Integrand size = 39, antiderivative size = 61 \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{b x^2+a x^6}}\right )}{\sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{b x^2+a x^6}}\right )}{\sqrt [4]{2}} \]
-1/2*arctan(2^(1/4)*x/(a*x^6+b*x^2)^(1/4))*2^(3/4)-1/2*arctanh(2^(1/4)*x/( a*x^6+b*x^2)^(1/4))*2^(3/4)
Time = 13.48 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2 \left (b+a x^4\right )}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2 \left (b+a x^4\right )}}\right )}{\sqrt [4]{2}} \]
-((ArcTan[(2^(1/4)*x)/(x^2*(b + a*x^4))^(1/4)] + ArcTanh[(2^(1/4)*x)/(x^2* (b + a*x^4))^(1/4)])/2^(1/4))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.54 (sec) , antiderivative size = 471, normalized size of antiderivative = 7.72, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {2467, 25, 2035, 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.
\(\displaystyle \int \frac {a x^4-b}{\left (a x^4+b-2 x^2\right ) \sqrt [4]{a x^6+b x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt [4]{a x^4+b} \int -\frac {b-a x^4}{\sqrt {x} \sqrt [4]{a x^4+b} \left (a x^4-2 x^2+b\right )}dx}{\sqrt [4]{a x^6+b x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^4+b} \int \frac {b-a x^4}{\sqrt {x} \sqrt [4]{a x^4+b} \left (a x^4-2 x^2+b\right )}dx}{\sqrt [4]{a x^6+b x^2}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^4+b} \int \frac {b-a x^4}{\sqrt [4]{a x^4+b} \left (a x^4-2 x^2+b\right )}d\sqrt {x}}{\sqrt [4]{a x^6+b x^2}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^4+b} \int \left (\frac {2 \left (b-x^2\right )}{\sqrt [4]{a x^4+b} \left (a x^4-2 x^2+b\right )}-\frac {1}{\sqrt [4]{a x^4+b}}\right )d\sqrt {x}}{\sqrt [4]{a x^6+b x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^4+b} \left (\frac {\sqrt {x} \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {a^2 x^4}{-a b-2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^4+b}}+\frac {\sqrt {x} \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {a^2 x^4}{-a b+2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^4+b}}+\frac {a x^{5/2} \left (1-\sqrt {1-a b}\right ) \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {a^2 x^4}{-a b-2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{5 \left (-a b-2 \sqrt {1-a b}+2\right ) \sqrt [4]{a x^4+b}}+\frac {a x^{5/2} \left (\sqrt {1-a b}+1\right ) \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {a^2 x^4}{-a b+2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{5 \left (-a b+2 \sqrt {1-a b}+2\right ) \sqrt [4]{a x^4+b}}-\frac {\sqrt {x} \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a x^6+b x^2}}\) |
(-2*Sqrt[x]*(b + a*x^4)^(1/4)*((Sqrt[x]*(1 + (a*x^4)/b)^(1/4)*AppellF1[1/8 , 1, 1/4, 9/8, (a^2*x^4)/(2 - a*b - 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(b + a*x^4)^(1/4) + (Sqrt[x]*(1 + (a*x^4)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, ( a^2*x^4)/(2 - a*b + 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(b + a*x^4)^(1/4) + ( a*(1 - Sqrt[1 - a*b])*x^(5/2)*(1 + (a*x^4)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (a^2*x^4)/(2 - a*b - 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(5*(2 - a*b - 2*Sqrt[1 - a*b])*(b + a*x^4)^(1/4)) + (a*(1 + Sqrt[1 - a*b])*x^(5/2)*(1 + (a*x^4)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (a^2*x^4)/(2 - a*b + 2*Sqrt[1 - a*b]), -((a*x^4)/b)])/(5*(2 - a*b + 2*Sqrt[1 - a*b])*(b + a*x^4)^(1/4)) - (Sqrt[x]*(1 + (a*x^4)/b)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, -((a*x^ 4)/b)])/(b + a*x^4)^(1/4)))/(b*x^2 + a*x^6)^(1/4)
3.9.6.3.1 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]
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 = 2.87 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (a \,x^{4}+b \right )\right )^{\frac {1}{4}}}{2 x}\right )-\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{4}+b \right )\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{4}+b \right )\right )^{\frac {1}{4}}}\right )\right )}{4}\) | \(76\) |
1/4*2^(3/4)*(2*arctan(1/2*2^(3/4)/x*(x^2*(a*x^4+b))^(1/4))-ln((2^(1/4)*x+( x^2*(a*x^4+b))^(1/4))/(-2^(1/4)*x+(x^2*(a*x^4+b))^(1/4))))
Timed out. \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\text {Timed out} \]
\[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int \frac {a x^{4} - b}{\sqrt [4]{x^{2} \left (a x^{4} + b\right )} \left (a x^{4} + b - 2 x^{2}\right )}\, dx \]
\[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int { \frac {a x^{4} - b}{{\left (a x^{6} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, x^{2} + b\right )}} \,d x } \]
\[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int { \frac {a x^{4} - b}{{\left (a x^{6} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, x^{2} + b\right )}} \,d x } \]
Timed out. \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int -\frac {b-a\,x^4}{{\left (a\,x^6+b\,x^2\right )}^{1/4}\,\left (a\,x^4-2\,x^2+b\right )} \,d x \]