Integrand size = 43, antiderivative size = 69 \[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\frac {4 \left (3+7 x^3+3 x^4\right ) \left (x+x^5\right )^{3/4}}{21 x^6}-4 \arctan \left (\frac {\left (x+x^5\right )^{3/4}}{1+x^4}\right )-4 \text {arctanh}\left (\frac {\left (x+x^5\right )^{3/4}}{1+x^4}\right ) \]
4/21*(3*x^4+7*x^3+3)*(x^5+x)^(3/4)/x^6-4*arctan((x^5+x)^(3/4)/(x^4+1))-4*a rctanh((x^5+x)^(3/4)/(x^4+1))
Time = 11.51 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\frac {4 \left (3+7 x^3+3 x^4\right ) \left (x+x^5\right )^{3/4}}{21 x^6}+2 \left (-2 \arctan \left (\frac {\left (x+x^5\right )^{3/4}}{1+x^4}\right )-2 \text {arctanh}\left (\frac {\left (x+x^5\right )^{3/4}}{1+x^4}\right )\right ) \]
(4*(3 + 7*x^3 + 3*x^4)*(x + x^5)^(3/4))/(21*x^6) + 2*(-2*ArcTan[(x + x^5)^ (3/4)/(1 + x^4)] - 2*ArcTanh[(x + x^5)^(3/4)/(1 + x^4)])
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.
| \(\displaystyle \int \frac {\left (x^4-3\right ) \left (x^8+x^6+2 x^4+1\right )}{x^6 \left (x^4-x^3+1\right ) \sqrt [4]{x^5+x}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{x^4+1} \int -\frac {\left (3-x^4\right ) \left (x^8+x^6+2 x^4+1\right )}{x^{25/4} \sqrt [4]{x^4+1} \left (x^4-x^3+1\right )}dx}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{x^4+1} \int \frac {\left (3-x^4\right ) \left (x^8+x^6+2 x^4+1\right )}{x^{25/4} \sqrt [4]{x^4+1} \left (x^4-x^3+1\right )}dx}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \int \frac {\left (3-x^4\right ) \left (x^8+x^6+2 x^4+1\right )}{x^{11/2} \sqrt [4]{x^4+1} \left (x^4-x^3+1\right )}d\sqrt [4]{x}}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \int \left (-\frac {x^{5/2}}{\sqrt [4]{x^4+1}}-\frac {x^{3/2}}{\sqrt [4]{x^4+1}}-\frac {2 \sqrt {x}}{\sqrt [4]{x^4+1}}-\frac {2 \left (x^3-4\right ) \sqrt {x}}{\sqrt [4]{x^4+1} \left (x^4-x^3+1\right )}+\frac {2}{\sqrt [4]{x^4+1} x^{3/2}}+\frac {3}{\sqrt [4]{x^4+1} x^{5/2}}+\frac {3}{\sqrt [4]{x^4+1} x^{11/2}}\right )d\sqrt [4]{x}}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \left (8 \int \frac {\sqrt {x}}{\sqrt [4]{x^4+1} \left (x^4-x^3+1\right )}d\sqrt [4]{x}-2 \int \frac {x^{7/2}}{\sqrt [4]{x^4+1} \left (x^4-x^3+1\right )}d\sqrt [4]{x}-\frac {1}{11} x^{11/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{16},\frac {27}{16},-x^4\right )-\frac {1}{7} x^{7/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{16},\frac {23}{16},-x^4\right )-\frac {2}{3} x^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{16},\frac {1}{4},\frac {19}{16},-x^4\right )-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{16},\frac {1}{4},\frac {11}{16},-x^4\right )}{5 x^{5/4}}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {9}{16},\frac {1}{4},\frac {7}{16},-x^4\right )}{3 x^{9/4}}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {21}{16},\frac {1}{4},-\frac {5}{16},-x^4\right )}{7 x^{21/4}}\right )}{\sqrt [4]{x^5+x}}\) |
3.10.16.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 4.83 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32
| method | result | size |
| pseudoelliptic | \(\frac {2 \ln \left (\frac {{\left (x \left (x^{4}+1\right )\right )}^{\frac {1}{4}}-x}{x}\right ) x^{6}-2 \ln \left (\frac {{\left (x \left (x^{4}+1\right )\right )}^{\frac {1}{4}}+x}{x}\right ) x^{6}+4 \arctan \left (\frac {{\left (x \left (x^{4}+1\right )\right )}^{\frac {1}{4}}}{x}\right ) x^{6}+\frac {4 \left (x^{4}+\frac {7}{3} x^{3}+1\right ) {\left (x \left (x^{4}+1\right )\right )}^{\frac {3}{4}}}{7}}{x^{6}}\) | \(91\) |
| trager | \(\frac {4 \left (3 x^{4}+7 x^{3}+3\right ) \left (x^{5}+x \right )^{\frac {3}{4}}}{21 x^{6}}-2 \ln \left (-\frac {x^{4}+2 \left (x^{5}+x \right )^{\frac {3}{4}}+2 x \sqrt {x^{5}+x}+2 \left (x^{5}+x \right )^{\frac {1}{4}} x^{2}+x^{3}+1}{x^{4}-x^{3}+1}\right )-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{5}+x}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{5}+x \right )^{\frac {3}{4}}-2 \left (x^{5}+x \right )^{\frac {1}{4}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{4}-x^{3}+1}\right )\) | \(168\) |
| risch | \(\frac {\frac {4}{7} x^{8}+\frac {8}{7} x^{4}+\frac {4}{7}+\frac {4}{3} x^{7}+\frac {4}{3} x^{3}}{x^{5} {\left (x \left (x^{4}+1\right )\right )}^{\frac {1}{4}}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{5}+x}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{5}+x \right )^{\frac {3}{4}}-2 \left (x^{5}+x \right )^{\frac {1}{4}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{4}-x^{3}+1}\right )+2 \ln \left (-\frac {-x^{4}+2 \left (x^{5}+x \right )^{\frac {3}{4}}-2 x \sqrt {x^{5}+x}+2 \left (x^{5}+x \right )^{\frac {1}{4}} x^{2}-x^{3}-1}{x^{4}-x^{3}+1}\right )\) | \(189\) |
2/7*(7*ln(((x*(x^4+1))^(1/4)-x)/x)*x^6-7*ln(((x*(x^4+1))^(1/4)+x)/x)*x^6+1 4*arctan((x*(x^4+1))^(1/4)/x)*x^6+2*(x^4+7/3*x^3+1)*(x*(x^4+1))^(3/4))/x^6
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (61) = 122\).
Time = 25.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.81 \[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=-\frac {2 \, {\left (21 \, x^{6} \arctan \left (\frac {{\left (x^{5} + x\right )}^{\frac {3}{4}} x - {\left (x^{5} + x\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}}{2 \, {\left (x^{5} + x\right )}}\right ) - 21 \, x^{6} \log \left (-\frac {x^{4} + x^{3} - 2 \, {\left (x^{5} + x\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{5} + x} x - 2 \, {\left (x^{5} + x\right )}^{\frac {3}{4}} + 1}{x^{4} - x^{3} + 1}\right ) - 2 \, {\left (x^{5} + x\right )}^{\frac {3}{4}} {\left (3 \, x^{4} + 7 \, x^{3} + 3\right )}\right )}}{21 \, x^{6}} \]
-2/21*(21*x^6*arctan(1/2*((x^5 + x)^(3/4)*x - (x^5 + x)^(1/4)*(x^4 + 1))/( x^5 + x)) - 21*x^6*log(-(x^4 + x^3 - 2*(x^5 + x)^(1/4)*x^2 + 2*sqrt(x^5 + x)*x - 2*(x^5 + x)^(3/4) + 1)/(x^4 - x^3 + 1)) - 2*(x^5 + x)^(3/4)*(3*x^4 + 7*x^3 + 3))/x^6
Timed out. \[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\int { \frac {{\left (x^{8} + x^{6} + 2 \, x^{4} + 1\right )} {\left (x^{4} - 3\right )}}{{\left (x^{5} + x\right )}^{\frac {1}{4}} {\left (x^{4} - x^{3} + 1\right )} x^{6}} \,d x } \]
\[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\int { \frac {{\left (x^{8} + x^{6} + 2 \, x^{4} + 1\right )} {\left (x^{4} - 3\right )}}{{\left (x^{5} + x\right )}^{\frac {1}{4}} {\left (x^{4} - x^{3} + 1\right )} x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (-3+x^4\right ) \left (1+2 x^4+x^6+x^8\right )}{x^6 \left (1-x^3+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\int \frac {\left (x^4-3\right )\,\left (x^8+x^6+2\,x^4+1\right )}{x^6\,{\left (x^5+x\right )}^{1/4}\,\left (x^4-x^3+1\right )} \,d x \]