\(\int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx\) [50]

Optimal result

Integrand size = 14, antiderivative size = 279 \[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\frac {d \cos \left (\pi (a+b x)^2\right )}{4 b^2 \pi ^2}+\frac {2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \operatorname {FresnelS}(a+b x)}{b^2 \pi }+\frac {d (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \operatorname {FresnelS}(a+b x)}{b^2 \pi }-\frac {d \operatorname {FresnelC}(a+b x) \operatorname {FresnelS}(a+b x)}{2 b^2 \pi }+\frac {(b c-a d) (a+b x) \operatorname {FresnelS}(a+b x)^2}{b^2}+\frac {d (a+b x)^2 \operatorname {FresnelS}(a+b x)^2}{2 b^2}-\frac {(b c-a d) \operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^2 \pi }+\frac {i d (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{8 b^2 \pi }-\frac {i d (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{8 b^2 \pi } \] Output:

1/4*d*cos(Pi*(b*x+a)^2)/b^2/Pi^2+2*(-a*d+b*c)*cos(1/2*Pi*(b*x+a)^2)*Fresne 
lS(b*x+a)/b^2/Pi+d*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)*FresnelS(b*x+a)/b^2/Pi-1/ 
2*d*FresnelC(b*x+a)*FresnelS(b*x+a)/b^2/Pi+(-a*d+b*c)*(b*x+a)*FresnelS(b*x 
+a)^2/b^2+1/2*d*(b*x+a)^2*FresnelS(b*x+a)^2/b^2-1/2*(-a*d+b*c)*FresnelS(2^ 
(1/2)*(b*x+a))*2^(1/2)/b^2/Pi+1/8*I*d*(b*x+a)^2*hypergeom([1, 1],[3/2, 2], 
-1/2*I*Pi*(b*x+a)^2)/b^2/Pi-1/8*I*d*(b*x+a)^2*hypergeom([1, 1],[3/2, 2],1/ 
2*I*Pi*(b*x+a)^2)/b^2/Pi
 

Mathematica [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx \] Input:

Integrate[(c + d*x)*FresnelS[a + b*x]^2,x]
 

Output:

Integrate[(c + d*x)*FresnelS[a + b*x]^2, x]
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6986, 2009}

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.

\(\displaystyle \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx\)

\(\Big \downarrow \) 6986

\(\displaystyle \frac {\int \left ((b c-a d) \operatorname {FresnelS}(a+b x)^2+d (a+b x) \operatorname {FresnelS}(a+b x)^2\right )d(a+b x)}{b^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {i d (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{8 \pi }-\frac {i d (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{8 \pi }+(a+b x) (b c-a d) \operatorname {FresnelS}(a+b x)^2-\frac {(b c-a d) \operatorname {FresnelS}\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi }+\frac {2 (b c-a d) \operatorname {FresnelS}(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {d \operatorname {FresnelC}(a+b x) \operatorname {FresnelS}(a+b x)}{2 \pi }+\frac {1}{2} d (a+b x)^2 \operatorname {FresnelS}(a+b x)^2+\frac {d (a+b x) \operatorname {FresnelS}(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }+\frac {d \cos \left (\pi (a+b x)^2\right )}{4 \pi ^2}}{b^2}\)

Input:

Int[(c + d*x)*FresnelS[a + b*x]^2,x]
 

Output:

((d*Cos[Pi*(a + b*x)^2])/(4*Pi^2) + (2*(b*c - a*d)*Cos[(Pi*(a + b*x)^2)/2] 
*FresnelS[a + b*x])/Pi + (d*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2]*FresnelS[a + 
 b*x])/Pi - (d*FresnelC[a + b*x]*FresnelS[a + b*x])/(2*Pi) + (b*c - a*d)*( 
a + b*x)*FresnelS[a + b*x]^2 + (d*(a + b*x)^2*FresnelS[a + b*x]^2)/2 - ((b 
*c - a*d)*FresnelS[Sqrt[2]*(a + b*x)])/(Sqrt[2]*Pi) + ((I/8)*d*(a + b*x)^2 
*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-1/2*I)*Pi*(a + b*x)^2])/Pi - ((I/8) 
*d*(a + b*x)^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*Pi*(a + b*x)^2])/ 
Pi)/b^2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6986

Int[FresnelS[(a_) + (b_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> 
Simp[1/b^(m + 1)   Subst[Int[ExpandIntegrand[FresnelS[x]^2, (b*c - a*d + d* 
x)^m, x], x], x, a + b*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0]
 

Maple [F]

\[\int \left (d x +c \right ) \operatorname {FresnelS}\left (b x +a \right )^{2}d x\]

Input:

int((d*x+c)*FresnelS(b*x+a)^2,x)
 

Output:

int((d*x+c)*FresnelS(b*x+a)^2,x)
 

Fricas [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*x+c)*fresnel_sin(b*x+a)^2,x, algorithm="fricas")
 

Output:

integral((d*x + c)*fresnel_sin(b*x + a)^2, x)
 

Sympy [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\int \left (c + d x\right ) S^{2}\left (a + b x\right )\, dx \] Input:

integrate((d*x+c)*fresnels(b*x+a)**2,x)
 

Output:

Integral((c + d*x)*fresnels(a + b*x)**2, x)
 

Maxima [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*x+c)*fresnel_sin(b*x+a)^2,x, algorithm="maxima")
 

Output:

integrate((d*x + c)*fresnel_sin(b*x + a)^2, x)
 

Giac [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*x+c)*fresnel_sin(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)*fresnel_sin(b*x + a)^2, x)
 

Mupad [F(-1)]

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

int(FresnelS(a + b*x)^2*(c + d*x),x)
 

Output:

int(FresnelS(a + b*x)^2*(c + d*x), x)
 

Reduce [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x)^2 \, dx=\int \left (d x +c \right ) \mathrm {FresnelS}\left (b x +a \right )^{2}d x \] Input:

int((d*x+c)*FresnelS(b*x+a)^2,x)
 

Output:

int((d*x+c)*FresnelS(b*x+a)^2,x)