3.112 \(\int x^5 C(b x) \, dx\)

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

[Out]

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

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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 = {6427, 3386, 3385, 3351} \[ -\frac {5 S(b x)}{2 \pi ^3 b^6}-\frac {x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 \pi b}+\frac {5 x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi ^3 b^5}-\frac {5 x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 \pi ^2 b^3}+\frac {1}{6} x^6 \text {FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6427

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 80, normalized size = 0.81 \[ \frac {\pi ^3 b^6 x^6 C(b x)+b x \left (15-\pi ^2 b^4 x^4\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )-5 \pi b^3 x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )-15 S(b x)}{6 \pi ^3 b^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{5} {\rm fresnelc}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^5*fresnelc(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^5*fresnelc(b*x), x)

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maple [A]  time = 0.00, size = 97, normalized size = 0.98 \[ \frac {\frac {b^{6} x^{6} \FresnelC \left (b x \right )}{6}-\frac {b^{5} x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }+\frac {-\frac {5 b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }+\frac {5 \left (\frac {3 b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \,\mathrm {S}\left (b x \right )}{\pi }\right )}{6 \pi }}{\pi }}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^6*(1/6*b^6*x^6*FresnelC(b*x)-1/6/Pi*b^5*x^5*sin(1/2*b^2*Pi*x^2)+5/6/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 \[ \int x^{5} {\rm fresnelc}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^5*fresnelc(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^5*FresnelC(b*x), x)

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sympy [A]  time = 0.97, size = 49, normalized size = 0.49 \[ \frac {b x^{7} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {7}{4} \\ \frac {1}{2}, \frac {5}{4}, \frac {11}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x**7*gamma(1/4)*gamma(7/4)*hyper((1/4, 7/4), (1/2, 5/4, 11/4), -pi**2*b**4*x**4/16)/(16*gamma(5/4)*gamma(11/
4))

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