Integrand size = 13, antiderivative size = 394 \[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=\frac {1}{6 x^2 \sqrt {2+x^6}}-\frac {5 \sqrt {2+x^6}}{24 x^2}+\frac {5 \sqrt {2+x^6}}{24 \left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )}-\frac {5 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{8\ 2^{5/6} 3^{3/4} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}}+\frac {5 \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right ),-7-4 \sqrt {3}\right )}{12 \sqrt [3]{2} \sqrt [4]{3} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}} \]
1/6/x^2/(x^6+2)^(1/2)-5/24*(x^6+2)^(1/2)/x^2+5/24*(x^6+2)^(1/2)/(x^2+2^(1/ 3)*(1+3^(1/2)))+5/72*2^(2/3)*(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)*((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)-5/48*2^(1/6)*3^(1/4)*(2^(1/3)+x^2)*Ellipti cE((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)/(x^6+2)^(1/2)/((2^(1/3)+x^2)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.07 \[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {3}{2},\frac {2}{3},-\frac {x^6}{2}\right )}{4 \sqrt {2} x^2} \]
Time = 0.44 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {807, 819, 847, 832, 759, 2416}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^3 \left (x^6+2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (x^6+2\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} \int \frac {1}{x^4 \sqrt {x^6+2}}dx^2+\frac {1}{3 x^2 \sqrt {x^6+2}}\right )\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {1}{4} \int \frac {x^2}{\sqrt {x^6+2}}dx^2-\frac {\sqrt {x^6+2}}{2 x^2}\right )+\frac {1}{3 x^2 \sqrt {x^6+2}}\right )\) |
\(\Big \downarrow \) 832 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {1}{4} \left (\int \frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{\sqrt {x^6+2}}dx^2-\sqrt [3]{2} \left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {x^6+2}}dx^2\right )-\frac {\sqrt {x^6+2}}{2 x^2}\right )+\frac {1}{3 x^2 \sqrt {x^6+2}}\right )\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {1}{4} \left (\int \frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{\sqrt {x^6+2}}dx^2-\frac {2 \sqrt [6]{2} \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}\right )-\frac {\sqrt {x^6+2}}{2 x^2}\right )+\frac {1}{3 x^2 \sqrt {x^6+2}}\right )\) |
\(\Big \downarrow \) 2416 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {1}{4} \left (-\frac {2 \sqrt [6]{2} \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}-\frac {\sqrt [6]{2} \sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} E\left (\arcsin \left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}+\frac {2 \sqrt {x^6+2}}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )-\frac {\sqrt {x^6+2}}{2 x^2}\right )+\frac {1}{3 x^2 \sqrt {x^6+2}}\right )\) |
(1/(3*x^2*Sqrt[2 + x^6]) + (5*(-1/2*Sqrt[2 + x^6]/x^2 + ((2*Sqrt[2 + x^6]) /(2^(1/3)*(1 + Sqrt[3]) + x^2) - (2^(1/6)*3^(1/4)*Sqrt[2 - Sqrt[3]]*(2^(1/ 3) + x^2)*Sqrt[(2^(2/3) - 2^(1/3)*x^2 + x^4)/(2^(1/3)*(1 + Sqrt[3]) + x^2) ^2]*EllipticE[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x ^2)^2]*Sqrt[2 + x^6]) - (2*2^(1/6)*(1 - Sqrt[3])*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]])/(3^(1/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[ 3]) + x^2)^2]*Sqrt[2 + x^6]))/4))/6)/2
3.15.31.3.1 Defintions of rubi rules used
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]*Sqrt[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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
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] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] ], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r) Int[1/Sqrt[a + b*x ^3], x], x] + Simp[1/r Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x ]] /; FreeQ[{a, b}, x] && PosQ[a]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) ]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( (1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt [3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) *s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 5.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.05
method | result | size |
meijerg | \(-\frac {\sqrt {2}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{3},\frac {3}{2};\frac {2}{3};-\frac {x^{6}}{2}\right )}{8 x^{2}}\) | \(20\) |
risch | \(-\frac {5 x^{6}+6}{24 x^{2} \sqrt {x^{6}+2}}+\frac {5 \sqrt {2}\, x^{4} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};-\frac {x^{6}}{2}\right )}{192}\) | \(40\) |
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.12 \[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=-\frac {5 \, {\left (x^{8} + 2 \, x^{2}\right )} {\rm weierstrassZeta}\left (0, -8, {\rm weierstrassPInverse}\left (0, -8, x^{2}\right )\right ) + {\left (5 \, x^{6} + 6\right )} \sqrt {x^{6} + 2}}{24 \, {\left (x^{8} + 2 \, x^{2}\right )}} \]
-1/24*(5*(x^8 + 2*x^2)*weierstrassZeta(0, -8, weierstrassPInverse(0, -8, x ^2)) + (5*x^6 + 6)*sqrt(x^6 + 2))/(x^8 + 2*x^2)
Time = 0.49 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.10 \[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=\frac {\sqrt {2} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {2}{3} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{24 x^{2} \Gamma \left (\frac {2}{3}\right )} \]
\[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{6} + 2\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
\[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{6} + 2\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{x^3 \left (2+x^6\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (x^6+2\right )}^{3/2}} \,d x \]