3.73 \(\int x^6 S(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=248 \[ \frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi ^3 b^5}-\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi ^3 b^5}-\frac {15 C(b x) S(b x)}{2 \pi ^3 b^7}-\frac {5 x^4}{8 \pi ^2 b^3}-\frac {x^5 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {11 \cos \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7}+\frac {15 x S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac {7 x^2 \sin \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^4 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]

[Out]

-5/8*x^4/b^3/Pi^2+11/2*cos(b^2*Pi*x^2)/b^7/Pi^4-1/4*x^4*cos(b^2*Pi*x^2)/b^3/Pi^2+15*x*cos(1/2*b^2*Pi*x^2)*Fres
nelS(b*x)/b^6/Pi^3-x^5*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^2/Pi-15/2*FresnelC(b*x)*FresnelS(b*x)/b^7/Pi^3+15/8
*I*x^2*HypergeometricPFQ([1, 1],[3/2, 2],-1/2*I*b^2*Pi*x^2)/b^5/Pi^3-15/8*I*x^2*HypergeometricPFQ([1, 1],[3/2,
2],1/2*I*b^2*Pi*x^2)/b^5/Pi^3+5*x^3*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^4/Pi^2+7/4*x^2*sin(b^2*Pi*x^2)/b^5/Pi
^3

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Rubi [A]  time = 0.25, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6454, 6462, 3379, 3309, 30, 3296, 2638, 6446} \[ \frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi ^3 b^5}-\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi ^3 b^5}-\frac {15 \text {FresnelC}(b x) S(b x)}{2 \pi ^3 b^7}+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^5 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {15 x S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac {5 x^4}{8 \pi ^2 b^3}+\frac {7 x^2 \sin \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}-\frac {x^4 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {11 \cos \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7} \]

Antiderivative was successfully verified.

[In]

Int[x^6*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

(-5*x^4)/(8*b^3*Pi^2) + (11*Cos[b^2*Pi*x^2])/(2*b^7*Pi^4) - (x^4*Cos[b^2*Pi*x^2])/(4*b^3*Pi^2) + (15*x*Cos[(b^
2*Pi*x^2)/2]*FresnelS[b*x])/(b^6*Pi^3) - (x^5*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi) - (15*FresnelC[b*x]*
FresnelS[b*x])/(2*b^7*Pi^3) + (((15*I)/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2])/(b^5*Pi^
3) - (((15*I)/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2])/(b^5*Pi^3) + (5*x^3*FresnelS[b*x]*
Sin[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + (7*x^2*Sin[b^2*Pi*x^2])/(4*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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 3309

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + ((f_.)*(x_))/2]^2, x_Symbol] :> Dist[1/2, Int[(c + d*x)^m, x], x] -
Dist[1/2, Int[(c + d*x)^m*Cos[2*e + f*x], x], x] /; FreeQ[{c, d, e, f, m}, x]

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 6446

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)], x_Symbol] :> Simp[(FresnelC[b*x]*FresnelS[b*x])/(2*b), x] + (-Simp
[(1*I*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, -((I*b^2*Pi*x^2)/2)])/8, x] + Simp[(1*I*b*x^2*HypergeometricPF
Q[{1, 1}, {3/2, 2}, (1*I*b^2*Pi*x^2)/2])/8, x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

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 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^6 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx &=-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {5 \int x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^2 \pi }+\frac {\int x^5 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {15 \int x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}-\frac {5 \int x^3 \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}+\frac {\operatorname {Subst}\left (\int x^2 \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi }\\ &=-\frac {x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {15 \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^6 \pi ^3}-\frac {15 \int x \sin \left (b^2 \pi x^2\right ) \, dx}{2 b^5 \pi ^3}+\frac {\operatorname {Subst}\left (\int x \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^3 \pi ^2}-\frac {5 \operatorname {Subst}\left (\int x \sin ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^3 \pi ^2}\\ &=-\frac {x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {15 C(b x) S(b x)}{2 b^7 \pi ^3}+\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {x^2 \sin \left (b^2 \pi x^2\right )}{2 b^5 \pi ^3}-\frac {\operatorname {Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^5 \pi ^3}-\frac {15 \operatorname {Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^5 \pi ^3}-\frac {5 \operatorname {Subst}\left (\int x \, dx,x,x^2\right )}{4 b^3 \pi ^2}+\frac {5 \operatorname {Subst}\left (\int x \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^3 \pi ^2}\\ &=-\frac {5 x^4}{8 b^3 \pi ^2}+\frac {17 \cos \left (b^2 \pi x^2\right )}{4 b^7 \pi ^4}-\frac {x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {15 C(b x) S(b x)}{2 b^7 \pi ^3}+\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {7 x^2 \sin \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}-\frac {5 \operatorname {Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^5 \pi ^3}\\ &=-\frac {5 x^4}{8 b^3 \pi ^2}+\frac {11 \cos \left (b^2 \pi x^2\right )}{2 b^7 \pi ^4}-\frac {x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac {15 C(b x) S(b x)}{2 b^7 \pi ^3}+\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {15 i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {5 x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {7 x^2 \sin \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}\\ \end {align*}

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Mathematica [F]  time = 0.53, size = 0, normalized size = 0.00 \[ \int x^6 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^6*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

Integrate[x^6*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2], x]

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{6} {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnels(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")

[Out]

integral(x^6*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 \[ \int x^{6} {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnels(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")

[Out]

integrate(x^6*fresnels(b*x)*sin(1/2*pi*b^2*x^2), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int x^{6} \mathrm {S}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)

[Out]

int(x^6*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{6} {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnels(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")

[Out]

integrate(x^6*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 \[ \int x^6\,\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*FresnelS(b*x)*sin((Pi*b^2*x^2)/2),x)

[Out]

int(x^6*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 \[ \int x^{6} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*fresnels(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Integral(x**6*sin(pi*b**2*x**2/2)*fresnels(b*x), x)

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