3.2.12 \(\int x^5 \text {FresnelC}(b x) \, dx\) [112]

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

[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 = 0.04, 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 = {6562, 3467, 3466, 3432} \begin {gather*} -\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) \end {gather*}

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 3432

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

Rule 3466

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 3467

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 6562

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/2)*b^2*x^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 = 0.04, size = 80, normalized size = 0.81 \begin {gather*} \frac {-5 b^3 \pi x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )+b^6 \pi ^3 x^6 \text {FresnelC}(b x)-15 S(b x)+b x \left (15-b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 b^6 \pi ^3} \end {gather*}

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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Maple [A]
time = 0.32, size = 97, normalized size = 0.98

method	result	size
meijerg	\(\frac {b \,x^{7} \hypergeom \left (\left [\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {1}{2}, \frac {5}{4}, \frac {11}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{7}\)	\(26\)
derivativedivides	\(\frac {\frac {\FresnelC \left (b x \right ) b^{6} x^{6}}{6}-\frac {b^{5} x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }+\frac {-\frac {5 b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }+\frac {5 \left (\frac {3 b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \,\mathrm {S}\left (b x \right )}{\pi }\right )}{6 \pi }}{\pi }}{b^{6}}\)	\(97\)
default	\(\frac {\frac {\FresnelC \left (b x \right ) b^{6} x^{6}}{6}-\frac {b^{5} x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }+\frac {-\frac {5 b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }+\frac {5 \left (\frac {3 b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \,\mathrm {S}\left (b x \right )}{\pi }\right )}{6 \pi }}{\pi }}{b^{6}}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*FresnelC(b*x),x,method=_RETURNVERBOSE)

[Out]

1/b^6*(1/6*FresnelC(b*x)*b^6*x^6-1/6/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 [C] Result contains complex when optimal does not.
time = 0.48, size = 110, normalized size = 1.11 \begin {gather*} \frac {1}{6} \, x^{6} \operatorname {C}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (20 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (15 i + 15\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (15 i - 15\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right ) + 4 \, {\left (\sqrt {\frac {1}{2}} \pi ^{3} b^{5} x^{5} - 15 \, \sqrt {\frac {1}{2}} \pi b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )}}{12 \, \pi ^{4} b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*fresnel_cos(b*x) - 1/12*sqrt(1/2)*(20*sqrt(1/2)*pi^2*b^3*x^3*cos(1/2*pi*b^2*x^2) + (15*I + 15)*(1/4)^(
1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) - (15*I - 15)*(1/4)^(1/4)*pi*erf(sqrt(-1/2*I*pi)*b*x) + 4*(sqrt(1/2)*pi^3*b^5*
x^5 - 15*sqrt(1/2)*pi*b*x)*sin(1/2*pi*b^2*x^2))/(pi^4*b^6)

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Fricas [A]
time = 0.35, size = 86, normalized size = 0.87 \begin {gather*} \frac {\pi ^{3} b^{7} x^{6} \operatorname {C}\left (b x\right ) - 5 \, \pi b^{4} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{6} x^{5} - 15 \, b^{2} x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 15 \, \sqrt {b^{2}} \operatorname {S}\left (\sqrt {b^{2}} x\right )}{6 \, \pi ^{3} b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(pi^3*b^7*x^6*fresnel_cos(b*x) - 5*pi*b^4*x^3*cos(1/2*pi*b^2*x^2) - (pi^2*b^6*x^5 - 15*b^2*x)*sin(1/2*pi*b
^2*x^2) - 15*sqrt(b^2)*fresnel_sin(sqrt(b^2)*x))/(pi^3*b^7)

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Sympy [A]
time = 0.53, size = 49, normalized size = 0.49 \begin {gather*} \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 )} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^5*fresnel_cos(b*x), x)

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

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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