Integrand size = 20, antiderivative size = 243 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx=\frac {e^2 \left (6 c^2 d^2+b^2 e^2-c e (4 b d+a e)\right ) x}{c^3}+\frac {e^3 (4 c d-b e) x^2}{2 c^2}+\frac {e^4 x^3}{3 c}-\frac {\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 c^4} \] Output:
e^2*(6*c^2*d^2+b^2*e^2-c*e*(a*e+4*b*d))*x/c^3+1/2*e^3*(-b*e+4*c*d)*x^2/c^2 +1/3*e^4*x^3/c-(2*c^4*d^4+b^4*e^4-4*b^2*c*e^3*(a*e+b*d)-4*c^3*d^2*e*(3*a*e +b*d)+2*c^2*e^2*(a^2*e^2+6*a*b*d*e+3*b^2*d^2))*arctanh((2*c*x+b)/(-4*a*c+b ^2)^(1/2))/c^4/(-4*a*c+b^2)^(1/2)+1/2*e*(-b*e+2*c*d)*(2*c^2*d^2+b^2*e^2-2* c*e*(a*e+b*d))*ln(c*x^2+b*x+a)/c^4
Time = 0.13 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx=\frac {6 c e^2 \left (6 c^2 d^2+b^2 e^2-c e (4 b d+a e)\right ) x+3 c^2 e^3 (4 c d-b e) x^2+2 c^3 e^4 x^3+\frac {6 \left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+3 e (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (a+x (b+c x))}{6 c^4} \] Input:
Integrate[(d + e*x)^4/(a + b*x + c*x^2),x]
Output:
(6*c*e^2*(6*c^2*d^2 + b^2*e^2 - c*e*(4*b*d + a*e))*x + 3*c^2*e^3*(4*c*d - b*e)*x^2 + 2*c^3*e^4*x^3 + (6*(2*c^4*d^4 + b^4*e^4 - 4*b^2*c*e^3*(b*d + a* e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a*b*d*e + a^2*e^ 2))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + 3*e*(2*c* d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*Log[a + x*(b + c*x)])/( 6*c^4)
Time = 0.57 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1143, 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.
\(\displaystyle \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1143 |
\(\displaystyle \int \left (\frac {e^2 \left (-c e (a e+4 b d)+b^2 e^2+6 c^2 d^2\right )}{c^3}+\frac {e x (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )-a b^2 e^4+a c e^3 (a e+4 b d)-6 a c^2 d^2 e^2+c^3 d^4}{c^3 \left (a+b x+c x^2\right )}+\frac {e^3 x (4 c d-b e)}{c^2}+\frac {e^4 x^2}{c}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^4}+\frac {e^2 x \left (-c e (a e+4 b d)+b^2 e^2+6 c^2 d^2\right )}{c^3}+\frac {e^3 x^2 (4 c d-b e)}{2 c^2}+\frac {e^4 x^3}{3 c}\) |
Input:
Int[(d + e*x)^4/(a + b*x + c*x^2),x]
Output:
(e^2*(6*c^2*d^2 + b^2*e^2 - c*e*(4*b*d + a*e))*x)/c^3 + (e^3*(4*c*d - b*e) *x^2)/(2*c^2) + (e^4*x^3)/(3*c) - ((2*c^4*d^4 + b^4*e^4 - 4*b^2*c*e^3*(b*d + a*e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a*b*d*e + a ^2*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^4*Sqrt[b^2 - 4*a*c]) + (e*(2*c*d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*Log[a + b*x + c*x^2])/(2*c^4)
Int[((d_.) + (e_.)*(x_))^(m_)/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 1]
Time = 1.12 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.22
method | result | size |
default | \(-\frac {e^{2} \left (-\frac {1}{3} c^{2} e^{2} x^{3}+\frac {1}{2} b c \,e^{2} x^{2}-2 c^{2} d e \,x^{2}+a c \,e^{2} x -b^{2} e^{2} x +4 b c d e x -6 c^{2} d^{2} x \right )}{c^{3}}+\frac {\frac {\left (2 a b c \,e^{4}-4 a \,c^{2} d \,e^{3}-e^{4} b^{3}+4 b^{2} c d \,e^{3}-6 d^{2} e^{2} b \,c^{2}+4 d^{3} e \,c^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (e^{4} a^{2} c -a \,b^{2} e^{4}+4 a b c d \,e^{3}-6 d^{2} e^{2} a \,c^{2}+d^{4} c^{3}-\frac {\left (2 a b c \,e^{4}-4 a \,c^{2} d \,e^{3}-e^{4} b^{3}+4 b^{2} c d \,e^{3}-6 d^{2} e^{2} b \,c^{2}+4 d^{3} e \,c^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{3}}\) | \(297\) |
risch | \(\text {Expression too large to display}\) | \(9468\) |
Input:
int((e*x+d)^4/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
-e^2/c^3*(-1/3*c^2*e^2*x^3+1/2*b*c*e^2*x^2-2*c^2*d*e*x^2+a*c*e^2*x-b^2*e^2 *x+4*b*c*d*e*x-6*c^2*d^2*x)+1/c^3*(1/2*(2*a*b*c*e^4-4*a*c^2*d*e^3-b^3*e^4+ 4*b^2*c*d*e^3-6*b*c^2*d^2*e^2+4*c^3*d^3*e)/c*ln(c*x^2+b*x+a)+2*(e^4*a^2*c- a*b^2*e^4+4*a*b*c*d*e^3-6*d^2*e^2*a*c^2+d^4*c^3-1/2*(2*a*b*c*e^4-4*a*c^2*d *e^3-b^3*e^4+4*b^2*c*d*e^3-6*b*c^2*d^2*e^2+4*c^3*d^3*e)*b/c)/(4*a*c-b^2)^( 1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))
Time = 0.12 (sec) , antiderivative size = 821, normalized size of antiderivative = 3.38 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx =\text {Too large to display} \] Input:
integrate((e*x+d)^4/(c*x^2+b*x+a),x, algorithm="fricas")
Output:
[1/6*(2*(b^2*c^3 - 4*a*c^4)*e^4*x^3 + 3*(4*(b^2*c^3 - 4*a*c^4)*d*e^3 - (b^ 3*c^2 - 4*a*b*c^3)*e^4)*x^2 + 3*(2*c^4*d^4 - 4*b*c^3*d^3*e + 6*(b^2*c^2 - 2*a*c^3)*d^2*e^2 - 4*(b^3*c - 3*a*b*c^2)*d*e^3 + (b^4 - 4*a*b^2*c + 2*a^2* c^2)*e^4)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt( b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 6*(6*(b^2*c^3 - 4*a*c^4)*d^ 2*e^2 - 4*(b^3*c^2 - 4*a*b*c^3)*d*e^3 + (b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)* e^4)*x + 3*(4*(b^2*c^3 - 4*a*c^4)*d^3*e - 6*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 + 4*(b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*d*e^3 - (b^5 - 6*a*b^3*c + 8*a^2*b*c ^2)*e^4)*log(c*x^2 + b*x + a))/(b^2*c^4 - 4*a*c^5), 1/6*(2*(b^2*c^3 - 4*a* c^4)*e^4*x^3 + 3*(4*(b^2*c^3 - 4*a*c^4)*d*e^3 - (b^3*c^2 - 4*a*b*c^3)*e^4) *x^2 - 6*(2*c^4*d^4 - 4*b*c^3*d^3*e + 6*(b^2*c^2 - 2*a*c^3)*d^2*e^2 - 4*(b ^3*c - 3*a*b*c^2)*d*e^3 + (b^4 - 4*a*b^2*c + 2*a^2*c^2)*e^4)*sqrt(-b^2 + 4 *a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 6*(6*(b^2*c^ 3 - 4*a*c^4)*d^2*e^2 - 4*(b^3*c^2 - 4*a*b*c^3)*d*e^3 + (b^4*c - 5*a*b^2*c^ 2 + 4*a^2*c^3)*e^4)*x + 3*(4*(b^2*c^3 - 4*a*c^4)*d^3*e - 6*(b^3*c^2 - 4*a* b*c^3)*d^2*e^2 + 4*(b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*d*e^3 - (b^5 - 6*a*b^ 3*c + 8*a^2*b*c^2)*e^4)*log(c*x^2 + b*x + a))/(b^2*c^4 - 4*a*c^5)]
Leaf count of result is larger than twice the leaf count of optimal. 1556 vs. \(2 (238) = 476\).
Time = 4.15 (sec) , antiderivative size = 1556, normalized size of antiderivative = 6.40 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)**4/(c*x**2+b*x+a),x)
Output:
x**2*(-b*e**4/(2*c**2) + 2*d*e**3/c) + x*(-a*e**4/c**2 + b**2*e**4/c**3 - 4*b*d*e**3/c**2 + 6*d**2*e**2/c) + (e*(b*e - 2*c*d)*(2*a*c*e**2 - b**2*e** 2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**4) - sqrt(-4*a*c + b**2)*(2*a**2*c**2*e **4 - 4*a*b**2*c*e**4 + 12*a*b*c**2*d*e**3 - 12*a*c**3*d**2*e**2 + b**4*e* *4 - 4*b**3*c*d*e**3 + 6*b**2*c**2*d**2*e**2 - 4*b*c**3*d**3*e + 2*c**4*d* *4)/(2*c**4*(4*a*c - b**2)))*log(x + (-3*a**2*b*c*e**4 + 8*a**2*c**2*d*e** 3 + a*b**3*e**4 - 4*a*b**2*c*d*e**3 + 6*a*b*c**2*d**2*e**2 + 4*a*c**4*(e*( b*e - 2*c*d)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**4) - sqrt(-4*a*c + b**2)*(2*a**2*c**2*e**4 - 4*a*b**2*c*e**4 + 12*a*b*c**2*d*e **3 - 12*a*c**3*d**2*e**2 + b**4*e**4 - 4*b**3*c*d*e**3 + 6*b**2*c**2*d**2 *e**2 - 4*b*c**3*d**3*e + 2*c**4*d**4)/(2*c**4*(4*a*c - b**2))) - 8*a*c**3 *d**3*e - b**2*c**3*(e*(b*e - 2*c*d)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**4) - sqrt(-4*a*c + b**2)*(2*a**2*c**2*e**4 - 4*a*b**2* c*e**4 + 12*a*b*c**2*d*e**3 - 12*a*c**3*d**2*e**2 + b**4*e**4 - 4*b**3*c*d *e**3 + 6*b**2*c**2*d**2*e**2 - 4*b*c**3*d**3*e + 2*c**4*d**4)/(2*c**4*(4* a*c - b**2))) + b*c**3*d**4)/(2*a**2*c**2*e**4 - 4*a*b**2*c*e**4 + 12*a*b* c**2*d*e**3 - 12*a*c**3*d**2*e**2 + b**4*e**4 - 4*b**3*c*d*e**3 + 6*b**2*c **2*d**2*e**2 - 4*b*c**3*d**3*e + 2*c**4*d**4)) + (e*(b*e - 2*c*d)*(2*a*c* e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**4) + sqrt(-4*a*c + b**2) *(2*a**2*c**2*e**4 - 4*a*b**2*c*e**4 + 12*a*b*c**2*d*e**3 - 12*a*c**3*d...
Exception generated. \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^4/(c*x^2+b*x+a),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
Time = 0.40 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx=\frac {2 \, c^{2} e^{4} x^{3} + 12 \, c^{2} d e^{3} x^{2} - 3 \, b c e^{4} x^{2} + 36 \, c^{2} d^{2} e^{2} x - 24 \, b c d e^{3} x + 6 \, b^{2} e^{4} x - 6 \, a c e^{4} x}{6 \, c^{3}} + \frac {{\left (4 \, c^{3} d^{3} e - 6 \, b c^{2} d^{2} e^{2} + 4 \, b^{2} c d e^{3} - 4 \, a c^{2} d e^{3} - b^{3} e^{4} + 2 \, a b c e^{4}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{4}} + \frac {{\left (2 \, c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 12 \, a c^{3} d^{2} e^{2} - 4 \, b^{3} c d e^{3} + 12 \, a b c^{2} d e^{3} + b^{4} e^{4} - 4 \, a b^{2} c e^{4} + 2 \, a^{2} c^{2} e^{4}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{4}} \] Input:
integrate((e*x+d)^4/(c*x^2+b*x+a),x, algorithm="giac")
Output:
1/6*(2*c^2*e^4*x^3 + 12*c^2*d*e^3*x^2 - 3*b*c*e^4*x^2 + 36*c^2*d^2*e^2*x - 24*b*c*d*e^3*x + 6*b^2*e^4*x - 6*a*c*e^4*x)/c^3 + 1/2*(4*c^3*d^3*e - 6*b* c^2*d^2*e^2 + 4*b^2*c*d*e^3 - 4*a*c^2*d*e^3 - b^3*e^4 + 2*a*b*c*e^4)*log(c *x^2 + b*x + a)/c^4 + (2*c^4*d^4 - 4*b*c^3*d^3*e + 6*b^2*c^2*d^2*e^2 - 12* a*c^3*d^2*e^2 - 4*b^3*c*d*e^3 + 12*a*b*c^2*d*e^3 + b^4*e^4 - 4*a*b^2*c*e^4 + 2*a^2*c^2*e^4)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a* c)*c^4)
Time = 5.82 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.51 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx=x\,\left (\frac {b\,\left (\frac {b\,e^4}{c^2}-\frac {4\,d\,e^3}{c}\right )}{c}-\frac {a\,e^4}{c^2}+\frac {6\,d^2\,e^2}{c}\right )-x^2\,\left (\frac {b\,e^4}{2\,c^2}-\frac {2\,d\,e^3}{c}\right )+\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (8\,a^2\,b\,c^2\,e^4-16\,a^2\,c^3\,d\,e^3-6\,a\,b^3\,c\,e^4+20\,a\,b^2\,c^2\,d\,e^3-24\,a\,b\,c^3\,d^2\,e^2+16\,a\,c^4\,d^3\,e+b^5\,e^4-4\,b^4\,c\,d\,e^3+6\,b^3\,c^2\,d^2\,e^2-4\,b^2\,c^3\,d^3\,e\right )}{2\,\left (4\,a\,c^5-b^2\,c^4\right )}+\frac {e^4\,x^3}{3\,c}+\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (2\,a^2\,c^2\,e^4-4\,a\,b^2\,c\,e^4+12\,a\,b\,c^2\,d\,e^3-12\,a\,c^3\,d^2\,e^2+b^4\,e^4-4\,b^3\,c\,d\,e^3+6\,b^2\,c^2\,d^2\,e^2-4\,b\,c^3\,d^3\,e+2\,c^4\,d^4\right )}{c^4\,\sqrt {4\,a\,c-b^2}} \] Input:
int((d + e*x)^4/(a + b*x + c*x^2),x)
Output:
x*((b*((b*e^4)/c^2 - (4*d*e^3)/c))/c - (a*e^4)/c^2 + (6*d^2*e^2)/c) - x^2* ((b*e^4)/(2*c^2) - (2*d*e^3)/c) + (log(a + b*x + c*x^2)*(b^5*e^4 + 8*a^2*b *c^2*e^4 - 16*a^2*c^3*d*e^3 - 4*b^2*c^3*d^3*e + 6*b^3*c^2*d^2*e^2 - 6*a*b^ 3*c*e^4 + 16*a*c^4*d^3*e - 4*b^4*c*d*e^3 - 24*a*b*c^3*d^2*e^2 + 20*a*b^2*c ^2*d*e^3))/(2*(4*a*c^5 - b^2*c^4)) + (e^4*x^3)/(3*c) + (atan(b/(4*a*c - b^ 2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2))*(b^4*e^4 + 2*c^4*d^4 + 2*a^2*c^2*e ^4 - 12*a*c^3*d^2*e^2 + 6*b^2*c^2*d^2*e^2 - 4*a*b^2*c*e^4 - 4*b*c^3*d^3*e - 4*b^3*c*d*e^3 + 12*a*b*c^2*d*e^3))/(c^4*(4*a*c - b^2)^(1/2))
Time = 0.19 (sec) , antiderivative size = 793, normalized size of antiderivative = 3.26 \[ \int \frac {(d+e x)^4}{a+b x+c x^2} \, dx =\text {Too large to display} \] Input:
int((e*x+d)^4/(c*x^2+b*x+a),x)
Output:
(12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*c**2*e**4 - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c*e** 4 + 72*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c**2*d* e**3 - 72*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c**3*d **2*e**2 + 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**4* e**4 - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*c*d *e**3 + 36*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c* *2*d**2*e**2 - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))* b*c**3*d**3*e + 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2)) *c**4*d**4 + 24*log(a + b*x + c*x**2)*a**2*b*c**2*e**4 - 48*log(a + b*x + c*x**2)*a**2*c**3*d*e**3 - 18*log(a + b*x + c*x**2)*a*b**3*c*e**4 + 60*log (a + b*x + c*x**2)*a*b**2*c**2*d*e**3 - 72*log(a + b*x + c*x**2)*a*b*c**3* d**2*e**2 + 48*log(a + b*x + c*x**2)*a*c**4*d**3*e + 3*log(a + b*x + c*x** 2)*b**5*e**4 - 12*log(a + b*x + c*x**2)*b**4*c*d*e**3 + 18*log(a + b*x + c *x**2)*b**3*c**2*d**2*e**2 - 12*log(a + b*x + c*x**2)*b**2*c**3*d**3*e - 2 4*a**2*c**3*e**4*x + 30*a*b**2*c**2*e**4*x - 96*a*b*c**3*d*e**3*x - 12*a*b *c**3*e**4*x**2 + 144*a*c**4*d**2*e**2*x + 48*a*c**4*d*e**3*x**2 + 8*a*c** 4*e**4*x**3 - 6*b**4*c*e**4*x + 24*b**3*c**2*d*e**3*x + 3*b**3*c**2*e**4*x **2 - 36*b**2*c**3*d**2*e**2*x - 12*b**2*c**3*d*e**3*x**2 - 2*b**2*c**3*e* *4*x**3)/(6*c**4*(4*a*c - b**2))