Integrand size = 55, antiderivative size = 1668 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx =\text {Too large to display} \]
k*x/c+1/2*l*x^2/c+1/3*m*x^3/c+1/6*(-b*m+c*j)*ln(c*x^6+b*x^3+a)/c^2+1/6*ln( 2^(1/3)*c^(1/3)*x+(b-(-4*a*c+b^2)^(1/2))^(1/3))*(g-b*k/c+(2*c^2*d+b^2*k-c* (2*a*k+b*g))/c/(-4*a*c+b^2)^(1/2))*2^(2/3)/c^(1/3)/(b-(-4*a*c+b^2)^(1/2))^ (2/3)-1/12*ln(2^(2/3)*c^(2/3)*x^2-2^(1/3)*c^(1/3)*x*(b-(-4*a*c+b^2)^(1/2)) ^(1/3)+(b-(-4*a*c+b^2)^(1/2))^(2/3))*(g-b*k/c+(2*c^2*d+b^2*k-c*(2*a*k+b*g) )/c/(-4*a*c+b^2)^(1/2))*2^(2/3)/c^(1/3)/(b-(-4*a*c+b^2)^(1/2))^(2/3)-1/6*a rctan(1/3*(1-2*2^(1/3)*c^(1/3)*x/(b-(-4*a*c+b^2)^(1/2))^(1/3))*3^(1/2))*(g -b*k/c+(2*c^2*d+b^2*k-c*(2*a*k+b*g))/c/(-4*a*c+b^2)^(1/2))*2^(2/3)/c^(1/3) *3^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(2/3)-1/6*ln(2^(1/3)*c^(1/3)*x+(b-(-4*a*c+ b^2)^(1/2))^(1/3))*(h-b*l/c+(2*c^2*e+b^2*l-c*(2*a*l+b*h))/c/(-4*a*c+b^2)^( 1/2))*2^(1/3)/c^(2/3)/(b-(-4*a*c+b^2)^(1/2))^(1/3)+1/12*ln(2^(2/3)*c^(2/3) *x^2-2^(1/3)*c^(1/3)*x*(b-(-4*a*c+b^2)^(1/2))^(1/3)+(b-(-4*a*c+b^2)^(1/2)) ^(2/3))*(h-b*l/c+(2*c^2*e+b^2*l-c*(2*a*l+b*h))/c/(-4*a*c+b^2)^(1/2))*2^(1/ 3)/c^(2/3)/(b-(-4*a*c+b^2)^(1/2))^(1/3)-1/6*arctan(1/3*(1-2*2^(1/3)*c^(1/3 )*x/(b-(-4*a*c+b^2)^(1/2))^(1/3))*3^(1/2))*(h-b*l/c+(2*c^2*e+b^2*l-c*(2*a* l+b*h))/c/(-4*a*c+b^2)^(1/2))*2^(1/3)/c^(2/3)*3^(1/2)/(b-(-4*a*c+b^2)^(1/2 ))^(1/3)-1/3*(-2*a*c*m+b^2*m-b*c*j+2*c^2*f)*arctanh((2*c*x^3+b)/(-4*a*c+b^ 2)^(1/2))/c^2/(-4*a*c+b^2)^(1/2)+1/6*ln(2^(1/3)*c^(1/3)*x+(b+(-4*a*c+b^2)^ (1/2))^(1/3))*(g-b*k/c+(2*a*c*k-b^2*k+b*c*g-2*c^2*d)/c/(-4*a*c+b^2)^(1/2)) *2^(2/3)/c^(1/3)/(b+(-4*a*c+b^2)^(1/2))^(2/3)-1/12*ln(2^(2/3)*c^(2/3)*x...
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.74 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.13 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx=\frac {6 k x+3 l x^2+2 m x^3-2 \text {RootSum}\left [a+b \text {$\#$1}^3+c \text {$\#$1}^6\&,\frac {-c d \log (x-\text {$\#$1})+a k \log (x-\text {$\#$1})-c e \log (x-\text {$\#$1}) \text {$\#$1}+a l \log (x-\text {$\#$1}) \text {$\#$1}-c f \log (x-\text {$\#$1}) \text {$\#$1}^2+a m \log (x-\text {$\#$1}) \text {$\#$1}^2-c g \log (x-\text {$\#$1}) \text {$\#$1}^3+b k \log (x-\text {$\#$1}) \text {$\#$1}^3-c h \log (x-\text {$\#$1}) \text {$\#$1}^4+b l \log (x-\text {$\#$1}) \text {$\#$1}^4-c j \log (x-\text {$\#$1}) \text {$\#$1}^5+b m \log (x-\text {$\#$1}) \text {$\#$1}^5}{b \text {$\#$1}^2+2 c \text {$\#$1}^5}\&\right ]}{6 c} \]
Integrate[(d + e*x + f*x^2 + g*x^3 + h*x^4 + j*x^5 + k*x^6 + l*x^7 + m*x^8 )/(a + b*x^3 + c*x^6),x]
(6*k*x + 3*l*x^2 + 2*m*x^3 - 2*RootSum[a + b*#1^3 + c*#1^6 & , (-(c*d*Log[ x - #1]) + a*k*Log[x - #1] - c*e*Log[x - #1]*#1 + a*l*Log[x - #1]*#1 - c*f *Log[x - #1]*#1^2 + a*m*Log[x - #1]*#1^2 - c*g*Log[x - #1]*#1^3 + b*k*Log[ x - #1]*#1^3 - c*h*Log[x - #1]*#1^4 + b*l*Log[x - #1]*#1^4 - c*j*Log[x - # 1]*#1^5 + b*m*Log[x - #1]*#1^5)/(b*#1^2 + 2*c*#1^5) & ])/(6*c)
Time = 3.72 (sec) , antiderivative size = 1668, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2322, 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.
| \(\displaystyle \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx\) |
| \(\Big \downarrow \) 2322 |
| \(\displaystyle \int \left (\frac {d+g x^3+k x^6}{a+b x^3+c x^6}+\frac {x \left (e+h x^3+l x^6\right )}{a+b x^3+c x^6}+\frac {x^2 \left (f+j x^3+m x^6\right )}{a+b x^3+c x^6}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {m x^3}{3 c}+\frac {l x^2}{2 c}+\frac {k x}{c}-\frac {\left (g-\frac {b k}{c}+\frac {k b^2+2 c^2 d-c (b g+2 a k)}{c \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \sqrt [3]{c} x}{\sqrt [3]{b-\sqrt {b^2-4 a c}}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} \sqrt [3]{c} \left (b-\sqrt {b^2-4 a c}\right )^{2/3}}-\frac {\left (h-\frac {b l}{c}+\frac {l b^2+2 c^2 e-c (b h+2 a l)}{c \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \sqrt [3]{c} x}{\sqrt [3]{b-\sqrt {b^2-4 a c}}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} c^{2/3} \sqrt [3]{b-\sqrt {b^2-4 a c}}}-\frac {\left (g-\frac {b k}{c}-\frac {k b^2-c g b+2 c^2 d-2 a c k}{c \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \sqrt [3]{c} x}{\sqrt [3]{b+\sqrt {b^2-4 a c}}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} \sqrt [3]{c} \left (b+\sqrt {b^2-4 a c}\right )^{2/3}}-\frac {\left (h-\frac {b l}{c}-\frac {l b^2-c h b+2 c^2 e-2 a c l}{c \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \sqrt [3]{c} x}{\sqrt [3]{b+\sqrt {b^2-4 a c}}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} c^{2/3} \sqrt [3]{b+\sqrt {b^2-4 a c}}}-\frac {\left (m b^2-c j b+2 c^2 f-2 a c m\right ) \text {arctanh}\left (\frac {2 c x^3+b}{\sqrt {b^2-4 a c}}\right )}{3 c^2 \sqrt {b^2-4 a c}}+\frac {\left (g-\frac {b k}{c}+\frac {k b^2+2 c^2 d-c (b g+2 a k)}{c \sqrt {b^2-4 a c}}\right ) \log \left (\sqrt [3]{2} \sqrt [3]{c} x+\sqrt [3]{b-\sqrt {b^2-4 a c}}\right )}{3 \sqrt [3]{2} \sqrt [3]{c} \left (b-\sqrt {b^2-4 a c}\right )^{2/3}}-\frac {\left (h-\frac {b l}{c}+\frac {l b^2+2 c^2 e-c (b h+2 a l)}{c \sqrt {b^2-4 a c}}\right ) \log \left (\sqrt [3]{2} \sqrt [3]{c} x+\sqrt [3]{b-\sqrt {b^2-4 a c}}\right )}{3\ 2^{2/3} c^{2/3} \sqrt [3]{b-\sqrt {b^2-4 a c}}}+\frac {\left (g-\frac {b k}{c}-\frac {k b^2-c g b+2 c^2 d-2 a c k}{c \sqrt {b^2-4 a c}}\right ) \log \left (\sqrt [3]{2} \sqrt [3]{c} x+\sqrt [3]{b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt [3]{2} \sqrt [3]{c} \left (b+\sqrt {b^2-4 a c}\right )^{2/3}}-\frac {\left (h-\frac {b l}{c}-\frac {l b^2-c h b+2 c^2 e-2 a c l}{c \sqrt {b^2-4 a c}}\right ) \log \left (\sqrt [3]{2} \sqrt [3]{c} x+\sqrt [3]{b+\sqrt {b^2-4 a c}}\right )}{3\ 2^{2/3} c^{2/3} \sqrt [3]{b+\sqrt {b^2-4 a c}}}-\frac {\left (g-\frac {b k}{c}+\frac {k b^2+2 c^2 d-c (b g+2 a k)}{c \sqrt {b^2-4 a c}}\right ) \log \left (2^{2/3} c^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{c} \sqrt [3]{b-\sqrt {b^2-4 a c}} x+\left (b-\sqrt {b^2-4 a c}\right )^{2/3}\right )}{6 \sqrt [3]{2} \sqrt [3]{c} \left (b-\sqrt {b^2-4 a c}\right )^{2/3}}+\frac {\left (h-\frac {b l}{c}+\frac {l b^2+2 c^2 e-c (b h+2 a l)}{c \sqrt {b^2-4 a c}}\right ) \log \left (2^{2/3} c^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{c} \sqrt [3]{b-\sqrt {b^2-4 a c}} x+\left (b-\sqrt {b^2-4 a c}\right )^{2/3}\right )}{6\ 2^{2/3} c^{2/3} \sqrt [3]{b-\sqrt {b^2-4 a c}}}-\frac {\left (g-\frac {b k}{c}-\frac {k b^2-c g b+2 c^2 d-2 a c k}{c \sqrt {b^2-4 a c}}\right ) \log \left (2^{2/3} c^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{c} \sqrt [3]{b+\sqrt {b^2-4 a c}} x+\left (b+\sqrt {b^2-4 a c}\right )^{2/3}\right )}{6 \sqrt [3]{2} \sqrt [3]{c} \left (b+\sqrt {b^2-4 a c}\right )^{2/3}}+\frac {\left (h-\frac {b l}{c}-\frac {l b^2-c h b+2 c^2 e-2 a c l}{c \sqrt {b^2-4 a c}}\right ) \log \left (2^{2/3} c^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{c} \sqrt [3]{b+\sqrt {b^2-4 a c}} x+\left (b+\sqrt {b^2-4 a c}\right )^{2/3}\right )}{6\ 2^{2/3} c^{2/3} \sqrt [3]{b+\sqrt {b^2-4 a c}}}+\frac {(c j-b m) \log \left (c x^6+b x^3+a\right )}{6 c^2}\) |
(k*x)/c + (l*x^2)/(2*c) + (m*x^3)/(3*c) - ((g - (b*k)/c + (2*c^2*d + b^2*k - c*(b*g + 2*a*k))/(c*Sqrt[b^2 - 4*a*c]))*ArcTan[(1 - (2*2^(1/3)*c^(1/3)* x)/(b - Sqrt[b^2 - 4*a*c])^(1/3))/Sqrt[3]])/(2^(1/3)*Sqrt[3]*c^(1/3)*(b - Sqrt[b^2 - 4*a*c])^(2/3)) - ((h - (b*l)/c + (2*c^2*e + b^2*l - c*(b*h + 2* a*l))/(c*Sqrt[b^2 - 4*a*c]))*ArcTan[(1 - (2*2^(1/3)*c^(1/3)*x)/(b - Sqrt[b ^2 - 4*a*c])^(1/3))/Sqrt[3]])/(2^(2/3)*Sqrt[3]*c^(2/3)*(b - Sqrt[b^2 - 4*a *c])^(1/3)) - ((g - (b*k)/c - (2*c^2*d - b*c*g + b^2*k - 2*a*c*k)/(c*Sqrt[ b^2 - 4*a*c]))*ArcTan[(1 - (2*2^(1/3)*c^(1/3)*x)/(b + Sqrt[b^2 - 4*a*c])^( 1/3))/Sqrt[3]])/(2^(1/3)*Sqrt[3]*c^(1/3)*(b + Sqrt[b^2 - 4*a*c])^(2/3)) - ((h - (b*l)/c - (2*c^2*e - b*c*h + b^2*l - 2*a*c*l)/(c*Sqrt[b^2 - 4*a*c])) *ArcTan[(1 - (2*2^(1/3)*c^(1/3)*x)/(b + Sqrt[b^2 - 4*a*c])^(1/3))/Sqrt[3]] )/(2^(2/3)*Sqrt[3]*c^(2/3)*(b + Sqrt[b^2 - 4*a*c])^(1/3)) - ((2*c^2*f - b* c*j + b^2*m - 2*a*c*m)*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*c^2*Sq rt[b^2 - 4*a*c]) + ((g - (b*k)/c + (2*c^2*d + b^2*k - c*(b*g + 2*a*k))/(c* Sqrt[b^2 - 4*a*c]))*Log[(b - Sqrt[b^2 - 4*a*c])^(1/3) + 2^(1/3)*c^(1/3)*x] )/(3*2^(1/3)*c^(1/3)*(b - Sqrt[b^2 - 4*a*c])^(2/3)) - ((h - (b*l)/c + (2*c ^2*e + b^2*l - c*(b*h + 2*a*l))/(c*Sqrt[b^2 - 4*a*c]))*Log[(b - Sqrt[b^2 - 4*a*c])^(1/3) + 2^(1/3)*c^(1/3)*x])/(3*2^(2/3)*c^(2/3)*(b - Sqrt[b^2 - 4* a*c])^(1/3)) + ((g - (b*k)/c - (2*c^2*d - b*c*g + b^2*k - 2*a*c*k)/(c*Sqrt [b^2 - 4*a*c]))*Log[(b + Sqrt[b^2 - 4*a*c])^(1/3) + 2^(1/3)*c^(1/3)*x])...
3.1.1.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Mo dule[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, x, j + k*n]*x^(k*n ), {k, 0, (q - j)/n + 1}]*(a + b*x^n + c*x^(2*n))^p, {j, 0, n - 1}], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && !PolyQ[Pq, x^n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 29.02 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.08
| method | result | size |
| default | \(\frac {\frac {1}{3} m \,x^{3}+\frac {1}{2} l \,x^{2}+k x}{c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6} c +\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (\left (-b m +c j \right ) \textit {\_R}^{5}+\left (-b l +c h \right ) \textit {\_R}^{4}+\left (-b k +g c \right ) \textit {\_R}^{3}+\left (-a m +c f \right ) \textit {\_R}^{2}+\left (-a l +e c \right ) \textit {\_R} -a k +c d \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{5} c +\textit {\_R}^{2} b}}{3 c}\) | \(130\) |
| risch | \(\frac {m \,x^{3}}{3 c}+\frac {l \,x^{2}}{2 c}+\frac {k x}{c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6} c +\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (\left (-b m +c j \right ) \textit {\_R}^{5}+\left (-b l +c h \right ) \textit {\_R}^{4}+\left (-b k +g c \right ) \textit {\_R}^{3}+\left (-a m +c f \right ) \textit {\_R}^{2}+\left (-a l +e c \right ) \textit {\_R} -a k +c d \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{5} c +\textit {\_R}^{2} b}}{3 c}\) | \(134\) |
int((m*x^8+l*x^7+k*x^6+j*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^6+b*x^3+a),x,me thod=_RETURNVERBOSE)
1/c*(1/3*m*x^3+1/2*l*x^2+k*x)+1/3/c*sum(((-b*m+c*j)*_R^5+(-b*l+c*h)*_R^4+( -b*k+c*g)*_R^3+(-a*m+c*f)*_R^2+(-a*l+c*e)*_R-a*k+c*d)/(2*_R^5*c+_R^2*b)*ln (x-_R),_R=RootOf(_Z^6*c+_Z^3*b+a))
Timed out. \[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx=\text {Timed out} \]
integrate((m*x^8+l*x^7+k*x^6+j*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^6+b*x^3+a ),x, algorithm="fricas")
Timed out. \[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx=\text {Timed out} \]
\[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx=\int { \frac {m x^{8} + l x^{7} + k x^{6} + j x^{5} + h x^{4} + g x^{3} + f x^{2} + e x + d}{c x^{6} + b x^{3} + a} \,d x } \]
integrate((m*x^8+l*x^7+k*x^6+j*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^6+b*x^3+a ),x, algorithm="maxima")
1/6*(2*m*x^3 + 3*l*x^2 + 6*k*x)/c - integrate(-((c*j - b*m)*x^5 + (c*h - b *l)*x^4 + (c*g - b*k)*x^3 + (c*f - a*m)*x^2 + c*d - a*k + (c*e - a*l)*x)/( c*x^6 + b*x^3 + a), x)/c
Timed out. \[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx=\text {Timed out} \]
integrate((m*x^8+l*x^7+k*x^6+j*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^6+b*x^3+a ),x, algorithm="giac")
Time = 52.38 (sec) , antiderivative size = 359169, normalized size of antiderivative = 215.33 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+j x^5+k x^6+l x^7+m x^8}{a+b x^3+c x^6} \, dx=\text {Too large to display} \]
symsum(log((x*(c^7*e^5 + c^7*d^4*j - a^5*c^2*l^5 - b^7*e^2*m^3 - a^2*b*c^4 *h^5 - a*c^6*e^2*g^3 - b*c^6*e^2*f^3 + 2*a*c^6*e^3*h^2 + b*c^6*d^3*h^2 + a ^2*c^5*e*h^4 + a^4*b^2*c*l^5 + 3*c^7*d^2*e*f^2 + 3*c^7*d^2*e^2*g + a^2*b^5 *d*m^4 - a^2*c^5*g^4*j + a^3*c^4*g*j^4 + 5*a^4*c^3*e*l^4 + 3*b^2*c^5*e^4*l + b^6*c*e^2*l^3 - a^3*b^4*g*m^4 - a^3*c^4*h^4*l - a^5*c^2*g*m^4 + a^4*c^3 *j*k^4 + a^4*b^3*k*m^4 + b^2*c^5*e^2*g^3 + 3*b^2*c^5*e^3*h^2 - b^3*c^4*e^2 *h^3 + a^2*c^5*e^2*j^3 + a^2*c^5*g^3*h^2 + b^4*c^3*e^2*j^3 + 10*a^2*c^5*e^ 3*l^2 - 10*a^3*c^4*e^2*l^3 + b^3*c^4*d^3*l^2 - b^5*c^2*e^2*k^3 - a^3*c^4*h ^2*j^3 + 3*b^4*c^3*e^3*l^2 - a^3*c^4*g^3*l^2 - a^2*b^5*h^2*m^3 - 2*a^4*c^3 *h^2*l^3 + a^4*c^3*j^3*l^2 - a^4*b^3*l^2*m^3 + b*c^6*d*f^4 - a*c^6*f^4*g - 3*b*c^6*e^4*h - 4*c^7*d*e^3*f - 2*c^7*d^3*e*h - 2*c^7*d^3*f*g - 5*a*c^6*e ^4*l - b*c^6*d^4*m + b^7*d*f*m^3 + a*b*c^5*f*g^4 - 2*a*c^6*d*f*g^3 + 2*a*c ^6*e*f^3*h + 3*b*c^6*e^3*f*g + 2*a*c^6*d*f^3*j + 3*b*c^6*d*e^3*j + 4*a*c^6 *d*e^3*m + 4*a*c^6*e^3*f*k + 4*a*c^6*e^3*g*j + 2*b*c^6*d^3*e*l + 2*b*c^6*d ^3*f*k - b*c^6*d^3*g*j - b^6*c*d*f*l^3 + 2*a*c^6*d^3*g*m + 2*a*c^6*d^3*h*l + 2*a*b^6*e*h*m^3 - a*b^6*f*g*m^3 - 4*a*c^6*d^3*j*k - a*b^6*d*j*m^3 - 2*a ^5*b*c*k*m^4 + 12*a^2*b^2*c^3*e^2*l^3 + a^2*b^2*c^3*h^2*j^3 - 10*a^2*b^3*c ^2*e^2*m^3 - a^2*b^3*c^2*h^2*k^3 - 3*a^2*b^3*c^2*h^3*l^2 + 3*a^3*b^2*c^2*h ^2*l^3 - 4*a*b*c^5*e^2*h^3 + 2*a*b^2*c^4*e*h^4 + a*b^3*c^3*d*j^4 - 2*a^2*b *c^4*d*j^4 - 3*b*c^6*d*e^2*g^2 - 2*a*b*c^5*d^3*l^2 + 3*a*c^6*e*f^2*g^2 ...