\(\int x^5 \sqrt {a+b x^2+c x^4} \, dx\) [949]

Optimal result

Integrand size = 20, antiderivative size = 153 \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\frac {\left (5 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{128 c^3}-\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{7/2}} \] Output:

1/128*(-4*a*c+5*b^2)*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c^3-5/48*b*(c*x^4+b 
*x^2+a)^(3/2)/c^2+1/8*x^2*(c*x^4+b*x^2+a)^(3/2)/c-1/256*(-4*a*c+b^2)*(-4*a 
*c+5*b^2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.86 \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\frac {\sqrt {a+b x^2+c x^4} \left (15 b^3-52 a b c-10 b^2 c x^2+24 a c^2 x^2+8 b c^2 x^4+48 c^3 x^6\right )}{384 c^3}+\frac {\left (5 b^4-24 a b^2 c+16 a^2 c^2\right ) \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{256 c^{7/2}} \] Input:

Integrate[x^5*Sqrt[a + b*x^2 + c*x^4],x]
 

Output:

(Sqrt[a + b*x^2 + c*x^4]*(15*b^3 - 52*a*b*c - 10*b^2*c*x^2 + 24*a*c^2*x^2 
+ 8*b*c^2*x^4 + 48*c^3*x^6))/(384*c^3) + ((5*b^4 - 24*a*b^2*c + 16*a^2*c^2 
)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(256*c^(7/2))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1434, 1166, 27, 1160, 1087, 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.

\(\displaystyle \int x^5 \sqrt {a+b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int x^4 \sqrt {c x^4+b x^2+a}dx^2\)

\(\Big \downarrow \) 1166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

Input:

Int[x^5*Sqrt[a + b*x^2 + c*x^4],x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

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 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

rule 1434

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {\left (-48 c^{3} x^{6}-8 b \,c^{2} x^{4}-24 a \,c^{2} x^{2}+10 b^{2} c \,x^{2}+52 a b c -15 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{384 c^{3}}-\frac {\left (16 a^{2} c^{2}-24 a \,b^{2} c +5 b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {7}{2}}}\)	\(122\)
pseudoelliptic	\(-\frac {\left (a^{2} c^{2}-\frac {3}{2} a \,b^{2} c +\frac {5}{16} b^{4}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (\frac {13 \left (\frac {5 b \,x^{2}}{26}+a \right ) b \,c^{\frac {3}{2}}}{6}+\left (-\frac {1}{3} b \,x^{4}-a \,x^{2}\right ) c^{\frac {5}{2}}-2 c^{\frac {7}{2}} x^{6}-\frac {5 \sqrt {c}\, b^{3}}{8}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}-\ln \left (2\right ) \left (a c -\frac {b^{2}}{4}\right ) \left (a c -\frac {5 b^{2}}{4}\right )}{16 c^{\frac {7}{2}}}\)	\(145\)
default	\(\frac {x^{2} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{8 c}-\frac {5 b \left (\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{16 c}-\frac {a \left (\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\right )}{8 c}\)	\(206\)
elliptic	\(\frac {x^{2} \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{8 c}-\frac {5 b \left (\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{16 c}-\frac {a \left (\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\right )}{8 c}\)	\(206\)

Input:

int(x^5*(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/384*(-48*c^3*x^6-8*b*c^2*x^4-24*a*c^2*x^2+10*b^2*c*x^2+52*a*b*c-15*b^3) 
*(c*x^4+b*x^2+a)^(1/2)/c^3-1/256*(16*a^2*c^2-24*a*b^2*c+5*b^4)/c^(7/2)*ln( 
(1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.98 \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{1536 \, c^{4}}, \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{768 \, c^{4}}\right ] \] Input:

integrate(x^5*(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/1536*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*sqrt(c)*log(-8*c^2*x^4 - 8*b* 
c*x^2 - b^2 + 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4 
*(48*c^4*x^6 + 8*b*c^3*x^4 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c 
^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/768*(3*(5*b^4 - 24*a*b^2*c + 16*a 
^2*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c) 
/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(48*c^4*x^6 + 8*b*c^3*x^4 + 15*b^3*c - 52* 
a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4]
 

Sympy [F]

\[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\int x^{5} \sqrt {a + b x^{2} + c x^{4}}\, dx \] Input:

integrate(x**5*(c*x**4+b*x**2+a)**(1/2),x)
 

Output:

Integral(x**5*sqrt(a + b*x**2 + c*x**4), x)
 

Maxima [F(-2)]

Exception generated. \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5*(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.86 \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\frac {1}{384} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} \] Input:

integrate(x^5*(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

1/384*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) + 1/256*(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))*sqrt(c) + b))/c^( 
7/2)
 

Mupad [B] (verification not implemented)

Time = 18.33 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.26 \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx=\frac {x^2\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}-\frac {a\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{8\,c}-\frac {5\,b\,\left (\frac {\left (8\,c\,\left (c\,x^4+a\right )-3\,b^2+2\,b\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^4+b\,x^2+a}+\frac {2\,c\,x^2+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{16\,c} \] Input:

int(x^5*(a + b*x^2 + c*x^4)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(x^2*(a + b*x^2 + c*x^4)^(3/2))/(8*c) - (a*((b/(4*c) + x^2/2)*(a + b*x^2 + 
 c*x^4)^(1/2) + (log((a + b*x^2 + c*x^4)^(1/2) + (b/2 + c*x^2)/c^(1/2))*(a 
*c - b^2/4))/(2*c^(3/2))))/(8*c) - (5*b*(((8*c*(a + c*x^4) - 3*b^2 + 2*b*c 
*x^2)*(a + b*x^2 + c*x^4)^(1/2))/(24*c^2) + (log(2*(a + b*x^2 + c*x^4)^(1/ 
2) + (b + 2*c*x^2)/c^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2))))/(16*c)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 2818, normalized size of antiderivative = 18.42 \[ \int x^5 \sqrt {a+b x^2+c x^4} \, dx =\text {Too large to display} \] Input:

int(x^5*(c*x^4+b*x^2+a)^(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)*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**3*c**4*x**2 + 1920*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 + 768*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**2*c**3*x**2 - 9216*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*s 
qrt(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 
)*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**5*c + 4800*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 + 13824*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**3*c**3*x**4 + 9216*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqr 
t(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 - 120*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 - 1200*sqrt(c)*sqrt(a...