3.1537 \(\int \frac{1}{x^4 \sqrt{1+x^8}} \, dx\)

Optimal. Leaf size=39 \[ \frac{1}{15} x^5 \, _2F_1\left (\frac{1}{2},\frac{5}{8};\frac{13}{8};-x^8\right )-\frac{\sqrt{x^8+1}}{3 x^3} \]

[Out]

-Sqrt[1 + x^8]/(3*x^3) + (x^5*Hypergeometric2F1[1/2, 5/8, 13/8, -x^8])/15

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Rubi [A]  time = 0.0084266, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {325, 364} \[ \frac{1}{15} x^5 \, _2F_1\left (\frac{1}{2},\frac{5}{8};\frac{13}{8};-x^8\right )-\frac{\sqrt{x^8+1}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[1 + x^8]),x]

[Out]

-Sqrt[1 + x^8]/(3*x^3) + (x^5*Hypergeometric2F1[1/2, 5/8, 13/8, -x^8])/15

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^4 \sqrt{1+x^8}} \, dx &=-\frac{\sqrt{1+x^8}}{3 x^3}+\frac{1}{3} \int \frac{x^4}{\sqrt{1+x^8}} \, dx\\ &=-\frac{\sqrt{1+x^8}}{3 x^3}+\frac{1}{15} x^5 \, _2F_1\left (\frac{1}{2},\frac{5}{8};\frac{13}{8};-x^8\right )\\ \end{align*}

Mathematica [A]  time = 0.0025476, size = 22, normalized size = 0.56 \[ -\frac{\, _2F_1\left (-\frac{3}{8},\frac{1}{2};\frac{5}{8};-x^8\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[1 + x^8]),x]

[Out]

-Hypergeometric2F1[-3/8, 1/2, 5/8, -x^8]/(3*x^3)

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Maple [A]  time = 0.025, size = 30, normalized size = 0.8 \begin{align*}{\frac{{x}^{5}}{15}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{5}{8}};\,{\frac{13}{8}};\,-{x}^{8})}}-{\frac{1}{3\,{x}^{3}}\sqrt{{x}^{8}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(x^8+1)^(1/2),x)

[Out]

1/15*x^5*hypergeom([1/2,5/8],[13/8],-x^8)-1/3*(x^8+1)^(1/2)/x^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{8} + 1} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^8 + 1)*x^4), 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^{12} + x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^8 + 1)/(x^12 + x^4), x)

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Sympy [C]  time = 0.665945, size = 32, normalized size = 0.82 \begin{align*} \frac{\Gamma \left (- \frac{3}{8}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{8}, \frac{1}{2} \\ \frac{5}{8} \end{matrix}\middle |{x^{8} e^{i \pi }} \right )}}{8 x^{3} \Gamma \left (\frac{5}{8}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(x**8+1)**(1/2),x)

[Out]

gamma(-3/8)*hyper((-3/8, 1/2), (5/8,), x**8*exp_polar(I*pi))/(8*x**3*gamma(5/8))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{8} + 1} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^8+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^8 + 1)*x^4), x)