Integrand size = 21, antiderivative size = 211 \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}+\frac {x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a-b x^4\right )^2}+\frac {x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a-b x^4\right )}+\frac {\left (77 \sqrt {b} c-15 \sqrt {a} e\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 a^{15/4} b^{3/4}}+\frac {\left (77 \sqrt {b} c+15 \sqrt {a} e\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 a^{15/4} b^{3/4}}+\frac {5 d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} \sqrt {b}} \]
1/12*x*(e*x^2+d*x+c)/a/(-b*x^4+a)^3+1/96*x*(9*e*x^2+10*d*x+11*c)/a^2/(-b*x ^4+a)^2+1/384*x*(45*e*x^2+60*d*x+77*c)/a^3/(-b*x^4+a)+5/32*d*arctanh(x^2*b ^(1/2)/a^(1/2))/a^(7/2)/b^(1/2)+1/256*arctan(b^(1/4)*x/a^(1/4))*(-15*e*a^( 1/2)+77*c*b^(1/2))/a^(15/4)/b^(3/4)+1/256*arctanh(b^(1/4)*x/a^(1/4))*(15*e *a^(1/2)+77*c*b^(1/2))/a^(15/4)/b^(3/4)
Time = 0.26 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.31 \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=\frac {\frac {128 a^3 x (c+x (d+e x))}{\left (a-b x^4\right )^3}+\frac {4 a x (77 c+15 x (4 d+3 e x))}{a-b x^4}+\frac {16 a^2 x (11 c+x (10 d+9 e x))}{\left (a-b x^4\right )^2}+\frac {6 \sqrt [4]{a} \left (77 \sqrt {b} c-15 \sqrt {a} e\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{b^{3/4}}-\frac {3 \left (77 \sqrt [4]{a} \sqrt {b} c+40 \sqrt {a} \sqrt [4]{b} d+15 a^{3/4} e\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right )}{b^{3/4}}+\frac {3 \left (77 \sqrt [4]{a} \sqrt {b} c-40 \sqrt {a} \sqrt [4]{b} d+15 a^{3/4} e\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )}{b^{3/4}}+\frac {120 \sqrt {a} d \log \left (\sqrt {a}+\sqrt {b} x^2\right )}{\sqrt {b}}}{1536 a^4} \]
((128*a^3*x*(c + x*(d + e*x)))/(a - b*x^4)^3 + (4*a*x*(77*c + 15*x*(4*d + 3*e*x)))/(a - b*x^4) + (16*a^2*x*(11*c + x*(10*d + 9*e*x)))/(a - b*x^4)^2 + (6*a^(1/4)*(77*Sqrt[b]*c - 15*Sqrt[a]*e)*ArcTan[(b^(1/4)*x)/a^(1/4)])/b^ (3/4) - (3*(77*a^(1/4)*Sqrt[b]*c + 40*Sqrt[a]*b^(1/4)*d + 15*a^(3/4)*e)*Lo g[a^(1/4) - b^(1/4)*x])/b^(3/4) + (3*(77*a^(1/4)*Sqrt[b]*c - 40*Sqrt[a]*b^ (1/4)*d + 15*a^(3/4)*e)*Log[a^(1/4) + b^(1/4)*x])/b^(3/4) + (120*Sqrt[a]*d *Log[Sqrt[a] + Sqrt[b]*x^2])/Sqrt[b])/(1536*a^4)
Time = 0.50 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2394, 25, 2394, 25, 2394, 27, 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.
| \(\displaystyle \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}-\frac {\int -\frac {9 e x^2+10 d x+11 c}{\left (a-b x^4\right )^3}dx}{12 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {9 e x^2+10 d x+11 c}{\left (a-b x^4\right )^3}dx}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {\frac {x \left (11 c+10 d x+9 e x^2\right )}{8 a \left (a-b x^4\right )^2}-\frac {\int -\frac {45 e x^2+60 d x+77 c}{\left (a-b x^4\right )^2}dx}{8 a}}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {45 e x^2+60 d x+77 c}{\left (a-b x^4\right )^2}dx}{8 a}+\frac {x \left (11 c+10 d x+9 e x^2\right )}{8 a \left (a-b x^4\right )^2}}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {\frac {\frac {x \left (77 c+60 d x+45 e x^2\right )}{4 a \left (a-b x^4\right )}-\frac {\int -\frac {3 \left (15 e x^2+40 d x+77 c\right )}{a-b x^4}dx}{4 a}}{8 a}+\frac {x \left (11 c+10 d x+9 e x^2\right )}{8 a \left (a-b x^4\right )^2}}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {15 e x^2+40 d x+77 c}{a-b x^4}dx}{4 a}+\frac {x \left (77 c+60 d x+45 e x^2\right )}{4 a \left (a-b x^4\right )}}{8 a}+\frac {x \left (11 c+10 d x+9 e x^2\right )}{8 a \left (a-b x^4\right )^2}}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
| \(\Big \downarrow \) 2415 |
| \(\displaystyle \frac {\frac {\frac {3 \int \left (\frac {40 d x}{a-b x^4}+\frac {15 e x^2+77 c}{a-b x^4}\right )dx}{4 a}+\frac {x \left (77 c+60 d x+45 e x^2\right )}{4 a \left (a-b x^4\right )}}{8 a}+\frac {x \left (11 c+10 d x+9 e x^2\right )}{8 a \left (a-b x^4\right )^2}}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (77 \sqrt {b} c-15 \sqrt {a} e\right )}{2 a^{3/4} b^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (15 \sqrt {a} e+77 \sqrt {b} c\right )}{2 a^{3/4} b^{3/4}}+\frac {20 d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}\right )}{4 a}+\frac {x \left (77 c+60 d x+45 e x^2\right )}{4 a \left (a-b x^4\right )}}{8 a}+\frac {x \left (11 c+10 d x+9 e x^2\right )}{8 a \left (a-b x^4\right )^2}}{12 a}+\frac {x \left (c+d x+e x^2\right )}{12 a \left (a-b x^4\right )^3}\) |
(x*(c + d*x + e*x^2))/(12*a*(a - b*x^4)^3) + ((x*(11*c + 10*d*x + 9*e*x^2) )/(8*a*(a - b*x^4)^2) + ((x*(77*c + 60*d*x + 45*e*x^2))/(4*a*(a - b*x^4)) + (3*(((77*Sqrt[b]*c - 15*Sqrt[a]*e)*ArcTan[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/ 4)*b^(3/4)) + ((77*Sqrt[b]*c + 15*Sqrt[a]*e)*ArcTanh[(b^(1/4)*x)/a^(1/4)]) /(2*a^(3/4)*b^(3/4)) + (20*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/(Sqrt[a]*Sqrt [b])))/(4*a))/(8*a))/(12*a)
3.2.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b *x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) Int[ExpandToSum[n *(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x ] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 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 = 1.50 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(\frac {\frac {15 e \,b^{2} x^{11}}{128 a^{3}}+\frac {5 d \,b^{2} x^{10}}{32 a^{3}}+\frac {77 c \,b^{2} x^{9}}{384 a^{3}}-\frac {21 b e \,x^{7}}{64 a^{2}}-\frac {5 b d \,x^{6}}{12 a^{2}}-\frac {33 b c \,x^{5}}{64 a^{2}}+\frac {113 e \,x^{3}}{384 a}+\frac {11 d \,x^{2}}{32 a}+\frac {51 c x}{128 a}}{\left (-b \,x^{4}+a \right )^{3}}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} b -a \right )}{\sum }\frac {\left (15 \textit {\_R}^{2} e +40 \textit {\_R} d +77 c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{512 a^{3} b}\) | \(150\) |
| default | \(\frac {\frac {15 e \,b^{2} x^{11}}{128 a^{3}}+\frac {5 d \,b^{2} x^{10}}{32 a^{3}}+\frac {77 c \,b^{2} x^{9}}{384 a^{3}}-\frac {21 b e \,x^{7}}{64 a^{2}}-\frac {5 b d \,x^{6}}{12 a^{2}}-\frac {33 b c \,x^{5}}{64 a^{2}}+\frac {113 e \,x^{3}}{384 a}+\frac {11 d \,x^{2}}{32 a}+\frac {51 c x}{128 a}}{\left (-b \,x^{4}+a \right )^{3}}+\frac {\frac {77 c \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {10 d \ln \left (\frac {a +x^{2} \sqrt {a b}}{a -x^{2} \sqrt {a b}}\right )}{\sqrt {a b}}-\frac {15 e \left (2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{128 a^{3}}\) | \(248\) |
(15/128*e/a^3*b^2*x^11+5/32*d/a^3*b^2*x^10+77/384*c/a^3*b^2*x^9-21/64*b*e/ a^2*x^7-5/12*b*d/a^2*x^6-33/64*b*c/a^2*x^5+113/384/a*e*x^3+11/32*d/a*x^2+5 1/128*c/a*x)/(-b*x^4+a)^3-1/512/a^3/b*sum((15*_R^2*e+40*_R*d+77*c)/_R^3*ln (x-_R),_R=RootOf(_Z^4*b-a))
Result contains complex when optimal does not.
Time = 9.18 (sec) , antiderivative size = 118903, normalized size of antiderivative = 563.52 \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.32 \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=-\frac {45 \, b^{2} e x^{11} + 60 \, b^{2} d x^{10} + 77 \, b^{2} c x^{9} - 126 \, a b e x^{7} - 160 \, a b d x^{6} - 198 \, a b c x^{5} + 113 \, a^{2} e x^{3} + 132 \, a^{2} d x^{2} + 153 \, a^{2} c x}{384 \, {\left (a^{3} b^{3} x^{12} - 3 \, a^{4} b^{2} x^{8} + 3 \, a^{5} b x^{4} - a^{6}\right )}} + \frac {\frac {40 \, d \log \left (\sqrt {b} x^{2} + \sqrt {a}\right )}{\sqrt {a} \sqrt {b}} - \frac {40 \, d \log \left (\sqrt {b} x^{2} - \sqrt {a}\right )}{\sqrt {a} \sqrt {b}} + \frac {2 \, {\left (77 \, \sqrt {b} c - 15 \, \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {b} x}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {{\left (77 \, \sqrt {b} c + 15 \, \sqrt {a} e\right )} \log \left (\frac {\sqrt {b} x - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} x + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}}{512 \, a^{3}} \]
-1/384*(45*b^2*e*x^11 + 60*b^2*d*x^10 + 77*b^2*c*x^9 - 126*a*b*e*x^7 - 160 *a*b*d*x^6 - 198*a*b*c*x^5 + 113*a^2*e*x^3 + 132*a^2*d*x^2 + 153*a^2*c*x)/ (a^3*b^3*x^12 - 3*a^4*b^2*x^8 + 3*a^5*b*x^4 - a^6) + 1/512*(40*d*log(sqrt( b)*x^2 + sqrt(a))/(sqrt(a)*sqrt(b)) - 40*d*log(sqrt(b)*x^2 - sqrt(a))/(sqr t(a)*sqrt(b)) + 2*(77*sqrt(b)*c - 15*sqrt(a)*e)*arctan(sqrt(b)*x/sqrt(sqrt (a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - (77*sqrt(b)*c + 15 *sqrt(a)*e)*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^3
Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (170) = 340\).
Time = 0.29 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.75 \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=-\frac {\sqrt {2} {\left (77 \, b^{2} c - 40 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} b d + 15 \, \sqrt {-a b} b e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{512 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{3}} - \frac {\sqrt {2} {\left (77 \, b^{2} c + 40 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} b d - 15 \, \sqrt {-a b} b e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{512 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{3}} - \frac {\sqrt {2} {\left (77 \, b^{2} c - 15 \, \sqrt {-a b} b e\right )} \log \left (x^{2} + \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{1024 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{3}} + \frac {\sqrt {2} {\left (77 \, b^{2} c - 15 \, \sqrt {-a b} b e\right )} \log \left (x^{2} - \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{1024 \, \left (-a b^{3}\right )^{\frac {3}{4}} a^{3}} - \frac {45 \, b^{2} e x^{11} + 60 \, b^{2} d x^{10} + 77 \, b^{2} c x^{9} - 126 \, a b e x^{7} - 160 \, a b d x^{6} - 198 \, a b c x^{5} + 113 \, a^{2} e x^{3} + 132 \, a^{2} d x^{2} + 153 \, a^{2} c x}{384 \, {\left (b x^{4} - a\right )}^{3} a^{3}} \]
-1/512*sqrt(2)*(77*b^2*c - 40*sqrt(2)*(-a*b^3)^(1/4)*b*d + 15*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^3) - 1/512*sqrt(2)*(77*b^2*c + 40*sqrt(2)*(-a*b^3)^(1/4)*b*d - 1 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^3) - 1/1024*sqrt(2)*(77*b^2*c - 15*sqrt(-a*b)*b*e) *log(x^2 + sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/((-a*b^3)^(3/4)*a^3) + 1/1 024*sqrt(2)*(77*b^2*c - 15*sqrt(-a*b)*b*e)*log(x^2 - sqrt(2)*x*(-a/b)^(1/4 ) + sqrt(-a/b))/((-a*b^3)^(3/4)*a^3) - 1/384*(45*b^2*e*x^11 + 60*b^2*d*x^1 0 + 77*b^2*c*x^9 - 126*a*b*e*x^7 - 160*a*b*d*x^6 - 198*a*b*c*x^5 + 113*a^2 *e*x^3 + 132*a^2*d*x^2 + 153*a^2*c*x)/((b*x^4 - a)^3*a^3)
Time = 9.95 (sec) , antiderivative size = 874, normalized size of antiderivative = 4.14 \[ \int \frac {c+d x+e x^2}{\left (a-b x^4\right )^4} \, dx=\frac {\frac {11\,d\,x^2}{32\,a}+\frac {113\,e\,x^3}{384\,a}+\frac {51\,c\,x}{128\,a}+\frac {77\,b^2\,c\,x^9}{384\,a^3}+\frac {5\,b^2\,d\,x^{10}}{32\,a^3}+\frac {15\,b^2\,e\,x^{11}}{128\,a^3}-\frac {33\,b\,c\,x^5}{64\,a^2}-\frac {5\,b\,d\,x^6}{12\,a^2}-\frac {21\,b\,e\,x^7}{64\,a^2}}{a^3-3\,a^2\,b\,x^4+3\,a\,b^2\,x^8-b^3\,x^{12}}+\left (\sum _{k=1}^4\ln \left (-\frac {b\,\left (3375\,a\,e^3+123200\,b\,c\,d^2-88935\,b\,c^2\,e+64000\,b\,d^3\,x+{\mathrm {root}\left (68719476736\,a^{15}\,b^3\,z^4-1211105280\,a^8\,b^2\,c\,e\,z^2-838860800\,a^8\,b^2\,d^2\,z^2+485703680\,a^4\,b^2\,c^2\,d\,z+18432000\,a^5\,b\,d\,e^2\,z-7392000\,a\,b\,c\,d^2\,e+2668050\,a\,b\,c^2\,e^2+2560000\,a\,b\,d^4-35153041\,b^2\,c^4-50625\,a^2\,e^4,z,k\right )}^2\,a^7\,b^2\,c\,20185088+\mathrm {root}\left (68719476736\,a^{15}\,b^3\,z^4-1211105280\,a^8\,b^2\,c\,e\,z^2-838860800\,a^8\,b^2\,d^2\,z^2+485703680\,a^4\,b^2\,c^2\,d\,z+18432000\,a^5\,b\,d\,e^2\,z-7392000\,a\,b\,c\,d^2\,e+2668050\,a\,b\,c^2\,e^2+2560000\,a\,b\,d^4-35153041\,b^2\,c^4-50625\,a^2\,e^4,z,k\right )\,a^4\,b\,e^2\,x\,115200-92400\,b\,c\,d\,e\,x+\mathrm {root}\left (68719476736\,a^{15}\,b^3\,z^4-1211105280\,a^8\,b^2\,c\,e\,z^2-838860800\,a^8\,b^2\,d^2\,z^2+485703680\,a^4\,b^2\,c^2\,d\,z+18432000\,a^5\,b\,d\,e^2\,z-7392000\,a\,b\,c\,d^2\,e+2668050\,a\,b\,c^2\,e^2+2560000\,a\,b\,d^4-35153041\,b^2\,c^4-50625\,a^2\,e^4,z,k\right )\,a^3\,b^2\,c^2\,x\,3035648-{\mathrm {root}\left (68719476736\,a^{15}\,b^3\,z^4-1211105280\,a^8\,b^2\,c\,e\,z^2-838860800\,a^8\,b^2\,d^2\,z^2+485703680\,a^4\,b^2\,c^2\,d\,z+18432000\,a^5\,b\,d\,e^2\,z-7392000\,a\,b\,c\,d^2\,e+2668050\,a\,b\,c^2\,e^2+2560000\,a\,b\,d^4-35153041\,b^2\,c^4-50625\,a^2\,e^4,z,k\right )}^2\,a^7\,b^2\,d\,x\,10485760-\mathrm {root}\left (68719476736\,a^{15}\,b^3\,z^4-1211105280\,a^8\,b^2\,c\,e\,z^2-838860800\,a^8\,b^2\,d^2\,z^2+485703680\,a^4\,b^2\,c^2\,d\,z+18432000\,a^5\,b\,d\,e^2\,z-7392000\,a\,b\,c\,d^2\,e+2668050\,a\,b\,c^2\,e^2+2560000\,a\,b\,d^4-35153041\,b^2\,c^4-50625\,a^2\,e^4,z,k\right )\,a^4\,b\,d\,e\,614400\right )}{a^9\,2097152}\right )\,\mathrm {root}\left (68719476736\,a^{15}\,b^3\,z^4-1211105280\,a^8\,b^2\,c\,e\,z^2-838860800\,a^8\,b^2\,d^2\,z^2+485703680\,a^4\,b^2\,c^2\,d\,z+18432000\,a^5\,b\,d\,e^2\,z-7392000\,a\,b\,c\,d^2\,e+2668050\,a\,b\,c^2\,e^2+2560000\,a\,b\,d^4-35153041\,b^2\,c^4-50625\,a^2\,e^4,z,k\right )\right ) \]
((11*d*x^2)/(32*a) + (113*e*x^3)/(384*a) + (51*c*x)/(128*a) + (77*b^2*c*x^ 9)/(384*a^3) + (5*b^2*d*x^10)/(32*a^3) + (15*b^2*e*x^11)/(128*a^3) - (33*b *c*x^5)/(64*a^2) - (5*b*d*x^6)/(12*a^2) - (21*b*e*x^7)/(64*a^2))/(a^3 - b^ 3*x^12 - 3*a^2*b*x^4 + 3*a*b^2*x^8) + symsum(log(-(b*(3375*a*e^3 + 123200* b*c*d^2 - 88935*b*c^2*e + 64000*b*d^3*x + 20185088*root(68719476736*a^15*b ^3*z^4 - 1211105280*a^8*b^2*c*e*z^2 - 838860800*a^8*b^2*d^2*z^2 + 48570368 0*a^4*b^2*c^2*d*z + 18432000*a^5*b*d*e^2*z - 7392000*a*b*c*d^2*e + 2668050 *a*b*c^2*e^2 + 2560000*a*b*d^4 - 35153041*b^2*c^4 - 50625*a^2*e^4, z, k)^2 *a^7*b^2*c + 115200*root(68719476736*a^15*b^3*z^4 - 1211105280*a^8*b^2*c*e *z^2 - 838860800*a^8*b^2*d^2*z^2 + 485703680*a^4*b^2*c^2*d*z + 18432000*a^ 5*b*d*e^2*z - 7392000*a*b*c*d^2*e + 2668050*a*b*c^2*e^2 + 2560000*a*b*d^4 - 35153041*b^2*c^4 - 50625*a^2*e^4, z, k)*a^4*b*e^2*x - 92400*b*c*d*e*x + 3035648*root(68719476736*a^15*b^3*z^4 - 1211105280*a^8*b^2*c*e*z^2 - 83886 0800*a^8*b^2*d^2*z^2 + 485703680*a^4*b^2*c^2*d*z + 18432000*a^5*b*d*e^2*z - 7392000*a*b*c*d^2*e + 2668050*a*b*c^2*e^2 + 2560000*a*b*d^4 - 35153041*b ^2*c^4 - 50625*a^2*e^4, z, k)*a^3*b^2*c^2*x - 10485760*root(68719476736*a^ 15*b^3*z^4 - 1211105280*a^8*b^2*c*e*z^2 - 838860800*a^8*b^2*d^2*z^2 + 4857 03680*a^4*b^2*c^2*d*z + 18432000*a^5*b*d*e^2*z - 7392000*a*b*c*d^2*e + 266 8050*a*b*c^2*e^2 + 2560000*a*b*d^4 - 35153041*b^2*c^4 - 50625*a^2*e^4, z, k)^2*a^7*b^2*d*x - 614400*root(68719476736*a^15*b^3*z^4 - 1211105280*a^...