Integrand size = 8, antiderivative size = 84 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)+\frac {8 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi } \] Output:
-4/5*x^2*cos(1/2*b^2*Pi*x^2)/b^3/Pi^2+1/5*x^5*FresnelC(b*x)+8/5*sin(1/2*b^ 2*Pi*x^2)/b^5/Pi^3-1/5*x^4*sin(1/2*b^2*Pi*x^2)/b/Pi
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {\left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3} \] Input:
Integrate[x^4*FresnelC[b*x],x]
Output:
(-4*x^2*Cos[(b^2*Pi*x^2)/2])/(5*b^3*Pi^2) + (x^5*FresnelC[b*x])/5 - ((-8 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3)
Time = 0.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {6981, 3861, 3042, 3777, 25, 3042, 3777, 3042, 3117}
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.
\(\displaystyle \int x^4 \operatorname {FresnelC}(b x) \, dx\) |
\(\Big \downarrow \) 6981 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{5} b \int x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx\) |
\(\Big \downarrow \) 3861 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \int x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \int x^4 \sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )dx^2\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \left (\frac {4 \int -x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}+\frac {2 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \left (\frac {2 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {4 \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \left (\frac {2 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {4 \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \left (\frac {2 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {4 \left (\frac {2 \int \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}-\frac {2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \left (\frac {2 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {4 \left (\frac {2 \int \sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )dx^2}{\pi b^2}-\frac {2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \left (\frac {2 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {4 \left (\frac {4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )\) |
Input:
Int[x^4*FresnelC[b*x],x]
Output:
(x^5*FresnelC[b*x])/5 - (b*((2*x^4*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) - (4*((-2 *x^2*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi) + (4*Sin[(b^2*Pi*x^2)/2])/(b^4*Pi^2)))/ (b^2*Pi)))/10
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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 = 0.31
method | result | size |
meijerg | \(\frac {b \,x^{6} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {1}{2}, \frac {5}{4}, \frac {5}{2}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{6}\) | \(26\) |
derivativedivides | \(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{5} x^{5}}{5}-\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {-\frac {4 b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {8 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi ^{2}}}{\pi }}{b^{5}}\) | \(81\) |
default | \(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{5} x^{5}}{5}-\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {-\frac {4 b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {8 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi ^{2}}}{\pi }}{b^{5}}\) | \(81\) |
parts | \(\frac {x^{5} \operatorname {FresnelC}\left (b x \right )}{5}-\frac {b \left (\frac {x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {4 \left (-\frac {x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{b^{2} \pi }\right )}{5}\) | \(83\) |
Input:
int(x^4*FresnelC(b*x),x,method=_RETURNVERBOSE)
Output:
1/6*b*x^6*hypergeom([1/4,3/2],[1/2,5/4,5/2],-1/16*x^4*Pi^2*b^4)
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {\pi ^{3} b^{5} x^{5} \operatorname {C}\left (b x\right ) - 4 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \] Input:
integrate(x^4*fresnel_cos(b*x),x, algorithm="fricas")
Output:
1/5*(pi^3*b^5*x^5*fresnel_cos(b*x) - 4*pi*b^2*x^2*cos(1/2*pi*b^2*x^2) - (p i^2*b^4*x^4 - 8)*sin(1/2*pi*b^2*x^2))/(pi^3*b^5)
Time = 0.73 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.38 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {x^{5} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{20 \Gamma \left (\frac {5}{4}\right )} - \frac {x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{20 \pi b \Gamma \left (\frac {5}{4}\right )} - \frac {x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {2 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(x**4*fresnelc(b*x),x)
Output:
x**5*fresnelc(b*x)*gamma(1/4)/(20*gamma(5/4)) - x**4*sin(pi*b**2*x**2/2)*g amma(1/4)/(20*pi*b*gamma(5/4)) - x**2*cos(pi*b**2*x**2/2)*gamma(1/4)/(5*pi **2*b**3*gamma(5/4)) + 2*sin(pi*b**2*x**2/2)*gamma(1/4)/(5*pi**3*b**5*gamm a(5/4))
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {C}\left (b x\right ) - \frac {4 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \] Input:
integrate(x^4*fresnel_cos(b*x),x, algorithm="maxima")
Output:
1/5*x^5*fresnel_cos(b*x) - 1/5*(4*pi*b^2*x^2*cos(1/2*pi*b^2*x^2) + (pi^2*b ^4*x^4 - 8)*sin(1/2*pi*b^2*x^2))/(pi^3*b^5)
\[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int { x^{4} \operatorname {C}\left (b x\right ) \,d x } \] Input:
integrate(x^4*fresnel_cos(b*x),x, algorithm="giac")
Output:
integrate(x^4*fresnel_cos(b*x), x)
Timed out. \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int x^4\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \] Input:
int(x^4*FresnelC(b*x),x)
Output:
int(x^4*FresnelC(b*x), x)
\[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int x^{4} \mathrm {FresnelC}\left (b x \right )d x \] Input:
int(x^4*FresnelC(b*x),x)
Output:
int(x^4*FresnelC(b*x),x)