b x +a \right ) \FresnelC \left (b x +a \right )-\frac {\sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm fresnelc}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x+a),x, algorithm="maxima")

[Out]

integrate(fresnelc(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {FresnelC}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(a + b*x),x)

[Out]

int(FresnelC(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int C\left (a + b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(fresnelc(a + b*x), x)

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