3.15.23 \(\int \frac {1}{x^5 (2+x^6)^{3/2}} \, dx\) [1423]

Optimal. Leaf size=202 \[ \frac {1}{6 x^4 \sqrt {2+x^6}}-\frac {7 \sqrt {2+x^6}}{48 x^4}-\frac {7 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{48 \sqrt [6]{2} \sqrt [4]{3} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}} \]

[Out]

1/6/x^4/(x^6+2)^(1/2)-7/48*(x^6+2)^(1/2)/x^4-7/288*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 = 0.08, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 296, 331, 224} \begin {gather*} -\frac {7 \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 (\text {ArcSin}\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 )}{48 \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}}-\frac {7 \sqrt {x^6+2}}{48 x^4}+\frac {1}{6 x^4 \sqrt {x^6+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(2 + x^6)^(3/2)),x]

[Out]

1/(6*x^4*Sqrt[2 + x^6]) - (7*Sqrt[2 + x^6])/(48*x^4) - (7*Sqrt[2 + Sqrt[3]]*(2^(1/3) + x^2)*Sqrt[(2^(2/3) - 2^
(1/3)*x^2 + x^4)/(2^(1/3)*(1 + Sqrt[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]])/(48*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 224

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

Rule 281

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 296

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

Rule 331

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 \left (2+x^6\right )^{3/2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^3 \left (2+x^3\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {1}{6 x^4 \sqrt {2+x^6}}+\frac {7}{12} \text {Subst}\left (\int \frac {1}{x^3 \sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=\frac {1}{6 x^4 \sqrt {2+x^6}}-\frac {7 \sqrt {2+x^6}}{48 x^4}-\frac {7}{96} \text {Subst}\left (\int \frac {1}{\sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=\frac {1}{6 x^4 \sqrt {2+x^6}}-\frac {7 \sqrt {2+x^6}}{48 x^4}-\frac {7 \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 )}{48 \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 29, normalized size = 0.14 \begin {gather*} -\frac {\, _2F_1\left (-\frac {2}{3},\frac {3}{2};\frac {1}{3};-\frac {x^6}{2}\right )}{8 \sqrt {2} x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(2 + x^6)^(3/2)),x]

[Out]

-1/8*Hypergeometric2F1[-2/3, 3/2, 1/3, -1/2*x^6]/(Sqrt[2]*x^4)

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Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.20, size = 20, normalized size = 0.10

method	result	size
meijerg	\(-\frac {\sqrt {2}\, \hypergeom \left (\left [-\frac {2}{3}, \frac {3}{2}\right ], \left [\frac {1}{3}\right ], -\frac {x^{6}}{2}\right )}{16 x^{4}}\)	\(20\)
risch	\(-\frac {7 x^{6}+6}{48 x^{4} \sqrt {x^{6}+2}}-\frac {7 \sqrt {2}\, x^{2} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], -\frac {x^{6}}{2}\right )}{192}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(x^6+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/16*2^(1/2)/x^4*hypergeom([-2/3,3/2],[1/3],-1/2*x^6)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^6+2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((x^6 + 2)^(3/2)*x^5), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.07, size = 46, normalized size = 0.23 \begin {gather*} -\frac {7 \, {\left (x^{10} + 2 \, x^{4}\right )} {\rm weierstrassPInverse}\left (0, -8, x^{2}\right ) + {\left (7 \, x^{6} + 6\right )} \sqrt {x^{6} + 2}}{48 \, {\left (x^{10} + 2 \, x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^6+2)^(3/2),x, algorithm="fricas")

[Out]

-1/48*(7*(x^10 + 2*x^4)*weierstrassPInverse(0, -8, x^2) + (7*x^6 + 6)*sqrt(x^6 + 2))/(x^10 + 2*x^4)

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Sympy [A]
time = 0.47, size = 39, normalized size = 0.19 \begin {gather*} \frac {\sqrt {2} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {3}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{24 x^{4} \Gamma \left (\frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(x**6+2)**(3/2),x)

[Out]

sqrt(2)*gamma(-2/3)*hyper((-2/3, 3/2), (1/3,), x**6*exp_polar(I*pi)/2)/(24*x**4*gamma(1/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^6+2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((x^6 + 2)^(3/2)*x^5), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^5\,{\left (x^6+2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(x^6 + 2)^(3/2)),x)

[Out]

int(1/(x^5*(x^6 + 2)^(3/2)), x)

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