Integrand size = 39, antiderivative size = 211 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=-a^{5/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+a^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )-\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a b \log (x)-a b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )-a^2 \log (x) \text {$\#$1}^4+b \log (x) \text {$\#$1}^4+a^2 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (4 a^{5/4} \left (-\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )+\text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a b \log (x)+2 a b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )+a^2 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4-2 a^2 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4+2 b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ]\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}} \]
((-(b*x^2) + a*x^4)^(1/4)*(4*a^(5/4)*(-ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^ 2)^(1/4)] + ArcTanh[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)]) + RootSum[b - a *#1^4 + #1^8 & , (-(a*b*Log[x]) + 2*a*b*Log[(-b + a*x^2)^(1/4) - Sqrt[x]*# 1] + a^2*Log[x]*#1^4 - b*Log[x]*#1^4 - 2*a^2*Log[(-b + a*x^2)^(1/4) - Sqrt [x]*#1]*#1^4 + 2*b*Log[(-b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1^4)/(a*#1^3 - 2* #1^7) & ]))/(4*Sqrt[x]*(-b + a*x^2)^(1/4))
Time = 0.61 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2467, 1592, 1852, 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.
\(\displaystyle \int \frac {\left (a x^2-b\right ) \sqrt [4]{a x^4-b x^2}}{-a x^2+b+x^4} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \int \frac {\sqrt {x} \left (a x^2-b\right )^{5/4}}{x^4-a x^2+b}dx}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 1592 |
\(\displaystyle \frac {2 \sqrt [4]{a x^4-b x^2} \int \frac {x \left (a x^2-b\right )^{5/4}}{x^4-a x^2+b}d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 1852 |
\(\displaystyle \frac {2 \sqrt [4]{a x^4-b x^2} \int \left (-\frac {2 x \left (a x^2-b\right )^{5/4}}{\sqrt {a^2-4 b} \left (-2 x^2+a+\sqrt {a^2-4 b}\right )}-\frac {2 x \left (a x^2-b\right )^{5/4}}{\sqrt {a^2-4 b} \left (2 x^2-a+\sqrt {a^2-4 b}\right )}\right )d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt [4]{a x^4-b x^2} \left (\frac {2 b x^{3/2} \sqrt [4]{a x^2-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a-\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (-a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {2 b x^{3/2} \sqrt [4]{a x^2-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a+\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
(2*(-(b*x^2) + a*x^4)^(1/4)*((2*b*x^(3/2)*(-b + a*x^2)^(1/4)*AppellF1[3/4, 1, -5/4, 7/4, (2*x^2)/(a - Sqrt[a^2 - 4*b]), (a*x^2)/b])/(3*(a^2 - a*Sqrt [a^2 - 4*b] - 4*b)*(1 - (a*x^2)/b)^(1/4)) + (2*b*x^(3/2)*(-b + a*x^2)^(1/4 )*AppellF1[3/4, 1, -5/4, 7/4, (2*x^2)/(a + Sqrt[a^2 - 4*b]), (a*x^2)/b])/( 3*(a^2 + a*Sqrt[a^2 - 4*b] - 4*b)*(1 - (a*x^2)/b)^(1/4))))/(Sqrt[x]*(-b + a*x^2)^(1/4))
3.26.26.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/f Subst[ Int[x^(k*(m + 1) - 1)*(d + e*(x^(2*k)/f^2))^q*(a + b*(x^(2*k)/f^k) + c*(x^( 4*k)/f^4))^p, x], x, (f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, p, q}, x ] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q, ( f*x)^m/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, f, q, n}, x ] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && !IntegerQ[q] && IntegerQ[m]
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 = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.73
method | result | size |
pseudoelliptic | \(a^{\frac {5}{4}} \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {a^{\frac {5}{4}} \ln \left (\frac {x \,a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{-x \,a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right )}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a \,\textit {\_Z}^{4}+b \right )}{\sum }\left (-\frac {\left (\left (a^{2}-b \right ) \textit {\_R}^{4}-a b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (-2 \textit {\_R}^{4}+a \right )}\right )\right )}{2}\) | \(153\) |
a^(5/4)*arctan(1/a^(1/4)/x*(x^2*(a*x^2-b))^(1/4))+1/2*a^(5/4)*ln((x*a^(1/4 )+(x^2*(a*x^2-b))^(1/4))/(-x*a^(1/4)+(x^2*(a*x^2-b))^(1/4)))+1/2*sum(-((a^ 2-b)*_R^4-a*b)*ln((-_R*x+(x^2*(a*x^2-b))^(1/4))/x)/_R^3/(-2*_R^4+a),_R=Roo tOf(_Z^8-_Z^4*a+b))
Timed out. \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\text {Timed out} \]
Not integrable
Time = 13.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.15 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{2} - b\right )}{- a x^{2} + b + x^{4}}\, dx \]
Not integrable
Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}{x^{4} - a x^{2} + b} \,d x } \]
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 7.97 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) \]
1/2*sqrt(2)*(-a)^(1/4)*a*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b /x^2)^(1/4))/(-a)^(1/4)) + 1/2*sqrt(2)*(-a)^(1/4)*a*arctan(-1/2*sqrt(2)*(s qrt(2)*(-a)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4)) + 1/4*sqrt(2)*(-a)^(1 /4)*a*log(sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2 )) - 1/4*sqrt(2)*(-a)^(1/4)*a*log(-sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))
Not integrable
Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int -\frac {\left (b-a\,x^2\right )\,{\left (a\,x^4-b\,x^2\right )}^{1/4}}{x^4-a\,x^2+b} \,d x \]