\(\int \frac {1}{(a+\frac {c}{x^2}+\frac {b}{x}) x^3 (d+e x)} \, dx\) [67]
Optimal result
Integrand size = 25, antiderivative size = 158 \[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=\frac {\left (a b d-b^2 e+2 a c e\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-b d e+c e^2\right )}-\frac {(a d-b e) \log \left (c+b x+a x^2\right )}{2 c \left (a d^2-e (b d-c e)\right )}
\]
[Out]
ln(x)/c/d-e^2*ln(e*x+d)/d/(a*d^2-b*d*e+c*e^2)-1/2*(a*d-b*e)*ln(a*x^2+b*x+c)/c/(a*d^2-e*(b*d-c*e))+(a*b*d+2*a*c
*e-b^2*e)*arctanh((2*a*x+b)/(-4*a*c+b^2)^(1/2))/c/(a*d^2-e*(b*d-c*e))/(-4*a*c+b^2)^(1/2)
Rubi [A] (verified)
Time = 0.17 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.01, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1583, 907, 648, 632, 212, 642}
\[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=\frac {\text {arctanh}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right ) \left (a b d+2 a c e+b^2 (-e)\right )}{c \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac {(a d-b e) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )}+\frac {\log (x)}{c d}
\]
[In]
Int[1/((a + c/x^2 + b/x)*x^3*(d + e*x)),x]
[Out]
((a*b*d - b^2*e + 2*a*c*e)*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e)
)) + Log[x]/(c*d) - (e^2*Log[d + e*x])/(d*(a*d^2 - e*(b*d - c*e))) - ((a*d - b*e)*Log[c + b*x + a*x^2])/(2*c*(
a*d^2 - e*(b*d - c*e)))
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 632
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 642
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rule 648
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && !NiceSqrtQ[b^2 - 4*a*c]
Rule 907
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))
Rule 1583
Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbo
l] :> 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] && E
qQ[mn, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {1}{x (d+e x) \left (c+b x+a x^2\right )} \, dx \\ & = \int \left (\frac {1}{c d x}+\frac {e^3}{d \left (-a d^2+e (b d-c e)\right ) (d+e x)}+\frac {b^2 e-a (b d+c e)-a (a d-b e) x}{c \left (a d^2-e (b d-c e)\right ) \left (c+b x+a x^2\right )}\right ) \, dx \\ & = \frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}+\frac {\int \frac {b^2 e-a (b d+c e)-a (a d-b e) x}{c+b x+a x^2} \, dx}{c \left (a d^2-b d e+c e^2\right )} \\ & = \frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}+\frac {\left (-a b d+b^2 e-2 a c e\right ) \int \frac {1}{c+b x+a x^2} \, dx}{2 c \left (a d^2-b d e+c e^2\right )}-\frac {(a d-b e) \int \frac {b+2 a x}{c+b x+a x^2} \, dx}{2 c \left (a d^2-e (b d-c e)\right )} \\ & = \frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac {(a d-b e) \log \left (c+b x+a x^2\right )}{2 c \left (a d^2-e (b d-c e)\right )}-\frac {\left (-a b d+b^2 e-2 a c e\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 a x\right )}{c \left (a d^2-b d e+c e^2\right )} \\ & = \frac {\left (a b d-b^2 e+2 a c e\right ) \tanh ^{-1}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (a d^2-b d e+c e^2\right )}+\frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac {(a d-b e) \log \left (c+b x+a x^2\right )}{2 c \left (a d^2-e (b d-c e)\right )} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.96
\[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=-\frac {2 d \left (a b d-b^2 e+2 a c e\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} \left (-2 \left (a d^2+e (-b d+c e)\right ) \log (x)+2 c e^2 \log (d+e x)+d (a d-b e) \log (c+x (b+a x))\right )}{2 c \sqrt {-b^2+4 a c} d \left (a d^2+e (-b d+c e)\right )}
\]
[In]
Integrate[1/((a + c/x^2 + b/x)*x^3*(d + e*x)),x]
[Out]
-1/2*(2*d*(a*b*d - b^2*e + 2*a*c*e)*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*(-2*(a*d^2 + e
*(-(b*d) + c*e))*Log[x] + 2*c*e^2*Log[d + e*x] + d*(a*d - b*e)*Log[c + x*(b + a*x)]))/(c*Sqrt[-b^2 + 4*a*c]*d*
(a*d^2 + e*(-(b*d) + c*e)))
Maple [A] (verified)
Time = 0.77 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01
| | |
| method | result | size |
| | |
| default |
\(\frac {\ln \left (x \right )}{c d}+\frac {\frac {\left (-d \,a^{2}+a b e \right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (-d a b -a c e +b^{2} e -\frac {\left (-d \,a^{2}+a b e \right ) b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right ) c}-\frac {e^{2} \ln \left (e x +d \right )}{d \left (a \,d^{2}-b d e +c \,e^{2}\right )}\) |
\(160\) |
| risch |
\(\text {Expression too large to display}\) |
\(19172\) |
| | |
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[In]
int(1/(a+c/x^2+b/x)/x^3/(e*x+d),x,method=_RETURNVERBOSE)
[Out]
ln(x)/c/d+1/(a*d^2-b*d*e+c*e^2)/c*(1/2*(-a^2*d+a*b*e)/a*ln(a*x^2+b*x+c)+2*(-d*a*b-a*c*e+b^2*e-1/2*(-a^2*d+a*b*
e)*b/a)/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2)))-e^2*ln(e*x+d)/d/(a*d^2-b*d*e+c*e^2)
Fricas [A] (verification not implemented)
none
Time = 184.15 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.19
\[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=\left [-\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} \log \left (e x + d\right ) - {\left (a b d^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right ) + {\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e\right )} \log \left (a x^{2} + b x + c\right ) - 2 \, {\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )} \log \left (x\right )}{2 \, {\left ({\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d^{3} - {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{2} e + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e^{2}\right )}}, -\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} \log \left (e x + d\right ) - 2 \, {\left (a b d^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e\right )} \log \left (a x^{2} + b x + c\right ) - 2 \, {\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )} \log \left (x\right )}{2 \, {\left ({\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d^{3} - {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{2} e + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e^{2}\right )}}\right ]
\]
[In]
integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d),x, algorithm="fricas")
[Out]
[-1/2*(2*(b^2*c - 4*a*c^2)*e^2*log(e*x + d) - (a*b*d^2 - (b^2 - 2*a*c)*d*e)*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)) + ((a*b^2 - 4*a^2*c)*d^2 - (b^3 - 4
*a*b*c)*d*e)*log(a*x^2 + b*x + c) - 2*((a*b^2 - 4*a^2*c)*d^2 - (b^3 - 4*a*b*c)*d*e + (b^2*c - 4*a*c^2)*e^2)*lo
g(x))/((a*b^2*c - 4*a^2*c^2)*d^3 - (b^3*c - 4*a*b*c^2)*d^2*e + (b^2*c^2 - 4*a*c^3)*d*e^2), -1/2*(2*(b^2*c - 4*
a*c^2)*e^2*log(e*x + d) - 2*(a*b*d^2 - (b^2 - 2*a*c)*d*e)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*a*x
+ b)/(b^2 - 4*a*c)) + ((a*b^2 - 4*a^2*c)*d^2 - (b^3 - 4*a*b*c)*d*e)*log(a*x^2 + b*x + c) - 2*((a*b^2 - 4*a^2*
c)*d^2 - (b^3 - 4*a*b*c)*d*e + (b^2*c - 4*a*c^2)*e^2)*log(x))/((a*b^2*c - 4*a^2*c^2)*d^3 - (b^3*c - 4*a*b*c^2)
*d^2*e + (b^2*c^2 - 4*a*c^3)*d*e^2)]
Sympy [F(-1)]
Timed out. \[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=\text {Timed out}
\]
[In]
integrate(1/(a+c/x**2+b/x)/x**3/(e*x+d),x)
[Out]
Timed out
Maxima [F(-2)]
Exception generated. \[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta
Giac [A] (verification not implemented)
none
Time = 0.36 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03
\[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=-\frac {e^{3} \log \left ({\left | e x + d \right |}\right )}{a d^{3} e - b d^{2} e^{2} + c d e^{3}} - \frac {{\left (a d - b e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a c d^{2} - b c d e + c^{2} e^{2}\right )}} - \frac {{\left (a b d - b^{2} e + 2 \, a c e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a c d^{2} - b c d e + c^{2} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {\log \left ({\left | x \right |}\right )}{c d}
\]
[In]
integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d),x, algorithm="giac")
[Out]
-e^3*log(abs(e*x + d))/(a*d^3*e - b*d^2*e^2 + c*d*e^3) - 1/2*(a*d - b*e)*log(a*x^2 + b*x + c)/(a*c*d^2 - b*c*d
*e + c^2*e^2) - (a*b*d - b^2*e + 2*a*c*e)*arctan((2*a*x + b)/sqrt(-b^2 + 4*a*c))/((a*c*d^2 - b*c*d*e + c^2*e^2
)*sqrt(-b^2 + 4*a*c)) + log(abs(x))/(c*d)
Mupad [B] (verification not implemented)
Time = 11.09 (sec) , antiderivative size = 2399, normalized size of antiderivative = 15.18
\[
\int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx=\text {Too large to display}
\]
[In]
int(1/(x^3*(d + e*x)*(a + b/x + c/x^2)),x)
[Out]
(log(b^3*c^3*e^5 - 6*a^4*c^2*d^5 + 2*a^3*b^2*c*d^5 + 8*a^2*c^4*d*e^4 - b^4*c^2*d*e^4 - 2*b^5*c*d^2*e^3 + 2*a^3
*b^3*d^5*x + 8*a^2*c^4*e^5*x + b^4*c^2*e^5*x - 2*b^6*d^2*e^3*x + b^2*c^3*e^5*(b^2 - 4*a*c)^(1/2) + 18*a^3*c^3*
d^3*e^2 - 4*a*b*c^4*e^5 - 4*a*c^4*e^5*(b^2 - 4*a*c)^(1/2) - 5*a^2*c^3*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 7*a^4*b*c*
d^5*x - b^5*c*d*e^4*x - 27*a^2*b^2*c^2*d^3*e^2 + 2*a^3*b*c*d^5*(b^2 - 4*a*c)^(1/2) - 3*a^4*c*d^5*x*(b^2 - 4*a*
c)^(1/2) + 2*a*b^2*c^3*d*e^4 + 6*a*b^4*c*d^3*e^2 - 6*a^2*b^3*c*d^4*e + 21*a^3*b*c^2*d^4*e - 6*a*b^2*c^3*e^5*x
+ 6*a*b^5*d^3*e^2*x - 6*a^2*b^4*d^4*e*x - 14*a^4*c^2*d^4*e*x + 7*a^3*c^2*d^4*e*(b^2 - 4*a*c)^(1/2) - b^3*c^2*d
*e^4*(b^2 - 4*a*c)^(1/2) - 2*b^4*c*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 2*a^3*b^2*d^5*x*(b^2 - 4*a*c)^(1/2) + b^3*c^2
*e^5*x*(b^2 - 4*a*c)^(1/2) - 2*b^5*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) + 13*a*b^3*c^2*d^2*e^3 - 21*a^2*b*c^3*d^2*e^3
+ 10*a^3*c^3*d^2*e^3*x + 6*a*b^3*c*d^3*e^2*(b^2 - 4*a*c)^(1/2) - 6*a^2*b^2*c*d^4*e*(b^2 - 4*a*c)^(1/2) + 6*a*
b^4*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) - 6*a^2*b^3*d^4*e*x*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^3*d*e^4*x*(b^2 - 4*a*c)^(1
/2) - 32*a^2*b^3*c*d^3*e^2*x + 35*a^3*b*c^2*d^3*e^2*x + 7*a*b^2*c^2*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2
*d^3*e^2*(b^2 - 4*a*c)^(1/2) + 9*a^3*c^2*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) - 27*a^2*b^2*c^2*d^2*e^3*x + 4*a*b*c^3*
d*e^4*(b^2 - 4*a*c)^(1/2) - 4*a*b*c^3*e^5*x*(b^2 - 4*a*c)^(1/2) - b^4*c*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 5*a*b^3*
c^2*d*e^4*x + 14*a*b^4*c*d^2*e^3*x - 4*a^2*b*c^3*d*e^4*x + 26*a^3*b^2*c*d^4*e*x + 14*a^3*b*c*d^4*e*x*(b^2 - 4*
a*c)^(1/2) + 3*a*b^2*c^2*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 10*a*b^3*c*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2
*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 20*a^2*b^2*c*d^3*e^2*x*(b^2 - 4*a*c)^(1/2))*(d*((a*b^2)/2 - 2*a^2*c + (a*b*(b
^2 - 4*a*c)^(1/2))/2) - (b^3*e)/2 - (b^2*e*(b^2 - 4*a*c)^(1/2))/2 + a*c*e*(b^2 - 4*a*c)^(1/2) + 2*a*b*c*e))/(4
*a*c^3*e^2 + 4*a^2*c^2*d^2 - b^2*c^2*e^2 + b^3*c*d*e - a*b^2*c*d^2 - 4*a*b*c^2*d*e) - (log(6*a^4*c^2*d^5 - b^3
*c^3*e^5 - 2*a^3*b^2*c*d^5 - 8*a^2*c^4*d*e^4 + b^4*c^2*d*e^4 + 2*b^5*c*d^2*e^3 - 2*a^3*b^3*d^5*x - 8*a^2*c^4*e
^5*x - b^4*c^2*e^5*x + 2*b^6*d^2*e^3*x + b^2*c^3*e^5*(b^2 - 4*a*c)^(1/2) - 18*a^3*c^3*d^3*e^2 + 4*a*b*c^4*e^5
- 4*a*c^4*e^5*(b^2 - 4*a*c)^(1/2) - 5*a^2*c^3*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 7*a^4*b*c*d^5*x + b^5*c*d*e^4*x +
27*a^2*b^2*c^2*d^3*e^2 + 2*a^3*b*c*d^5*(b^2 - 4*a*c)^(1/2) - 3*a^4*c*d^5*x*(b^2 - 4*a*c)^(1/2) - 2*a*b^2*c^3*d
*e^4 - 6*a*b^4*c*d^3*e^2 + 6*a^2*b^3*c*d^4*e - 21*a^3*b*c^2*d^4*e + 6*a*b^2*c^3*e^5*x - 6*a*b^5*d^3*e^2*x + 6*
a^2*b^4*d^4*e*x + 14*a^4*c^2*d^4*e*x + 7*a^3*c^2*d^4*e*(b^2 - 4*a*c)^(1/2) - b^3*c^2*d*e^4*(b^2 - 4*a*c)^(1/2)
- 2*b^4*c*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 2*a^3*b^2*d^5*x*(b^2 - 4*a*c)^(1/2) + b^3*c^2*e^5*x*(b^2 - 4*a*c)^(1/
2) - 2*b^5*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 13*a*b^3*c^2*d^2*e^3 + 21*a^2*b*c^3*d^2*e^3 - 10*a^3*c^3*d^2*e^3*x
+ 6*a*b^3*c*d^3*e^2*(b^2 - 4*a*c)^(1/2) - 6*a^2*b^2*c*d^4*e*(b^2 - 4*a*c)^(1/2) + 6*a*b^4*d^3*e^2*x*(b^2 - 4*a
*c)^(1/2) - 6*a^2*b^3*d^4*e*x*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^3*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 32*a^2*b^3*c*d^3*e
^2*x - 35*a^3*b*c^2*d^3*e^2*x + 7*a*b^2*c^2*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2*d^3*e^2*(b^2 - 4*a*c)^(
1/2) + 9*a^3*c^2*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) + 27*a^2*b^2*c^2*d^2*e^3*x + 4*a*b*c^3*d*e^4*(b^2 - 4*a*c)^(1/2
) - 4*a*b*c^3*e^5*x*(b^2 - 4*a*c)^(1/2) - b^4*c*d*e^4*x*(b^2 - 4*a*c)^(1/2) - 5*a*b^3*c^2*d*e^4*x - 14*a*b^4*c
*d^2*e^3*x + 4*a^2*b*c^3*d*e^4*x - 26*a^3*b^2*c*d^4*e*x + 14*a^3*b*c*d^4*e*x*(b^2 - 4*a*c)^(1/2) + 3*a*b^2*c^2
*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 10*a*b^3*c*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2*d^2*e^3*x*(b^2 - 4*a*c)
^(1/2) - 20*a^2*b^2*c*d^3*e^2*x*(b^2 - 4*a*c)^(1/2))*((b^3*e)/2 + d*(2*a^2*c - (a*b^2)/2 + (a*b*(b^2 - 4*a*c)^
(1/2))/2) - (b^2*e*(b^2 - 4*a*c)^(1/2))/2 + a*c*e*(b^2 - 4*a*c)^(1/2) - 2*a*b*c*e))/(4*a*c^3*e^2 + 4*a^2*c^2*d
^2 - b^2*c^2*e^2 + b^3*c*d*e - a*b^2*c*d^2 - 4*a*b*c^2*d*e) - (e^2*log(d + e*x))/(a*d^3 - b*d^2*e + c*d*e^2) +
log(x)/(c*d)