3.1398 \(\int \frac {1}{x^5 \sqrt {2+x^6}} \, dx\)

Optimal. Leaf size=186 \[ -\frac {\sqrt {x^6+2}}{8 x^4}-\frac {\sqrt {2+\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} F\left (\sin ^{-1}\left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{8 \sqrt [6]{2} \sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}} \]

[Out]

-1/8*(x^6+2)^(1/2)/x^4-1/48*2^(5/6)*(2^(1/3)+x^2)*EllipticF((x^2+2^(1/3)*(1-3^(1/2)))/(x^2+2^(1/3)*(1+3^(1/2))
),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((2^(2/3)-2^(1/3)*x^2+x^4)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/
4)/(x^6+2)^(1/2)/((2^(1/3)+x^2)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {275, 325, 218} \[ -\frac {\sqrt {x^6+2}}{8 x^4}-\frac {\sqrt {2+\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} F\left (\sin ^{-1}\left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{8 \sqrt [6]{2} \sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[2 + x^6]),x]

[Out]

-Sqrt[2 + x^6]/(8*x^4) - (Sqrt[2 + Sqrt[3]]*(2^(1/3) + x^2)*Sqrt[(2^(2/3) - 2^(1/3)*x^2 + x^4)/(2^(1/3)*(1 + S
qrt[3]) + x^2)^2]*EllipticF[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3
]])/(8*2^(1/6)*3^(1/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[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]

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]

Rubi steps

\begin {align*} \int \frac {1}{x^5 \sqrt {2+x^6}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {2+x^6}}{8 x^4}-\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {2+x^6}}{8 x^4}-\frac {\sqrt {2+\sqrt {3}} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{8 \sqrt [6]{2} \sqrt [4]{3} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.16 \[ -\frac {\, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};-\frac {x^6}{2}\right )}{4 \sqrt {2} x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[2 + x^6]),x]

[Out]

-1/4*Hypergeometric2F1[-2/3, 1/2, 1/3, -1/2*x^6]/(Sqrt[2]*x^4)

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{6} + 2}}{x^{11} + 2 \, x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^6+2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^6 + 2)/(x^11 + 2*x^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{6} + 2} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^6+2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^6 + 2)*x^5), x)

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maple [C]  time = 0.14, size = 33, normalized size = 0.18 \[ -\frac {\sqrt {2}\, x^{2} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], -\frac {x^{6}}{2}\right )}{32}-\frac {\sqrt {x^{6}+2}}{8 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(x^6+2)^(1/2),x)

[Out]

-1/8*(x^6+2)^(1/2)/x^4-1/32*2^(1/2)*x^2*hypergeom([1/3,1/2],[4/3],-1/2*x^6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{6} + 2} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^6+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^6 + 2)*x^5), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^5\,\sqrt {x^6+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(x^6 + 2)^(1/2)),x)

[Out]

int(1/(x^5*(x^6 + 2)^(1/2)), x)

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sympy [A]  time = 0.95, size = 39, normalized size = 0.21 \[ \frac {\sqrt {2} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{12 x^{4} \Gamma \left (\frac {1}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(x**6+2)**(1/2),x)

[Out]

sqrt(2)*gamma(-2/3)*hyper((-2/3, 1/2), (1/3,), x**6*exp_polar(I*pi)/2)/(12*x**4*gamma(1/3))

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