Integrand size = 17, antiderivative size = 124 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=-\frac {5 d}{3 (b c-a d)^2 (c+d x)^{3/2}}-\frac {1}{(b c-a d) (a+b x) (c+d x)^{3/2}}-\frac {5 b d}{(b c-a d)^3 \sqrt {c+d x}}+\frac {5 b^{3/2} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{7/2}} \]
-5/3*d/(-a*d+b*c)^2/(d*x+c)^(3/2)-1/(-a*d+b*c)/(b*x+a)/(d*x+c)^(3/2)+5*b^( 3/2)*d*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(7/2)-5* b*d/(-a*d+b*c)^3/(d*x+c)^(1/2)
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {2 a^2 d^2-2 a b d (7 c+5 d x)-b^2 \left (3 c^2+20 c d x+15 d^2 x^2\right )}{3 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}+\frac {5 b^{3/2} d \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}} \]
(2*a^2*d^2 - 2*a*b*d*(7*c + 5*d*x) - b^2*(3*c^2 + 20*c*d*x + 15*d^2*x^2))/ (3*(b*c - a*d)^3*(a + b*x)*(c + d*x)^(3/2)) + (5*b^(3/2)*d*ArcTan[(Sqrt[b] *Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2)
Time = 0.23 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {52, 61, 61, 73, 221}
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.
\(\displaystyle \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5 d \int \frac {1}{(a+b x) (c+d x)^{5/2}}dx}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 d \left (\frac {b \int \frac {1}{(a+b x) (c+d x)^{3/2}}dx}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 d \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{b c-a d}+\frac {2}{\sqrt {c+d x} (b c-a d)}\right )}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 d \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d (b c-a d)}+\frac {2}{\sqrt {c+d x} (b c-a d)}\right )}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {5 d \left (\frac {b \left (\frac {2}{\sqrt {c+d x} (b c-a d)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
-(1/((b*c - a*d)*(a + b*x)*(c + d*x)^(3/2))) - (5*d*(2/(3*(b*c - a*d)*(c + d*x)^(3/2)) + (b*(2/((b*c - a*d)*Sqrt[c + d*x]) - (2*Sqrt[b]*ArcTanh[(Sqr t[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2)))/(b*c - a*d)))/(2 *(b*c - a*d))
3.15.41.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 0.60 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(2 d \left (-\frac {1}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{3} \sqrt {d x +c}}+\frac {b^{2} \left (\frac {\sqrt {d x +c}}{2 \left (d x +c \right ) b +2 a d -2 b c}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\) | \(121\) |
default | \(2 d \left (-\frac {1}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{3} \sqrt {d x +c}}+\frac {b^{2} \left (\frac {\sqrt {d x +c}}{2 \left (d x +c \right ) b +2 a d -2 b c}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\) | \(121\) |
pseudoelliptic | \(d \left (\frac {\sqrt {d x +c}\, b^{2}}{d \left (b x +a \right ) \left (a d -b c \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) b^{2}}{\sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{3}}-\frac {2}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b}{\left (a d -b c \right )^{3} \sqrt {d x +c}}\right )\) | \(123\) |
2*d*(-1/3/(a*d-b*c)^2/(d*x+c)^(3/2)+2/(a*d-b*c)^3*b/(d*x+c)^(1/2)+1/(a*d-b *c)^3*b^2*(1/2*(d*x+c)^(1/2)/((d*x+c)*b+a*d-b*c)+5/2/((a*d-b*c)*b)^(1/2)*a rctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (106) = 212\).
Time = 0.24 (sec) , antiderivative size = 782, normalized size of antiderivative = 6.31 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}}, \frac {15 \, {\left (b^{2} d^{3} x^{3} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - {\left (15 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}}\right ] \]
[-1/6*(15*(b^2*d^3*x^3 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^2 + (b^2*c^ 2*d + 2*a*b*c*d^2)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d - 2*(b* c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) + 2*(15*b^2*d^2*x^2 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x)*sqrt(d *x + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b ^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c ^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4 )*x), 1/3*(15*(b^2*d^3*x^3 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^2 + (b^ 2*c^2*d + 2*a*b*c*d^2)*x)*sqrt(-b/(b*c - a*d))*arctan(-(b*c - a*d)*sqrt(d* x + c)*sqrt(-b/(b*c - a*d))/(b*d*x + b*c)) - (15*b^2*d^2*x^2 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x)*sqrt(d*x + c))/(a*b^ 3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3 *a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3 *c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b ^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)]
\[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{2} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (106) = 212\).
Time = 0.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.74 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=-\frac {5 \, b^{2} d \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {\sqrt {d x + c} b^{2} d}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} - \frac {2 \, {\left (6 \, {\left (d x + c\right )} b d + b c d - a d^{2}\right )}}{3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} \]
-5*b^2*d*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2* c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b^2*c + a*b*d)) - sqrt(d*x + c)*b^2 *d/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c + a*d)) - 2/3*(6*(d*x + c)*b*d + b*c*d - a*d^2)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x + c)^(3/2))
Time = 0.36 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {\frac {10\,b\,d\,\left (c+d\,x\right )}{3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,d}{3\,\left (a\,d-b\,c\right )}+\frac {5\,b^2\,d\,{\left (c+d\,x\right )}^2}{{\left (a\,d-b\,c\right )}^3}}{b\,{\left (c+d\,x\right )}^{5/2}+\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}+\frac {5\,b^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{{\left (a\,d-b\,c\right )}^{7/2}} \]
((10*b*d*(c + d*x))/(3*(a*d - b*c)^2) - (2*d)/(3*(a*d - b*c)) + (5*b^2*d*( c + d*x)^2)/(a*d - b*c)^3)/(b*(c + d*x)^(5/2) + (a*d - b*c)*(c + d*x)^(3/2 )) + (5*b^(3/2)*d*atan((b^(1/2)*(c + d*x)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b ^2*c^2*d - 3*a^2*b*c*d^2))/(a*d - b*c)^(7/2)))/(a*d - b*c)^(7/2)
Time = 0.00 (sec) , antiderivative size = 493, normalized size of antiderivative = 3.98 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b c d +15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b \,d^{2} x +15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} c d x +15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} d^{2} x^{2}-2 a^{3} d^{3}+16 a^{2} b c \,d^{2}+10 a^{2} b \,d^{3} x -11 a \,b^{2} c^{2} d +10 a \,b^{2} c \,d^{2} x +15 a \,b^{2} d^{3} x^{2}-3 b^{3} c^{3}-20 b^{3} c^{2} d x -15 b^{3} c \,d^{2} x^{2}}{3 \sqrt {d x +c}\, \left (a^{4} b \,d^{5} x^{2}-4 a^{3} b^{2} c \,d^{4} x^{2}+6 a^{2} b^{3} c^{2} d^{3} x^{2}-4 a \,b^{4} c^{3} d^{2} x^{2}+b^{5} c^{4} d \,x^{2}+a^{5} d^{5} x -3 a^{4} b c \,d^{4} x +2 a^{3} b^{2} c^{2} d^{3} x +2 a^{2} b^{3} c^{3} d^{2} x -3 a \,b^{4} c^{4} d x +b^{5} c^{5} x +a^{5} c \,d^{4}-4 a^{4} b \,c^{2} d^{3}+6 a^{3} b^{2} c^{3} d^{2}-4 a^{2} b^{3} c^{4} d +a \,b^{4} c^{5}\right )} \]
int(1/(sqrt(c + d*x)*(a**2*c**2 + 2*a**2*c*d*x + a**2*d**2*x**2 + 2*a*b*c* *2*x + 4*a*b*c*d*x**2 + 2*a*b*d**2*x**3 + b**2*c**2*x**2 + 2*b**2*c*d*x**3 + b**2*d**2*x**4)),x)
(15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)* sqrt(a*d - b*c)))*a*b*c*d + 15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan( (sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b*d**2*x + 15*sqrt(b)*sqrt( c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c))) *b**2*c*d*x + 15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x) *b)/(sqrt(b)*sqrt(a*d - b*c)))*b**2*d**2*x**2 - 2*a**3*d**3 + 16*a**2*b*c* d**2 + 10*a**2*b*d**3*x - 11*a*b**2*c**2*d + 10*a*b**2*c*d**2*x + 15*a*b** 2*d**3*x**2 - 3*b**3*c**3 - 20*b**3*c**2*d*x - 15*b**3*c*d**2*x**2)/(3*sqr t(c + d*x)*(a**5*c*d**4 + a**5*d**5*x - 4*a**4*b*c**2*d**3 - 3*a**4*b*c*d* *4*x + a**4*b*d**5*x**2 + 6*a**3*b**2*c**3*d**2 + 2*a**3*b**2*c**2*d**3*x - 4*a**3*b**2*c*d**4*x**2 - 4*a**2*b**3*c**4*d + 2*a**2*b**3*c**3*d**2*x + 6*a**2*b**3*c**2*d**3*x**2 + a*b**4*c**5 - 3*a*b**4*c**4*d*x - 4*a*b**4*c **3*d**2*x**2 + b**5*c**5*x + b**5*c**4*d*x**2))