3.92 \(\int x^7 \cos (\frac{1}{2} b^2 \pi x^2) S(b x) \, dx\)

Optimal. Leaf size=217 \[ \frac{x^6 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac{24 x^2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac{6 x^4 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{48 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^4 b^8}+\frac{531 S\left (\sqrt{2} b x\right )}{16 \sqrt{2} \pi ^4 b^8}+\frac{4 x^3}{\pi ^3 b^5}+\frac{x^5 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac{147 x \sin \left (\pi b^2 x^2\right )}{16 \pi ^4 b^7}+\frac{17 x^3 \cos \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{x^7}{14 \pi b} \]

[Out]

(4*x^3)/(b^5*Pi^3) - x^7/(14*b*Pi) + (17*x^3*Cos[b^2*Pi*x^2])/(8*b^5*Pi^3) - (48*Cos[(b^2*Pi*x^2)/2]*FresnelS[
b*x])/(b^8*Pi^4) + (6*x^4*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^4*Pi^2) + (531*FresnelS[Sqrt[2]*b*x])/(16*Sqrt
[2]*b^8*Pi^4) - (24*x^2*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^6*Pi^3) + (x^6*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2]
)/(b^2*Pi) - (147*x*Sin[b^2*Pi*x^2])/(16*b^7*Pi^4) + (x^5*Sin[b^2*Pi*x^2])/(4*b^3*Pi^2)

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Rubi [A]  time = 0.271429, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6462, 3391, 30, 3386, 3385, 3351, 6454, 6452} \[ \frac{x^6 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac{24 x^2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac{6 x^4 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{48 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^4 b^8}+\frac{531 S\left (\sqrt{2} b x\right )}{16 \sqrt{2} \pi ^4 b^8}+\frac{4 x^3}{\pi ^3 b^5}+\frac{x^5 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac{147 x \sin \left (\pi b^2 x^2\right )}{16 \pi ^4 b^7}+\frac{17 x^3 \cos \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{x^7}{14 \pi b} \]

Antiderivative was successfully verified.

[In]

Int[x^7*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

(4*x^3)/(b^5*Pi^3) - x^7/(14*b*Pi) + (17*x^3*Cos[b^2*Pi*x^2])/(8*b^5*Pi^3) - (48*Cos[(b^2*Pi*x^2)/2]*FresnelS[
b*x])/(b^8*Pi^4) + (6*x^4*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^4*Pi^2) + (531*FresnelS[Sqrt[2]*b*x])/(16*Sqrt
[2]*b^8*Pi^4) - (24*x^2*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^6*Pi^3) + (x^6*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2]
)/(b^2*Pi) - (147*x*Sin[b^2*Pi*x^2])/(16*b^7*Pi^4) + (x^5*Sin[b^2*Pi*x^2])/(4*b^3*Pi^2)

Rule 6462

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*Sin[d*x^2]*FresnelS[b*x])/(
2*d), x] + (-Dist[1/(Pi*b), Int[x^(m - 1)*Sin[d*x^2]^2, x], x] - Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*
FresnelS[b*x], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rule 3391

Int[(x_)^(m_.)*Sin[(a_.) + ((b_.)*(x_)^(n_))/2]^2, x_Symbol] :> Dist[1/2, Int[x^m, x], x] - Dist[1/2, Int[x^m*
Cos[2*a + b*x^n], x], x] /; FreeQ[{a, b, m, n}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[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 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 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 6454

Int[FresnelS[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(x^(m - 1)*Cos[d*x^2]*FresnelS[b*x])/
(2*d), x] + (Dist[(m - 1)/(2*d), Int[x^(m - 2)*Cos[d*x^2]*FresnelS[b*x], x], x] + Dist[1/(2*b*Pi), Int[x^(m -
1)*Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rule 6452

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(Cos[d*x^2]*FresnelS[b*x])/(2*d), x] + Dis
t[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rubi steps

\begin{align*} \int x^7 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx &=\frac{x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{6 \int x^5 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac{\int x^6 \sin ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=\frac{6 x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}+\frac{x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{24 \int x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^4 \pi ^2}-\frac{3 \int x^4 \sin \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}-\frac{\int x^6 \, dx}{2 b \pi }+\frac{\int x^6 \cos \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac{x^7}{14 b \pi }+\frac{3 x^3 \cos \left (b^2 \pi x^2\right )}{2 b^5 \pi ^3}+\frac{6 x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac{24 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac{x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac{x^5 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{48 \int x S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^6 \pi ^3}-\frac{9 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{2 b^5 \pi ^3}+\frac{24 \int x^2 \sin ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}-\frac{5 \int x^4 \sin \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}\\ &=-\frac{x^7}{14 b \pi }+\frac{17 x^3 \cos \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{48 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^8 \pi ^4}+\frac{6 x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac{24 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac{x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{9 x \sin \left (b^2 \pi x^2\right )}{4 b^7 \pi ^4}+\frac{x^5 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{9 \int \sin \left (b^2 \pi x^2\right ) \, dx}{4 b^7 \pi ^4}+\frac{24 \int \sin \left (b^2 \pi x^2\right ) \, dx}{b^7 \pi ^4}-\frac{15 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{8 b^5 \pi ^3}+\frac{12 \int x^2 \, dx}{b^5 \pi ^3}-\frac{12 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}\\ &=\frac{4 x^3}{b^5 \pi ^3}-\frac{x^7}{14 b \pi }+\frac{17 x^3 \cos \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{48 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^8 \pi ^4}+\frac{6 x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}+\frac{9 S\left (\sqrt{2} b x\right )}{4 \sqrt{2} b^8 \pi ^4}+\frac{12 \sqrt{2} S\left (\sqrt{2} b x\right )}{b^8 \pi ^4}-\frac{24 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac{x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{147 x \sin \left (b^2 \pi x^2\right )}{16 b^7 \pi ^4}+\frac{x^5 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{15 \int \sin \left (b^2 \pi x^2\right ) \, dx}{16 b^7 \pi ^4}+\frac{6 \int \sin \left (b^2 \pi x^2\right ) \, dx}{b^7 \pi ^4}\\ &=\frac{4 x^3}{b^5 \pi ^3}-\frac{x^7}{14 b \pi }+\frac{17 x^3 \cos \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{48 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^8 \pi ^4}+\frac{6 x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}+\frac{51 S\left (\sqrt{2} b x\right )}{16 \sqrt{2} b^8 \pi ^4}+\frac{15 \sqrt{2} S\left (\sqrt{2} b x\right )}{b^8 \pi ^4}-\frac{24 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac{x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{147 x \sin \left (b^2 \pi x^2\right )}{16 b^7 \pi ^4}+\frac{x^5 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end{align*}

Mathematica [A]  time = 0.175812, size = 163, normalized size = 0.75 \[ \frac{224 S(b x) \left (\pi b^2 x^2 \left (\pi ^2 b^4 x^4-24\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )+6 \left (\pi ^2 b^4 x^4-8\right ) \cos \left (\frac{1}{2} \pi b^2 x^2\right )\right )-16 \pi ^3 b^7 x^7+896 \pi b^3 x^3+56 \pi ^2 b^5 x^5 \sin \left (\pi b^2 x^2\right )-2058 b x \sin \left (\pi b^2 x^2\right )+476 \pi b^3 x^3 \cos \left (\pi b^2 x^2\right )+3717 \sqrt{2} S\left (\sqrt{2} b x\right )}{224 \pi ^4 b^8} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

(896*b^3*Pi*x^3 - 16*b^7*Pi^3*x^7 + 476*b^3*Pi*x^3*Cos[b^2*Pi*x^2] + 3717*Sqrt[2]*FresnelS[Sqrt[2]*b*x] + 224*
FresnelS[b*x]*(6*(-8 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + b^2*Pi*x^2*(-24 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2]
) - 2058*b*x*Sin[b^2*Pi*x^2] + 56*b^5*Pi^2*x^5*Sin[b^2*Pi*x^2])/(224*b^8*Pi^4)

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Maple [A]  time = 0.084, size = 321, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ({\frac{{\it FresnelS} \left ( bx \right ) }{{b}^{7}} \left ({\frac{{b}^{6}{x}^{6}}{\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-6\,{\frac{1}{\pi } \left ( -{\frac{{x}^{4}{b}^{4}\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}+4\,{\frac{1}{\pi } \left ({\frac{{b}^{2}{x}^{2}\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}+2\,{\frac{\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) } \right ) } \right ) }-{\frac{1}{{b}^{7}} \left ({\frac{1}{2\,{\pi }^{3}} \left ({\frac{{\pi }^{2}{b}^{7}{x}^{7}}{7}}-8\,{x}^{3}{b}^{3} \right ) }+3\,{\frac{1}{{\pi }^{4}} \left ( -1/2\,\pi \,{b}^{3}{x}^{3}\cos \left ({b}^{2}\pi \,{x}^{2} \right ) +3/2\,\pi \, \left ( 1/2\,{\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{\pi }}-1/4\,{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{\pi }} \right ) -4\,\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) \right ) }-{\frac{1}{2\,{\pi }^{3}} \left ({\frac{\pi \,{b}^{5}{x}^{5}\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{2}}-{\frac{5\,\pi }{2} \left ( -{\frac{{x}^{3}{b}^{3}\cos \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}+{\frac{3}{2\,\pi } \left ({\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}-{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{4\,\pi }} \right ) } \right ) }-12\,{\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{\pi }}+6\,{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{\pi }} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x)

[Out]

(FresnelS(b*x)/b^7*(1/Pi*b^6*x^6*sin(1/2*b^2*Pi*x^2)-6/Pi*(-1/Pi*b^4*x^4*cos(1/2*b^2*Pi*x^2)+4/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))))-1/b^7*(1/2/Pi^3*(1/7*Pi^2*b^7*x^7-8*x^3*b^3)+3/Pi^4*(-1/2*
Pi*b^3*x^3*cos(b^2*Pi*x^2)+3/2*Pi*(1/2/Pi*b*x*sin(b^2*Pi*x^2)-1/4/Pi*2^(1/2)*FresnelS(b*x*2^(1/2)))-4*2^(1/2)*
FresnelS(b*x*2^(1/2)))-1/2/Pi^3*(1/2*Pi*b^5*x^5*sin(b^2*Pi*x^2)-5/2*Pi*(-1/2/Pi*b^3*x^3*cos(b^2*Pi*x^2)+3/2/Pi
*(1/2/Pi*b*x*sin(b^2*Pi*x^2)-1/4/Pi*2^(1/2)*FresnelS(b*x*2^(1/2))))-12/Pi*b*x*sin(b^2*Pi*x^2)+6/Pi*2^(1/2)*Fre
snelS(b*x*2^(1/2)))))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{7} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^7*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{7} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnels}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^7*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*cos(1/2*b**2*pi*x**2)*fresnels(b*x),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{7} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^7*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)