\(\int x \operatorname {FresnelS}(a+b x) \, dx\) [27]

Optimal result

Integrand size = 8, antiderivative size = 96 \[ \int x \operatorname {FresnelS}(a+b x) \, dx=-\frac {a \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^2 \pi }+\frac {(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }-\frac {\operatorname {FresnelC}(a+b x)}{2 b^2 \pi }-\frac {a^2 \operatorname {FresnelS}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \operatorname {FresnelS}(a+b x) \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.53 \[ \int x \operatorname {FresnelS}(a+b x) \, dx=-\frac {\operatorname {FresnelC}(a+b x)+(a-b x) \left (\cos \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi (a+b x) \operatorname {FresnelS}(a+b x)\right )}{2 b^2 \pi } \] Input:

Integrate[x*FresnelS[a + b*x],x]
 

Output:

-1/2*(FresnelC[a + b*x] + (a - b*x)*(Cos[(Pi*(a + b*x)^2)/2] + Pi*(a + b*x 
)*FresnelS[a + b*x]))/(b^2*Pi)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6982, 3914, 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.

\(\displaystyle \int x \operatorname {FresnelS}(a+b x) \, dx\)

\(\Big \downarrow \) 6982

\(\displaystyle \frac {1}{2} x^2 \operatorname {FresnelS}(a+b x)-\frac {1}{2} b \int x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )dx\)

\(\Big \downarrow \) 3914

\(\displaystyle \frac {1}{2} x^2 \operatorname {FresnelS}(a+b x)-\frac {\int \left (\sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a^2-2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a+(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )\right )d(a+b x)}{2 b^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} x^2 \operatorname {FresnelS}(a+b x)-\frac {a^2 \operatorname {FresnelS}(a+b x)+\frac {\operatorname {FresnelC}(a+b x)}{\pi }+\frac {2 a \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }}{2 b^2}\)

Input:

Int[x*FresnelS[a + b*x],x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 3914

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat 
or[n], 1]}, Simp[k/f^(m + 1)   Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x 
^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x 
]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
 

rule 6982

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\operatorname {FresnelS}\left (b x +a \right ) \left (\frac {\left (b x +a \right )^{2}}{2}-\left (b x +a \right ) a \right )-\frac {a \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }-\frac {\operatorname {FresnelC}\left (b x +a \right )}{2 \pi }}{b^{2}}\)	\(80\)
default	\(\frac {\operatorname {FresnelS}\left (b x +a \right ) \left (\frac {\left (b x +a \right )^{2}}{2}-\left (b x +a \right ) a \right )-\frac {a \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }-\frac {\operatorname {FresnelC}\left (b x +a \right )}{2 \pi }}{b^{2}}\)	\(80\)
parts	\(\frac {x^{2} \operatorname {FresnelS}\left (b x +a \right )}{2}-\frac {b \left (-\frac {x \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}+\frac {\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{2}\)	\(167\)

Input:

int(x*FresnelS(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08 \[ \int x \operatorname {FresnelS}(a+b x) \, dx=\frac {\pi b^{3} x^{2} \operatorname {S}\left (b x + a\right ) - \pi a^{2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (b^{2} x - a b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{2 \, \pi b^{3}} \] Input:

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

Output:

1/2*(pi*b^3*x^2*fresnel_sin(b*x + a) - pi*a^2*sqrt(b^2)*fresnel_sin(sqrt(b 
^2)*(b*x + a)/b) + (b^2*x - a*b)*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^ 
2) - sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b))/(pi*b^3)
 

Sympy [F]

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

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

Output:

Integral(x*fresnels(a + b*x), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 307, normalized size of antiderivative = 3.20 \[ \int x \operatorname {FresnelS}(a+b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {S}\left (b x + a\right ) - \frac {{\left (8 \, {\left (\pi e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + \pi e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a b x + 8 \, {\left (\pi e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + \pi e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{2} - \sqrt {2 \, \pi b^{2} x^{2} + 4 \, \pi a b x + 2 \, \pi a^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \pi ^{\frac {3}{2}} {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \pi ^{\frac {3}{2}} {\left (\operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )}\right )} a^{2} - \left (2 i - 2\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + \left (2 i + 2\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )}\right )} b}{16 \, {\left (\pi ^{2} b^{4} x + \pi ^{2} a b^{3}\right )}} \] Input:

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

Output:

1/2*x^2*fresnel_sin(b*x + a) - 1/16*(8*(pi*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b* 
x + 1/2*I*pi*a^2) + pi*e^(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))* 
a*b*x + 8*(pi*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi*e^(-1/ 
2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 - sqrt(2*pi*b^2*x^2 + 4*p 
i*a*b*x + 2*pi*a^2)*((-(I + 1)*sqrt(2)*pi^(3/2)*(erf(sqrt(1/2*I*pi*b^2*x^2 
 + I*pi*a*b*x + 1/2*I*pi*a^2)) - 1) + (I - 1)*sqrt(2)*pi^(3/2)*(erf(sqrt(- 
1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - 1))*a^2 - (2*I - 2)*sqrt( 
2)*gamma(3/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + (2*I + 2)*sq 
rt(2)*gamma(3/2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)))*b/(pi^2* 
b^4*x + pi^2*a*b^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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