\(\int \frac {x}{(a+\frac {c}{x^2}+\frac {b}{x}) (d+e x)} \, dx\) [202]

Optimal result

Integrand size = 23, antiderivative size = 176 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\frac {x}{a e}+\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a c^2 e\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )} \] Output:

x/a/e+(-3*a*b*c*d+2*a*c^2*e+b^3*d-b^2*c*e)*arctanh((2*a*x+b)/(-4*a*c+b^2)^ 
(1/2))/a^2/(-4*a*c+b^2)^(1/2)/(a*d^2-e*(b*d-c*e))-d^3*ln(e*x+d)/e^2/(a*d^2 
-e*(b*d-c*e))+1/2*(-a*c*d+b^2*d-b*c*e)*ln(a*x^2+b*x+c)/a^2/(a*d^2-e*(b*d-c 
*e))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.01 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\frac {x}{a e}+\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a c^2 e\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{a^2 \sqrt {-b^2+4 a c} \left (-a d^2+b d e-c e^2\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-b d e+c e^2\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-b d e+c e^2\right )} \] Input:

Integrate[x/((a + c/x^2 + b/x)*(d + e*x)),x]
 

Output:

x/(a*e) + ((b^3*d - 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*ArcTan[(b + 2*a*x)/Sq 
rt[-b^2 + 4*a*c]])/(a^2*Sqrt[-b^2 + 4*a*c]*(-(a*d^2) + b*d*e - c*e^2)) - ( 
d^3*Log[d + e*x])/(e^2*(a*d^2 - b*d*e + c*e^2)) + ((b^2*d - a*c*d - b*c*e) 
*Log[c + b*x + a*x^2])/(2*a^2*(a*d^2 - b*d*e + c*e^2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1893, 1200, 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.

\(\displaystyle \int \frac {x}{(d+e x) \left (a+\frac {b}{x}+\frac {c}{x^2}\right )} \, dx\)

\(\Big \downarrow \) 1893

\(\displaystyle \int \frac {x^3}{(d+e x) \left (a x^2+b x+c\right )}dx\)

\(\Big \downarrow \) 1200

\(\displaystyle \int \left (\frac {x \left (-a c d+b^2 d-b c e\right )+c (b d-c e)}{a \left (a x^2+b x+c\right ) \left (a d^2-e (b d-c e)\right )}+\frac {d^3}{e (d+e x) \left (e (b d-c e)-a d^2\right )}+\frac {1}{a e}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (-3 a b c d+2 a c^2 e+b^3 d-b^2 c e\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {x}{a e}\)

Input:

Int[x/((a + c/x^2 + b/x)*(d + e*x)),x]
 

Output:

x/(a*e) + ((b^3*d - 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*ArcTanh[(b + 2*a*x)/S 
qrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))) - (d^3* 
Log[d + e*x])/(e^2*(a*d^2 - e*(b*d - c*e))) + ((b^2*d - a*c*d - b*c*e)*Log 
[c + b*x + a*x^2])/(2*a^2*(a*d^2 - e*(b*d - c*e)))
 

Defintions of rubi rules used

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1893

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) 
 + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c 
 + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && EqQ[mn 
, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {x}{a e}+\frac {\frac {\left (-a c d +d \,b^{2}-b c e \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (c b d -c^{2} e -\frac {\left (-a c d +d \,b^{2}-b c e \right ) b}{2 a}\right ) \arctan \left (\frac {2 x a +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right ) a}-\frac {d^{3} \ln \left (e x +d \right )}{e^{2} \left (a \,d^{2}-b d e +c \,e^{2}\right )}\)	\(164\)
risch	\(\frac {x}{a e}-\frac {d^{3} \ln \left (e x +d \right )}{e^{2} \left (a \,d^{2}-b d e +c \,e^{2}\right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c \,d^{2}-b^{2} d^{2} a^{2}-4 a^{2} b c d e +4 a^{2} c^{2} e^{2}+a \,b^{3} d e -a \,b^{2} c \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 a^{2} c^{2} d e -5 a \,b^{2} c d e +4 c^{2} e^{2} a b +d e \,b^{4}-c \,e^{2} b^{3}\right ) \textit {\_Z} +a \,c^{3} e^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-2 a^{3} d^{2} e +2 a^{2} b d \,e^{2}+6 a^{2} c \,e^{3}-2 a \,b^{2} e^{3}\right ) \textit {\_R}^{2}+\left (-2 a^{3} d^{3}+3 a^{2} c d \,e^{2}+6 a b c \,e^{3}-2 b^{3} e^{3}\right ) \textit {\_R} -a^{2} d^{2} e c +a \,b^{2} d^{2} e +a b c d \,e^{2}+a \,c^{2} e^{3}\right ) x +\left (-a^{2} b \,d^{2} e +8 a^{2} c d \,e^{2}-a \,b^{2} d \,e^{2}-a b c \,e^{3}\right ) \textit {\_R}^{2}+\left (-a^{2} b \,d^{3}+4 a^{2} d^{2} e c -a \,b^{2} d^{2} e +5 a b c d \,e^{2}+a \,c^{2} e^{3}-b^{3} d \,e^{2}-b^{2} c \,e^{3}\right ) \textit {\_R} +a b c \,d^{2} e +a \,c^{2} d \,e^{2}\right )}{a e}\)	\(419\)

Input:

int(x/(a+c/x^2+b/x)/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

x/a/e+1/(a*d^2-b*d*e+c*e^2)/a*(1/2*(-a*c*d+b^2*d-b*c*e)/a*ln(a*x^2+b*x+c)+ 
2*(c*b*d-c^2*e-1/2*(-a*c*d+b^2*d-b*c*e)*b/a)/(4*a*c-b^2)^(1/2)*arctan((2*a 
*x+b)/(4*a*c-b^2)^(1/2)))-1/e^2*d^3/(a*d^2-b*d*e+c*e^2)*ln(e*x+d)
 

Fricas [A] (verification not implemented)

Time = 2.91 (sec) , antiderivative size = 596, normalized size of antiderivative = 3.39 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\left [-\frac {2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - {\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} - {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right ) - 2 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )} x - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} - {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )}}, -\frac {2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - 2 \, {\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} - {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )} x - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} - {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )}}\right ] \] Input:

integrate(x/(a+c/x^2+b/x)/(e*x+d),x, algorithm="fricas")
 

Output:

[-1/2*(2*(a^2*b^2 - 4*a^3*c)*d^3*log(e*x + d) - ((b^3 - 3*a*b*c)*d*e^2 - ( 
b^2*c - 2*a*c^2)*e^3)*sqrt(b^2 - 4*a*c)*log((2*a^2*x^2 + 2*a*b*x + b^2 - 2 
*a*c + sqrt(b^2 - 4*a*c)*(2*a*x + b))/(a*x^2 + b*x + c)) - 2*((a^2*b^2 - 4 
*a^3*c)*d^2*e - (a*b^3 - 4*a^2*b*c)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3)*x - 
 ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*e^2 - (b^3*c - 4*a*b*c^2)*e^3)*log(a*x^2 
 + b*x + c))/((a^3*b^2 - 4*a^4*c)*d^2*e^2 - (a^2*b^3 - 4*a^3*b*c)*d*e^3 + 
(a^2*b^2*c - 4*a^3*c^2)*e^4), -1/2*(2*(a^2*b^2 - 4*a^3*c)*d^3*log(e*x + d) 
 - 2*((b^3 - 3*a*b*c)*d*e^2 - (b^2*c - 2*a*c^2)*e^3)*sqrt(-b^2 + 4*a*c)*ar 
ctan(-sqrt(-b^2 + 4*a*c)*(2*a*x + b)/(b^2 - 4*a*c)) - 2*((a^2*b^2 - 4*a^3* 
c)*d^2*e - (a*b^3 - 4*a^2*b*c)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3)*x - ((b^ 
4 - 5*a*b^2*c + 4*a^2*c^2)*d*e^2 - (b^3*c - 4*a*b*c^2)*e^3)*log(a*x^2 + b* 
x + c))/((a^3*b^2 - 4*a^4*c)*d^2*e^2 - (a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^2* 
b^2*c - 4*a^3*c^2)*e^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\text {Timed out} \] Input:

integrate(x/(a+c/x**2+b/x)/(e*x+d),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x/(a+c/x^2+b/x)/(e*x+d),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=-\frac {d^{3} \log \left ({\left | e x + d \right |}\right )}{a d^{2} e^{2} - b d e^{3} + c e^{4}} + \frac {{\left (b^{2} d - a c d - b c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )}} - \frac {{\left (b^{3} d - 3 \, a b c d - b^{2} c e + 2 \, a c^{2} e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {x}{a e} \] Input:

integrate(x/(a+c/x^2+b/x)/(e*x+d),x, algorithm="giac")
 

Output:

-d^3*log(abs(e*x + d))/(a*d^2*e^2 - b*d*e^3 + c*e^4) + 1/2*(b^2*d - a*c*d 
- b*c*e)*log(a*x^2 + b*x + c)/(a^3*d^2 - a^2*b*d*e + a^2*c*e^2) - (b^3*d - 
 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*arctan((2*a*x + b)/sqrt(-b^2 + 4*a*c))/( 
(a^3*d^2 - a^2*b*d*e + a^2*c*e^2)*sqrt(-b^2 + 4*a*c)) + x/(a*e)
 

Mupad [B] (verification not implemented)

Time = 24.75 (sec) , antiderivative size = 1367, normalized size of antiderivative = 7.77 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx =\text {Too large to display} \] Input:

int(x/((d + e*x)*(a + b/x + c/x^2)),x)
 

Output:

x/(a*e) - (log(c^3*e^5*(b^2 - 4*a*c)^(1/2) - b*c^3*e^5 - 4*a^3*c*d^5 + a^2 
*b^2*d^5 + b^4*d^3*e^2 + 3*b^2*c^2*d*e^4 - 3*b^3*c*d^2*e^3 - b^3*d^3*e^2*( 
b^2 - 4*a*c)^(1/2) + 6*a^2*c^2*d^3*e^2 - 6*a*c^3*d*e^4 - 2*a*c^3*e^5*x - a 
^2*b*d^5*(b^2 - 4*a*c)^(1/2) - 2*a^3*d^5*x*(b^2 - 4*a*c)^(1/2) - 8*a^3*c*d 
^4*e*x + 4*a^2*c*d^4*e*(b^2 - 4*a*c)^(1/2) - 3*b*c^2*d*e^4*(b^2 - 4*a*c)^( 
1/2) + 9*a*b*c^2*d^2*e^3 - 5*a*b^2*c*d^3*e^2 + 2*a^2*b^2*d^4*e*x - 3*a*c^2 
*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 3*b^2*c*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 6*a^2 
*c^2*d^2*e^3*x - 2*a*b^2*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) + 3*a^2*c*d^3*e^2*x 
*(b^2 - 4*a*c)^(1/2) + 3*a*b*c^2*d*e^4*x + a*b*c*d^3*e^2*(b^2 - 4*a*c)^(1/ 
2) + 2*a^2*b*d^4*e*x*(b^2 - 4*a*c)^(1/2) - 3*a*c^2*d*e^4*x*(b^2 - 4*a*c)^( 
1/2) - 3*a*b^2*c*d^2*e^3*x + a^2*b*c*d^3*e^2*x + 3*a*b*c*d^2*e^3*x*(b^2 - 
4*a*c)^(1/2))*(b^4*d - b^3*d*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*d - b^3*c*e - 
 5*a*b^2*c*d + 4*a*b*c^2*e - 2*a*c^2*e*(b^2 - 4*a*c)^(1/2) + b^2*c*e*(b^2 
- 4*a*c)^(1/2) + 3*a*b*c*d*(b^2 - 4*a*c)^(1/2)))/(2*(4*a^4*c*d^2 - a^3*b^2 
*d^2 + 4*a^3*c^2*e^2 - a^2*b^2*c*e^2 + a^2*b^3*d*e - 4*a^3*b*c*d*e)) - (lo 
g(a^2*b^2*d^5 - b*c^3*e^5 - c^3*e^5*(b^2 - 4*a*c)^(1/2) - 4*a^3*c*d^5 + b^ 
4*d^3*e^2 + 3*b^2*c^2*d*e^4 - 3*b^3*c*d^2*e^3 + b^3*d^3*e^2*(b^2 - 4*a*c)^ 
(1/2) + 6*a^2*c^2*d^3*e^2 - 6*a*c^3*d*e^4 - 2*a*c^3*e^5*x + a^2*b*d^5*(b^2 
 - 4*a*c)^(1/2) + 2*a^3*d^5*x*(b^2 - 4*a*c)^(1/2) - 8*a^3*c*d^4*e*x - 4*a^ 
2*c*d^4*e*(b^2 - 4*a*c)^(1/2) + 3*b*c^2*d*e^4*(b^2 - 4*a*c)^(1/2) + 9*a...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.45 \[ \int \frac {x}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx=\frac {6 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) a b c d \,e^{2}-4 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) a \,c^{2} e^{3}-2 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{3} d \,e^{2}+2 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c \,e^{3}-4 \,\mathrm {log}\left (a \,x^{2}+b x +c \right ) a^{2} c^{2} d \,e^{2}+5 \,\mathrm {log}\left (a \,x^{2}+b x +c \right ) a \,b^{2} c d \,e^{2}-4 \,\mathrm {log}\left (a \,x^{2}+b x +c \right ) a b \,c^{2} e^{3}-\mathrm {log}\left (a \,x^{2}+b x +c \right ) b^{4} d \,e^{2}+\mathrm {log}\left (a \,x^{2}+b x +c \right ) b^{3} c \,e^{3}-8 \,\mathrm {log}\left (e x +d \right ) a^{3} c \,d^{3}+2 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{2} d^{3}+8 a^{3} c \,d^{2} e x -2 a^{2} b^{2} d^{2} e x -8 a^{2} b c d \,e^{2} x +8 a^{2} c^{2} e^{3} x +2 a \,b^{3} d \,e^{2} x -2 a \,b^{2} c \,e^{3} x}{2 a^{2} e^{2} \left (4 a^{2} c \,d^{2}-a \,b^{2} d^{2}-4 a b c d e +4 a \,c^{2} e^{2}+b^{3} d e -b^{2} c \,e^{2}\right )} \] Input:

int(x/(a+c/x^2+b/x)/(e*x+d),x)
 

Output:

(6*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b*c*d*e**2 - 
4*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*c**2*e**3 - 2* 
sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b**3*d*e**2 + 2*sq 
rt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b**2*c*e**3 - 4*log( 
a*x**2 + b*x + c)*a**2*c**2*d*e**2 + 5*log(a*x**2 + b*x + c)*a*b**2*c*d*e* 
*2 - 4*log(a*x**2 + b*x + c)*a*b*c**2*e**3 - log(a*x**2 + b*x + c)*b**4*d* 
e**2 + log(a*x**2 + b*x + c)*b**3*c*e**3 - 8*log(d + e*x)*a**3*c*d**3 + 2* 
log(d + e*x)*a**2*b**2*d**3 + 8*a**3*c*d**2*e*x - 2*a**2*b**2*d**2*e*x - 8 
*a**2*b*c*d*e**2*x + 8*a**2*c**2*e**3*x + 2*a*b**3*d*e**2*x - 2*a*b**2*c*e 
**3*x)/(2*a**2*e**2*(4*a**2*c*d**2 - a*b**2*d**2 - 4*a*b*c*d*e + 4*a*c**2* 
e**2 + b**3*d*e - b**2*c*e**2))