3.1512 \(\int \frac{\sqrt{1+x^8}}{x^3} \, dx\)

Optimal. Leaf size=125 \[ \frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (x^2\right ),\frac{1}{2}\right )}{2 \sqrt{x^8+1}}+\frac{\sqrt{x^8+1} x^2}{x^4+1}-\frac{\sqrt{x^8+1}}{2 x^2}-\frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac{1}{2}\right )}{\sqrt{x^8+1}} \]

[Out]

-Sqrt[1 + x^8]/(2*x^2) + (x^2*Sqrt[1 + x^8])/(1 + x^4) - ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticE[2*Ar
cTan[x^2], 1/2])/Sqrt[1 + x^8] + ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticF[2*ArcTan[x^2], 1/2])/(2*Sqrt
[1 + x^8])

________________________________________________________________________________________

Rubi [A]  time = 0.0546132, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {275, 277, 305, 220, 1196} \[ \frac{\sqrt{x^8+1} x^2}{x^4+1}-\frac{\sqrt{x^8+1}}{2 x^2}+\frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac{1}{2}\right )}{2 \sqrt{x^8+1}}-\frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac{1}{2}\right )}{\sqrt{x^8+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^8]/x^3,x]

[Out]

-Sqrt[1 + x^8]/(2*x^2) + (x^2*Sqrt[1 + x^8])/(1 + x^4) - ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticE[2*Ar
cTan[x^2], 1/2])/Sqrt[1 + x^8] + ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticF[2*ArcTan[x^2], 1/2])/(2*Sqrt
[1 + x^8])

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]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

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

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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 0.0030187, size = 22, normalized size = 0.18 \[ -\frac{\, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-x^8\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x^8]/x^3,x]

[Out]

-Hypergeometric2F1[-1/2, -1/4, 3/4, -x^8]/(2*x^2)

________________________________________________________________________________________

Maple [C]  time = 0.03, size = 30, normalized size = 0.2 \begin{align*} -{\frac{1}{2\,{x}^{2}}\sqrt{{x}^{8}+1}}+{\frac{{x}^{6}}{3}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{3}{4}};\,{\frac{7}{4}};\,-{x}^{8})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8+1)^(1/2)/x^3,x)

[Out]

-1/2*(x^8+1)^(1/2)/x^2+1/3*x^6*hypergeom([1/2,3/4],[7/4],-x^8)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{8} + 1}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+1)^(1/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sqrt(x^8 + 1)/x^3, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{8} + 1}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+1)^(1/2)/x^3,x, algorithm="fricas")

[Out]

integral(sqrt(x^8 + 1)/x^3, x)

________________________________________________________________________________________

Sympy [C]  time = 0.586596, size = 34, normalized size = 0.27 \begin{align*} \frac{\Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{x^{8} e^{i \pi }} \right )}}{8 x^{2} \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**8+1)**(1/2)/x**3,x)

[Out]

gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), x**8*exp_polar(I*pi))/(8*x**2*gamma(3/4))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{8} + 1}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+1)^(1/2)/x^3,x, algorithm="giac")

[Out]

integrate(sqrt(x^8 + 1)/x^3, x)