Integrand size = 38, antiderivative size = 49 \[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\frac {2 \sqrt {x+x^5}}{1+x^4}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {x+x^5}}{\sqrt {a} \left (1+x^4\right )}\right ) \]
\[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx \]
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.
\(\displaystyle \int \frac {3 x^5-x}{\left (x^4+1\right ) \sqrt {x^5+x} \left (a x^4+a-x\right )} \, dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {x \left (3 x^4-1\right )}{\left (x^4+1\right ) \sqrt {x^5+x} \left (a x^4+a-x\right )}dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {x^4+1} \int -\frac {\sqrt {x} \left (1-3 x^4\right )}{\left (x^4+1\right )^{3/2} \left (a x^4-x+a\right )}dx}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {x^4+1} \int \frac {\sqrt {x} \left (1-3 x^4\right )}{\left (x^4+1\right )^{3/2} \left (a x^4-x+a\right )}dx}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^4+1} \int \frac {x \left (1-3 x^4\right )}{\left (x^4+1\right )^{3/2} \left (a x^4-x+a\right )}d\sqrt {x}}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^4+1} \int \left (\frac {(4 a-3 x) x}{a \left (x^4+1\right )^{3/2} \left (a x^4-x+a\right )}-\frac {3 x}{a \left (x^4+1\right )^{3/2}}\right )d\sqrt {x}}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^4+1} \left (4 \int \frac {x}{\left (x^4+1\right )^{3/2} \left (a x^4-x+a\right )}d\sqrt {x}-\frac {3 \int \frac {x^2}{\left (x^4+1\right )^{3/2} \left (a x^4-x+a\right )}d\sqrt {x}}{a}+\frac {3 \sqrt {\frac {(x+1)^2}{x}} \sqrt {-\frac {x^4+1}{x^2}} x^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} x^2-2 x+\sqrt {2}}{x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{8 \sqrt {2+\sqrt {2}} a (x+1) \sqrt {x^4+1}}+\frac {3 \sqrt {-\frac {(1-x)^2}{x}} \sqrt {-\frac {x^4+1}{x^2}} x^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} x^2+2 x+\sqrt {2}}{x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{8 \sqrt {2+\sqrt {2}} a (1-x) \sqrt {x^4+1}}-\frac {3 x^{3/2}}{4 a \sqrt {x^4+1}}\right )}{\sqrt {x^5+x}}\) |
3.7.26.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
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 = 1.38 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92
method | result | size |
pseudoelliptic | \(\frac {-2 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{4}+1\right )}\, \sqrt {a}}{x}\right ) \sqrt {x \left (x^{4}+1\right )}+2 x}{\sqrt {x \left (x^{4}+1\right )}}\) | \(45\) |
Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 3.67 \[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\left [\frac {{\left (x^{4} + 1\right )} \sqrt {a} \log \left (\frac {a^{2} x^{8} + 2 \, a^{2} x^{4} + 6 \, a x^{5} - 4 \, {\left (a x^{4} + a + x\right )} \sqrt {x^{5} + x} \sqrt {a} + a^{2} + 6 \, a x + x^{2}}{a^{2} x^{8} + 2 \, a^{2} x^{4} - 2 \, a x^{5} + a^{2} - 2 \, a x + x^{2}}\right ) + 4 \, \sqrt {x^{5} + x}}{2 \, {\left (x^{4} + 1\right )}}, \frac {{\left (x^{4} + 1\right )} \sqrt {-a} \arctan \left (\frac {{\left (a x^{4} + a + x\right )} \sqrt {x^{5} + x} \sqrt {-a}}{2 \, {\left (a x^{5} + a x\right )}}\right ) + 2 \, \sqrt {x^{5} + x}}{x^{4} + 1}\right ] \]
[1/2*((x^4 + 1)*sqrt(a)*log((a^2*x^8 + 2*a^2*x^4 + 6*a*x^5 - 4*(a*x^4 + a + x)*sqrt(x^5 + x)*sqrt(a) + a^2 + 6*a*x + x^2)/(a^2*x^8 + 2*a^2*x^4 - 2*a *x^5 + a^2 - 2*a*x + x^2)) + 4*sqrt(x^5 + x))/(x^4 + 1), ((x^4 + 1)*sqrt(- a)*arctan(1/2*(a*x^4 + a + x)*sqrt(x^5 + x)*sqrt(-a)/(a*x^5 + a*x)) + 2*sq rt(x^5 + x))/(x^4 + 1)]
Timed out. \[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\text {Timed out} \]
\[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\int { \frac {3 \, x^{5} - x}{{\left (a x^{4} + a - x\right )} \sqrt {x^{5} + x} {\left (x^{4} + 1\right )}} \,d x } \]
\[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\int { \frac {3 \, x^{5} - x}{{\left (a x^{4} + a - x\right )} \sqrt {x^{5} + x} {\left (x^{4} + 1\right )}} \,d x } \]
Time = 6.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12 \[ \int \frac {-x+3 x^5}{\left (1+x^4\right ) \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\frac {2\,\sqrt {x^5+x}}{x^4+1}+\sqrt {a}\,\ln \left (\frac {a+x-2\,\sqrt {a}\,\sqrt {x^5+x}+a\,x^4}{a\,x^4-x+a}\right ) \]