\(\int \operatorname {FresnelC}(a+b x)^2 \, dx\) [160]

Optimal result

Integrand size = 8, antiderivative size = 69 \[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\frac {(a+b x) \operatorname {FresnelC}(a+b x)^2}{b}+\frac {\operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b \pi }-\frac {2 \operatorname {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b \pi } \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6556, 6588, 3432} \[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\frac {(a+b x) \operatorname {FresnelC}(a+b x)^2}{b}-\frac {2 \operatorname {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b}+\frac {\operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b} \]

[In]

Int[FresnelC[a + b*x]^2,x]

[Out]

((a + b*x)*FresnelC[a + b*x]^2)/b + FresnelS[Sqrt[2]*(a + b*x)]/(Sqrt[2]*b*Pi) - (2*FresnelC[a + b*x]*Sin[(Pi*
(a + b*x)^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 6556

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

Rule 6588

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\frac {2 \pi (a+b x) \operatorname {FresnelC}(a+b x)^2+\sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )-4 \operatorname {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b \pi } \]

[In]

Integrate[FresnelC[a + b*x]^2,x]

[Out]

(2*Pi*(a + b*x)*FresnelC[a + b*x]^2 + Sqrt[2]*FresnelS[Sqrt[2]*(a + b*x)] - 4*FresnelC[a + b*x]*Sin[(Pi*(a + b
*x)^2)/2])/(2*b*Pi)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\operatorname {FresnelC}\left (b x +a \right )^{2} \left (b x +a \right )-\frac {2 \,\operatorname {FresnelC}\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\sqrt {2}\, \operatorname {FresnelS}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b}\)	\(60\)
default	\(\frac {\operatorname {FresnelC}\left (b x +a \right )^{2} \left (b x +a \right )-\frac {2 \,\operatorname {FresnelC}\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\sqrt {2}\, \operatorname {FresnelS}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b}\)	\(60\)

[In]

int(FresnelC(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28 \[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\frac {2 \, {\left (\pi b^{2} x + \pi a b\right )} \operatorname {C}\left (b x + a\right )^{2} - 4 \, b \operatorname {C}\left (b x + a\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{2 \, \pi b^{2}} \]

[In]

integrate(fresnel_cos(b*x+a)^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\int C^{2}\left (a + b x\right )\, dx \]

[In]

integrate(fresnelc(b*x+a)**2,x)

[Out]

Integral(fresnelc(a + b*x)**2, x)

Maxima [F]

\[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\int { \operatorname {C}\left (b x + a\right )^{2} \,d x } \]

[In]

integrate(fresnel_cos(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate(fresnel_cos(b*x + a)^2, x)

Giac [F]

\[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\int { \operatorname {C}\left (b x + a\right )^{2} \,d x } \]

[In]

integrate(fresnel_cos(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \operatorname {FresnelC}(a+b x)^2 \, dx=\int {\mathrm {FresnelC}\left (a+b\,x\right )}^2 \,d x \]

[In]

int(FresnelC(a + b*x)^2,x)

[Out]

int(FresnelC(a + b*x)^2, x)