Integrand size = 13, antiderivative size = 156 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\frac {1}{2 x^5 \sqrt {1+x^4}}-\frac {7 \sqrt {1+x^4}}{10 x^5}+\frac {21 \sqrt {1+x^4}}{10 x}-\frac {21 x \sqrt {1+x^4}}{10 \left (1+x^2\right )}+\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{10 \sqrt {1+x^4}}-\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{20 \sqrt {1+x^4}} \]
1/2/x^5/(x^4+1)^(1/2)-7/10*(x^4+1)^(1/2)/x^5+21/10*(x^4+1)^(1/2)/x-21/10*x *(x^4+1)^(1/2)/(x^2+1)+21/10*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arct an(x))*EllipticE(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/( x^4+1)^(1/2)-21/20*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*Ell ipticF(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/ 2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.14 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},-x^4\right )}{5 x^5} \]
Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {819, 847, 847, 834, 761, 1510}
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.
\(\displaystyle \int \frac {1}{x^6 \left (x^4+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {7}{2} \int \frac {1}{x^6 \sqrt {x^4+1}}dx+\frac {1}{2 x^5 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {7}{2} \left (-\frac {3}{5} \int \frac {1}{x^2 \sqrt {x^4+1}}dx-\frac {\sqrt {x^4+1}}{5 x^5}\right )+\frac {1}{2 x^5 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {7}{2} \left (-\frac {3}{5} \left (\int \frac {x^2}{\sqrt {x^4+1}}dx-\frac {\sqrt {x^4+1}}{x}\right )-\frac {\sqrt {x^4+1}}{5 x^5}\right )+\frac {1}{2 x^5 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {7}{2} \left (-\frac {3}{5} \left (\int \frac {1}{\sqrt {x^4+1}}dx-\int \frac {1-x^2}{\sqrt {x^4+1}}dx-\frac {\sqrt {x^4+1}}{x}\right )-\frac {\sqrt {x^4+1}}{5 x^5}\right )+\frac {1}{2 x^5 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {7}{2} \left (-\frac {3}{5} \left (-\int \frac {1-x^2}{\sqrt {x^4+1}}dx+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\sqrt {x^4+1}}{x}\right )-\frac {\sqrt {x^4+1}}{5 x^5}\right )+\frac {1}{2 x^5 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {7}{2} \left (-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+1}}-\frac {\sqrt {x^4+1}}{x}+\frac {\sqrt {x^4+1} x}{x^2+1}\right )-\frac {\sqrt {x^4+1}}{5 x^5}\right )+\frac {1}{2 x^5 \sqrt {x^4+1}}\) |
1/(2*x^5*Sqrt[1 + x^4]) + (7*(-1/5*Sqrt[1 + x^4]/x^5 - (3*(-(Sqrt[1 + x^4] /x) + (x*Sqrt[1 + x^4])/(1 + x^2) - ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2] *EllipticE[2*ArcTan[x], 1/2])/Sqrt[1 + x^4] + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4])))/5))/2
3.10.58.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/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*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_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 4.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11
method | result | size |
meijerg | \(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {5}{4},\frac {3}{2};-\frac {1}{4};-x^{4}\right )}{5 x^{5}}\) | \(17\) |
risch | \(\frac {21 x^{8}+14 x^{4}-2}{10 x^{5} \sqrt {x^{4}+1}}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
default | \(\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{5 x^{5}}+\frac {8 \sqrt {x^{4}+1}}{5 x}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(119\) |
elliptic | \(\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {x^{4}+1}}{5 x^{5}}+\frac {8 \sqrt {x^{4}+1}}{5 x}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(119\) |
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.52 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=-\frac {21 \, \sqrt {i} {\left (-i \, x^{9} - i \, x^{5}\right )} E(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + 21 \, \sqrt {i} {\left (i \, x^{9} + i \, x^{5}\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) - {\left (21 \, x^{8} + 14 \, x^{4} - 2\right )} \sqrt {x^{4} + 1}}{10 \, {\left (x^{9} + x^{5}\right )}} \]
-1/10*(21*sqrt(I)*(-I*x^9 - I*x^5)*elliptic_e(arcsin(sqrt(I)*x), -1) + 21* sqrt(I)*(I*x^9 + I*x^5)*elliptic_f(arcsin(sqrt(I)*x), -1) - (21*x^8 + 14*x ^4 - 2)*sqrt(x^4 + 1))/(x^9 + x^5)
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \]
\[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{6}} \,d x } \]
\[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{6}} \,d x } \]
Timed out. \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/2}} \, dx=\int \frac {1}{x^6\,{\left (x^4+1\right )}^{3/2}} \,d x \]