Integrand size = 17, antiderivative size = 393 \[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=-\frac {1}{2 \left (a+b c^3\right ) x^2}+\frac {3 b c^2 d}{\left (a+b c^3\right )^2 x}+\frac {b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \left (a-3 a^{2/3} \sqrt [3]{b} c+b c^3\right ) d^2 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \left (a+b c^3\right )^3}-\frac {3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}-\frac {b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac {b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )^3}+\frac {b c \left (a-2 b c^3\right ) d^2 \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3} \]
-1/2/(b*c^3+a)/x^2+3*b*c^2*d/(b*c^3+a)^2/x-3*b*c*(-2*b*c^3+a)*d^2*ln(x)/(b *c^3+a)^3-1/3*b^(2/3)*(a^2+6*a^(4/3)*b^(2/3)*c^2-7*a*b*c^3-3*a^(1/3)*b^(5/ 3)*c^5+b^2*c^6)*d^2*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(2/3)/(b*c^3+a)^3+1/6*b^ (2/3)*(a^2+6*a^(4/3)*b^(2/3)*c^2-7*a*b*c^3-3*a^(1/3)*b^(5/3)*c^5+b^2*c^6)* d^2*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(2/3)/(b*c^3+a )^3+b*c*(-2*b*c^3+a)*d^2*ln(a+b*(d*x+c)^3)/(b*c^3+a)^3+1/3*b^(2/3)*(a^(1/3 )+b^(1/3)*c)^3*(a-3*a^(2/3)*b^(1/3)*c+b*c^3)*d^2*arctan(1/3*(a^(1/3)-2*b^( 1/3)*(d*x+c))/a^(1/3)*3^(1/2))/a^(2/3)/(b*c^3+a)^3*3^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.10 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=-\frac {3 \left (a+b c^3\right ) \left (a+b c^2 (c-6 d x)\right )+18 b c \left (a-2 b c^3\right ) d^2 x^2 \log (x)+2 d^2 x^2 \text {RootSum}\left [a+b c^3+3 b c^2 d \text {$\#$1}+3 b c d^2 \text {$\#$1}^2+b d^3 \text {$\#$1}^3\&,\frac {a^2 \log (x-\text {$\#$1})-16 a b c^3 \log (x-\text {$\#$1})+10 b^2 c^6 \log (x-\text {$\#$1})-12 a b c^2 d \log (x-\text {$\#$1}) \text {$\#$1}+15 b^2 c^5 d \log (x-\text {$\#$1}) \text {$\#$1}-3 a b c d^2 \log (x-\text {$\#$1}) \text {$\#$1}^2+6 b^2 c^4 d^2 \log (x-\text {$\#$1}) \text {$\#$1}^2}{c^2+2 c d \text {$\#$1}+d^2 \text {$\#$1}^2}\&\right ]}{6 \left (a+b c^3\right )^3 x^2} \]
-1/6*(3*(a + b*c^3)*(a + b*c^2*(c - 6*d*x)) + 18*b*c*(a - 2*b*c^3)*d^2*x^2 *Log[x] + 2*d^2*x^2*RootSum[a + b*c^3 + 3*b*c^2*d*#1 + 3*b*c*d^2*#1^2 + b* d^3*#1^3 & , (a^2*Log[x - #1] - 16*a*b*c^3*Log[x - #1] + 10*b^2*c^6*Log[x - #1] - 12*a*b*c^2*d*Log[x - #1]*#1 + 15*b^2*c^5*d*Log[x - #1]*#1 - 3*a*b* c*d^2*Log[x - #1]*#1^2 + 6*b^2*c^4*d^2*Log[x - #1]*#1^2)/(c^2 + 2*c*d*#1 + d^2*#1^2) & ])/((a + b*c^3)^3*x^2)
Time = 0.82 (sec) , antiderivative size = 380, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {896, 25, 7276, 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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle d^2 \int \frac {1}{d^3 x^3 \left (b (c+d x)^3+a\right )}d(c+d x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -d^2 \int -\frac {1}{d^3 x^3 \left (b (c+d x)^3+a\right )}d(c+d x)\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -d^2 \int \left (\frac {3 b c^2}{\left (b c^3+a\right )^2 d^2 x^2}-\frac {3 b \left (2 b c^3-a\right ) c}{\left (b c^3+a\right )^3 d x}+\frac {b \left (b^2 c^6-7 a b c^3-3 b \left (2 a-b c^3\right ) (c+d x) c^2-3 b \left (a-2 b c^3\right ) (c+d x)^2 c+a^2\right )}{\left (b c^3+a\right )^3 \left (b (c+d x)^3+a\right )}-\frac {1}{\left (b c^3+a\right ) d^3 x^3}\right )d(c+d x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle d^2 \left (\frac {b^{2/3} \left (-3 a^{2/3} \sqrt [3]{b} c+a+b c^3\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \left (a+b c^3\right )^3}-\frac {b^{2/3} \left (a^2+3 \sqrt [3]{a} b^{2/3} c^2 \left (2 a-b c^3\right )-7 a b c^3+b^2 c^6\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac {b^{2/3} \left (a^2+3 \sqrt [3]{a} b^{2/3} c^2 \left (2 a-b c^3\right )-7 a b c^3+b^2 c^6\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )^3}-\frac {1}{2 d^2 x^2 \left (a+b c^3\right )}-\frac {3 b c \left (a-2 b c^3\right ) \log (-d x)}{\left (a+b c^3\right )^3}+\frac {b c \left (a-2 b c^3\right ) \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}+\frac {3 b c^2}{d x \left (a+b c^3\right )^2}\right )\) |
d^2*(-1/2*1/((a + b*c^3)*d^2*x^2) + (3*b*c^2)/((a + b*c^3)^2*d*x) + (b^(2/ 3)*(a^(1/3) + b^(1/3)*c)^3*(a - 3*a^(2/3)*b^(1/3)*c + b*c^3)*ArcTan[(a^(1/ 3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*(a + b*c^3) ^3) - (3*b*c*(a - 2*b*c^3)*Log[-(d*x)])/(a + b*c^3)^3 - (b^(2/3)*(a^2 - 7* a*b*c^3 + b^2*c^6 + 3*a^(1/3)*b^(2/3)*c^2*(2*a - b*c^3))*Log[a^(1/3) + b^( 1/3)*(c + d*x)])/(3*a^(2/3)*(a + b*c^3)^3) + (b^(2/3)*(a^2 - 7*a*b*c^3 + b ^2*c^6 + 3*a^(1/3)*b^(2/3)*c^2*(2*a - b*c^3))*Log[a^(2/3) - a^(1/3)*b^(1/3 )*(c + d*x) + b^(2/3)*(c + d*x)^2])/(6*a^(2/3)*(a + b*c^3)^3) + (b*c*(a - 2*b*c^3)*Log[a + b*(c + d*x)^3])/(a + b*c^3)^3)
3.2.9.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.15 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.51
method | result | size |
default | \(-\frac {1}{2 \left (b \,c^{3}+a \right ) x^{2}}+\frac {3 b \,c^{2} d}{\left (b \,c^{3}+a \right )^{2} x}-\frac {3 b c \left (-2 b \,c^{3}+a \right ) d^{2} \ln \left (x \right )}{\left (b \,c^{3}+a \right )^{3}}-\frac {d^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +b \,c^{3}+a \right )}{\sum }\frac {\left (6 \textit {\_R}^{2} b^{2} c^{4} d^{2}+15 \textit {\_R} \,b^{2} c^{5} d +10 c^{6} b^{2}-3 \textit {\_R}^{2} a b c \,d^{2}-12 \textit {\_R} a b \,c^{2} d -16 a b \,c^{3}+a^{2}\right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{3 \left (b \,c^{3}+a \right )^{3}}\) | \(200\) |
risch | \(\frac {\frac {3 b \,c^{2} d x}{c^{6} b^{2}+2 a b \,c^{3}+a^{2}}-\frac {1}{2 \left (b \,c^{3}+a \right )}}{x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{2} b^{3} c^{9}+3 a^{3} b^{2} c^{6}+3 a^{4} b \,c^{3}+a^{5}\right ) \textit {\_Z}^{3}+\left (18 d^{2} c^{4} b^{2} a^{2}-9 a^{3} b c \,d^{2}\right ) \textit {\_Z}^{2}+9 a \,b^{2} c^{2} d^{4} \textit {\_Z} +b^{2} d^{6}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (2 a \,b^{5} c^{15} d +4 a^{2} b^{4} c^{12} d -4 a^{3} b^{3} c^{9} d -16 a^{4} b^{2} c^{6} d -14 a^{5} b \,c^{3} d -4 a^{6} d \right ) \textit {\_R}^{3}+\left (b^{5} c^{13} d^{3}-29 a \,b^{4} c^{10} d^{3}-39 a^{2} b^{3} c^{7} d^{3}+13 a^{3} b^{2} c^{4} d^{3}+22 a^{4} b c \,d^{3}\right ) \textit {\_R}^{2}+\left (-18 b^{4} c^{8} d^{5}-90 a \,b^{3} c^{5} d^{5}+9 a^{2} b^{2} c^{2} d^{5}\right ) \textit {\_R} -3 b^{3} c^{3} d^{7}-3 a \,b^{2} d^{7}\right ) x +\left (a \,b^{5} c^{16}+5 a^{2} b^{4} c^{13}+10 a^{3} b^{3} c^{10}+10 a^{4} b^{2} c^{7}+5 a^{5} b \,c^{4}+c \,a^{6}\right ) \textit {\_R}^{3}+\left (b^{5} c^{14} d^{2}-14 a \,b^{4} c^{11} d^{2}-21 a^{2} b^{3} c^{8} d^{2}+4 a^{3} b^{2} c^{5} d^{2}+10 a^{4} b \,c^{2} d^{2}\right ) \textit {\_R}^{2}+\left (-19 b^{4} c^{9} d^{4}-12 a \,b^{3} c^{6} d^{4}+6 a^{2} b^{2} c^{3} d^{4}-a^{3} b \,d^{4}\right ) \textit {\_R} +18 b^{3} c^{4} d^{6}-9 a \,b^{2} c \,d^{6}\right )\right )}{3}+\frac {6 b^{2} c^{4} d^{2} \ln \left (-x \right )}{b^{3} c^{9}+3 a \,b^{2} c^{6}+3 a^{2} b \,c^{3}+a^{3}}-\frac {3 b c \,d^{2} \ln \left (-x \right ) a}{b^{3} c^{9}+3 a \,b^{2} c^{6}+3 a^{2} b \,c^{3}+a^{3}}\) | \(625\) |
-1/2/(b*c^3+a)/x^2+3*b*c^2*d/(b*c^3+a)^2/x-3*b*c*(-2*b*c^3+a)*d^2*ln(x)/(b *c^3+a)^3-1/3*d^2*sum((6*_R^2*b^2*c^4*d^2+15*_R*b^2*c^5*d+10*b^2*c^6-3*_R^ 2*a*b*c*d^2-12*_R*a*b*c^2*d-16*a*b*c^3+a^2)/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_ R),_R=RootOf(_Z^3*b*d^3+3*_Z^2*b*c*d^2+3*_Z*b*c^2*d+b*c^3+a))/(b*c^3+a)^3
Result contains complex when optimal does not.
Time = 3.88 (sec) , antiderivative size = 14765, normalized size of antiderivative = 37.57 \[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )} x^{3}} \,d x } \]
-b*d^3*integrate((10*b^2*c^6 - 16*a*b*c^3 + 3*(2*b^2*c^4 - a*b*c)*d^2*x^2 + 3*(5*b^2*c^5 - 4*a*b*c^2)*d*x + a^2)/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^ 2*d*x + b*c^3 + a), x)/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3) + 3*(2* b^2*c^4 - a*b*c)*d^2*log(x)/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3) + 1/2*(6*b*c^2*d*x - b*c^3 - a)/((b^2*c^6 + 2*a*b*c^3 + a^2)*x^2)
\[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )} x^{3}} \,d x } \]
Time = 9.27 (sec) , antiderivative size = 1328, normalized size of antiderivative = 3.38 \[ \int \frac {1}{x^3 \left (a+b (c+d x)^3\right )} \, dx=\left (\sum _{k=1}^3\ln \left (\frac {6\,b^6\,c^4\,d^{14}-3\,a\,b^5\,c\,d^{14}}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}-\mathrm {root}\left (81\,a^3\,b^2\,c^6\,z^3+27\,a^2\,b^3\,c^9\,z^3+81\,a^4\,b\,c^3\,z^3+27\,a^5\,z^3-81\,a^3\,b\,c\,d^2\,z^2+162\,a^2\,b^2\,c^4\,d^2\,z^2+27\,a\,b^2\,c^2\,d^4\,z+b^2\,d^6,z,k\right )\,\left (\frac {a^3\,b^4\,d^{12}-6\,a^2\,b^5\,c^3\,d^{12}+12\,a\,b^6\,c^6\,d^{12}+19\,b^7\,c^9\,d^{12}}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}-\mathrm {root}\left (81\,a^3\,b^2\,c^6\,z^3+27\,a^2\,b^3\,c^9\,z^3+81\,a^4\,b\,c^3\,z^3+27\,a^5\,z^3-81\,a^3\,b\,c\,d^2\,z^2+162\,a^2\,b^2\,c^4\,d^2\,z^2+27\,a\,b^2\,c^2\,d^4\,z+b^2\,d^6,z,k\right )\,\left (\mathrm {root}\left (81\,a^3\,b^2\,c^6\,z^3+27\,a^2\,b^3\,c^9\,z^3+81\,a^4\,b\,c^3\,z^3+27\,a^5\,z^3-81\,a^3\,b\,c\,d^2\,z^2+162\,a^2\,b^2\,c^4\,d^2\,z^2+27\,a\,b^2\,c^2\,d^4\,z+b^2\,d^6,z,k\right )\,\left (\frac {9\,a^6\,b^3\,c\,d^8+45\,a^5\,b^4\,c^4\,d^8+90\,a^4\,b^5\,c^7\,d^8+90\,a^3\,b^6\,c^{10}\,d^8+45\,a^2\,b^7\,c^{13}\,d^8+9\,a\,b^8\,c^{16}\,d^8}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}-\frac {x\,\left (36\,a^6\,b^3\,d^9+126\,a^5\,b^4\,c^3\,d^9+144\,a^4\,b^5\,c^6\,d^9+36\,a^3\,b^6\,c^9\,d^9-36\,a^2\,b^7\,c^{12}\,d^9-18\,a\,b^8\,c^{15}\,d^9\right )}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}\right )+\frac {30\,a^4\,b^4\,c^2\,d^{10}+12\,a^3\,b^5\,c^5\,d^{10}-63\,a^2\,b^6\,c^8\,d^{10}-42\,a\,b^7\,c^{11}\,d^{10}+3\,b^8\,c^{14}\,d^{10}}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}+\frac {x\,\left (66\,a^4\,b^4\,c\,d^{11}+39\,a^3\,b^5\,c^4\,d^{11}-117\,a^2\,b^6\,c^7\,d^{11}-87\,a\,b^7\,c^{10}\,d^{11}+3\,b^8\,c^{13}\,d^{11}\right )}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}\right )+\frac {x\,\left (-9\,a^2\,b^5\,c^2\,d^{13}+90\,a\,b^6\,c^5\,d^{13}+18\,b^7\,c^8\,d^{13}\right )}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}\right )-\frac {x\,\left (b^6\,c^3\,d^{15}+a\,b^5\,d^{15}\right )}{a^4+4\,a^3\,b\,c^3+6\,a^2\,b^2\,c^6+4\,a\,b^3\,c^9+b^4\,c^{12}}\right )\,\mathrm {root}\left (81\,a^3\,b^2\,c^6\,z^3+27\,a^2\,b^3\,c^9\,z^3+81\,a^4\,b\,c^3\,z^3+27\,a^5\,z^3-81\,a^3\,b\,c\,d^2\,z^2+162\,a^2\,b^2\,c^4\,d^2\,z^2+27\,a\,b^2\,c^2\,d^4\,z+b^2\,d^6,z,k\right )\right )-\frac {1}{2\,\left (b\,c^3\,x^2+a\,x^2\right )}+\frac {3\,b\,c^2\,d}{x\,a^2+2\,x\,a\,b\,c^3+x\,b^2\,c^6}+\frac {6\,b^2\,c^4\,d^2\,\ln \left (x\right )}{a^3+3\,a^2\,b\,c^3+3\,a\,b^2\,c^6+b^3\,c^9}-\frac {3\,a\,b\,c\,d^2\,\ln \left (x\right )}{a^3+3\,a^2\,b\,c^3+3\,a\,b^2\,c^6+b^3\,c^9} \]
symsum(log((6*b^6*c^4*d^14 - 3*a*b^5*c*d^14)/(a^4 + b^4*c^12 + 4*a^3*b*c^3 + 4*a*b^3*c^9 + 6*a^2*b^2*c^6) - root(81*a^3*b^2*c^6*z^3 + 27*a^2*b^3*c^9 *z^3 + 81*a^4*b*c^3*z^3 + 27*a^5*z^3 - 81*a^3*b*c*d^2*z^2 + 162*a^2*b^2*c^ 4*d^2*z^2 + 27*a*b^2*c^2*d^4*z + b^2*d^6, z, k)*((a^3*b^4*d^12 + 19*b^7*c^ 9*d^12 + 12*a*b^6*c^6*d^12 - 6*a^2*b^5*c^3*d^12)/(a^4 + b^4*c^12 + 4*a^3*b *c^3 + 4*a*b^3*c^9 + 6*a^2*b^2*c^6) - root(81*a^3*b^2*c^6*z^3 + 27*a^2*b^3 *c^9*z^3 + 81*a^4*b*c^3*z^3 + 27*a^5*z^3 - 81*a^3*b*c*d^2*z^2 + 162*a^2*b^ 2*c^4*d^2*z^2 + 27*a*b^2*c^2*d^4*z + b^2*d^6, z, k)*(root(81*a^3*b^2*c^6*z ^3 + 27*a^2*b^3*c^9*z^3 + 81*a^4*b*c^3*z^3 + 27*a^5*z^3 - 81*a^3*b*c*d^2*z ^2 + 162*a^2*b^2*c^4*d^2*z^2 + 27*a*b^2*c^2*d^4*z + b^2*d^6, z, k)*((9*a^6 *b^3*c*d^8 + 9*a*b^8*c^16*d^8 + 45*a^5*b^4*c^4*d^8 + 90*a^4*b^5*c^7*d^8 + 90*a^3*b^6*c^10*d^8 + 45*a^2*b^7*c^13*d^8)/(a^4 + b^4*c^12 + 4*a^3*b*c^3 + 4*a*b^3*c^9 + 6*a^2*b^2*c^6) - (x*(36*a^6*b^3*d^9 - 18*a*b^8*c^15*d^9 + 1 26*a^5*b^4*c^3*d^9 + 144*a^4*b^5*c^6*d^9 + 36*a^3*b^6*c^9*d^9 - 36*a^2*b^7 *c^12*d^9))/(a^4 + b^4*c^12 + 4*a^3*b*c^3 + 4*a*b^3*c^9 + 6*a^2*b^2*c^6)) + (3*b^8*c^14*d^10 - 42*a*b^7*c^11*d^10 + 30*a^4*b^4*c^2*d^10 + 12*a^3*b^5 *c^5*d^10 - 63*a^2*b^6*c^8*d^10)/(a^4 + b^4*c^12 + 4*a^3*b*c^3 + 4*a*b^3*c ^9 + 6*a^2*b^2*c^6) + (x*(3*b^8*c^13*d^11 + 66*a^4*b^4*c*d^11 - 87*a*b^7*c ^10*d^11 + 39*a^3*b^5*c^4*d^11 - 117*a^2*b^6*c^7*d^11))/(a^4 + b^4*c^12 + 4*a^3*b*c^3 + 4*a*b^3*c^9 + 6*a^2*b^2*c^6)) + (x*(18*b^7*c^8*d^13 + 90*...