3.1528 \(\int \frac{1}{x^3 \sqrt{1+x^8}} \, dx\)

Optimal. Leaf size=130 \[ \frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (x^2\right ),\frac{1}{2}\right )}{4 \sqrt{x^8+1}}+\frac{\sqrt{x^8+1} x^2}{2 \left (x^4+1\right )}-\frac{\sqrt{x^8+1}}{2 x^2}-\frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac{1}{2}\right )}{2 \sqrt{x^8+1}} \]

[Out]

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

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Rubi [A]  time = 0.0457453, antiderivative size = 130, 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, 325, 305, 220, 1196} \[ \frac{\sqrt{x^8+1} x^2}{2 \left (x^4+1\right )}-\frac{\sqrt{x^8+1}}{2 x^2}+\frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac{1}{2}\right )}{4 \sqrt{x^8+1}}-\frac{\left (x^4+1\right ) \sqrt{\frac{x^8+1}{\left (x^4+1\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac{1}{2}\right )}{2 \sqrt{x^8+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[1 + x^8]),x]

[Out]

-Sqrt[1 + x^8]/(2*x^2) + (x^2*Sqrt[1 + x^8])/(2*(1 + x^4)) - ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticE[
2*ArcTan[x^2], 1/2])/(2*Sqrt[1 + x^8]) + ((1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticF[2*ArcTan[x^2], 1/2])
/(4*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{1}{x^3 \sqrt{1+x^8}} \, dx &=\frac{1}{2} \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{1}{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}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,x^2\right )-\frac{1}{2} \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}}{2 \left (1+x^4\right )}-\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 )}{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 )}{4 \sqrt{1+x^8}}\\ \end{align*}

Mathematica [C]  time = 0.0037449, size = 22, normalized size = 0.17 \[ -\frac{\, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-x^8\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[1 + x^8]),x]

[Out]

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

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Maple [C]  time = 0.013, size = 30, normalized size = 0.2 \begin{align*} -{\frac{1}{2\,{x}^{2}}\sqrt{{x}^{8}+1}}+{\frac{{x}^{6}}{6}{\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(1/x^3/(x^8+1)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x^8 + 1)*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{8} + 1}}{x^{11} + x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^8 + 1)/(x^11 + x^3), x)

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Sympy [C]  time = 0.618386, size = 32, normalized size = 0.25 \begin{align*} \frac{\Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \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(1/x**3/(x**8+1)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x^8 + 1)*x^3), x)