Integrand size = 36, antiderivative size = 241 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=\frac {x \left (b c+a g+(b d+a h) x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac {4 a f+x \left (7 b c-a g+2 (3 b d-a h) x+5 b e x^2\right )}{32 a^2 b \left (a-b x^4\right )}+\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e-3 a g\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 a^{11/4} b^{5/4}}+\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e-3 a g\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 a^{11/4} b^{5/4}}+\frac {(3 b d-a h) \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}} \]
1/8*x*(b*c+a*g+(a*h+b*d)*x+b*e*x^2+b*f*x^3)/a/b/(-b*x^4+a)^2+1/32*(4*a*f+x *(7*b*c-a*g+2*(-a*h+3*b*d)*x+5*b*e*x^2))/a^2/b/(-b*x^4+a)+1/16*(-a*h+3*b*d )*arctanh(x^2*b^(1/2)/a^(1/2))/a^(5/2)/b^(3/2)+1/64*arctan(b^(1/4)*x/a^(1/ 4))*(21*b*c-3*a*g-5*e*a^(1/2)*b^(1/2))/a^(11/4)/b^(5/4)+1/64*arctanh(b^(1/ 4)*x/a^(1/4))*(21*b*c-3*a*g+5*e*a^(1/2)*b^(1/2))/a^(11/4)/b^(5/4)
Time = 0.27 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.28 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=\frac {\frac {4 a^{3/4} \sqrt {b} x (7 b c+b x (6 d+5 e x)-a (g+2 h x))}{a-b x^4}+\frac {16 a^{7/4} \sqrt {b} (b x (c+x (d+e x))+a (f+x (g+h x)))}{\left (a-b x^4\right )^2}+2 \sqrt [4]{b} \left (21 b c-5 \sqrt {a} \sqrt {b} e-3 a g\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )+\left (-21 b^{5/4} c-12 \sqrt [4]{a} b d-5 \sqrt {a} b^{3/4} e+3 a \sqrt [4]{b} g+4 a^{5/4} h\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right )+\left (21 b^{5/4} c-12 \sqrt [4]{a} b d+5 \sqrt {a} b^{3/4} e-3 a \sqrt [4]{b} g+4 a^{5/4} h\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )-4 \sqrt [4]{a} (-3 b d+a h) \log \left (\sqrt {a}+\sqrt {b} x^2\right )}{128 a^{11/4} b^{3/2}} \]
((4*a^(3/4)*Sqrt[b]*x*(7*b*c + b*x*(6*d + 5*e*x) - a*(g + 2*h*x)))/(a - b* x^4) + (16*a^(7/4)*Sqrt[b]*(b*x*(c + x*(d + e*x)) + a*(f + x*(g + h*x))))/ (a - b*x^4)^2 + 2*b^(1/4)*(21*b*c - 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTan[(b ^(1/4)*x)/a^(1/4)] + (-21*b^(5/4)*c - 12*a^(1/4)*b*d - 5*Sqrt[a]*b^(3/4)*e + 3*a*b^(1/4)*g + 4*a^(5/4)*h)*Log[a^(1/4) - b^(1/4)*x] + (21*b^(5/4)*c - 12*a^(1/4)*b*d + 5*Sqrt[a]*b^(3/4)*e - 3*a*b^(1/4)*g + 4*a^(5/4)*h)*Log[a ^(1/4) + b^(1/4)*x] - 4*a^(1/4)*(-3*b*d + a*h)*Log[Sqrt[a] + Sqrt[b]*x^2]) /(128*a^(11/4)*b^(3/2))
Time = 0.62 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2397, 25, 2393, 25, 2415, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx\) |
| \(\Big \downarrow \) 2397 |
| \(\displaystyle \frac {x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}-\frac {\int -\frac {4 b^2 f x^3+5 b^2 e x^2+2 b (3 b d-a h) x+b (7 b c-a g)}{\left (a-b x^4\right )^2}dx}{8 a b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {4 b^2 f x^3+5 b^2 e x^2+2 b (3 b d-a h) x+b (7 b c-a g)}{\left (a-b x^4\right )^2}dx}{8 a b^2}+\frac {x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}\) |
| \(\Big \downarrow \) 2393 |
| \(\displaystyle \frac {\frac {x \left (b (7 b c-a g)+2 b x (3 b d-a h)+5 b^2 e x^2\right )+4 a b f}{4 a \left (a-b x^4\right )}-\frac {\int -\frac {5 b^2 e x^2+4 b (3 b d-a h) x+3 b (7 b c-a g)}{a-b x^4}dx}{4 a}}{8 a b^2}+\frac {x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {5 b^2 e x^2+4 b (3 b d-a h) x+3 b (7 b c-a g)}{a-b x^4}dx}{4 a}+\frac {x \left (b (7 b c-a g)+2 b x (3 b d-a h)+5 b^2 e x^2\right )+4 a b f}{4 a \left (a-b x^4\right )}}{8 a b^2}+\frac {x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}\) |
| \(\Big \downarrow \) 2415 |
| \(\displaystyle \frac {\frac {\int \left (\frac {4 b (3 b d-a h) x}{a-b x^4}+\frac {5 b^2 e x^2+3 b (7 b c-a g)}{a-b x^4}\right )dx}{4 a}+\frac {x \left (b (7 b c-a g)+2 b x (3 b d-a h)+5 b^2 e x^2\right )+4 a b f}{4 a \left (a-b x^4\right )}}{8 a b^2}+\frac {x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {b^{3/4} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-5 \sqrt {a} \sqrt {b} e-3 a g+21 b c\right )}{2 a^{3/4}}+\frac {b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (5 \sqrt {a} \sqrt {b} e-3 a g+21 b c\right )}{2 a^{3/4}}+\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ) (3 b d-a h)}{\sqrt {a}}}{4 a}+\frac {x \left (b (7 b c-a g)+2 b x (3 b d-a h)+5 b^2 e x^2\right )+4 a b f}{4 a \left (a-b x^4\right )}}{8 a b^2}+\frac {x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}\) |
(x*(b*c + a*g + (b*d + a*h)*x + b*e*x^2 + b*f*x^3))/(8*a*b*(a - b*x^4)^2) + ((4*a*b*f + x*(b*(7*b*c - a*g) + 2*b*(3*b*d - a*h)*x + 5*b^2*e*x^2))/(4* a*(a - b*x^4)) + ((b^(3/4)*(21*b*c - 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTan[( b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)) + (b^(3/4)*(21*b*c + 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)) + (2*Sqrt[b]*(3*b*d - a *h)*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/Sqrt[a])/(4*a))/(8*a*b^2)
3.2.98.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a*Coeff[Pq, x, q] - b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q , x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) In t[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^( p + 1), x], x] /; q == n - 1] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n , 0] && LtQ[p, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x]}, S imp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) Int[(a + b*x^n)^(p + 1)* ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff [Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1 }]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.54 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.72
| method | result | size |
| risch | \(\frac {-\frac {5 b e \,x^{7}}{32 a^{2}}+\frac {\left (a h -3 b d \right ) x^{6}}{16 a^{2}}+\frac {\left (a g -7 b c \right ) x^{5}}{32 a^{2}}+\frac {9 e \,x^{3}}{32 a}+\frac {\left (a h +5 b d \right ) x^{2}}{16 a b}+\frac {\left (3 a g +11 b c \right ) x}{32 a b}+\frac {f}{8 b}}{\left (-b \,x^{4}+a \right )^{2}}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} b -a \right )}{\sum }\frac {\left (5 \textit {\_R}^{2} e -\frac {4 \left (a h -3 b d \right ) \textit {\_R}}{b}-\frac {3 \left (a g -7 b c \right )}{b}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{128 a^{2} b}\) | \(173\) |
| default | \(\frac {-\frac {5 b e \,x^{7}}{32 a^{2}}+\frac {\left (a h -3 b d \right ) x^{6}}{16 a^{2}}+\frac {\left (a g -7 b c \right ) x^{5}}{32 a^{2}}+\frac {9 e \,x^{3}}{32 a}+\frac {\left (a h +5 b d \right ) x^{2}}{16 a b}+\frac {\left (3 a g +11 b c \right ) x}{32 a b}+\frac {f}{8 b}}{\left (-b \,x^{4}+a \right )^{2}}+\frac {\frac {\left (-3 a g +21 b c \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {\left (-4 a h +12 b d \right ) \ln \left (\frac {a +x^{2} \sqrt {a b}}{a -x^{2} \sqrt {a b}}\right )}{4 \sqrt {a b}}-\frac {5 e \left (2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{32 a^{2} b}\) | \(267\) |
(-5/32*b*e/a^2*x^7+1/16*(a*h-3*b*d)/a^2*x^6+1/32*(a*g-7*b*c)/a^2*x^5+9/32/ a*e*x^3+1/16*(a*h+5*b*d)/a/b*x^2+1/32*(3*a*g+11*b*c)/a/b*x+1/8*f/b)/(-b*x^ 4+a)^2-1/128/a^2/b*sum((5*_R^2*e-4/b*(a*h-3*b*d)*_R-3/b*(a*g-7*b*c))/_R^3* ln(x-_R),_R=RootOf(_Z^4*b-a))
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.31 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=-\frac {5 \, b^{2} e x^{7} + 2 \, {\left (3 \, b^{2} d - a b h\right )} x^{6} - 9 \, a b e x^{3} + {\left (7 \, b^{2} c - a b g\right )} x^{5} - 4 \, a^{2} f - 2 \, {\left (5 \, a b d + a^{2} h\right )} x^{2} - {\left (11 \, a b c + 3 \, a^{2} g\right )} x}{32 \, {\left (a^{2} b^{3} x^{8} - 2 \, a^{3} b^{2} x^{4} + a^{4} b\right )}} + \frac {\frac {4 \, {\left (3 \, b d - a h\right )} \log \left (\sqrt {b} x^{2} + \sqrt {a}\right )}{\sqrt {a} \sqrt {b}} - \frac {4 \, {\left (3 \, b d - a h\right )} \log \left (\sqrt {b} x^{2} - \sqrt {a}\right )}{\sqrt {a} \sqrt {b}} + \frac {2 \, {\left (21 \, b^{\frac {3}{2}} c - 5 \, \sqrt {a} b e - 3 \, a \sqrt {b} g\right )} \arctan \left (\frac {\sqrt {b} x}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {{\left (21 \, b^{\frac {3}{2}} c + 5 \, \sqrt {a} b e - 3 \, a \sqrt {b} g\right )} \log \left (\frac {\sqrt {b} x - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} x + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}}{128 \, a^{2} b} \]
-1/32*(5*b^2*e*x^7 + 2*(3*b^2*d - a*b*h)*x^6 - 9*a*b*e*x^3 + (7*b^2*c - a* b*g)*x^5 - 4*a^2*f - 2*(5*a*b*d + a^2*h)*x^2 - (11*a*b*c + 3*a^2*g)*x)/(a^ 2*b^3*x^8 - 2*a^3*b^2*x^4 + a^4*b) + 1/128*(4*(3*b*d - a*h)*log(sqrt(b)*x^ 2 + sqrt(a))/(sqrt(a)*sqrt(b)) - 4*(3*b*d - a*h)*log(sqrt(b)*x^2 - sqrt(a) )/(sqrt(a)*sqrt(b)) + 2*(21*b^(3/2)*c - 5*sqrt(a)*b*e - 3*a*sqrt(b)*g)*arc tan(sqrt(b)*x/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b )) - (21*b^(3/2)*c + 5*sqrt(a)*b*e - 3*a*sqrt(b)*g)*log((sqrt(b)*x - sqrt( sqrt(a)*sqrt(b)))/(sqrt(b)*x + sqrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqrt( a)*sqrt(b))*sqrt(b)))/(a^2*b)
Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (201) = 402\).
Time = 0.28 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.80 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=-\frac {\sqrt {2} {\left (21 \, b^{2} c - 3 \, a b g - 12 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} b d + 4 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} a h + 5 \, \sqrt {-a b} b e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{128 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{2}} - \frac {\sqrt {2} {\left (21 \, b^{2} c - 3 \, a b g + 12 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} b d - 4 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} a h - 5 \, \sqrt {-a b} b e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{128 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{2}} - \frac {\sqrt {2} {\left (21 \, b^{2} c - 3 \, a b g - 5 \, \sqrt {-a b} b e\right )} \log \left (x^{2} + \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{256 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{2}} + \frac {\sqrt {2} {\left (21 \, b^{2} c - 3 \, a b g - 5 \, \sqrt {-a b} b e\right )} \log \left (x^{2} - \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{256 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{2}} - \frac {5 \, b^{2} e x^{7} + 6 \, b^{2} d x^{6} - 2 \, a b h x^{6} + 7 \, b^{2} c x^{5} - a b g x^{5} - 9 \, a b e x^{3} - 10 \, a b d x^{2} - 2 \, a^{2} h x^{2} - 11 \, a b c x - 3 \, a^{2} g x - 4 \, a^{2} f}{32 \, {\left (b x^{4} - a\right )}^{2} a^{2} b} \]
-1/128*sqrt(2)*(21*b^2*c - 3*a*b*g - 12*sqrt(2)*(-a*b^3)^(1/4)*b*d + 4*sqr t(2)*(-a*b^3)^(1/4)*a*h + 5*sqrt(-a*b)*b*e)*arctan(1/2*sqrt(2)*(2*x + sqrt (2)*(-a/b)^(1/4))/(-a/b)^(1/4))/((-a*b^3)^(3/4)*a^2) - 1/128*sqrt(2)*(21*b ^2*c - 3*a*b*g + 12*sqrt(2)*(-a*b^3)^(1/4)*b*d - 4*sqrt(2)*(-a*b^3)^(1/4)* a*h - 5*sqrt(-a*b)*b*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(-a/b)^(1/4))/(- a/b)^(1/4))/((-a*b^3)^(3/4)*a^2) - 1/256*sqrt(2)*(21*b^2*c - 3*a*b*g - 5*s qrt(-a*b)*b*e)*log(x^2 + sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/((-a*b^3)^(3 /4)*a^2) + 1/256*sqrt(2)*(21*b^2*c - 3*a*b*g - 5*sqrt(-a*b)*b*e)*log(x^2 - sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/((-a*b^3)^(3/4)*a^2) - 1/32*(5*b^2*e *x^7 + 6*b^2*d*x^6 - 2*a*b*h*x^6 + 7*b^2*c*x^5 - a*b*g*x^5 - 9*a*b*e*x^3 - 10*a*b*d*x^2 - 2*a^2*h*x^2 - 11*a*b*c*x - 3*a^2*g*x - 4*a^2*f)/((b*x^4 - a)^2*a^2*b)
Time = 9.89 (sec) , antiderivative size = 1687, normalized size of antiderivative = 7.00 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^3} \, dx=\text {Too large to display} \]
(f/(8*b) + (9*e*x^3)/(32*a) - (x^5*(7*b*c - a*g))/(32*a^2) - (x^6*(3*b*d - a*h))/(16*a^2) + (x*(11*b*c + 3*a*g))/(32*a*b) + (x^2*(5*b*d + a*h))/(16* a*b) - (5*b*e*x^7)/(32*a^2))/(a^2 + b^2*x^8 - 2*a*b*x^4) + symsum(log(- ro ot(268435456*a^11*b^6*z^4 + 3145728*a^7*b^4*d*h*z^2 + 983040*a^7*b^4*e*g*z ^2 - 6881280*a^6*b^5*c*e*z^2 - 524288*a^8*b^3*h^2*z^2 - 4718592*a^6*b^5*d^ 2*z^2 + 258048*a^5*b^3*c*g*h*z - 774144*a^4*b^4*c*d*g*z - 18432*a^6*b^2*g^ 2*h*z - 51200*a^5*b^3*e^2*h*z - 903168*a^4*b^4*c^2*h*z + 55296*a^5*b^3*d*g ^2*z + 153600*a^4*b^4*d*e^2*z + 2709504*a^3*b^5*c^2*d*z - 5760*a^3*b^2*d*e *g*h + 40320*a^2*b^3*c*d*e*h + 8640*a^2*b^3*d^2*e*g - 6720*a^3*b^2*c*e*h^2 - 6300*a^2*b^3*c*e^2*g + 960*a^4*b*e*g*h^2 - 60480*a*b^4*c*d^2*e - 3072*a ^4*b*d*h^3 + 111132*a*b^4*c^3*g + 13824*a^3*b^2*d^2*h^2 + 450*a^3*b^2*e^2* g^2 - 23814*a^2*b^3*c^2*g^2 - 27648*a^2*b^3*d^3*h + 2268*a^3*b^2*c*g^3 + 2 2050*a*b^4*c^2*e^2 - 625*a^2*b^3*e^4 - 81*a^4*b*g^4 + 20736*a*b^4*d^4 + 25 6*a^5*h^4 - 194481*b^5*c^4, z, k)*(root(268435456*a^11*b^6*z^4 + 3145728*a ^7*b^4*d*h*z^2 + 983040*a^7*b^4*e*g*z^2 - 6881280*a^6*b^5*c*e*z^2 - 524288 *a^8*b^3*h^2*z^2 - 4718592*a^6*b^5*d^2*z^2 + 258048*a^5*b^3*c*g*h*z - 7741 44*a^4*b^4*c*d*g*z - 18432*a^6*b^2*g^2*h*z - 51200*a^5*b^3*e^2*h*z - 90316 8*a^4*b^4*c^2*h*z + 55296*a^5*b^3*d*g^2*z + 153600*a^4*b^4*d*e^2*z + 27095 04*a^3*b^5*c^2*d*z - 5760*a^3*b^2*d*e*g*h + 40320*a^2*b^3*c*d*e*h + 8640*a ^2*b^3*d^2*e*g - 6720*a^3*b^2*c*e*h^2 - 6300*a^2*b^3*c*e^2*g + 960*a^4*...