3.1510 \(\int x \sqrt {1+x^8} \, dx\)

Optimal. Leaf size=62 \[ \frac {1}{6} \sqrt {x^8+1} x^2+\frac {\left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{6 \sqrt {x^8+1}} \]

[Out]

1/6*x^2*(x^8+1)^(1/2)+1/6*(x^4+1)*(cos(2*arctan(x^2))^2)^(1/2)/cos(2*arctan(x^2))*EllipticF(sin(2*arctan(x^2))
,1/2*2^(1/2))*((x^8+1)/(x^4+1)^2)^(1/2)/(x^8+1)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {275, 195, 220} \[ \frac {1}{6} \sqrt {x^8+1} x^2+\frac {\left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{6 \sqrt {x^8+1}} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[1 + x^8],x]

[Out]

(x^2*Sqrt[1 + x^8])/6 + ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticF[2*ArcTan[x^2], 1/2])/(6*Sqrt[1 + x^8]
)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int x \sqrt {1+x^8} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \sqrt {1+x^4} \, dx,x,x^2\right )\\ &=\frac {1}{6} x^2 \sqrt {1+x^8}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,x^2\right )\\ &=\frac {1}{6} x^2 \sqrt {1+x^8}+\frac {\left (1+x^4\right ) \sqrt {\frac {1+x^8}{\left (1+x^4\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{6 \sqrt {1+x^8}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 22, normalized size = 0.35 \[ \frac {1}{2} x^2 \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-x^8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[1 + x^8],x]

[Out]

(x^2*Hypergeometric2F1[-1/2, 1/4, 5/4, -x^8])/2

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {x^{8} + 1} x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^8 + 1)*x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x^{8} + 1} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^8+1)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(x^8 + 1)*x, x)

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maple [C]  time = 0.15, size = 30, normalized size = 0.48 \[ \frac {x^{2} \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {5}{4}\right ], -x^{8}\right )}{3}+\frac {\sqrt {x^{8}+1}\, x^{2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^8+1)^(1/2),x)

[Out]

1/6*x^2*(x^8+1)^(1/2)+1/3*x^2*hypergeom([1/4,1/2],[5/4],-x^8)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x^{8} + 1} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^8 + 1)*x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\sqrt {x^8+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^8 + 1)^(1/2),x)

[Out]

int(x*(x^8 + 1)^(1/2), x)

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sympy [C]  time = 0.76, size = 31, normalized size = 0.50 \[ \frac {x^{2} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{8} e^{i \pi }} \right )}}{8 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x**8+1)**(1/2),x)

[Out]

x**2*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**8*exp_polar(I*pi))/(8*gamma(5/4))

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