Integrand size = 8, antiderivative size = 59 \[ \int x^2 \operatorname {FresnelC}(b x) \, dx=-\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi } \] Output:
-2/3*cos(1/2*b^2*Pi*x^2)/b^3/Pi^2+1/3*x^3*FresnelC(b*x)-1/3*x^2*sin(1/2*b^ 2*Pi*x^2)/b/Pi
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelC}(b x) \, dx=-\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi } \] Input:
Integrate[x^2*FresnelC[b*x],x]
Output:
(-2*Cos[(b^2*Pi*x^2)/2])/(3*b^3*Pi^2) + (x^3*FresnelC[b*x])/3 - (x^2*Sin[( b^2*Pi*x^2)/2])/(3*b*Pi)
Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6981, 3861, 3042, 3777, 25, 3042, 3118}
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.
\(\displaystyle \int x^2 \operatorname {FresnelC}(b x) \, dx\) |
\(\Big \downarrow \) 6981 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{3} b \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx\) |
\(\Big \downarrow \) 3861 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{6} b \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{6} b \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )dx^2\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{6} b \left (\frac {2 \int -\sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}+\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{6} b \left (\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {2 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{6} b \left (\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {2 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelC}(b x)-\frac {1}{6} b \left (\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}\right )\) |
Input:
Int[x^2*FresnelC[b*x],x]
Output:
(x^3*FresnelC[b*x])/3 - (b*((4*Cos[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + (2*x^2*Si n[(b^2*Pi*x^2)/2])/(b^2*Pi)))/6
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(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[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/n], 0]))
Int[FresnelC[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1 )*(FresnelC[b*x]/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[(d*x)^(m + 1)* Cos[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.44
method | result | size |
meijerg | \(\frac {b \,x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{4}, 1\right ], \left [\frac {1}{2}, \frac {5}{4}, 2\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{4}\) | \(26\) |
parts | \(\frac {x^{3} \operatorname {FresnelC}\left (b x \right )}{3}-\frac {b \left (\frac {x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{3}\) | \(53\) |
derivativedivides | \(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{3} x^{3}}{3}-\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\) | \(54\) |
default | \(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{3} x^{3}}{3}-\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\) | \(54\) |
Input:
int(x^2*FresnelC(b*x),x,method=_RETURNVERBOSE)
Output:
1/4*b*x^4*hypergeom([1/4,1],[1/2,5/4,2],-1/16*x^4*Pi^2*b^4)
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int x^2 \operatorname {FresnelC}(b x) \, dx=\frac {\pi ^{2} b^{3} x^{3} \operatorname {C}\left (b x\right ) - \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \] Input:
integrate(x^2*fresnel_cos(b*x),x, algorithm="fricas")
Output:
1/3*(pi^2*b^3*x^3*fresnel_cos(b*x) - pi*b^2*x^2*sin(1/2*pi*b^2*x^2) - 2*co s(1/2*pi*b^2*x^2))/(pi^2*b^3)
Time = 0.68 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int x^2 \operatorname {FresnelC}(b x) \, dx=\frac {x^{3} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{12 \Gamma \left (\frac {5}{4}\right )} - \frac {x^{2} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{12 \pi b \Gamma \left (\frac {5}{4}\right )} - \frac {\cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{6 \pi ^{2} b^{3} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(x**2*fresnelc(b*x),x)
Output:
x**3*fresnelc(b*x)*gamma(1/4)/(12*gamma(5/4)) - x**2*sin(pi*b**2*x**2/2)*g amma(1/4)/(12*pi*b*gamma(5/4)) - cos(pi*b**2*x**2/2)*gamma(1/4)/(6*pi**2*b **3*gamma(5/4))
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int x^2 \operatorname {FresnelC}(b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {C}\left (b x\right ) - \frac {\pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \] Input:
integrate(x^2*fresnel_cos(b*x),x, algorithm="maxima")
Output:
1/3*x^3*fresnel_cos(b*x) - 1/3*(pi*b^2*x^2*sin(1/2*pi*b^2*x^2) + 2*cos(1/2 *pi*b^2*x^2))/(pi^2*b^3)
\[ \int x^2 \operatorname {FresnelC}(b x) \, dx=\int { x^{2} \operatorname {C}\left (b x\right ) \,d x } \] Input:
integrate(x^2*fresnel_cos(b*x),x, algorithm="giac")
Output:
integrate(x^2*fresnel_cos(b*x), x)
Timed out. \[ \int x^2 \operatorname {FresnelC}(b x) \, dx=\int x^2\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \] Input:
int(x^2*FresnelC(b*x),x)
Output:
int(x^2*FresnelC(b*x), x)
\[ \int x^2 \operatorname {FresnelC}(b x) \, dx=\int x^{2} \mathrm {FresnelC}\left (b x \right )d x \] Input:
int(x^2*FresnelC(b*x),x)
Output:
int(x^2*FresnelC(b*x),x)