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]

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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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.00, 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]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{8} + 1}}{x^{12} + x^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{8} + 1} x^{4}}\,{d x} \]

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)

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maple [A]  time = 0.15, size = 30, normalized size = 0.77 \[ \frac {x^{5} \hypergeom \left (\left [\frac {1}{2}, \frac {5}{8}\right ], \left [\frac {13}{8}\right ], -x^{8}\right )}{15}-\frac {\sqrt {x^{8}+1}}{3 x^{3}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{8} + 1} x^{4}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{x^4\,\sqrt {x^8+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(x^8 + 1)^(1/2)),x)

[Out]

int(1/(x^4*(x^8 + 1)^(1/2)), x)

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sympy [C]  time = 0.84, size = 32, normalized size = 0.82 \[ \frac {\Gamma \left (- \frac {3}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{8}, \frac {1}{2} \\ \frac {5}{8} \end {matrix}\middle | {x^{8} e^{i \pi }} \right )}}{8 x^{3} \Gamma \left (\frac {5}{8}\right )} \]

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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