Integrand size = 8, antiderivative size = 70 \[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \operatorname {FresnelS}(a+b x)}{b \pi }+\frac {(a+b x) \operatorname {FresnelS}(a+b x)^2}{b}-\frac {\operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b \pi } \] Output:
2*cos(1/2*Pi*(b*x+a)^2)*FresnelS(b*x+a)/b/Pi+(b*x+a)*FresnelS(b*x+a)^2/b-1 /2*FresnelS(2^(1/2)*(b*x+a))*2^(1/2)/b/Pi
Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\frac {4 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \operatorname {FresnelS}(a+b x)+2 \pi (a+b x) \operatorname {FresnelS}(a+b x)^2-\sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{2 b \pi } \] Input:
Integrate[FresnelS[a + b*x]^2,x]
Output:
(4*Cos[(Pi*(a + b*x)^2)/2]*FresnelS[a + b*x] + 2*Pi*(a + b*x)*FresnelS[a + b*x]^2 - Sqrt[2]*FresnelS[Sqrt[2]*(a + b*x)])/(2*b*Pi)
Time = 0.49 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6974, 7281, 7006, 3832}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \operatorname {FresnelS}(a+b x)^2 \, dx\) |
\(\Big \downarrow \) 6974 |
\(\displaystyle \frac {(a+b x) \operatorname {FresnelS}(a+b x)^2}{b}-2 \int (a+b x) \operatorname {FresnelS}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {(a+b x) \operatorname {FresnelS}(a+b x)^2}{b}-\frac {2 \int (a+b x) \operatorname {FresnelS}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )d(a+b x)}{b}\) |
\(\Big \downarrow \) 7006 |
\(\displaystyle \frac {(a+b x) \operatorname {FresnelS}(a+b x)^2}{b}-\frac {2 \left (\frac {\int \sin \left (\pi (a+b x)^2\right )d(a+b x)}{2 \pi }-\frac {\operatorname {FresnelS}(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }\right )}{b}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle \frac {(a+b x) \operatorname {FresnelS}(a+b x)^2}{b}-\frac {2 \left (\frac {\operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{2 \sqrt {2} \pi }-\frac {\operatorname {FresnelS}(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }\right )}{b}\) |
Input:
Int[FresnelS[a + b*x]^2,x]
Output:
((a + b*x)*FresnelS[a + b*x]^2)/b - (2*(-((Cos[(Pi*(a + b*x)^2)/2]*Fresnel S[a + b*x])/Pi) + FresnelS[Sqrt[2]*(a + b*x)]/(2*Sqrt[2]*Pi)))/b
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]
Int[FresnelS[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(FresnelS[a + b*x]^2/b), x] - Simp[2 Int[(a + b*x)*Sin[(Pi/2)*(a + b*x)^2]*FresnelS[ a + b*x], x], x] /; FreeQ[{a, b}, x]
Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-Cos[d* x^2])*(FresnelS[b*x]/(2*d)), x] + Simp[1/(2*b*Pi) Int[Sin[2*d*x^2], x], x ] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Time = 0.45 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {\operatorname {FresnelS}\left (b x +a \right )^{2} \left (b x +a \right )+\frac {2 \,\operatorname {FresnelS}\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (\sqrt {2}\, \left (b x +a \right )\right )}{2 \pi }}{b}\) | \(60\) |
default | \(\frac {\operatorname {FresnelS}\left (b x +a \right )^{2} \left (b x +a \right )+\frac {2 \,\operatorname {FresnelS}\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (\sqrt {2}\, \left (b x +a \right )\right )}{2 \pi }}{b}\) | \(60\) |
Input:
int(FresnelS(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(FresnelS(b*x+a)^2*(b*x+a)+2*FresnelS(b*x+a)/Pi*cos(1/2*Pi*(b*x+a)^2)- 1/2/Pi*2^(1/2)*FresnelS(2^(1/2)*(b*x+a)))
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.27 \[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\frac {4 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) \operatorname {S}\left (b x + a\right ) + 2 \, {\left (\pi b^{2} x + \pi a b\right )} \operatorname {S}\left (b x + a\right )^{2} - \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{2 \, \pi b^{2}} \] Input:
integrate(fresnel_sin(b*x+a)^2,x, algorithm="fricas")
Output:
1/2*(4*b*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2)*fresnel_sin(b*x + a) + 2*(pi*b^2*x + pi*a*b)*fresnel_sin(b*x + a)^2 - sqrt(2)*sqrt(b^2)*fresnel _sin(sqrt(2)*sqrt(b^2)*(b*x + a)/b))/(pi*b^2)
\[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\int S^{2}\left (a + b x\right )\, dx \] Input:
integrate(fresnels(b*x+a)**2,x)
Output:
Integral(fresnels(a + b*x)**2, x)
\[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\int { \operatorname {S}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(fresnel_sin(b*x+a)^2,x, algorithm="maxima")
Output:
integrate(fresnel_sin(b*x + a)^2, x)
\[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\int { \operatorname {S}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(fresnel_sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate(fresnel_sin(b*x + a)^2, x)
Timed out. \[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\int {\mathrm {FresnelS}\left (a+b\,x\right )}^2 \,d x \] Input:
int(FresnelS(a + b*x)^2,x)
Output:
int(FresnelS(a + b*x)^2, x)
\[ \int \operatorname {FresnelS}(a+b x)^2 \, dx=\int \mathrm {FresnelS}\left (b x +a \right )^{2}d x \] Input:
int(FresnelS(b*x+a)^2,x)
Output:
int(FresnelS(b*x+a)^2,x)