3.113 \(\int x^4 C(b x) \, dx\)

Optimal. Leaf size=84 \[ -\frac {x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac {8 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac {4 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 C(b x) \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6427, 3380, 3296, 2637} \[ -\frac {x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac {8 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac {4 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 \text {FresnelC}(b x) \]

Antiderivative was successfully verified.

[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 2637

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

Rule 3296

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 3380

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 6427

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*b^2*x^2)/2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^4 C(b x) \, dx &=\frac {1}{5} x^5 C(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 C(b x)-\frac {1}{10} b \operatorname {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 C(b x)-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {2 \operatorname {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 C(b x)-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {4 \operatorname {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 C(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*}

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Mathematica [A]  time = 0.06, size = 71, normalized size = 0.85 \[ -\frac {4 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}-\frac {\left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}+\frac {1}{5} x^5 C(b x) \]

Antiderivative was successfully verified.

[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)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{4} {\rm fresnelc}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4*fresnelc(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} {\rm fresnelc}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnelc(b*x), x)

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maple [A]  time = 0.00, size = 81, normalized size = 0.96 \[ \frac {\frac {b^{5} x^{5} \FresnelC \left (b x \right )}{5}-\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {-\frac {4 b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {8 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi ^{2}}}{\pi }}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^5*(1/5*b^5*x^5*FresnelC(b*x)-1/5/Pi*b^4*x^4*sin(1/2*b^2*Pi*x^2)+4/5/Pi*(-1/Pi*b^2*x^2*cos(1/2*b^2*Pi*x^2)+
2/Pi^2*sin(1/2*b^2*Pi*x^2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} {\rm fresnelc}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnelc(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4*FresnelC(b*x), x)

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sympy [A]  time = 1.20, size = 116, normalized size = 1.38 \[ \frac {x^{5} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{20 \Gamma \left (\frac {5}{4}\right )} - \frac {x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{20 \pi b \Gamma \left (\frac {5}{4}\right )} - \frac {x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {2 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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