Integrand size = 30, antiderivative size = 474 \[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\frac {(b c-a d) (e x)^{3/2} \sqrt {c-d x^2}}{2 a b e \left (a-b x^2\right )}-\frac {c^{3/4} \sqrt [4]{d} (b c-5 a d) \sqrt {e} \sqrt {1-\frac {d x^2}{c}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{2 a b^2 \sqrt {c-d x^2}}+\frac {c^{3/4} \sqrt [4]{d} (b c-5 a d) \sqrt {e} \sqrt {1-\frac {d x^2}{c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),-1\right )}{2 a b^2 \sqrt {c-d x^2}}-\frac {\sqrt [4]{c} \left (b^2 c^2+4 a b c d-5 a^2 d^2\right ) \sqrt {e} \sqrt {1-\frac {d x^2}{c}} \operatorname {EllipticPi}\left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}},\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),-1\right )}{4 a^{3/2} b^{5/2} \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} \left (b^2 c^2+4 a b c d-5 a^2 d^2\right ) \sqrt {e} \sqrt {1-\frac {d x^2}{c}} \operatorname {EllipticPi}\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}},\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),-1\right )}{4 a^{3/2} b^{5/2} \sqrt [4]{d} \sqrt {c-d x^2}} \] Output:
1/2*(-a*d+b*c)*(e*x)^(3/2)*(-d*x^2+c)^(1/2)/a/b/e/(-b*x^2+a)-1/2*c^(3/4)*d ^(1/4)*(-5*a*d+b*c)*e^(1/2)*(1-d*x^2/c)^(1/2)*EllipticE(d^(1/4)*(e*x)^(1/2 )/c^(1/4)/e^(1/2),I)/a/b^2/(-d*x^2+c)^(1/2)+1/2*c^(3/4)*d^(1/4)*(-5*a*d+b* c)*e^(1/2)*(1-d*x^2/c)^(1/2)*EllipticF(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2) ,I)/a/b^2/(-d*x^2+c)^(1/2)-1/4*c^(1/4)*(-5*a^2*d^2+4*a*b*c*d+b^2*c^2)*e^(1 /2)*(1-d*x^2/c)^(1/2)*EllipticPi(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2),-b^(1 /2)*c^(1/2)/a^(1/2)/d^(1/2),I)/a^(3/2)/b^(5/2)/d^(1/4)/(-d*x^2+c)^(1/2)+1/ 4*c^(1/4)*(-5*a^2*d^2+4*a*b*c*d+b^2*c^2)*e^(1/2)*(1-d*x^2/c)^(1/2)*Ellipti cPi(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2),b^(1/2)*c^(1/2)/a^(1/2)/d^(1/2),I) /a^(3/2)/b^(5/2)/d^(1/4)/(-d*x^2+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 11.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.40 \[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\frac {\sqrt {e x} \left (21 a (-b c+a d) x \left (c-d x^2\right )+7 c (b c+3 a d) x \left (-a+b x^2\right ) \sqrt {1-\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\frac {d x^2}{c},\frac {b x^2}{a}\right )+3 d (b c-5 a d) x^3 \left (-a+b x^2\right ) \sqrt {1-\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},1,\frac {11}{4},\frac {d x^2}{c},\frac {b x^2}{a}\right )\right )}{42 a^2 b \left (-a+b x^2\right ) \sqrt {c-d x^2}} \] Input:
Integrate[(Sqrt[e*x]*(c - d*x^2)^(3/2))/(a - b*x^2)^2,x]
Output:
(Sqrt[e*x]*(21*a*(-(b*c) + a*d)*x*(c - d*x^2) + 7*c*(b*c + 3*a*d)*x*(-a + b*x^2)*Sqrt[1 - (d*x^2)/c]*AppellF1[3/4, 1/2, 1, 7/4, (d*x^2)/c, (b*x^2)/a ] + 3*d*(b*c - 5*a*d)*x^3*(-a + b*x^2)*Sqrt[1 - (d*x^2)/c]*AppellF1[7/4, 1 /2, 1, 11/4, (d*x^2)/c, (b*x^2)/a]))/(42*a^2*b*(-a + b*x^2)*Sqrt[c - d*x^2 ])
Time = 0.81 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {368, 27, 968, 27, 1054, 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.
\(\displaystyle \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 368 |
\(\displaystyle \frac {2 \int \frac {e^5 x \left (c-d x^2\right )^{3/2}}{\left (a e^2-b e^2 x^2\right )^2}d\sqrt {e x}}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 e^3 \int \frac {e x \left (c-d x^2\right )^{3/2}}{\left (a e^2-b e^2 x^2\right )^2}d\sqrt {e x}\) |
\(\Big \downarrow \) 968 |
\(\displaystyle 2 e^3 \left (\frac {\int \frac {x \left (d (b c-5 a d) x^2 e^2+c (b c+3 a d) e^2\right )}{e \sqrt {c-d x^2} \left (a e^2-b e^2 x^2\right )}d\sqrt {e x}}{4 a b e^2}+\frac {(e x)^{3/2} \sqrt {c-d x^2} (b c-a d)}{4 a b e^2 \left (a e^2-b e^2 x^2\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 e^3 \left (\frac {\int \frac {e x \left (d (b c-5 a d) x^2 e^2+c (b c+3 a d) e^2\right )}{\sqrt {c-d x^2} \left (a e^2-b e^2 x^2\right )}d\sqrt {e x}}{4 a b e^4}+\frac {(e x)^{3/2} \sqrt {c-d x^2} (b c-a d)}{4 a b e^2 \left (a e^2-b e^2 x^2\right )}\right )\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle 2 e^3 \left (\frac {\int \left (\frac {e \left (b^2 c^2 e^2-5 a^2 d^2 e^2+4 a b c d e^2\right ) x}{b \sqrt {c-d x^2} \left (a e^2-b e^2 x^2\right )}-\frac {d (b c-5 a d) e x}{b \sqrt {c-d x^2}}\right )d\sqrt {e x}}{4 a b e^4}+\frac {(e x)^{3/2} \sqrt {c-d x^2} (b c-a d)}{4 a b e^2 \left (a e^2-b e^2 x^2\right )}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e^3 \left (\frac {-\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} \left (-5 a^2 d^2+4 a b c d+b^2 c^2\right ) \operatorname {EllipticPi}\left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}},\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),-1\right )}{2 \sqrt {a} b^{3/2} \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} \left (-5 a^2 d^2+4 a b c d+b^2 c^2\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}},\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),-1\right )}{2 \sqrt {a} b^{3/2} \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {c^{3/4} \sqrt [4]{d} e^{3/2} \sqrt {1-\frac {d x^2}{c}} (b c-5 a d) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),-1\right )}{b \sqrt {c-d x^2}}-\frac {c^{3/4} \sqrt [4]{d} e^{3/2} \sqrt {1-\frac {d x^2}{c}} (b c-5 a d) E\left (\left .\arcsin \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt {c-d x^2}}}{4 a b e^4}+\frac {(e x)^{3/2} \sqrt {c-d x^2} (b c-a d)}{4 a b e^2 \left (a e^2-b e^2 x^2\right )}\right )\) |
Input:
Int[(Sqrt[e*x]*(c - d*x^2)^(3/2))/(a - b*x^2)^2,x]
Output:
2*e^3*(((b*c - a*d)*(e*x)^(3/2)*Sqrt[c - d*x^2])/(4*a*b*e^2*(a*e^2 - b*e^2 *x^2)) + (-((c^(3/4)*d^(1/4)*(b*c - 5*a*d)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*Ell ipticE[ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(b*Sqrt[c - d*x ^2])) + (c^(3/4)*d^(1/4)*(b*c - 5*a*d)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*Ellipti cF[ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(b*Sqrt[c - d*x^2]) - (c^(1/4)*(b^2*c^2 + 4*a*b*c*d - 5*a^2*d^2)*e^(3/2)*Sqrt[1 - (d*x^2)/c]* EllipticPi[-((Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d])), ArcSin[(d^(1/4)*Sqrt[e* x])/(c^(1/4)*Sqrt[e])], -1])/(2*Sqrt[a]*b^(3/2)*d^(1/4)*Sqrt[c - d*x^2]) + (c^(1/4)*(b^2*c^2 + 4*a*b*c*d - 5*a^2*d^2)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*El lipticPi[(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d]), ArcSin[(d^(1/4)*Sqrt[e*x])/( c^(1/4)*Sqrt[e])], -1])/(2*Sqrt[a]*b^(3/2)*d^(1/4)*Sqrt[c - d*x^2]))/(4*a* b*e^4))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) *((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1)) Int [(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c *b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1297\) vs. \(2(364)=728\).
Time = 1.19 (sec) , antiderivative size = 1298, normalized size of antiderivative = 2.74
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(1298\) |
default | \(\text {Expression too large to display}\) | \(3846\) |
Input:
int((e*x)^(1/2)*(-d*x^2+c)^(3/2)/(-b*x^2+a)^2,x,method=_RETURNVERBOSE)
Output:
1/e/x*(e*x)^(1/2)/(-d*x^2+c)^(1/2)*((-d*x^2+c)*e*x)^(1/2)*(-1/2*(a*d-b*c)/ b/a*x*(-d*e*x^3+c*e*x)^(1/2)/(-b*x^2+a)-5/2*c*d*(1+x*d/(c*d)^(1/2))^(1/2)* (2-2*x*d/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^(1/2))^(1/2)/(-d*e*x^3+c*e*x)^(1/2 )*e/b^2*EllipticE(((x+1/d*(c*d)^(1/2))*d/(c*d)^(1/2))^(1/2),1/2*2^(1/2))+5 /4*c*d*(1+x*d/(c*d)^(1/2))^(1/2)*(2-2*x*d/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^( 1/2))^(1/2)/(-d*e*x^3+c*e*x)^(1/2)*e/b^2*EllipticF(((x+1/d*(c*d)^(1/2))*d/ (c*d)^(1/2))^(1/2),1/2*2^(1/2))+1/2*c^2*(1+x*d/(c*d)^(1/2))^(1/2)*(2-2*x*d /(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^(1/2))^(1/2)/(-d*e*x^3+c*e*x)^(1/2)/b*e/a* EllipticE(((x+1/d*(c*d)^(1/2))*d/(c*d)^(1/2))^(1/2),1/2*2^(1/2))-1/4*c^2*( 1+x*d/(c*d)^(1/2))^(1/2)*(2-2*x*d/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^(1/2))^(1 /2)/(-d*e*x^3+c*e*x)^(1/2)/b*e/a*EllipticF(((x+1/d*(c*d)^(1/2))*d/(c*d)^(1 /2))^(1/2),1/2*2^(1/2))+5/8*e*a/b^3*d*(c*d)^(1/2)*(1+x*d/(c*d)^(1/2))^(1/2 )*(2-2*x*d/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^(1/2))^(1/2)/(-d*e*x^3+c*e*x)^(1 /2)/(-1/d*(c*d)^(1/2)-1/b*(a*b)^(1/2))*EllipticPi(((x+1/d*(c*d)^(1/2))*d/( c*d)^(1/2))^(1/2),-1/d*(c*d)^(1/2)/(-1/d*(c*d)^(1/2)-1/b*(a*b)^(1/2)),1/2* 2^(1/2))-1/2*e/b^2*(c*d)^(1/2)*(1+x*d/(c*d)^(1/2))^(1/2)*(2-2*x*d/(c*d)^(1 /2))^(1/2)*(-x*d/(c*d)^(1/2))^(1/2)/(-d*e*x^3+c*e*x)^(1/2)/(-1/d*(c*d)^(1/ 2)-1/b*(a*b)^(1/2))*EllipticPi(((x+1/d*(c*d)^(1/2))*d/(c*d)^(1/2))^(1/2),- 1/d*(c*d)^(1/2)/(-1/d*(c*d)^(1/2)-1/b*(a*b)^(1/2)),1/2*2^(1/2))*c-1/8*e/a/ b/d*(c*d)^(1/2)*(1+x*d/(c*d)^(1/2))^(1/2)*(2-2*x*d/(c*d)^(1/2))^(1/2)*(...
Timed out. \[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((e*x)^(1/2)*(-d*x^2+c)^(3/2)/(-b*x^2+a)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\int \frac {\sqrt {e x} \left (c - d x^{2}\right )^{\frac {3}{2}}}{\left (- a + b x^{2}\right )^{2}}\, dx \] Input:
integrate((e*x)**(1/2)*(-d*x**2+c)**(3/2)/(-b*x**2+a)**2,x)
Output:
Integral(sqrt(e*x)*(c - d*x**2)**(3/2)/(-a + b*x**2)**2, x)
\[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\int { \frac {{\left (-d x^{2} + c\right )}^{\frac {3}{2}} \sqrt {e x}}{{\left (b x^{2} - a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^(1/2)*(-d*x^2+c)^(3/2)/(-b*x^2+a)^2,x, algorithm="maxima")
Output:
integrate((-d*x^2 + c)^(3/2)*sqrt(e*x)/(b*x^2 - a)^2, x)
\[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\int { \frac {{\left (-d x^{2} + c\right )}^{\frac {3}{2}} \sqrt {e x}}{{\left (b x^{2} - a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^(1/2)*(-d*x^2+c)^(3/2)/(-b*x^2+a)^2,x, algorithm="giac")
Output:
integrate((-d*x^2 + c)^(3/2)*sqrt(e*x)/(b*x^2 - a)^2, x)
Timed out. \[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx=\int \frac {\sqrt {e\,x}\,{\left (c-d\,x^2\right )}^{3/2}}{{\left (a-b\,x^2\right )}^2} \,d x \] Input:
int(((e*x)^(1/2)*(c - d*x^2)^(3/2))/(a - b*x^2)^2,x)
Output:
int(((e*x)^(1/2)*(c - d*x^2)^(3/2))/(a - b*x^2)^2, x)
\[ \int \frac {\sqrt {e x} \left (c-d x^2\right )^{3/2}}{\left (a-b x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int((e*x)^(1/2)*(-d*x^2+c)^(3/2)/(-b*x^2+a)^2,x)
Output:
(sqrt(e)*(4*sqrt(x)*sqrt(c - d*x**2)*c*d*x + 25*int((sqrt(x)*sqrt(c - d*x* *2)*x**4)/(5*a**3*c*d - 5*a**3*d**2*x**2 - a**2*b*c**2 - 9*a**2*b*c*d*x**2 + 10*a**2*b*d**2*x**4 + 2*a*b**2*c**2*x**2 + 3*a*b**2*c*d*x**4 - 5*a*b**2 *d**2*x**6 - b**3*c**2*x**4 + b**3*c*d*x**6),x)*a**3*d**4 - 20*int((sqrt(x )*sqrt(c - d*x**2)*x**4)/(5*a**3*c*d - 5*a**3*d**2*x**2 - a**2*b*c**2 - 9* a**2*b*c*d*x**2 + 10*a**2*b*d**2*x**4 + 2*a*b**2*c**2*x**2 + 3*a*b**2*c*d* x**4 - 5*a*b**2*d**2*x**6 - b**3*c**2*x**4 + b**3*c*d*x**6),x)*a**2*b*c*d* *3 - 25*int((sqrt(x)*sqrt(c - d*x**2)*x**4)/(5*a**3*c*d - 5*a**3*d**2*x**2 - a**2*b*c**2 - 9*a**2*b*c*d*x**2 + 10*a**2*b*d**2*x**4 + 2*a*b**2*c**2*x **2 + 3*a*b**2*c*d*x**4 - 5*a*b**2*d**2*x**6 - b**3*c**2*x**4 + b**3*c*d*x **6),x)*a**2*b*d**4*x**2 + 3*int((sqrt(x)*sqrt(c - d*x**2)*x**4)/(5*a**3*c *d - 5*a**3*d**2*x**2 - a**2*b*c**2 - 9*a**2*b*c*d*x**2 + 10*a**2*b*d**2*x **4 + 2*a*b**2*c**2*x**2 + 3*a*b**2*c*d*x**4 - 5*a*b**2*d**2*x**6 - b**3*c **2*x**4 + b**3*c*d*x**6),x)*a*b**2*c**2*d**2 + 20*int((sqrt(x)*sqrt(c - d *x**2)*x**4)/(5*a**3*c*d - 5*a**3*d**2*x**2 - a**2*b*c**2 - 9*a**2*b*c*d*x **2 + 10*a**2*b*d**2*x**4 + 2*a*b**2*c**2*x**2 + 3*a*b**2*c*d*x**4 - 5*a*b **2*d**2*x**6 - b**3*c**2*x**4 + b**3*c*d*x**6),x)*a*b**2*c*d**3*x**2 - 3* int((sqrt(x)*sqrt(c - d*x**2)*x**4)/(5*a**3*c*d - 5*a**3*d**2*x**2 - a**2* b*c**2 - 9*a**2*b*c*d*x**2 + 10*a**2*b*d**2*x**4 + 2*a*b**2*c**2*x**2 + 3* a*b**2*c*d*x**4 - 5*a*b**2*d**2*x**6 - b**3*c**2*x**4 + b**3*c*d*x**6),...