Integrand size = 22, antiderivative size = 209 \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {c \left (d-e x^2\right )}{4 a \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac {\sqrt {c} e^3 \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (c d^2+a e^2\right )^2}-\frac {\sqrt {c} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \left (c d^2+a e^2\right )}+\frac {\log (x)}{a^2 d}-\frac {e^4 \log \left (d+e x^2\right )}{2 d \left (c d^2+a e^2\right )^2}-\frac {c d \left (c d^2+2 a e^2\right ) \log \left (a+c x^4\right )}{4 a^2 \left (c d^2+a e^2\right )^2} \]
1/4*c*(-e*x^2+d)/a/(a*e^2+c*d^2)/(c*x^4+a)+ln(x)/a^2/d-1/2*e^4*ln(e*x^2+d) /d/(a*e^2+c*d^2)^2-1/4*c*d*(2*a*e^2+c*d^2)*ln(c*x^4+a)/a^2/(a*e^2+c*d^2)^2 -1/4*e*arctan(x^2*c^(1/2)/a^(1/2))*c^(1/2)/a^(3/2)/(a*e^2+c*d^2)-1/2*e^3*a rctan(x^2*c^(1/2)/a^(1/2))*c^(1/2)/(a*e^2+c*d^2)^2/a^(1/2)
Time = 0.13 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {a c d \left (c d^2+a e^2\right ) \left (d-e x^2\right )+\sqrt {a} \sqrt {c} d e \left (c d^2+3 a e^2\right ) \left (a+c x^4\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+\sqrt {a} \sqrt {c} d e \left (c d^2+3 a e^2\right ) \left (a+c x^4\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+4 \left (c d^2+a e^2\right )^2 \left (a+c x^4\right ) \log (x)-2 a^2 e^4 \left (a+c x^4\right ) \log \left (d+e x^2\right )-c d^2 \left (c d^2+2 a e^2\right ) \left (a+c x^4\right ) \log \left (a+c x^4\right )}{4 a^2 d \left (c d^2+a e^2\right )^2 \left (a+c x^4\right )} \]
(a*c*d*(c*d^2 + a*e^2)*(d - e*x^2) + Sqrt[a]*Sqrt[c]*d*e*(c*d^2 + 3*a*e^2) *(a + c*x^4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + Sqrt[a]*Sqrt[c]*d*e *(c*d^2 + 3*a*e^2)*(a + c*x^4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 4 *(c*d^2 + a*e^2)^2*(a + c*x^4)*Log[x] - 2*a^2*e^4*(a + c*x^4)*Log[d + e*x^ 2] - c*d^2*(c*d^2 + 2*a*e^2)*(a + c*x^4)*Log[a + c*x^4])/(4*a^2*d*(c*d^2 + a*e^2)^2*(a + c*x^4))
Time = 0.40 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1579, 615, 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.
| \(\displaystyle \int \frac {1}{x \left (a+c x^4\right )^2 \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1579 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (e x^2+d\right ) \left (c x^4+a\right )^2}dx^2\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {1}{2} \int \left (-\frac {e^5}{d \left (c d^2+a e^2\right )^2 \left (e x^2+d\right )}+\frac {c \left (-a^2 e^3-c d \left (c d^2+2 a e^2\right ) x^2\right )}{a^2 \left (c d^2+a e^2\right )^2 \left (c x^4+a\right )}+\frac {1}{a^2 d x^2}-\frac {c \left (c d x^2+a e\right )}{a \left (c d^2+a e^2\right ) \left (c x^4+a\right )^2}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\sqrt {c} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )}-\frac {c d \left (2 a e^2+c d^2\right ) \log \left (a+c x^4\right )}{2 a^2 \left (a e^2+c d^2\right )^2}+\frac {\log \left (x^2\right )}{a^2 d}-\frac {\sqrt {c} e^3 \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2}+\frac {c \left (d-e x^2\right )}{2 a \left (a+c x^4\right ) \left (a e^2+c d^2\right )}-\frac {e^4 \log \left (d+e x^2\right )}{d \left (a e^2+c d^2\right )^2}\right )\) |
((c*(d - e*x^2))/(2*a*(c*d^2 + a*e^2)*(a + c*x^4)) - (Sqrt[c]*e^3*ArcTan[( Sqrt[c]*x^2)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*e^2)^2) - (Sqrt[c]*e*ArcTan[(Sq rt[c]*x^2)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)) + Log[x^2]/(a^2*d) - (e^4 *Log[d + e*x^2])/(d*(c*d^2 + a*e^2)^2) - (c*d*(c*d^2 + 2*a*e^2)*Log[a + c* x^4])/(2*a^2*(c*d^2 + a*e^2)^2))/2
3.3.49.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
Time = 0.46 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.82
| method | result | size |
| default | \(\frac {\ln \left (x \right )}{a^{2} d}-\frac {c \left (\frac {\left (\frac {1}{2} e^{3} a^{2}+\frac {1}{2} a c \,d^{2} e \right ) x^{2}-\frac {a d \left (a \,e^{2}+c \,d^{2}\right )}{2}}{c \,x^{4}+a}+\frac {\left (4 a c d \,e^{2}+2 c^{2} d^{3}\right ) \ln \left (c \,x^{4}+a \right )}{4 c}+\frac {\left (3 e^{3} a^{2}+a c \,d^{2} e \right ) \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} a^{2}}-\frac {e^{4} \ln \left (e \,x^{2}+d \right )}{2 d \left (a \,e^{2}+c \,d^{2}\right )^{2}}\) | \(171\) |
| risch | \(\frac {-\frac {e c \,x^{2}}{4 a \left (a \,e^{2}+c \,d^{2}\right )}+\frac {c d}{4 a \left (a \,e^{2}+c \,d^{2}\right )}}{c \,x^{4}+a}+\frac {\ln \left (x \right )}{a^{2} d}-\frac {e^{4} \ln \left (e \,x^{2}+d \right )}{2 d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{6} e^{4}+2 d^{2} a^{5} c \,e^{2}+a^{4} d^{4} c^{2}\right ) \textit {\_Z}^{2}+\left (8 a^{3} c d \,e^{2}+4 a^{2} d^{3} c^{2}\right ) \textit {\_Z} +9 e^{2} a c +4 c^{2} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-6 a^{8} e^{8}-19 a^{7} c \,d^{2} e^{6}-25 a^{6} c^{2} d^{4} e^{4}-17 a^{5} c^{3} d^{6} e^{2}-5 a^{4} c^{4} d^{8}\right ) \textit {\_R}^{3}+\left (-52 a^{5} c d \,e^{6}-102 a^{4} c^{2} d^{3} e^{4}-60 a^{3} c^{3} d^{5} e^{2}-10 a^{2} c^{4} d^{7}\right ) \textit {\_R}^{2}+\left (-86 a^{3} c \,e^{6}-56 a^{2} c^{2} d^{2} e^{4}-36 a \,c^{3} d^{4} e^{2}\right ) \textit {\_R} +64 c^{2} d \,e^{4}\right ) x^{2}+\left (-2 a^{8} d \,e^{7}-2 a^{7} c \,d^{3} e^{5}+2 a^{6} c^{2} d^{5} e^{3}+2 d^{7} a^{5} c^{3} e \right ) \textit {\_R}^{3}+\left (16 a^{6} e^{7}+11 a^{5} c \,d^{2} e^{5}-18 a^{4} c^{2} d^{4} e^{3}-13 a^{3} d^{6} c^{3} e \right ) \textit {\_R}^{2}+\left (34 a^{3} c d \,e^{5}+8 a^{2} c^{2} d^{3} e^{3}-24 a \,c^{3} d^{5} e \right ) \textit {\_R} +48 a c \,e^{5}+64 c^{2} d^{2} e^{3}\right )\right )}{8}\) | \(520\) |
ln(x)/a^2/d-1/2*c/(a*e^2+c*d^2)^2/a^2*(((1/2*e^3*a^2+1/2*a*c*d^2*e)*x^2-1/ 2*a*d*(a*e^2+c*d^2))/(c*x^4+a)+1/4*(4*a*c*d*e^2+2*c^2*d^3)/c*ln(c*x^4+a)+1 /2*(3*a^2*e^3+a*c*d^2*e)/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2)))-1/2*e^4*ln (e*x^2+d)/d/(a*e^2+c*d^2)^2
Time = 74.98 (sec) , antiderivative size = 686, normalized size of antiderivative = 3.28 \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\left [\frac {2 \, a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} - 2 \, {\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{2} + {\left (a^{2} c d^{3} e + 3 \, a^{3} d e^{3} + {\left (a c^{2} d^{3} e + 3 \, a^{2} c d e^{3}\right )} x^{4}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) - 2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2}\right )} x^{4}\right )} \log \left (c x^{4} + a\right ) - 4 \, {\left (a^{2} c e^{4} x^{4} + a^{3} e^{4}\right )} \log \left (e x^{2} + d\right ) + 8 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} + {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{4}\right )} \log \left (x\right )}{8 \, {\left (a^{3} c^{2} d^{5} + 2 \, a^{4} c d^{3} e^{2} + a^{5} d e^{4} + {\left (a^{2} c^{3} d^{5} + 2 \, a^{3} c^{2} d^{3} e^{2} + a^{4} c d e^{4}\right )} x^{4}\right )}}, \frac {a c^{2} d^{4} + a^{2} c d^{2} e^{2} - {\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{2} + {\left (a^{2} c d^{3} e + 3 \, a^{3} d e^{3} + {\left (a c^{2} d^{3} e + 3 \, a^{2} c d e^{3}\right )} x^{4}\right )} \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) - {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2}\right )} x^{4}\right )} \log \left (c x^{4} + a\right ) - 2 \, {\left (a^{2} c e^{4} x^{4} + a^{3} e^{4}\right )} \log \left (e x^{2} + d\right ) + 4 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} + {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{3} c^{2} d^{5} + 2 \, a^{4} c d^{3} e^{2} + a^{5} d e^{4} + {\left (a^{2} c^{3} d^{5} + 2 \, a^{3} c^{2} d^{3} e^{2} + a^{4} c d e^{4}\right )} x^{4}\right )}}\right ] \]
[1/8*(2*a*c^2*d^4 + 2*a^2*c*d^2*e^2 - 2*(a*c^2*d^3*e + a^2*c*d*e^3)*x^2 + (a^2*c*d^3*e + 3*a^3*d*e^3 + (a*c^2*d^3*e + 3*a^2*c*d*e^3)*x^4)*sqrt(-c/a) *log((c*x^4 - 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) - 2*(a*c^2*d^4 + 2*a^2* c*d^2*e^2 + (c^3*d^4 + 2*a*c^2*d^2*e^2)*x^4)*log(c*x^4 + a) - 4*(a^2*c*e^4 *x^4 + a^3*e^4)*log(e*x^2 + d) + 8*(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4 + (c^3*d^4 + 2*a*c^2*d^2*e^2 + a^2*c*e^4)*x^4)*log(x))/(a^3*c^2*d^5 + 2*a^ 4*c*d^3*e^2 + a^5*d*e^4 + (a^2*c^3*d^5 + 2*a^3*c^2*d^3*e^2 + a^4*c*d*e^4)* x^4), 1/4*(a*c^2*d^4 + a^2*c*d^2*e^2 - (a*c^2*d^3*e + a^2*c*d*e^3)*x^2 + ( a^2*c*d^3*e + 3*a^3*d*e^3 + (a*c^2*d^3*e + 3*a^2*c*d*e^3)*x^4)*sqrt(c/a)*a rctan(a*sqrt(c/a)/(c*x^2)) - (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + (c^3*d^4 + 2*a *c^2*d^2*e^2)*x^4)*log(c*x^4 + a) - 2*(a^2*c*e^4*x^4 + a^3*e^4)*log(e*x^2 + d) + 4*(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4 + (c^3*d^4 + 2*a*c^2*d^2*e ^2 + a^2*c*e^4)*x^4)*log(x))/(a^3*c^2*d^5 + 2*a^4*c*d^3*e^2 + a^5*d*e^4 + (a^2*c^3*d^5 + 2*a^3*c^2*d^3*e^2 + a^4*c*d*e^4)*x^4)]
Timed out. \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {e^{4} \log \left (e x^{2} + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4}\right )}} - \frac {{\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )}} - \frac {{\left (c^{2} d^{2} e + 3 \, a c e^{3}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} - \frac {c e x^{2} - c d}{4 \, {\left (a^{2} c d^{2} + a^{3} e^{2} + {\left (a c^{2} d^{2} + a^{2} c e^{2}\right )} x^{4}\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2} d} \]
-1/2*e^4*log(e*x^2 + d)/(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4) - 1/4*(c^2*d ^3 + 2*a*c*d*e^2)*log(c*x^4 + a)/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4) - 1/4*(c^2*d^2*e + 3*a*c*e^3)*arctan(c*x^2/sqrt(a*c))/((a*c^2*d^4 + 2*a^2 *c*d^2*e^2 + a^3*e^4)*sqrt(a*c)) - 1/4*(c*e*x^2 - c*d)/(a^2*c*d^2 + a^3*e^ 2 + (a*c^2*d^2 + a^2*c*e^2)*x^4) + 1/2*log(x^2)/(a^2*d)
Time = 0.28 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.38 \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {e^{5} \log \left ({\left | e x^{2} + d \right |}\right )}{2 \, {\left (c^{2} d^{5} e + 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )}} - \frac {{\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )}} - \frac {{\left (c^{2} d^{2} e + 3 \, a c e^{3}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} + \frac {c^{3} d^{3} x^{4} + 2 \, a c^{2} d e^{2} x^{4} - a c^{2} d^{2} e x^{2} - a^{2} c e^{3} x^{2} + 2 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} {\left (c x^{4} + a\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2} d} \]
-1/2*e^5*log(abs(e*x^2 + d))/(c^2*d^5*e + 2*a*c*d^3*e^3 + a^2*d*e^5) - 1/4 *(c^2*d^3 + 2*a*c*d*e^2)*log(c*x^4 + a)/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a ^4*e^4) - 1/4*(c^2*d^2*e + 3*a*c*e^3)*arctan(c*x^2/sqrt(a*c))/((a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(a*c)) + 1/4*(c^3*d^3*x^4 + 2*a*c^2*d*e^2 *x^4 - a*c^2*d^2*e*x^2 - a^2*c*e^3*x^2 + 2*a*c^2*d^3 + 3*a^2*c*d*e^2)/((a^ 2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*(c*x^4 + a)) + 1/2*log(x^2)/(a^2*d)
Time = 9.05 (sec) , antiderivative size = 1082, normalized size of antiderivative = 5.18 \[ \int \frac {1}{x \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {\frac {c\,d}{4\,a\,\left (c\,d^2+a\,e^2\right )}-\frac {c\,e\,x^2}{4\,a\,\left (c\,d^2+a\,e^2\right )}}{c\,x^4+a}-\frac {\ln \left (400\,a^9\,c^{12}\,d^{20}\,x^2-10481\,d^4\,e^{16}\,{\left (-a^5\,c\right )}^{7/2}-1024\,a^{12}\,e^{20}\,{\left (-a^5\,c\right )}^{3/2}+1024\,a^{19}\,c^2\,e^{20}\,x^2-400\,a^2\,c^{10}\,d^{20}\,{\left (-a^5\,c\right )}^{3/2}+5840\,a^6\,d^2\,e^{18}\,{\left (-a^5\,c\right )}^{5/2}+33710\,c^6\,d^{14}\,e^6\,{\left (-a^5\,c\right )}^{5/2}+4104\,a^{10}\,c^{11}\,d^{18}\,e^2\,x^2+16689\,a^{11}\,c^{10}\,d^{16}\,e^4\,x^2+33710\,a^{12}\,c^9\,d^{14}\,e^6\,x^2+33391\,a^{13}\,c^8\,d^{12}\,e^8\,x^2+10748\,a^{14}\,c^7\,d^{10}\,e^{10}\,x^2-3585\,a^{15}\,c^6\,d^8\,e^{12}\,x^2+3998\,a^{16}\,c^5\,d^6\,e^{14}\,x^2+10481\,a^{17}\,c^4\,d^4\,e^{16}\,x^2+5840\,a^{18}\,c^3\,d^2\,e^{18}\,x^2+10748\,a^2\,c^4\,d^{10}\,e^{10}\,{\left (-a^5\,c\right )}^{5/2}-3585\,a^3\,c^3\,d^8\,e^{12}\,{\left (-a^5\,c\right )}^{5/2}+3998\,a^4\,c^2\,d^6\,e^{14}\,{\left (-a^5\,c\right )}^{5/2}-4104\,a^3\,c^9\,d^{18}\,e^2\,{\left (-a^5\,c\right )}^{3/2}-16689\,a^4\,c^8\,d^{16}\,e^4\,{\left (-a^5\,c\right )}^{3/2}+33391\,a\,c^5\,d^{12}\,e^8\,{\left (-a^5\,c\right )}^{5/2}\right )\,\left (3\,a\,e^3\,\sqrt {-a^5\,c}+2\,a^2\,c^2\,d^3+4\,a^3\,c\,d\,e^2+c\,d^2\,e\,\sqrt {-a^5\,c}\right )}{8\,\left (a^6\,e^4+2\,a^5\,c\,d^2\,e^2+a^4\,c^2\,d^4\right )}+\frac {\ln \left (1024\,a^{12}\,e^{20}\,{\left (-a^5\,c\right )}^{3/2}+10481\,d^4\,e^{16}\,{\left (-a^5\,c\right )}^{7/2}+400\,a^9\,c^{12}\,d^{20}\,x^2+1024\,a^{19}\,c^2\,e^{20}\,x^2+400\,a^2\,c^{10}\,d^{20}\,{\left (-a^5\,c\right )}^{3/2}-5840\,a^6\,d^2\,e^{18}\,{\left (-a^5\,c\right )}^{5/2}-33710\,c^6\,d^{14}\,e^6\,{\left (-a^5\,c\right )}^{5/2}+4104\,a^{10}\,c^{11}\,d^{18}\,e^2\,x^2+16689\,a^{11}\,c^{10}\,d^{16}\,e^4\,x^2+33710\,a^{12}\,c^9\,d^{14}\,e^6\,x^2+33391\,a^{13}\,c^8\,d^{12}\,e^8\,x^2+10748\,a^{14}\,c^7\,d^{10}\,e^{10}\,x^2-3585\,a^{15}\,c^6\,d^8\,e^{12}\,x^2+3998\,a^{16}\,c^5\,d^6\,e^{14}\,x^2+10481\,a^{17}\,c^4\,d^4\,e^{16}\,x^2+5840\,a^{18}\,c^3\,d^2\,e^{18}\,x^2-10748\,a^2\,c^4\,d^{10}\,e^{10}\,{\left (-a^5\,c\right )}^{5/2}+3585\,a^3\,c^3\,d^8\,e^{12}\,{\left (-a^5\,c\right )}^{5/2}-3998\,a^4\,c^2\,d^6\,e^{14}\,{\left (-a^5\,c\right )}^{5/2}+4104\,a^3\,c^9\,d^{18}\,e^2\,{\left (-a^5\,c\right )}^{3/2}+16689\,a^4\,c^8\,d^{16}\,e^4\,{\left (-a^5\,c\right )}^{3/2}-33391\,a\,c^5\,d^{12}\,e^8\,{\left (-a^5\,c\right )}^{5/2}\right )\,\left (3\,a\,e^3\,\sqrt {-a^5\,c}-2\,a^2\,c^2\,d^3-4\,a^3\,c\,d\,e^2+c\,d^2\,e\,\sqrt {-a^5\,c}\right )}{8\,\left (a^6\,e^4+2\,a^5\,c\,d^2\,e^2+a^4\,c^2\,d^4\right )}-\frac {e^4\,\ln \left (e\,x^2+d\right )}{2\,a^2\,d\,e^4+4\,a\,c\,d^3\,e^2+2\,c^2\,d^5}+\frac {\ln \left (x\right )}{a^2\,d} \]
((c*d)/(4*a*(a*e^2 + c*d^2)) - (c*e*x^2)/(4*a*(a*e^2 + c*d^2)))/(a + c*x^4 ) - (log(400*a^9*c^12*d^20*x^2 - 10481*d^4*e^16*(-a^5*c)^(7/2) - 1024*a^12 *e^20*(-a^5*c)^(3/2) + 1024*a^19*c^2*e^20*x^2 - 400*a^2*c^10*d^20*(-a^5*c) ^(3/2) + 5840*a^6*d^2*e^18*(-a^5*c)^(5/2) + 33710*c^6*d^14*e^6*(-a^5*c)^(5 /2) + 4104*a^10*c^11*d^18*e^2*x^2 + 16689*a^11*c^10*d^16*e^4*x^2 + 33710*a ^12*c^9*d^14*e^6*x^2 + 33391*a^13*c^8*d^12*e^8*x^2 + 10748*a^14*c^7*d^10*e ^10*x^2 - 3585*a^15*c^6*d^8*e^12*x^2 + 3998*a^16*c^5*d^6*e^14*x^2 + 10481* a^17*c^4*d^4*e^16*x^2 + 5840*a^18*c^3*d^2*e^18*x^2 + 10748*a^2*c^4*d^10*e^ 10*(-a^5*c)^(5/2) - 3585*a^3*c^3*d^8*e^12*(-a^5*c)^(5/2) + 3998*a^4*c^2*d^ 6*e^14*(-a^5*c)^(5/2) - 4104*a^3*c^9*d^18*e^2*(-a^5*c)^(3/2) - 16689*a^4*c ^8*d^16*e^4*(-a^5*c)^(3/2) + 33391*a*c^5*d^12*e^8*(-a^5*c)^(5/2))*(3*a*e^3 *(-a^5*c)^(1/2) + 2*a^2*c^2*d^3 + 4*a^3*c*d*e^2 + c*d^2*e*(-a^5*c)^(1/2))) /(8*(a^6*e^4 + a^4*c^2*d^4 + 2*a^5*c*d^2*e^2)) + (log(1024*a^12*e^20*(-a^5 *c)^(3/2) + 10481*d^4*e^16*(-a^5*c)^(7/2) + 400*a^9*c^12*d^20*x^2 + 1024*a ^19*c^2*e^20*x^2 + 400*a^2*c^10*d^20*(-a^5*c)^(3/2) - 5840*a^6*d^2*e^18*(- a^5*c)^(5/2) - 33710*c^6*d^14*e^6*(-a^5*c)^(5/2) + 4104*a^10*c^11*d^18*e^2 *x^2 + 16689*a^11*c^10*d^16*e^4*x^2 + 33710*a^12*c^9*d^14*e^6*x^2 + 33391* a^13*c^8*d^12*e^8*x^2 + 10748*a^14*c^7*d^10*e^10*x^2 - 3585*a^15*c^6*d^8*e ^12*x^2 + 3998*a^16*c^5*d^6*e^14*x^2 + 10481*a^17*c^4*d^4*e^16*x^2 + 5840* a^18*c^3*d^2*e^18*x^2 - 10748*a^2*c^4*d^10*e^10*(-a^5*c)^(5/2) + 3585*a...