Integrand size = 41, antiderivative size = 60 \[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=\frac {\sqrt {1-2 x^8} \left (-1-3 x^4+2 x^8\right )}{6 x^6}-\frac {1}{2} \text {arctanh}\left (\frac {x^2 \sqrt {1-2 x^8}}{-1+2 x^8}\right ) \]
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.24 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=-\frac {2+6 x^4-8 x^8-12 x^{12}+8 x^{16}-3\ 2^{3/4} x^6 \sqrt {1-2 x^8} \operatorname {EllipticPi}\left (-\frac {1}{\sqrt {2}},\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )-3\ 2^{3/4} x^6 \sqrt {1-2 x^8} \operatorname {EllipticPi}\left (\sqrt {2},\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )+6 x^8 \sqrt {1-2 x^8} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},2 x^8\right )}{12 x^6 \sqrt {1-2 x^8}} \]
-1/12*(2 + 6*x^4 - 8*x^8 - 12*x^12 + 8*x^16 - 3*2^(3/4)*x^6*Sqrt[1 - 2*x^8 ]*EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*x^2], -1] - 3*2^(3/4)*x^6*Sqrt[1 - 2*x^8]*EllipticPi[Sqrt[2], ArcSin[2^(1/4)*x^2], -1] + 6*x^8*Sqrt[1 - 2* x^8]*Hypergeometric2F1[1/4, 1/2, 5/4, 2*x^8])/(x^6*Sqrt[1 - 2*x^8])
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.06 (sec) , antiderivative size = 209, normalized size of antiderivative = 3.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {281, 25, 7279, 2009}
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.
\(\displaystyle \int \frac {\sqrt {1-2 x^8} \left (2 x^8-1\right ) \left (2 x^8+1\right )}{x^7 \left (2 x^8+x^4-1\right )} \, dx\) |
\(\Big \downarrow \) 281 |
\(\displaystyle -\int -\frac {\left (1-2 x^8\right )^{3/2} \left (2 x^8+1\right )}{x^7 \left (-2 x^8-x^4+1\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\left (1-2 x^8\right )^{3/2} \left (2 x^8+1\right )}{x^7 \left (-2 x^8-x^4+1\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {\left (1-2 x^8\right )^{3/2}}{x^7}+\frac {x \left (1-2 x^8\right )^{3/2}}{x^4+1}-\frac {4 x \left (1-2 x^8\right )^{3/2}}{2 x^4-1}+\frac {\left (1-2 x^8\right )^{3/2}}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (9+5 \sqrt {2}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )}{15\ 2^{3/4}}+\frac {\left (35-27 \sqrt {2}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )}{30 \sqrt [4]{2}}-\frac {2}{3} 2^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )+\frac {6}{5} \sqrt [4]{2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )+\frac {\operatorname {EllipticPi}\left (-\frac {1}{\sqrt {2}},\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )}{2 \sqrt [4]{2}}+\frac {\operatorname {EllipticPi}\left (\sqrt {2},\arcsin \left (\sqrt [4]{2} x^2\right ),-1\right )}{2 \sqrt [4]{2}}-\sqrt {1-2 x^8} x^6-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}\) |
-(x^6*Sqrt[1 - 2*x^8]) - (1 - 2*x^8)^(3/2)/(6*x^6) - (1 - 2*x^8)^(3/2)/(2* x^2) + (6*2^(1/4)*EllipticF[ArcSin[2^(1/4)*x^2], -1])/5 - (2*2^(3/4)*Ellip ticF[ArcSin[2^(1/4)*x^2], -1])/3 + ((35 - 27*Sqrt[2])*EllipticF[ArcSin[2^( 1/4)*x^2], -1])/(30*2^(1/4)) - ((9 + 5*Sqrt[2])*EllipticF[ArcSin[2^(1/4)*x ^2], -1])/(15*2^(3/4)) + EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*x^2], -1] /(2*2^(1/4)) + EllipticPi[Sqrt[2], ArcSin[2^(1/4)*x^2], -1]/(2*2^(1/4))
3.8.89.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 4.90 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20
method | result | size |
trager | \(\frac {\sqrt {-2 x^{8}+1}\, \left (2 x^{8}-3 x^{4}-1\right )}{6 x^{6}}+\frac {\ln \left (\frac {-2 x^{8}+x^{4}+2 \sqrt {-2 x^{8}+1}\, x^{2}+1}{\left (x^{4}+1\right ) \left (2 x^{4}-1\right )}\right )}{4}\) | \(72\) |
risch | \(-\frac {4 x^{16}-6 x^{12}-4 x^{8}+3 x^{4}+1}{6 x^{6} \sqrt {-2 x^{8}+1}}+\frac {\ln \left (-\frac {-2 x^{8}+x^{4}+2 \sqrt {-2 x^{8}+1}\, x^{2}+1}{\left (x^{4}+1\right ) \left (2 x^{4}-1\right )}\right )}{4}\) | \(83\) |
pseudoelliptic | \(\frac {\left (4 x^{8}-6 x^{4}-2\right ) \sqrt {\frac {-2 x^{8}+1}{x^{2}}}-3 x^{5} \left (\operatorname {arctanh}\left (\frac {\left (x^{4}+2 i x^{2}-1\right ) \sqrt {2}-2 i x^{4}-i}{\sqrt {\frac {-2 x^{8}+1}{x^{2}}}\, x}\right )+\operatorname {arctanh}\left (\frac {-\sqrt {2}\, x^{4}+2 i x^{4}+2 i \sqrt {2}\, x^{2}+i+\sqrt {2}}{x \sqrt {\frac {-2 x^{8}+1}{x^{2}}}}\right )\right )}{12 x^{5}}\) | \(128\) |
1/6*(-2*x^8+1)^(1/2)*(2*x^8-3*x^4-1)/x^6+1/4*ln((-2*x^8+x^4+2*(-2*x^8+1)^( 1/2)*x^2+1)/(x^4+1)/(2*x^4-1))
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=\frac {3 \, x^{6} \log \left (-\frac {2 \, x^{8} - x^{4} - 2 \, \sqrt {-2 \, x^{8} + 1} x^{2} - 1}{2 \, x^{8} + x^{4} - 1}\right ) + 2 \, {\left (2 \, x^{8} - 3 \, x^{4} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{12 \, x^{6}} \]
1/12*(3*x^6*log(-(2*x^8 - x^4 - 2*sqrt(-2*x^8 + 1)*x^2 - 1)/(2*x^8 + x^4 - 1)) + 2*(2*x^8 - 3*x^4 - 1)*sqrt(-2*x^8 + 1))/x^6
\[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=\int \frac {\sqrt {1 - 2 x^{8}} \cdot \left (2 x^{8} - 1\right ) \left (2 x^{8} + 1\right )}{x^{7} \left (x^{4} + 1\right ) \left (2 x^{4} - 1\right )}\, dx \]
\[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=\int { \frac {{\left (2 \, x^{8} + 1\right )} {\left (2 \, x^{8} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{{\left (2 \, x^{8} + x^{4} - 1\right )} x^{7}} \,d x } \]
\[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=\int { \frac {{\left (2 \, x^{8} + 1\right )} {\left (2 \, x^{8} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{{\left (2 \, x^{8} + x^{4} - 1\right )} x^{7}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx=\int -\frac {{\left (1-2\,x^8\right )}^{3/2}\,\left (2\,x^8+1\right )}{x^7\,\left (2\,x^8+x^4-1\right )} \,d x \]