3.3 \(\int x^5 S(b x) \, dx\)

Optimal. Leaf size=99 \[ \frac {5 C(b x)}{2 \pi ^3 b^6}+\frac {x^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 \pi b}-\frac {5 x \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi ^3 b^5}-\frac {5 x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 \pi ^2 b^3}+\frac {1}{6} x^6 S(b x) \]

[Out]

-5/2*x*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/6*x^5*cos(1/2*b^2*Pi*x^2)/b/Pi+5/2*FresnelC(b*x)/b^6/Pi^3+1/6*x^6*Fresne
lS(b*x)-5/6*x^3*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2

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Rubi [A]  time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6426, 3385, 3386, 3352} \[ \frac {5 \text {FresnelC}(b x)}{2 \pi ^3 b^6}-\frac {5 x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 \pi ^2 b^3}+\frac {x^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 \pi b}-\frac {5 x \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi ^3 b^5}+\frac {1}{6} x^6 S(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^5*FresnelS[b*x],x]

[Out]

(-5*x*Cos[(b^2*Pi*x^2)/2])/(2*b^5*Pi^3) + (x^5*Cos[(b^2*Pi*x^2)/2])/(6*b*Pi) + (5*FresnelC[b*x])/(2*b^6*Pi^3)
+ (x^6*FresnelS[b*x])/6 - (5*x^3*Sin[(b^2*Pi*x^2)/2])/(6*b^3*Pi^2)

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

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Mathematica [A]  time = 0.10, size = 79, normalized size = 0.80 \[ \frac {\pi ^3 b^6 x^6 S(b x)+b x \left (\pi ^2 b^4 x^4-15\right ) \cos \left (\frac {1}{2} \pi b^2 x^2\right )-5 \pi b^3 x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )+15 C(b x)}{6 \pi ^3 b^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*FresnelS[b*x],x]

[Out]

(b*x*(-15 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + 15*FresnelC[b*x] + b^6*Pi^3*x^6*FresnelS[b*x] - 5*b^3*Pi*x^3*S
in[(b^2*Pi*x^2)/2])/(6*b^6*Pi^3)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{5} {\rm fresnels}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^5*fresnels(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{5} {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^5*fresnels(b*x), x)

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maple [A]  time = 0.02, size = 96, normalized size = 0.97 \[ \frac {\frac {b^{6} x^{6} \mathrm {S}\left (b x \right )}{6}+\frac {b^{5} x^{5} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }-\frac {5 \left (\frac {b^{3} x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \left (-\frac {b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\FresnelC \left (b x \right )}{\pi }\right )}{\pi }\right )}{6 \pi }}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*FresnelS(b*x),x)

[Out]

1/b^6*(1/6*b^6*x^6*FresnelS(b*x)+1/6/Pi*b^5*x^5*cos(1/2*b^2*Pi*x^2)-5/6/Pi*(1/Pi*b^3*x^3*sin(1/2*b^2*Pi*x^2)-3
/Pi*(-1/Pi*b*x*cos(1/2*b^2*Pi*x^2)+1/Pi*FresnelC(b*x))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{5} {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^5*fresnels(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^5\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*FresnelS(b*x),x)

[Out]

int(x^5*FresnelS(b*x), x)

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sympy [A]  time = 0.85, size = 53, normalized size = 0.54 \[ \frac {\pi b^{3} x^{9} \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {13}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma \left (\frac {7}{4}\right ) \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*fresnels(b*x),x)

[Out]

pi*b**3*x**9*gamma(3/4)*gamma(9/4)*hyper((3/4, 9/4), (3/2, 7/4, 13/4), -pi**2*b**4*x**4/16)/(32*gamma(7/4)*gam
ma(13/4))

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