\(\int \frac {(a x+b x^3+c x^5)^{3/2}}{\sqrt {x}} \, dx\) [58]

Optimal result

Integrand size = 24, antiderivative size = 177 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=-\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {3 \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2} \sqrt {a x+b x^3+c x^5}} \] Output:

-3/128*(-4*a*c+b^2)*(2*c*x^2+b)*(c*x^5+b*x^3+a*x)^(1/2)/c^2/x^(1/2)+1/16*( 
2*c*x^2+b)*(c*x^5+b*x^3+a*x)^(3/2)/c/x^(3/2)+3/256*(-4*a*c+b^2)^2*x^(1/2)* 
(c*x^4+b*x^2+a)^(1/2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2 
))/c^(5/2)/(c*x^5+b*x^3+a*x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\frac {\left (x \left (a+b x^2+c x^4\right )\right )^{3/2} \left (\frac {\sqrt {c} \left (b+2 c x^2\right ) \left (-3 b^2+8 b c x^2+4 c \left (5 a+2 c x^4\right )\right )}{a+b x^2+c x^4}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2+c x^4}}\right )}{\left (a+b x^2+c x^4\right )^{3/2}}\right )}{128 c^{5/2} x^{3/2}} \] Input:

Integrate[(a*x + b*x^3 + c*x^5)^(3/2)/Sqrt[x],x]
 

Output:

((x*(a + b*x^2 + c*x^4))^(3/2)*((Sqrt[c]*(b + 2*c*x^2)*(-3*b^2 + 8*b*c*x^2 
 + 4*c*(5*a + 2*c*x^4)))/(a + b*x^2 + c*x^4) + (3*(b^2 - 4*a*c)^2*ArcTanh[ 
(Sqrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 + c*x^4])])/(a + b*x^2 + c*x^4)^( 
3/2)))/(128*c^(5/2)*x^(3/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1965, 1965, 1961, 1432, 1092, 219}

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.

\(\displaystyle \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx\)

\(\Big \downarrow \) 1965

\(\displaystyle \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {x} \sqrt {c x^5+b x^3+a x}dx}{16 c}\)

\(\Big \downarrow \) 1965

\(\displaystyle \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (b^2-4 a c\right ) \int \frac {x^{3/2}}{\sqrt {c x^5+b x^3+a x}}dx}{8 c}\right )}{16 c}\)

\(\Big \downarrow \) 1961

\(\displaystyle \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \int \frac {x}{\sqrt {c x^4+b x^2+a}}dx}{8 c \sqrt {a x+b x^3+c x^5}}\right )}{16 c}\)

\(\Big \downarrow \) 1432

\(\displaystyle \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{16 c \sqrt {a x+b x^3+c x^5}}\right )}{16 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{8 c \sqrt {a x+b x^3+c x^5}}\right )}{16 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} \sqrt {a x+b x^3+c x^5}}\right )}{16 c}\)

Input:

Int[(a*x + b*x^3 + c*x^5)^(3/2)/Sqrt[x],x]
 

Output:

((b + 2*c*x^2)*(a*x + b*x^3 + c*x^5)^(3/2))/(16*c*x^(3/2)) - (3*(b^2 - 4*a 
*c)*(((b + 2*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(8*c*Sqrt[x]) - ((b^2 - 4*a 
*c)*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[ 
a + b*x^2 + c*x^4])])/(16*c^(3/2)*Sqrt[a*x + b*x^3 + c*x^5])))/(16*c)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1432

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

rule 1961

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)] 
, x_Symbol] :> Simp[x^(q/2)*(Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]/Sqrt[a 
*x^q + b*x^n + c*x^(2*n - q)])   Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) + c*x 
^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && 
 PosQ[n - q] && ((EqQ[m, 1] && EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || 
 EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q, 1]))
 

rule 1965

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m - n + q + 1)*(b + 2*c*x^(n - q))*((a*x^q + b*x^n 
+ c*x^(2*n - q))^p/(2*c*(n - q)*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2* 
c*(2*p + 1)))   Int[x^(m + q)*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], 
x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p 
] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && E 
qQ[m + p*q + 1, n - q]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {\left (16 c^{3} x^{6}+24 b \,c^{2} x^{4}+40 a \,c^{2} x^{2}+2 b^{2} c \,x^{2}+20 a b c -3 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {x}}{128 c^{2} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{256 c^{\frac {5}{2}} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(170\)
default	\(\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \left (32 c^{\frac {7}{2}} x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}+48 b \,c^{\frac {5}{2}} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}+80 a \,c^{\frac {5}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+4 b^{2} c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+48 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a^{2} c^{2}-24 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a \,b^{2} c +3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) b^{4}+40 a b \,c^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}-6 b^{3} \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}} \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\)	\(295\)

Input:

int((c*x^5+b*x^3+a*x)^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/128*(16*c^3*x^6+24*b*c^2*x^4+40*a*c^2*x^2+2*b^2*c*x^2+20*a*b*c-3*b^3)/c^ 
2*(c*x^4+b*x^2+a)*x^(1/2)/(x*(c*x^4+b*x^2+a))^(1/2)+3/256*(16*a^2*c^2-8*a* 
b^2*c+b^4)/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*(c*x^4+ 
b*x^2+a)^(1/2)*x^(1/2)/(x*(c*x^4+b*x^2+a))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.88 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{512 \, c^{3} x}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{256 \, c^{3} x}\right ] \] Input:

integrate((c*x^5+b*x^3+a*x)^(3/2)/x^(1/2),x, algorithm="fricas")
 

Output:

[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*x*log(-(8*c^2*x^5 + 8*b*c 
*x^3 + 4*sqrt(c*x^5 + b*x^3 + a*x)*(2*c*x^2 + b)*sqrt(c)*sqrt(x) + (b^2 + 
4*a*c)*x)/x) + 4*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^ 
2*c^2 + 20*a*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x))/(c^3*x), -1/256* 
(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^5 + b*x^3 
 + a*x)*(2*c*x^2 + b)*sqrt(-c)*sqrt(x)/(c^2*x^5 + b*c*x^3 + a*c*x)) - 2*(1 
6*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x 
^2)*sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x))/(c^3*x)]
 

Sympy [F]

\[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\int \frac {\left (x \left (a + b x^{2} + c x^{4}\right )\right )^{\frac {3}{2}}}{\sqrt {x}}\, dx \] Input:

integrate((c*x**5+b*x**3+a*x)**(3/2)/x**(1/2),x)
 

Output:

Integral((x*(a + b*x**2 + c*x**4))**(3/2)/sqrt(x), x)
 

Maxima [F]

\[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\int { \frac {{\left (c x^{5} + b x^{3} + a x\right )}^{\frac {3}{2}}}{\sqrt {x}} \,d x } \] Input:

integrate((c*x^5+b*x^3+a*x)^(3/2)/x^(1/2),x, algorithm="maxima")
 

Output:

integrate((c*x^5 + b*x^3 + a*x)^(3/2)/sqrt(x), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (149) = 298\).

Time = 0.15 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}}\right )} b + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}}\right )} c \] Input:

integrate((c*x^5+b*x^3+a*x)^(3/2)/x^(1/2),x, algorithm="giac")
 

Output:

1/16*(2*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + (b^2 - 4*a*c)*log(abs(2*(s 
qrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(3/2))*a + 1/96*(2*s 
qrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) - 3*(b^ 
3 - 4*a*b*c)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b 
))/c^(5/2))*b + 1/768*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b/c)*x^2 - 
 (5*b^2*c - 12*a*c^2)/c^3)*x^2 + (15*b^3 - 52*a*b*c)/c^3) + 3*(5*b^4 - 24* 
a*b^2*c + 16*a^2*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sq 
rt(c) + b))/c^(7/2))*c
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\int \frac {{\left (c\,x^5+b\,x^3+a\,x\right )}^{3/2}}{\sqrt {x}} \,d x \] Input:

int((a*x + b*x^3 + c*x^5)^(3/2)/x^(1/2),x)
 

Output:

int((a*x + b*x^3 + c*x^5)^(3/2)/x^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 2869, normalized size of antiderivative = 16.21 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx =\text {Too large to display} \] Input:

int((c*x^5+b*x^3+a*x)^(3/2)/x^(1/2),x)
 

Output:

(1536*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c 
*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**3*b*c**3 + 3072*sqrt(c)*sqrt 
(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x 
**2)/sqrt(4*a*c - b**2))*a**3*c**4*x**2 - 384*sqrt(c)*sqrt(a + b*x**2 + c* 
x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c 
- b**2))*a**2*b**3*c**2 + 2304*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sq 
rt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b 
**2*c**3*x**2 + 9216*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt 
(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b*c**4*x**4 
 + 6144*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + 
 c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*c**5*x**6 - 96*sqrt(c)*s 
qrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2* 
c*x**2)/sqrt(4*a*c - b**2))*a*b**5*c - 1728*sqrt(c)*sqrt(a + b*x**2 + c*x* 
*4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - 
b**2))*a*b**4*c**2*x**2 - 4608*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sq 
rt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a*b**3 
*c**3*x**4 - 3072*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a 
+ b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a*b**2*c**4*x**6 + 
24*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x* 
*4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**7 + 240*sqrt(c)*sqrt(a + b*x...