Integrand size = 23, antiderivative size = 356 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx=\frac {(2 c d-b e) n}{6 e \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac {\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) n}{3 e \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {\sqrt {b^2-4 a c} \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{3 \left (c d^2-b d e+a e^2\right )^3}-\frac {(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) n \log (d+e x)}{3 e \left (c d^2-b d e+a e^2\right )^3}+\frac {(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) n \log \left (a+b x+c x^2\right )}{6 e \left (c d^2-b d e+a e^2\right )^3}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 e (d+e x)^3} \] Output:
1/6*(-b*e+2*c*d)*n/e/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2+1/3*(2*c^2*d^2+b^2*e^2- 2*c*e*(a*e+b*d))*n/e/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)+1/3*(-4*a*c+b^2)^(1/2)* (3*c^2*d^2+b^2*e^2-c*e*(a*e+3*b*d))*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2) )/(a*e^2-b*d*e+c*d^2)^3-1/3*(-b*e+2*c*d)*(c^2*d^2+b^2*e^2-c*e*(3*a*e+b*d)) *n*ln(e*x+d)/e/(a*e^2-b*d*e+c*d^2)^3+1/6*(-b*e+2*c*d)*(c^2*d^2+b^2*e^2-c*e *(3*a*e+b*d))*n*ln(c*x^2+b*x+a)/e/(a*e^2-b*d*e+c*d^2)^3-1/3*ln(d*(c*x^2+b* x+a)^n)/e/(e*x+d)^3
Time = 1.13 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx=\frac {\frac {n (d+e x) \left ((2 c d-b e) \left (c d^2+e (-b d+a e)\right )^2+2 \left (c d^2+e (-b d+a e)\right ) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) (d+e x)+2 \sqrt {b^2-4 a c} e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) (d+e x)^2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )-2 (2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) (d+e x)^2 \log (d+e x)+(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) (d+e x)^2 \log (a+x (b+c x))\right )}{\left (c d^2+e (-b d+a e)\right )^3}-2 \log \left (d (a+x (b+c x))^n\right )}{6 e (d+e x)^3} \] Input:
Integrate[Log[d*(a + b*x + c*x^2)^n]/(d + e*x)^4,x]
Output:
((n*(d + e*x)*((2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))^2 + 2*(c*d^2 + e*( -(b*d) + a*e))*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*(d + e*x) + 2*Sqr t[b^2 - 4*a*c]*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*(d + e*x)^2*Arc Tanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]] - 2*(2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e))*(d + e*x)^2*Log[d + e*x] + (2*c*d - b*e)*(c^2*d^2 + b^ 2*e^2 - c*e*(b*d + 3*a*e))*(d + e*x)^2*Log[a + x*(b + c*x)]))/(c*d^2 + e*( -(b*d) + a*e))^3 - 2*Log[d*(a + x*(b + c*x))^n])/(6*e*(d + e*x)^3)
Time = 0.91 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3005, 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.
| \(\displaystyle \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 3005 |
| \(\displaystyle \frac {n \int \frac {b+2 c x}{(d+e x)^3 \left (c x^2+b x+a\right )}dx}{3 e}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {n \int \left (\frac {e (b e-2 c d)}{\left (c d^2-b e d+a e^2\right ) (d+e x)^3}+\frac {e (2 c d-b e) \left (-c^2 d^2-b^2 e^2+c e (b d+3 a e)\right )}{\left (c d^2-b e d+a e^2\right )^3 (d+e x)}+\frac {-e^3 b^4+3 c d e^2 b^3-c e \left (3 c d^2-4 a e^2\right ) b^2+c^2 d \left (c d^2-9 a e^2\right ) b+2 a c^2 e \left (3 c d^2-a e^2\right )+c (2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right ) x}{\left (c d^2-b e d+a e^2\right )^3 \left (c x^2+b x+a\right )}+\frac {e \left (-2 c^2 d^2-b^2 e^2+2 c e (b d+a e)\right )}{\left (c d^2-b e d+a e^2\right )^2 (d+e x)^2}\right )dx}{3 e}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {n \left (\frac {e \sqrt {b^2-4 a c} \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^3}+\frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^3}+\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{(d+e x) \left (a e^2-b d e+c d^2\right )^2}-\frac {(2 c d-b e) \log (d+e x) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^3}+\frac {2 c d-b e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}\right )}{3 e}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 e (d+e x)^3}\) |
Input:
Int[Log[d*(a + b*x + c*x^2)^n]/(d + e*x)^4,x]
Output:
(n*((2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + (2*c^2*d^2 + b ^2*e^2 - 2*c*e*(b*d + a*e))/((c*d^2 - b*d*e + a*e^2)^2*(d + e*x)) + (Sqrt[ b^2 - 4*a*c]*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*ArcTanh[(b + 2*c* x)/Sqrt[b^2 - 4*a*c]])/(c*d^2 - b*d*e + a*e^2)^3 - ((2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e))*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^3 + ((2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e))*Log[a + b*x + c*x^2 ])/(2*(c*d^2 - b*d*e + a*e^2)^3)))/(3*e) - Log[d*(a + b*x + c*x^2)^n]/(3*e *(d + e*x)^3)
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]
Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_. ), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))) , x] - Simp[b*n*(p/(e*(m + 1))) Int[SimplifyIntegrand[(d + e*x)^(m + 1)*( a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]
Time = 4.46 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.37
| method | result | size |
| parts | \(-\frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )}{3 e \left (e x +d \right )^{3}}+\frac {n \left (-\frac {2 a c \,e^{2}-b^{2} e^{2}+2 b c d e -2 c^{2} d^{2}}{\left (a \,e^{2}-b d e +d^{2} c \right )^{2} \left (e x +d \right )}-\frac {\left (3 a b c \,e^{3}-6 a \,c^{2} d \,e^{2}-b^{3} e^{3}+3 b^{2} c d \,e^{2}-3 b \,c^{2} d^{2} e +2 c^{3} d^{3}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +d^{2} c \right )^{3}}-\frac {b e -2 c d}{2 \left (a \,e^{2}-b d e +d^{2} c \right ) \left (e x +d \right )^{2}}+\frac {\frac {\left (3 a b \,c^{2} e^{3}-6 a \,c^{3} e^{2} d -b^{3} c \,e^{3}+3 b^{2} c^{2} d \,e^{2}-3 b \,c^{3} d^{2} e +2 c^{4} d^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-2 a^{2} c^{2} e^{3}+4 a \,b^{2} c \,e^{3}-9 a b \,c^{2} d \,e^{2}+6 a \,c^{3} d^{2} e -b^{4} e^{3}+3 b^{3} c d \,e^{2}-3 b^{2} c^{2} d^{2} e +b \,c^{3} d^{3}-\frac {\left (3 a b \,c^{2} e^{3}-6 a \,c^{3} e^{2} d -b^{3} c \,e^{3}+3 b^{2} c^{2} d \,e^{2}-3 b \,c^{3} d^{2} e +2 c^{4} d^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,e^{2}-b d e +d^{2} c \right )^{3}}\right )}{3 e}\) | \(486\) |
| risch | \(\text {Expression too large to display}\) | \(72038\) |
Input:
int(ln(d*(c*x^2+b*x+a)^n)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
-1/3*ln(d*(c*x^2+b*x+a)^n)/e/(e*x+d)^3+1/3/e*n*(-(2*a*c*e^2-b^2*e^2+2*b*c* d*e-2*c^2*d^2)/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)-(3*a*b*c*e^3-6*a*c^2*d*e^2-b^ 3*e^3+3*b^2*c*d*e^2-3*b*c^2*d^2*e+2*c^3*d^3)/(a*e^2-b*d*e+c*d^2)^3*ln(e*x+ d)-1/2*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2+1/(a*e^2-b*d*e+c*d^2)^3*( 1/2*(3*a*b*c^2*e^3-6*a*c^3*d*e^2-b^3*c*e^3+3*b^2*c^2*d*e^2-3*b*c^3*d^2*e+2 *c^4*d^3)/c*ln(c*x^2+b*x+a)+2*(-2*a^2*c^2*e^3+4*a*b^2*c*e^3-9*a*b*c^2*d*e^ 2+6*a*c^3*d^2*e-b^4*e^3+3*b^3*c*d*e^2-3*b^2*c^2*d^2*e+b*c^3*d^3-1/2*(3*a*b *c^2*e^3-6*a*c^3*d*e^2-b^3*c*e^3+3*b^2*c^2*d*e^2-3*b*c^3*d^2*e+2*c^4*d^3)* b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 1496 vs. \(2 (340) = 680\).
Time = 8.13 (sec) , antiderivative size = 3013, normalized size of antiderivative = 8.46 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx=\text {Too large to display} \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/(e*x+d)^4,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx=\text {Timed out} \] Input:
integrate(ln(d*(c*x**2+b*x+a)**n)/(e*x+d)**4,x)
Output:
Timed out
Exception generated. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/(e*x+d)^4,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
Leaf count of result is larger than twice the leaf count of optimal. 1128 vs. \(2 (340) = 680\).
Time = 0.21 (sec) , antiderivative size = 1128, normalized size of antiderivative = 3.17 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx =\text {Too large to display} \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/(e*x+d)^4,x, algorithm="giac")
Output:
1/6*(2*c^3*d^3*n - 3*b*c^2*d^2*e*n + 3*b^2*c*d*e^2*n - 6*a*c^2*d*e^2*n - b ^3*e^3*n + 3*a*b*c*e^3*n)*log(c*x^2 + b*x + a)/(c^3*d^6*e - 3*b*c^2*d^5*e^ 2 + 3*b^2*c*d^4*e^3 + 3*a*c^2*d^4*e^3 - b^3*d^3*e^4 - 6*a*b*c*d^3*e^4 + 3* a*b^2*d^2*e^5 + 3*a^2*c*d^2*e^5 - 3*a^2*b*d*e^6 + a^3*e^7) - 1/3*n*log(c*x ^2 + b*x + a)/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e) - 1/3*(2*c^3*d ^3*n - 3*b*c^2*d^2*e*n + 3*b^2*c*d*e^2*n - 6*a*c^2*d*e^2*n - b^3*e^3*n + 3 *a*b*c*e^3*n)*log(e*x + d)/(c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e^3 + 3*a*c^2*d^4*e^3 - b^3*d^3*e^4 - 6*a*b*c*d^3*e^4 + 3*a*b^2*d^2*e^5 + 3*a^ 2*c*d^2*e^5 - 3*a^2*b*d*e^6 + a^3*e^7) - 1/3*(3*b^2*c^2*d^2*n - 12*a*c^3*d ^2*n - 3*b^3*c*d*e*n + 12*a*b*c^2*d*e*n + b^4*e^2*n - 5*a*b^2*c*e^2*n + 4* a^2*c^2*e^2*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^3*d^6 - 3*b*c^2* d^5*e + 3*b^2*c*d^4*e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3 - 6*a*b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 + a^3*e^6)*sqrt(-b^2 + 4*a*c)) + 1/6*(4*c^2*d^2*e^2*n*x^2 - 4*b*c*d*e^3*n*x^2 + 2*b^2*e^4*n*x^2 - 4*a*c*e^4*n*x^2 + 10*c^2*d^3*e*n*x - 11*b*c*d^2*e^2*n*x + 5*b^2*d*e^3*n* x - 6*a*c*d*e^3*n*x - a*b*e^4*n*x + 6*c^2*d^4*n - 7*b*c*d^3*e*n + 3*b^2*d^ 2*e^2*n - 2*a*c*d^2*e^2*n - a*b*d*e^3*n - 2*c^2*d^4*log(d) + 4*b*c*d^3*e*l og(d) - 2*b^2*d^2*e^2*log(d) - 4*a*c*d^2*e^2*log(d) + 4*a*b*d*e^3*log(d) - 2*a^2*e^4*log(d))/(c^2*d^4*e^4*x^3 - 2*b*c*d^3*e^5*x^3 + b^2*d^2*e^6*x^3 + 2*a*c*d^2*e^6*x^3 - 2*a*b*d*e^7*x^3 + a^2*e^8*x^3 + 3*c^2*d^5*e^3*x^2...
Time = 43.24 (sec) , antiderivative size = 2707, normalized size of antiderivative = 7.60 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx=\text {Too large to display} \] Input:
int(log(d*(a + b*x + c*x^2)^n)/(d + e*x)^4,x)
Output:
(log(d + e*x)*(e^3*(b^3*n - 3*a*b*c*n) + e^2*(6*a*c^2*d*n - 3*b^2*c*d*n) - 2*c^3*d^3*n + 3*b*c^2*d^2*e*n))/(3*a^3*e^7 + 3*c^3*d^6*e - 3*b^3*d^3*e^4 + 9*a*b^2*d^2*e^5 + 9*a*c^2*d^4*e^3 + 9*a^2*c*d^2*e^5 - 9*b*c^2*d^5*e^2 + 9*b^2*c*d^4*e^3 - 9*a^2*b*d*e^6 - 18*a*b*c*d^3*e^4) - (log(32*a*b^5*e^5 - 2*a*e^5*(b^2 - 4*a*c)^(5/2) - 192*a*c^5*d^5 + 32*b^6*e^5*x + 48*b^2*c^4*d^ 5 - 18*b^3*e^5*x*(b^2 - 4*a*c)^(3/2) - 3*b^5*e^5*x*(b^2 - 4*a*c)^(1/2) + 9 6*c^5*d^5*x*(b^2 - 4*a*c)^(1/2) - 208*a^2*b^3*c*e^5 + 320*a^3*b*c^2*e^5 - 704*a^3*c^3*d*e^4 - 48*b^3*c^3*d^4*e - 16*b^5*c*d^2*e^3 - 64*a^3*c^3*e^5*x + 1152*a^2*c^4*d^3*e^2 + 48*b^4*c^2*d^3*e^2 - 33*b*d*e^4*(b^2 - 4*a*c)^(5 /2) - 11*b*e^5*x*(b^2 - 4*a*c)^(5/2) - 24*a*b^2*e^5*(b^2 - 4*a*c)^(3/2) - 6*a*b^4*e^5*(b^2 - 4*a*c)^(1/2) + 48*b*c^4*d^5*(b^2 - 4*a*c)^(1/2) + 18*b^ 3*d*e^4*(b^2 - 4*a*c)^(3/2) + 15*b^5*d*e^4*(b^2 - 4*a*c)^(1/2) + 44*c*d^2* e^3*(b^2 - 4*a*c)^(5/2) + 72*c^3*d^4*e*(b^2 - 4*a*c)^(3/2) + 22*c*d*e^4*x* (b^2 - 4*a*c)^(5/2) + 192*a*b*c^4*d^4*e - 128*a*b^4*c*d*e^4 + 120*b^3*c^2* d^3*e^2*(b^2 - 4*a*c)^(1/2) - 224*a*b^4*c*e^5*x - 576*a*c^5*d^4*e*x - 160* b^5*c*d*e^4*x + 144*b^2*c^4*d^4*e*x - 72*b*c^2*d^3*e^2*(b^2 - 4*a*c)^(3/2) - 120*b^2*c^3*d^4*e*(b^2 - 4*a*c)^(1/2) - 60*b^4*c*d^2*e^3*(b^2 - 4*a*c)^ (1/2) + 144*c^3*d^3*e^2*x*(b^2 - 4*a*c)^(3/2) - 480*a*b^2*c^3*d^3*e^2 + 32 0*a*b^3*c^2*d^2*e^3 - 1024*a^2*b*c^3*d^2*e^3 + 688*a^2*b^2*c^2*d*e^4 + 400 *a^2*b^2*c^2*e^5*x + 1408*a^2*c^4*d^2*e^3*x - 288*b^3*c^3*d^3*e^2*x + 3...
Time = 0.17 (sec) , antiderivative size = 3818, normalized size of antiderivative = 10.72 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^4} \, dx =\text {Too large to display} \] Input:
int(log(d*(c*x^2+b*x+a)^n)/(e*x+d)^4,x)
Output:
( - 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c*d**6*e** 3*n - 18*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c*d**5* e**4*n*x - 18*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c* d**4*e**5*n*x**2 - 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2 ))*a*c*d**3*e**6*n*x**3 + 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*d**6*e**3*n + 18*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4 *a*c - b**2))*b**2*d**5*e**4*n*x + 18*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/ sqrt(4*a*c - b**2))*b**2*d**4*e**5*n*x**2 + 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*d**3*e**6*n*x**3 - 18*sqrt(4*a*c - b**2)* atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c*d**7*e**2*n - 54*sqrt(4*a*c - b** 2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c*d**6*e**3*n*x - 54*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c*d**5*e**4*n*x**2 - 18*sqr t(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c*d**4*e**5*n*x**3 + 18*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**2*d**8*e*n + 54*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**2*d**7*e* *2*n*x + 54*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**2*d **6*e**3*n*x**2 + 18*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2 ))*c**2*d**5*e**4*n*x**3 - 6*log(a + b*x + c*x**2)*a**3*d**3*e**6*n - 18*l og(a + b*x + c*x**2)*a**3*d**2*e**7*n*x - 18*log(a + b*x + c*x**2)*a**3*d* e**8*n*x**2 - 6*log(a + b*x + c*x**2)*a**3*e**9*n*x**3 + 18*log(a + b*x...