\(\int x^2 \operatorname {FresnelC}(a+b x) \, dx\) [135]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [C] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 148 \[ \int x^2 \operatorname {FresnelC}(a+b x) \, dx=-\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}+\frac {a^3 \operatorname {FresnelC}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)-\frac {a \operatorname {FresnelS}(a+b x)}{b^3 \pi }-\frac {a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi } \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6564, 3515, 3433, 3461, 2717, 3467, 3432, 3377, 2718} \[ \int x^2 \operatorname {FresnelC}(a+b x) \, dx=\frac {a^3 \operatorname {FresnelC}(a+b x)}{3 b^3}-\frac {a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {a \operatorname {FresnelS}(a+b x)}{\pi b^3}+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 \operatorname {FresnelC}(a+b x) \]

[In]

Int[x^2*FresnelC[a + b*x],x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3467

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sin[c + d*
x^n]/(d*n)), x] - Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3515

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Cos[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 6564

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)-\frac {1}{3} b \int x^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \, dx \\ & = \frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)-\frac {\text {Subst}\left (\int \left (-a^3 \cos \left (\frac {\pi x^2}{2}\right )+3 a^2 x \cos \left (\frac {\pi x^2}{2}\right )-3 a x^2 \cos \left (\frac {\pi x^2}{2}\right )+x^3 \cos \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{3 b^3} \\ & = \frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)-\frac {\text {Subst}\left (\int x^3 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}+\frac {a \text {Subst}\left (\int x^2 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac {a^2 \text {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}+\frac {a^3 \text {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3} \\ & = \frac {a^3 \operatorname {FresnelC}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {\text {Subst}\left (\int x \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{6 b^3}-\frac {a^2 \text {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^3}-\frac {a \text {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3 \pi } \\ & = \frac {a^3 \operatorname {FresnelC}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)-\frac {a \operatorname {FresnelS}(a+b x)}{b^3 \pi }-\frac {a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {\text {Subst}\left (\int \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{3 b^3 \pi } \\ & = -\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}+\frac {a^3 \operatorname {FresnelC}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \operatorname {FresnelC}(a+b x)-\frac {a \operatorname {FresnelS}(a+b x)}{b^3 \pi }-\frac {a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi } \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78 \[ \int x^2 \operatorname {FresnelC}(a+b x) \, dx=-\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-\pi ^2 \left (a^3+b^3 x^3\right ) \operatorname {FresnelC}(a+b x)+3 a \pi \operatorname {FresnelS}(a+b x)+a^2 \pi \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-a b \pi x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+b^2 \pi x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2} \]

[In]

Integrate[x^2*FresnelC[a + b*x],x]

[Out]

-1/3*(2*Cos[(Pi*(a + b*x)^2)/2] - Pi^2*(a^3 + b^3*x^3)*FresnelC[a + b*x] + 3*a*Pi*FresnelS[a + b*x] + a^2*Pi*S
in[(Pi*(a + b*x)^2)/2] - a*b*Pi*x*Sin[(Pi*(a + b*x)^2)/2] + b^2*Pi*x^2*Sin[(Pi*(a + b*x)^2)/2])/(b^3*Pi^2)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82

method result size
derivativedivides \(\frac {\frac {\operatorname {FresnelC}\left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \operatorname {FresnelC}\left (b x +a \right )}{3}-\frac {a^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \,\operatorname {FresnelS}\left (b x +a \right )}{\pi }-\frac {\left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\) \(122\)
default \(\frac {\frac {\operatorname {FresnelC}\left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \operatorname {FresnelC}\left (b x +a \right )}{3}-\frac {a^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \,\operatorname {FresnelS}\left (b x +a \right )}{\pi }-\frac {\left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\) \(122\)
parts \(\frac {x^{3} \operatorname {FresnelC}\left (b x +a \right )}{3}-\frac {b \left (\frac {x^{2} \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (\frac {x \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (\frac {\sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}-\frac {\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b}-\frac {2 \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b^{2} \pi }\right )}{3}\) \(286\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelC}(a+b x) \, dx=\frac {\pi ^{2} b^{4} x^{3} \operatorname {C}\left (b x + a\right ) + \pi ^{2} a^{3} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - 3 \, \pi a \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - 2 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (\pi b^{3} x^{2} - \pi a b^{2} x + \pi a^{2} b\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{3 \, \pi ^{2} b^{4}} \]

[In]

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

[Out]

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

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

[In]

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

[Out]

Integral(x**2*fresnelc(a + b*x), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.86 \[ \int x^2 \operatorname {FresnelC}(a+b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {C}\left (b x + a\right ) - \frac {{\left (12 \, {\left (-i \, \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 )} + i \, \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 (-i \, \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 )} + i \, \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 \, \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 \, \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 (\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 ) + \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 )}} \]

[In]

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

[Out]

1/3*x^3*fresnel_cos(b*x + a) - 1/24*(12*(-I*pi*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + I*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*(-I*pi*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2)
 + I*pi*e^(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 + 2*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/
2*I*pi*a^2) + 2*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*(gamma(2, 1/2*I*pi*b^2*x^2
+ I*pi*a*b*x + 1/2*I*pi*a^2) + 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)*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^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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^2*FresnelC(a + b*x),x)

[Out]

int(x^2*FresnelC(a + b*x), x)