\(\int x^4 \operatorname {FresnelC}(b x) \, dx\) [113]

Optimal result

Integrand size = 8, antiderivative size = 84 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)+\frac {8 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi } \]

[Out]

-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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6562, 3461, 3377, 2717} \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac {8 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac {4 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x) \]

[In]

Int[x^4*FresnelC[b*x],x]

[Out]

(-4*x^2*Cos[(b^2*Pi*x^2)/2])/(5*b^3*Pi^2) + (x^5*FresnelC[b*x])/5 + (8*Sin[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3) - (x^
4*Sin[(b^2*Pi*x^2)/2])/(5*b*Pi)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 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 6562

Int[FresnelC[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(FresnelC[b*x]/(d*(m + 1))), x] -
Dist[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]

Rubi steps \begin{align*} \text {integral}& = \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 \\ & = \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \text {Subst}\left (\int x^2 \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {2 \text {Subst}\left (\int x \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b \pi } \\ & = -\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 {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {4 \text {Subst}\left (\int \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b^3 \pi ^2} \\ & = -\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 } \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {\left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3} \]

[In]

Integrate[x^4*FresnelC[b*x],x]

[Out]

(-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)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.40 (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	\(\frac {x^{5} \operatorname {FresnelC}\left (b x \right )}{5}-\frac {b \left (\frac {x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {4 \left (-\frac {x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{b^{2} \pi }\right )}{5}\)	\(83\)

[In]

int(x^4*FresnelC(b*x),x,method=_RETURNVERBOSE)

[Out]

1/6*b*x^6*hypergeom([1/4,3/2],[1/2,5/4,5/2],-1/16*x^4*Pi^2*b^4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {\pi ^{3} b^{5} x^{5} \operatorname {C}\left (b x\right ) - 4 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \]

[In]

integrate(x^4*fresnel_cos(b*x),x, algorithm="fricas")

[Out]

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) - (pi^2*b^4*x^4 - 8)*sin(1/2*pi*b^2*x^2)
)/(pi^3*b^5)

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.38 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {x^{5} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{20 \Gamma \left (\frac {5}{4}\right )} - \frac {x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{20 \pi b \Gamma \left (\frac {5}{4}\right )} - \frac {x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {2 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(x**4*fresnelc(b*x),x)

[Out]

x**5*fresnelc(b*x)*gamma(1/4)/(20*gamma(5/4)) - x**4*sin(pi*b**2*x**2/2)*gamma(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*ga
mma(5/4))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {C}\left (b x\right ) - \frac {4 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \]

[In]

integrate(x^4*fresnel_cos(b*x),x, algorithm="maxima")

[Out]

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)

Giac [F]

\[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int { x^{4} \operatorname {C}\left (b x\right ) \,d x } \]

[In]

integrate(x^4*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x^4*fresnel_cos(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int x^4\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \]

[In]

int(x^4*FresnelC(b*x),x)

[Out]

int(x^4*FresnelC(b*x), x)