\(\int x^2 \operatorname {FresnelS}(a+b x) \, dx\) [26]

Optimal result

Integrand size = 10, antiderivative size = 147 \[ \int x^2 \operatorname {FresnelS}(a+b x) \, dx=\frac {a^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {(a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {a \operatorname {FresnelC}(a+b x)}{b^3 \pi }+\frac {a^3 \operatorname {FresnelS}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \operatorname {FresnelS}(a+b x)-\frac {2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2} \] Output:

a^2*cos(1/2*Pi*(b*x+a)^2)/b^3/Pi-a*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)/b^3/Pi+1/ 
3*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)/b^3/Pi+a*FresnelC(b*x+a)/b^3/Pi+1/3*a^3* 
FresnelS(b*x+a)/b^3+1/3*x^3*FresnelS(b*x+a)-2/3*sin(1/2*Pi*(b*x+a)^2)/b^3/ 
Pi^2
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.78 \[ \int x^2 \operatorname {FresnelS}(a+b x) \, dx=\frac {a^2 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+3 a \pi \operatorname {FresnelC}(a+b x)+\pi ^2 \left (a^3+b^3 x^3\right ) \operatorname {FresnelS}(a+b x)-2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2} \] Input:

Integrate[x^2*FresnelS[a + b*x],x]
 

Output:

(a^2*Pi*Cos[(Pi*(a + b*x)^2)/2] - a*b*Pi*x*Cos[(Pi*(a + b*x)^2)/2] + b^2*P 
i*x^2*Cos[(Pi*(a + b*x)^2)/2] + 3*a*Pi*FresnelC[a + b*x] + Pi^2*(a^3 + b^3 
*x^3)*FresnelS[a + b*x] - 2*Sin[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi^2)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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.

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

\(\Big \downarrow \) 6982

\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(a+b x)-\frac {1}{3} b \int x^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )dx\)

\(\Big \downarrow \) 3914

\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(a+b x)-\frac {\int \left (-\sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a^3+3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a^2-3 (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a+(a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )\right )d(a+b x)}{3 b^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(a+b x)-\frac {a^3 (-\operatorname {FresnelS}(a+b x))-\frac {3 a^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {3 a \operatorname {FresnelC}(a+b x)}{\pi }+\frac {2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2}+\frac {3 a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {(a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }}{3 b^3}\)

Input:

Int[x^2*FresnelS[a + b*x],x]
 

Output:

(x^3*FresnelS[a + b*x])/3 - ((-3*a^2*Cos[(Pi*(a + b*x)^2)/2])/Pi + (3*a*(a 
 + b*x)*Cos[(Pi*(a + b*x)^2)/2])/Pi - ((a + b*x)^2*Cos[(Pi*(a + b*x)^2)/2] 
)/Pi - (3*a*FresnelC[a + b*x])/Pi - a^3*FresnelS[a + b*x] + (2*Sin[(Pi*(a 
+ b*x)^2)/2])/Pi^2)/(3*b^3)
 

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 = 121, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \operatorname {FresnelS}\left (b x +a \right )}{3}+\frac {a^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a \,\operatorname {FresnelC}\left (b x +a \right )}{\pi }+\frac {\left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(121\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \operatorname {FresnelS}\left (b x +a \right )}{3}+\frac {a^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a \,\operatorname {FresnelC}\left (b x +a \right )}{\pi }+\frac {\left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(121\)
parts	\(\frac {x^{3} \operatorname {FresnelS}\left (b x +a \right )}{3}-\frac {b \left (-\frac {x^{2} \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {x \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}+\frac {\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b}+\frac {\frac {2 \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {2 \sqrt {\pi }\, a \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}}{b^{2} \pi }\right )}{3}\)	\(287\)

Input:

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

Output:

1/b^3*(1/3*FresnelS(b*x+a)*b^3*x^3+1/3*a^3*FresnelS(b*x+a)+a^2/Pi*cos(1/2* 
Pi*(b*x+a)^2)-a/Pi*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)+a/Pi*FresnelC(b*x+a)+1/3/ 
Pi*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)-2/3/Pi^2*sin(1/2*Pi*(b*x+a)^2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelS}(a+b x) \, dx=\frac {\pi ^{2} b^{4} x^{3} \operatorname {S}\left (b x + a\right ) + \pi ^{2} a^{3} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + 3 \, \pi a \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi b^{3} x^{2} - \pi a b^{2} x + \pi a^{2} b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - 2 \, b \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{3 \, \pi ^{2} b^{4}} \] Input:

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

Output:

1/3*(pi^2*b^4*x^3*fresnel_sin(b*x + a) + pi^2*a^3*sqrt(b^2)*fresnel_sin(sq 
rt(b^2)*(b*x + a)/b) + 3*pi*a*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b) 
 + (pi*b^3*x^2 - pi*a*b^2*x + pi*a^2*b)*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/ 
2*pi*a^2) - 2*b*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2))/(pi^2*b^4)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.88 \[ \int x^2 \operatorname {FresnelS}(a+b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {S}\left (b x + a\right ) + \frac {{\left (12 \, {\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^{3} + 4 \, {\left (3 \, {\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} - 2 i \, \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + 2 i \, \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} b x + 8 \, a {\left (-i \, \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + i \, \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} - \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^{3} - 6 \, {\left (\left (i - 1\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 (i + 1\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 )} a\right )}\right )} b}{24 \, {\left (\pi ^{2} b^{5} x + \pi ^{2} a b^{4}\right )}} \] Input:

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

Output:

1/3*x^3*fresnel_sin(b*x + a) + 1/24*(12*(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^3 + 4*(3*(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 - 2*I*gamma(2, 1/2*I*pi 
*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + 2*I*gamma(2, -1/2*I*pi*b^2*x^2 - I 
*pi*a*b*x - 1/2*I*pi*a^2))*b*x + 8*a*(-I*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi* 
a*b*x + 1/2*I*pi*a^2) + I*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I* 
pi*a^2)) - sqrt(2*pi*b^2*x^2 + 4*pi*a*b*x + 2*pi*a^2)*((-(I + 1)*sqrt(2)*p 
i^(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^3 - 6*((I - 1)*sqrt(2)*gamma(3/2, 1/2*I*pi*b^2*x^2 + I*pi*a 
*b*x + 1/2*I*pi*a^2) - (I + 1)*sqrt(2)*gamma(3/2, -1/2*I*pi*b^2*x^2 - I*pi 
*a*b*x - 1/2*I*pi*a^2))*a))*b/(pi^2*b^5*x + pi^2*a*b^4)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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