3.1.4 \(\int x^4 S(b x) \, dx\) [4]

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

[Out]

-8/5*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/5*x^4*cos(1/2*b^2*Pi*x^2)/b/Pi+1/5*x^5*FresnelS(b*x)-4/5*x^2*sin(1/2*b^2*P
i*x^2)/b^3/Pi^2

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Rubi [A]
time = 0.05, 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 = {6561, 3460, 3377, 2718} \begin {gather*} \frac {x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi b}-\frac {8 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac {4 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 S(b x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*Cos[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3) + (x^4*Cos[(b^2*Pi*x^2)/2])/(5*b*Pi) + (x^5*FresnelS[b*x])/5 - (4*x^2*Si
n[(b^2*Pi*x^2)/2])/(5*b^3*Pi^2)

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 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[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 6561

Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(FresnelS[b*x]/(d*(m + 1))), x] -
Dist[b/(d*(m + 1)), Int[(d*x)^(m + 1)*Sin[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^4 S(b x) \, dx &=\frac {1}{5} x^5 S(b x)-\frac {1}{5} b \int x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {1}{5} x^5 S(b x)-\frac {1}{10} b \text {Subst}\left (\int x^2 \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )\\ &=\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 S(b x)-\frac {2 \text {Subst}\left (\int x \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b \pi }\\ &=\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 S(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {4 \text {Subst}\left (\int \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b^3 \pi ^2}\\ &=-\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 S(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 71, normalized size = 0.85 \begin {gather*} \frac {\left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {1}{5} x^5 S(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.33, size = 80, normalized size = 0.95

method	result	size
meijerg	\(\frac {\pi \,b^{3} x^{8} \hypergeom \left (\left [\frac {3}{4}, 2\right ], \left [\frac {3}{2}, \frac {7}{4}, 3\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{48}\)	\(29\)
derivativedivides	\(\frac {\frac {\mathrm {S}\left (b x \right ) b^{5} x^{5}}{5}+\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }}{b^{5}}\)	\(80\)
default	\(\frac {\frac {\mathrm {S}\left (b x \right ) b^{5} x^{5}}{5}+\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }}{b^{5}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 62, normalized size = 0.74 \begin {gather*} \frac {1}{5} \, x^{5} \operatorname {S}\left (b x\right ) - \frac {4 \, \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 65, normalized size = 0.77 \begin {gather*} \frac {\pi ^{3} b^{5} x^{5} \operatorname {S}\left (b x\right ) - 4 \, \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.89, size = 121, normalized size = 1.44 \begin {gather*} \frac {3 x^{5} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{20 \Gamma \left (\frac {7}{4}\right )} + \frac {3 x^{4} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{20 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {3 x^{2} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {6 \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**5*fresnels(b*x)*gamma(3/4)/(20*gamma(7/4)) + 3*x**4*cos(pi*b**2*x**2/2)*gamma(3/4)/(20*pi*b*gamma(7/4)) -
3*x**2*sin(pi*b**2*x**2/2)*gamma(3/4)/(5*pi**2*b**3*gamma(7/4)) - 6*cos(pi*b**2*x**2/2)*gamma(3/4)/(5*pi**3*b
**5*gamma(7/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnel_sin(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4*FresnelS(b*x), x)

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