3.2 \(\int x^6 S(b x) \, dx\)

Optimal. Leaf size=109 \[ \frac {x^6 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi b}+\frac {48 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^4 b^7}-\frac {24 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^3 b^5}-\frac {6 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^2 b^3}+\frac {1}{7} x^7 S(b x) \]

[Out]

-24/7*x^2*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/7*x^6*cos(1/2*b^2*Pi*x^2)/b/Pi+1/7*x^7*FresnelS(b*x)+48/7*sin(1/2*b^2
*Pi*x^2)/b^7/Pi^4-6/7*x^4*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2

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Rubi [A]  time = 0.11, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6426, 3379, 3296, 2637} \[ -\frac {6 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^2 b^3}+\frac {48 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^4 b^7}+\frac {x^6 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi b}-\frac {24 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^3 b^5}+\frac {1}{7} x^7 S(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^6*FresnelS[b*x],x]

[Out]

(-24*x^2*Cos[(b^2*Pi*x^2)/2])/(7*b^5*Pi^3) + (x^6*Cos[(b^2*Pi*x^2)/2])/(7*b*Pi) + (x^7*FresnelS[b*x])/7 + (48*
Sin[(b^2*Pi*x^2)/2])/(7*b^7*Pi^4) - (6*x^4*Sin[(b^2*Pi*x^2)/2])/(7*b^3*Pi^2)

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 3379

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 6426

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

Rubi steps

\begin {align*} \int x^6 S(b x) \, dx &=\frac {1}{7} x^7 S(b x)-\frac {1}{7} b \int x^7 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {1}{7} x^7 S(b x)-\frac {1}{14} b \operatorname {Subst}\left (\int x^3 \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )\\ &=\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 S(b x)-\frac {3 \operatorname {Subst}\left (\int x^2 \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{7 b \pi }\\ &=\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 S(b x)-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2}+\frac {12 \operatorname {Subst}\left (\int x \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{7 b^3 \pi ^2}\\ &=-\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^5 \pi ^3}+\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 S(b x)-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2}+\frac {24 \operatorname {Subst}\left (\int \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{7 b^5 \pi ^3}\\ &=-\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^5 \pi ^3}+\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 S(b x)+\frac {48 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^7 \pi ^4}-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 83, normalized size = 0.76 \[ -\frac {6 \left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^4 b^7}+\frac {x^2 \left (\pi ^2 b^4 x^4-24\right ) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^3 b^5}+\frac {1}{7} x^7 S(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6*FresnelS[b*x],x]

[Out]

(x^2*(-24 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2])/(7*b^5*Pi^3) + (x^7*FresnelS[b*x])/7 - (6*(-8 + b^4*Pi^2*x^4)*S
in[(b^2*Pi*x^2)/2])/(7*b^7*Pi^4)

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fricas [F]  time = 0.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{6} {\rm fresnels}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^6*fresnels(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{6} {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^6*fresnels(b*x), x)

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maple [A]  time = 0.02, size = 107, normalized size = 0.98 \[ \frac {\frac {b^{7} x^{7} \mathrm {S}\left (b x \right )}{7}+\frac {b^{6} x^{6} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{7 \pi }-\frac {6 \left (\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {4 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{\pi }\right )}{7 \pi }}{b^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*FresnelS(b*x),x)

[Out]

1/b^7*(1/7*b^7*x^7*FresnelS(b*x)+1/7/Pi*b^6*x^6*cos(1/2*b^2*Pi*x^2)-6/7/Pi*(1/Pi*b^4*x^4*sin(1/2*b^2*Pi*x^2)-4
/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 \[ \int x^{6} {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^6*fresnels(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^6\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*FresnelS(b*x),x)

[Out]

int(x^6*FresnelS(b*x), x)

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sympy [A]  time = 1.50, size = 156, normalized size = 1.43 \[ \frac {3 x^{7} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{28 \Gamma \left (\frac {7}{4}\right )} + \frac {3 x^{6} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{28 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {9 x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{14 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {18 x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{7 \pi ^{3} b^{5} \Gamma \left (\frac {7}{4}\right )} + \frac {36 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{7 \pi ^{4} b^{7} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*fresnels(b*x),x)

[Out]

3*x**7*fresnels(b*x)*gamma(3/4)/(28*gamma(7/4)) + 3*x**6*cos(pi*b**2*x**2/2)*gamma(3/4)/(28*pi*b*gamma(7/4)) -
9*x**4*sin(pi*b**2*x**2/2)*gamma(3/4)/(14*pi**2*b**3*gamma(7/4)) - 18*x**2*cos(pi*b**2*x**2/2)*gamma(3/4)/(7*
pi**3*b**5*gamma(7/4)) + 36*sin(pi*b**2*x**2/2)*gamma(3/4)/(7*pi**4*b**7*gamma(7/4))

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