3.1 \(\int x^7 S(b x) \, dx\)

Optimal. Leaf size=124 \[ -\frac {105 S(b x)}{8 \pi ^4 b^8}+\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi b}+\frac {105 x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^4 b^7}-\frac {35 x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac {7 x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^2 b^3}+\frac {1}{8} x^8 S(b x) \]

[Out]

-35/8*x^3*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/8*x^7*cos(1/2*b^2*Pi*x^2)/b/Pi-105/8*FresnelS(b*x)/b^8/Pi^4+1/8*x^8*F
resnelS(b*x)+105/8*x*sin(1/2*b^2*Pi*x^2)/b^7/Pi^4-7/8*x^5*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2

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Rubi [A]  time = 0.09, antiderivative size = 124, 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, 3385, 3386, 3351} \[ -\frac {105 S(b x)}{8 \pi ^4 b^8}-\frac {7 x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^2 b^3}+\frac {105 x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^4 b^7}+\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi b}-\frac {35 x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^3 b^5}+\frac {1}{8} x^8 S(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^7*FresnelS[b*x],x]

[Out]

(-35*x^3*Cos[(b^2*Pi*x^2)/2])/(8*b^5*Pi^3) + (x^7*Cos[(b^2*Pi*x^2)/2])/(8*b*Pi) - (105*FresnelS[b*x])/(8*b^8*P
i^4) + (x^8*FresnelS[b*x])/8 + (105*x*Sin[(b^2*Pi*x^2)/2])/(8*b^7*Pi^4) - (7*x^5*Sin[(b^2*Pi*x^2)/2])/(8*b^3*P
i^2)

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

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^7 S(b x) \, dx &=\frac {1}{8} x^8 S(b x)-\frac {1}{8} b \int x^8 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {x^7 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b \pi }+\frac {1}{8} x^8 S(b x)-\frac {7 \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{8 b \pi }\\ &=\frac {x^7 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b \pi }+\frac {1}{8} x^8 S(b x)-\frac {7 x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac {35 \int x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{8 b^3 \pi ^2}\\ &=-\frac {35 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {x^7 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b \pi }+\frac {1}{8} x^8 S(b x)-\frac {7 x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac {105 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{8 b^5 \pi ^3}\\ &=-\frac {35 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {x^7 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b \pi }+\frac {1}{8} x^8 S(b x)+\frac {105 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^7 \pi ^4}-\frac {7 x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}-\frac {105 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{8 b^7 \pi ^4}\\ &=-\frac {35 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {x^7 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b \pi }-\frac {105 S(b x)}{8 b^8 \pi ^4}+\frac {1}{8} x^8 S(b x)+\frac {105 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^7 \pi ^4}-\frac {7 x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 88, normalized size = 0.71 \[ \frac {\left (\pi ^4 b^8 x^8-105\right ) S(b x)-7 b x \left (\pi ^2 b^4 x^4-15\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )+\pi b^3 x^3 \left (\pi ^2 b^4 x^4-35\right ) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{8 \pi ^4 b^8} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*FresnelS[b*x],x]

[Out]

(b^3*Pi*x^3*(-35 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + (-105 + b^8*Pi^4*x^8)*FresnelS[b*x] - 7*b*x*(-15 + b^4*
Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2])/(8*b^8*Pi^4)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{7} {\rm fresnels}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^7*fresnels(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{7} {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^7*fresnels(b*x), x)

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maple [A]  time = 0.03, size = 123, normalized size = 0.99 \[ \frac {\frac {\mathrm {S}\left (b x \right ) b^{8} x^{8}}{8}+\frac {b^{7} x^{7} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{8 \pi }-\frac {7 \left (\frac {b^{5} x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {5 \left (-\frac {b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\frac {3 b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \,\mathrm {S}\left (b x \right )}{\pi }}{\pi }\right )}{\pi }\right )}{8 \pi }}{b^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*FresnelS(b*x),x)

[Out]

1/b^8*(1/8*FresnelS(b*x)*b^8*x^8+1/8/Pi*b^7*x^7*cos(1/2*b^2*Pi*x^2)-7/8/Pi*(1/Pi*b^5*x^5*sin(1/2*b^2*Pi*x^2)-5
/Pi*(-1/Pi*b^3*x^3*cos(1/2*b^2*Pi*x^2)+3/Pi*(1/Pi*b*x*sin(1/2*b^2*Pi*x^2)-1/Pi*FresnelS(b*x)))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{7} {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^7*fresnels(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^7\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*FresnelS(b*x),x)

[Out]

int(x^7*FresnelS(b*x), x)

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sympy [A]  time = 1.98, size = 184, normalized size = 1.48 \[ \frac {231 x^{8} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{512 \Gamma \left (\frac {15}{4}\right )} + \frac {231 x^{7} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi b \Gamma \left (\frac {15}{4}\right )} - \frac {1617 x^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{2} b^{3} \Gamma \left (\frac {15}{4}\right )} - \frac {8085 x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{3} b^{5} \Gamma \left (\frac {15}{4}\right )} + \frac {24255 x \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{4} b^{7} \Gamma \left (\frac {15}{4}\right )} - \frac {24255 S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{4} b^{8} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*fresnels(b*x),x)

[Out]

231*x**8*fresnels(b*x)*gamma(3/4)/(512*gamma(15/4)) + 231*x**7*cos(pi*b**2*x**2/2)*gamma(3/4)/(512*pi*b*gamma(
15/4)) - 1617*x**5*sin(pi*b**2*x**2/2)*gamma(3/4)/(512*pi**2*b**3*gamma(15/4)) - 8085*x**3*cos(pi*b**2*x**2/2)
*gamma(3/4)/(512*pi**3*b**5*gamma(15/4)) + 24255*x*sin(pi*b**2*x**2/2)*gamma(3/4)/(512*pi**4*b**7*gamma(15/4))
- 24255*fresnels(b*x)*gamma(3/4)/(512*pi**4*b**8*gamma(15/4))

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