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

Optimal. Leaf size=216 \[ -\frac {531 C\left (\sqrt {2} b x\right )}{16 \sqrt {2} \pi ^4 b^8}+\frac {24 x}{\pi ^4 b^7}-\frac {3 x^5}{5 \pi ^2 b^3}-\frac {x^6 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {48 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^4 b^8}+\frac {147 x \cos \left (\pi b^2 x^2\right )}{16 \pi ^4 b^7}+\frac {24 x^2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac {17 x^3 \sin \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^5 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]

[Out]

24*x/b^7/Pi^4-3/5*x^5/b^3/Pi^2+147/16*x*cos(b^2*Pi*x^2)/b^7/Pi^4-1/4*x^5*cos(b^2*Pi*x^2)/b^3/Pi^2+24*x^2*cos(1
/2*b^2*Pi*x^2)*FresnelS(b*x)/b^6/Pi^3-x^6*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^2/Pi-48*FresnelS(b*x)*sin(1/2*b^
2*Pi*x^2)/b^8/Pi^4+6*x^4*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^4/Pi^2+17/8*x^3*sin(b^2*Pi*x^2)/b^5/Pi^3-531/32*F
resnelC(b*x*2^(1/2))/b^8/Pi^4*2^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6454, 6462, 3391, 30, 3386, 3385, 3352, 6460, 3357} \[ -\frac {531 \text {FresnelC}\left (\sqrt {2} b x\right )}{16 \sqrt {2} \pi ^4 b^8}+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {48 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^4 b^8}-\frac {x^6 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {24 x^2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac {3 x^5}{5 \pi ^2 b^3}+\frac {17 x^3 \sin \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac {x^5 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {147 x \cos \left (\pi b^2 x^2\right )}{16 \pi ^4 b^7}+\frac {24 x}{\pi ^4 b^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

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

Rule 3352

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

Rule 3357

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 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 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 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 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 6460

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

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]

Rubi steps

\begin {align*} \int x^7 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx &=-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {6 \int x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^2 \pi }+\frac {\int x^6 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac {x^5 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {24 \int x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}+\frac {5 \int x^4 \cos \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}-\frac {6 \int x^4 \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}\\ &=-\frac {x^5 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {5 x^3 \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {48 \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^6 \pi ^3}-\frac {15 \int x^2 \sin \left (b^2 \pi x^2\right ) \, dx}{8 b^5 \pi ^3}-\frac {12 \int x^2 \sin \left (b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}-\frac {3 \int x^4 \, dx}{b^3 \pi ^2}+\frac {3 \int x^4 \cos \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}\\ &=-\frac {3 x^5}{5 b^3 \pi ^2}+\frac {111 x \cos \left (b^2 \pi x^2\right )}{16 b^7 \pi ^4}-\frac {x^5 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {48 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^8 \pi ^4}+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {17 x^3 \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {15 \int \cos \left (b^2 \pi x^2\right ) \, dx}{16 b^7 \pi ^4}-\frac {6 \int \cos \left (b^2 \pi x^2\right ) \, dx}{b^7 \pi ^4}+\frac {48 \int \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^7 \pi ^4}-\frac {9 \int x^2 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b^5 \pi ^3}\\ &=-\frac {3 x^5}{5 b^3 \pi ^2}+\frac {147 x \cos \left (b^2 \pi x^2\right )}{16 b^7 \pi ^4}-\frac {x^5 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {15 C\left (\sqrt {2} b x\right )}{16 \sqrt {2} b^8 \pi ^4}-\frac {3 \sqrt {2} C\left (\sqrt {2} b x\right )}{b^8 \pi ^4}+\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {48 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^8 \pi ^4}+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {17 x^3 \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {9 \int \cos \left (b^2 \pi x^2\right ) \, dx}{4 b^7 \pi ^4}+\frac {48 \int \left (\frac {1}{2}-\frac {1}{2} \cos \left (b^2 \pi x^2\right )\right ) \, dx}{b^7 \pi ^4}\\ &=\frac {24 x}{b^7 \pi ^4}-\frac {3 x^5}{5 b^3 \pi ^2}+\frac {147 x \cos \left (b^2 \pi x^2\right )}{16 b^7 \pi ^4}-\frac {x^5 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {51 C\left (\sqrt {2} b x\right )}{16 \sqrt {2} b^8 \pi ^4}-\frac {3 \sqrt {2} C\left (\sqrt {2} b x\right )}{b^8 \pi ^4}+\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {48 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^8 \pi ^4}+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {17 x^3 \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {24 \int \cos \left (b^2 \pi x^2\right ) \, dx}{b^7 \pi ^4}\\ &=\frac {24 x}{b^7 \pi ^4}-\frac {3 x^5}{5 b^3 \pi ^2}+\frac {147 x \cos \left (b^2 \pi x^2\right )}{16 b^7 \pi ^4}-\frac {x^5 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {51 C\left (\sqrt {2} b x\right )}{16 \sqrt {2} b^8 \pi ^4}-\frac {15 \sqrt {2} C\left (\sqrt {2} b x\right )}{b^8 \pi ^4}+\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {48 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^8 \pi ^4}+\frac {6 x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {17 x^3 \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 153, normalized size = 0.71 \[ \frac {-160 S(b x) \left (\pi b^2 x^2 \left (\pi ^2 b^4 x^4-24\right ) \cos \left (\frac {1}{2} \pi b^2 x^2\right )-6 \left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )\right )+2 b x \left (2 \left (-24 \pi ^2 b^4 x^4+85 \pi b^2 x^2 \sin \left (\pi b^2 x^2\right )+960\right )+\left (735-20 \pi ^2 b^4 x^4\right ) \cos \left (\pi b^2 x^2\right )\right )-2655 \sqrt {2} C\left (\sqrt {2} b x\right )}{160 \pi ^4 b^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2655*Sqrt[2]*FresnelC[Sqrt[2]*b*x] - 160*FresnelS[b*x]*(b^2*Pi*x^2*(-24 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2]
- 6*(-8 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2]) + 2*b*x*((735 - 20*b^4*Pi^2*x^4)*Cos[b^2*Pi*x^2] + 2*(960 - 24*b^
4*Pi^2*x^4 + 85*b^2*Pi*x^2*Sin[b^2*Pi*x^2])))/(160*b^8*Pi^4)

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{7} {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{7} {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 318, normalized size = 1.47 \[ \frac {\frac {\mathrm {S}\left (b x \right ) \left (-\frac {b^{6} x^{6} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\frac {6 b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {24 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{\pi }}{\pi }\right )}{b^{7}}-\frac {\frac {\frac {3}{5} \pi ^{2} b^{5} x^{5}-24 b x}{\pi ^{4}}-\frac {3 \left (\frac {\pi \,b^{3} x^{3} \sin \left (b^{2} \pi \,x^{2}\right )}{2}-\frac {3 \pi \left (-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4 \pi }\right )}{2}-4 \sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )\right )}{\pi ^{4}}-\frac {-\frac {\pi \,b^{5} x^{5} \cos \left (b^{2} \pi \,x^{2}\right )}{2}+\frac {5 \pi \left (\frac {b^{3} x^{3} \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {3 \left (-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4 \pi }\right )}{2 \pi }\right )}{2}+\frac {12 b x \cos \left (b^{2} \pi \,x^{2}\right )}{\pi }-\frac {6 \sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{\pi }}{2 \pi ^{3}}}{b^{7}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(FresnelS(b*x)/b^7*(-1/Pi*b^6*x^6*cos(1/2*b^2*Pi*x^2)+6/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))))-1/b^7*(3/Pi^4*(1/5*Pi^2*b^5*x^5-8*b*x)-3/Pi^4*(1/2*Pi*b^3
*x^3*sin(b^2*Pi*x^2)-3/2*Pi*(-1/2/Pi*b*x*cos(b^2*Pi*x^2)+1/4/Pi*2^(1/2)*FresnelC(b*x*2^(1/2)))-4*2^(1/2)*Fresn
elC(b*x*2^(1/2)))-1/2/Pi^3*(-1/2*Pi*b^5*x^5*cos(b^2*Pi*x^2)+5/2*Pi*(1/2/Pi*b^3*x^3*sin(b^2*Pi*x^2)-3/2/Pi*(-1/
2/Pi*b*x*cos(b^2*Pi*x^2)+1/4/Pi*2^(1/2)*FresnelC(b*x*2^(1/2))))+12/Pi*b*x*cos(b^2*Pi*x^2)-6/Pi*2^(1/2)*Fresnel
C(b*x*2^(1/2)))))/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{7} {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^7\,\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{7} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**7*sin(pi*b**2*x**2/2)*fresnels(b*x), x)

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