Integrand size = 19, antiderivative size = 336 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \left (c d^2+a e^2\right )}-\frac {\sqrt [4]{c} \left (\sqrt {c} d-\sqrt {a} e\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )}+\frac {\sqrt [4]{c} \left (\sqrt {c} d-\sqrt {a} e\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )}-\frac {\sqrt [4]{c} \left (\sqrt {c} d+\sqrt {a} e\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )}+\frac {\sqrt [4]{c} \left (\sqrt {c} d+\sqrt {a} e\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )} \]
1/4*c^(1/4)*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))*(-e*a^(1/2)+d*c^(1/2))/a^ (3/4)/(a*e^2+c*d^2)*2^(1/2)+1/4*c^(1/4)*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4) )*(-e*a^(1/2)+d*c^(1/2))/a^(3/4)/(a*e^2+c*d^2)*2^(1/2)-1/8*c^(1/4)*ln(-a^( 1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(e*a^(1/2)+d*c^(1/2))/a^(3/4)/ (a*e^2+c*d^2)*2^(1/2)+1/8*c^(1/4)*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2 *c^(1/2))*(e*a^(1/2)+d*c^(1/2))/a^(3/4)/(a*e^2+c*d^2)*2^(1/2)+e^(3/2)*arct an(x*e^(1/2)/d^(1/2))/(a*e^2+c*d^2)/d^(1/2)
Time = 0.10 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {8 a^{3/4} e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+\sqrt {2} \sqrt [4]{c} \sqrt {d} \left (\left (-2 \sqrt {c} d+2 \sqrt {a} e\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \left (\sqrt {c} d-\sqrt {a} e\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-\left (\sqrt {c} d+\sqrt {a} e\right ) \left (\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )-\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )\right )\right )}{8 a^{3/4} \sqrt {d} \left (c d^2+a e^2\right )} \]
(8*a^(3/4)*e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]] + Sqrt[2]*c^(1/4)*Sqrt[d]*( (-2*Sqrt[c]*d + 2*Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 2*( Sqrt[c]*d - Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - (Sqrt[c]* d + Sqrt[a]*e)*(Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] - L og[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])))/(8*a^(3/4)*Sqrt[d ]*(c*d^2 + a*e^2))
Time = 0.45 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1485, 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.
| \(\displaystyle \int \frac {1}{\left (a+c x^4\right ) \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1485 |
| \(\displaystyle \int \left (\frac {e^2}{\left (d+e x^2\right ) \left (a e^2+c d^2\right )}+\frac {c \left (d-e x^2\right )}{\left (a+c x^4\right ) \left (a e^2+c d^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (\sqrt {c} d-\sqrt {a} e\right )}{2 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )}+\frac {\sqrt [4]{c} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt {c} d-\sqrt {a} e\right )}{2 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )}-\frac {\sqrt [4]{c} \left (\sqrt {a} e+\sqrt {c} d\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )}+\frac {\sqrt [4]{c} \left (\sqrt {a} e+\sqrt {c} d\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \left (a e^2+c d^2\right )}\) |
(e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*(c*d^2 + a*e^2)) - (c^(1/4) *(Sqrt[c]*d - Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[ 2]*a^(3/4)*(c*d^2 + a*e^2)) + (c^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*(c*d^2 + a*e^2)) - (c^(1/ 4)*(Sqrt[c]*d + Sqrt[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[ c]*x^2])/(4*Sqrt[2]*a^(3/4)*(c*d^2 + a*e^2)) + (c^(1/4)*(Sqrt[c]*d + Sqrt[ a]*e)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a ^(3/4)*(c*d^2 + a*e^2))
3.3.41.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[Expa ndIntegrand[(d + e*x^2)^q/(a + c*x^4), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]
Time = 0.33 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {c \left (\frac {d \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}-\frac {e \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{a \,e^{2}+c \,d^{2}}+\frac {e^{2} \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {e d}}\) | \(253\) |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{5} e^{4}+2 a^{4} c \,d^{2} e^{2}+a^{3} c^{2} d^{4}\right ) \textit {\_Z}^{4}-4 a^{2} c d e \,\textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (-2 a^{5} e^{7}-2 a^{4} c \,d^{2} e^{5}+2 a^{3} c^{2} d^{4} e^{3}+2 a^{2} c^{3} d^{6} e \right ) \textit {\_R}^{4}+\left (15 a^{2} c d \,e^{4}-2 a \,c^{2} d^{3} e^{2}-c^{3} d^{5}\right ) \textit {\_R}^{2}-6 e^{3} c \right ) x +\left (4 a^{4} e^{6}+7 a^{3} c \,d^{2} e^{4}+2 a^{2} c^{2} d^{4} e^{2}-a \,c^{3} d^{6}\right ) \textit {\_R}^{3}+\left (-5 a c d \,e^{3}-c^{2} d^{3} e \right ) \textit {\_R} \right )\right )}{4}+\frac {\sqrt {-e d}\, e \ln \left (\left (-16 a^{2} d \,e^{5}-a c \,d^{3} e^{3}-c^{2} d^{5} e \right ) x +16 \left (-e d \right )^{\frac {3}{2}} a^{2} e^{3}-4 \left (-e d \right )^{\frac {3}{2}} a c \,d^{2} e -5 a c \,e^{2} \sqrt {-e d}\, d^{3}-c^{2} d^{5} \sqrt {-e d}\right )}{2 d \left (a \,e^{2}+c \,d^{2}\right )}-\frac {\sqrt {-e d}\, e \ln \left (\left (-16 a^{2} d \,e^{5}-a c \,d^{3} e^{3}-c^{2} d^{5} e \right ) x -16 \left (-e d \right )^{\frac {3}{2}} a^{2} e^{3}+4 \left (-e d \right )^{\frac {3}{2}} a c \,d^{2} e +5 a c \,e^{2} \sqrt {-e d}\, d^{3}+c^{2} d^{5} \sqrt {-e d}\right )}{2 d \left (a \,e^{2}+c \,d^{2}\right )}\) | \(453\) |
c/(a*e^2+c*d^2)*(1/8*d*(a/c)^(1/4)/a*2^(1/2)*(ln((x^2+(a/c)^(1/4)*x*2^(1/2 )+(a/c)^(1/2))/(x^2-(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2)))+2*arctan(2^(1/2)/( a/c)^(1/4)*x+1)+2*arctan(2^(1/2)/(a/c)^(1/4)*x-1))-1/8*e/c/(a/c)^(1/4)*2^( 1/2)*(ln((x^2-(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2))/(x^2+(a/c)^(1/4)*x*2^(1/2 )+(a/c)^(1/2)))+2*arctan(2^(1/2)/(a/c)^(1/4)*x+1)+2*arctan(2^(1/2)/(a/c)^( 1/4)*x-1)))+e^2/(a*e^2+c*d^2)/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 2030 vs. \(2 (247) = 494\).
Time = 0.68 (sec) , antiderivative size = 4084, normalized size of antiderivative = 12.15 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\text {Too large to display} \]
[-1/4*((c*d^2 + a*e^2)*sqrt((2*c*d*e + (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3* e^4)*sqrt(-(c^3*d^4 - 2*a*c^2*d^2*e^2 + a^2*c*e^4)/(a^3*c^4*d^8 + 4*a^4*c^ 3*d^6*e^2 + 6*a^5*c^2*d^4*e^4 + 4*a^6*c*d^2*e^6 + a^7*e^8)))/(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4))*log(-(c^2*d^2 - a*c*e^2)*x + (a*c^2*d^3 - a^2* c*d*e^2 + (a^3*c^2*d^4*e + 2*a^4*c*d^2*e^3 + a^5*e^5)*sqrt(-(c^3*d^4 - 2*a *c^2*d^2*e^2 + a^2*c*e^4)/(a^3*c^4*d^8 + 4*a^4*c^3*d^6*e^2 + 6*a^5*c^2*d^4 *e^4 + 4*a^6*c*d^2*e^6 + a^7*e^8)))*sqrt((2*c*d*e + (a*c^2*d^4 + 2*a^2*c*d ^2*e^2 + a^3*e^4)*sqrt(-(c^3*d^4 - 2*a*c^2*d^2*e^2 + a^2*c*e^4)/(a^3*c^4*d ^8 + 4*a^4*c^3*d^6*e^2 + 6*a^5*c^2*d^4*e^4 + 4*a^6*c*d^2*e^6 + a^7*e^8)))/ (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4))) - (c*d^2 + a*e^2)*sqrt((2*c*d*e + (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(-(c^3*d^4 - 2*a*c^2*d^2*e^2 + a^2*c*e^4)/(a^3*c^4*d^8 + 4*a^4*c^3*d^6*e^2 + 6*a^5*c^2*d^4*e^4 + 4*a^6 *c*d^2*e^6 + a^7*e^8)))/(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4))*log(-(c^2 *d^2 - a*c*e^2)*x - (a*c^2*d^3 - a^2*c*d*e^2 + (a^3*c^2*d^4*e + 2*a^4*c*d^ 2*e^3 + a^5*e^5)*sqrt(-(c^3*d^4 - 2*a*c^2*d^2*e^2 + a^2*c*e^4)/(a^3*c^4*d^ 8 + 4*a^4*c^3*d^6*e^2 + 6*a^5*c^2*d^4*e^4 + 4*a^6*c*d^2*e^6 + a^7*e^8)))*s qrt((2*c*d*e + (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(-(c^3*d^4 - 2* a*c^2*d^2*e^2 + a^2*c*e^4)/(a^3*c^4*d^8 + 4*a^4*c^3*d^6*e^2 + 6*a^5*c^2*d^ 4*e^4 + 4*a^6*c*d^2*e^6 + a^7*e^8)))/(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^ 4))) + (c*d^2 + a*e^2)*sqrt((2*c*d*e - (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a...
Timed out. \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.31 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {d e}} + \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d - \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a c^{3} d^{2} + \sqrt {2} a^{2} c^{2} e^{2}\right )}} + \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d - \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a c^{3} d^{2} + \sqrt {2} a^{2} c^{2} e^{2}\right )}} + \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d + \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{4 \, {\left (\sqrt {2} a c^{3} d^{2} + \sqrt {2} a^{2} c^{2} e^{2}\right )}} - \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d + \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{4 \, {\left (\sqrt {2} a c^{3} d^{2} + \sqrt {2} a^{2} c^{2} e^{2}\right )}} \]
e^2*arctan(e*x/sqrt(d*e))/((c*d^2 + a*e^2)*sqrt(d*e)) + 1/2*((a*c^3)^(1/4) *c^2*d - (a*c^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/( a/c)^(1/4))/(sqrt(2)*a*c^3*d^2 + sqrt(2)*a^2*c^2*e^2) + 1/2*((a*c^3)^(1/4) *c^2*d - (a*c^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/( a/c)^(1/4))/(sqrt(2)*a*c^3*d^2 + sqrt(2)*a^2*c^2*e^2) + 1/4*((a*c^3)^(1/4) *c^2*d + (a*c^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sq rt(2)*a*c^3*d^2 + sqrt(2)*a^2*c^2*e^2) - 1/4*((a*c^3)^(1/4)*c^2*d + (a*c^3 )^(3/4)*e)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a*c^3*d^2 + sqrt(2)*a^2*c^2*e^2)
Time = 8.52 (sec) , antiderivative size = 4802, normalized size of antiderivative = 14.29 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\text {Too large to display} \]
atan(((((a*e^2*(-a^3*c)^(1/2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*(a ^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2)))^(1/2)*(4*c^6*d^3*e^3 - (((a*e^2* (-a^3*c)^(1/2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*(a^5*e^4 + a^3*c^ 2*d^4 + 2*a^4*c*d^2*e^2)))^(1/2)*(256*a^4*c^4*e^8 + x*((a*e^2*(-a^3*c)^(1/ 2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*(a^5*e^4 + a^3*c^2*d^4 + 2*a^ 4*c*d^2*e^2)))^(1/2)*(512*a^5*c^4*e^9 - 512*a^2*c^7*d^6*e^3 - 512*a^3*c^6* d^4*e^5 + 512*a^4*c^5*d^2*e^7) - 64*a*c^7*d^6*e^2 + 128*a^2*c^6*d^4*e^4 + 448*a^3*c^5*d^2*e^6) + x*(16*c^7*d^5*e^2 + 32*a*c^6*d^3*e^4 - 240*a^2*c^5* d*e^6))*((a*e^2*(-a^3*c)^(1/2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*( a^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2)))^(1/2) + 20*a*c^5*d*e^5) - 6*c^5 *e^5*x)*((a*e^2*(-a^3*c)^(1/2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*( a^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2)))^(1/2)*1i - (((a*e^2*(-a^3*c)^(1 /2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*(a^5*e^4 + a^3*c^2*d^4 + 2*a ^4*c*d^2*e^2)))^(1/2)*(4*c^6*d^3*e^3 - (((a*e^2*(-a^3*c)^(1/2) - c*d^2*(-a ^3*c)^(1/2) + 2*a^2*c*d*e)/(16*(a^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2))) ^(1/2)*(256*a^4*c^4*e^8 - x*((a*e^2*(-a^3*c)^(1/2) - c*d^2*(-a^3*c)^(1/2) + 2*a^2*c*d*e)/(16*(a^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2)))^(1/2)*(512* a^5*c^4*e^9 - 512*a^2*c^7*d^6*e^3 - 512*a^3*c^6*d^4*e^5 + 512*a^4*c^5*d^2* e^7) - 64*a*c^7*d^6*e^2 + 128*a^2*c^6*d^4*e^4 + 448*a^3*c^5*d^2*e^6) - x*( 16*c^7*d^5*e^2 + 32*a*c^6*d^3*e^4 - 240*a^2*c^5*d*e^6))*((a*e^2*(-a^3*c...