\(\int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx\) [989]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 145 \[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=-\frac {\sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {5 b \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {b \left (5 b^2-12 a c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}} \] Output:

-1/6*(c*x^4+b*x^2+a)^(1/2)/a/x^6+5/24*b*(c*x^4+b*x^2+a)^(1/2)/a^2/x^4-1/48 
*(-16*a*c+15*b^2)*(c*x^4+b*x^2+a)^(1/2)/a^3/x^2+1/32*b*(-12*a*c+5*b^2)*arc 
tanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=\frac {\sqrt {a+b x^2+c x^4} \left (-8 a^2+10 a b x^2-15 b^2 x^4+16 a c x^4\right )}{48 a^3 x^6}+\frac {\left (-5 b^3+12 a b c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{16 a^{7/2}} \] Input:

Integrate[1/(x^7*Sqrt[a + b*x^2 + c*x^4]),x]
 

Output:

(Sqrt[a + b*x^2 + c*x^4]*(-8*a^2 + 10*a*b*x^2 - 15*b^2*x^4 + 16*a*c*x^4))/ 
(48*a^3*x^6) + ((-5*b^3 + 12*a*b*c)*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 
+ c*x^4])/Sqrt[a]])/(16*a^(7/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1434, 1167, 27, 1237, 27, 1228, 1154, 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.

\(\displaystyle \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {1}{x^8 \sqrt {c x^4+b x^2+a}}dx^2\)

\(\Big \downarrow \) 1167

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {4 c x^2+5 b}{2 x^6 \sqrt {c x^4+b x^2+a}}dx^2}{3 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {4 c x^2+5 b}{x^6 \sqrt {c x^4+b x^2+a}}dx^2}{6 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

\(\Big \downarrow \) 1237

\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {\int \frac {15 b^2+10 c x^2 b-16 a c}{2 x^4 \sqrt {c x^4+b x^2+a}}dx^2}{2 a}-\frac {5 b \sqrt {a+b x^2+c x^4}}{2 a x^4}}{6 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {\int \frac {15 b^2+10 c x^2 b-16 a c}{x^4 \sqrt {c x^4+b x^2+a}}dx^2}{4 a}-\frac {5 b \sqrt {a+b x^2+c x^4}}{2 a x^4}}{6 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

\(\Big \downarrow \) 1228

\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {-\frac {3 b \left (5 b^2-12 a c\right ) \int \frac {1}{x^2 \sqrt {c x^4+b x^2+a}}dx^2}{2 a}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{a x^2}}{4 a}-\frac {5 b \sqrt {a+b x^2+c x^4}}{2 a x^4}}{6 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {\frac {3 b \left (5 b^2-12 a c\right ) \int \frac {1}{4 a-x^4}d\frac {b x^2+2 a}{\sqrt {c x^4+b x^2+a}}}{a}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{a x^2}}{4 a}-\frac {5 b \sqrt {a+b x^2+c x^4}}{2 a x^4}}{6 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {\frac {3 b \left (5 b^2-12 a c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 a^{3/2}}-\frac {\left (15 b^2-16 a c\right ) \sqrt {a+b x^2+c x^4}}{a x^2}}{4 a}-\frac {5 b \sqrt {a+b x^2+c x^4}}{2 a x^4}}{6 a}-\frac {\sqrt {a+b x^2+c x^4}}{3 a x^6}\right )\)

Input:

Int[1/(x^7*Sqrt[a + b*x^2 + c*x^4]),x]
 

Output:

(-1/3*Sqrt[a + b*x^2 + c*x^4]/(a*x^6) - ((-5*b*Sqrt[a + b*x^2 + c*x^4])/(2 
*a*x^4) - (-(((15*b^2 - 16*a*c)*Sqrt[a + b*x^2 + c*x^4])/(a*x^2)) + (3*b*( 
5*b^2 - 12*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])] 
)/(2*a^(3/2)))/(4*a))/(6*a))/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 1154
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1167
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)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2))   Int[ 
(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, 
 x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m 
, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp 
lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
 

rule 1228
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^ 
(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 1237
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* 
x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) 
*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ 
(c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m 
+ 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 
] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 = 101, normalized size of antiderivative = 0.70

method result size
risch \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (-16 a c \,x^{4}+15 b^{2} x^{4}-10 a b \,x^{2}+8 a^{2}\right )}{48 a^{3} x^{6}}-\frac {b \left (12 a c -5 b^{2}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {7}{2}}}\) \(101\)
pseudoelliptic \(-\frac {3 \left (b \,x^{6} \left (a c -\frac {5 b^{2}}{12}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )+\frac {5 \left (-\frac {2 \left (\frac {8 c \,x^{2}}{5}+b \right ) x^{2} a^{\frac {3}{2}}}{3}+\sqrt {a}\, b^{2} x^{4}+\frac {8 a^{\frac {5}{2}}}{15}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{6}\right )}{8 a^{\frac {7}{2}} x^{6}}\) \(105\)
default \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{6 a \,x^{6}}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{24 a^{2} x^{4}}-\frac {5 b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 a^{3} x^{2}}+\frac {5 b^{3} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {7}{2}}}-\frac {3 b c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{8 a^{\frac {5}{2}}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{3 a^{2} x^{2}}\) \(176\)
elliptic \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{6 a \,x^{6}}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{24 a^{2} x^{4}}-\frac {5 b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 a^{3} x^{2}}+\frac {5 b^{3} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {7}{2}}}-\frac {3 b c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{8 a^{\frac {5}{2}}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{3 a^{2} x^{2}}\) \(176\)

Input:

int(1/x^7/(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48*(c*x^4+b*x^2+a)^(1/2)*(-16*a*c*x^4+15*b^2*x^4-10*a*b*x^2+8*a^2)/a^3/ 
x^6-1/32*b*(12*a*c-5*b^2)/a^(7/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^ 
(1/2))/x^2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.83 \[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} - 12 \, a b c\right )} \sqrt {a} x^{6} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left (10 \, a^{2} b x^{2} - {\left (15 \, a b^{2} - 16 \, a^{2} c\right )} x^{4} - 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, a^{4} x^{6}}, -\frac {3 \, {\left (5 \, b^{3} - 12 \, a b c\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 2 \, {\left (10 \, a^{2} b x^{2} - {\left (15 \, a b^{2} - 16 \, a^{2} c\right )} x^{4} - 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, a^{4} x^{6}}\right ] \] Input:

integrate(1/x^7/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/192*(3*(5*b^3 - 12*a*b*c)*sqrt(a)*x^6*log(-((b^2 + 4*a*c)*x^4 + 8*a*b* 
x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) - 4*(1 
0*a^2*b*x^2 - (15*a*b^2 - 16*a^2*c)*x^4 - 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/ 
(a^4*x^6), -1/96*(3*(5*b^3 - 12*a*b*c)*sqrt(-a)*x^6*arctan(1/2*sqrt(c*x^4 
+ b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) - 2*(10*a^2 
*b*x^2 - (15*a*b^2 - 16*a^2*c)*x^4 - 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(a^4* 
x^6)]
 

Sympy [F]

\[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {1}{x^{7} \sqrt {a + b x^{2} + c x^{4}}}\, dx \] Input:

integrate(1/x**7/(c*x**4+b*x**2+a)**(1/2),x)
 

Output:

Integral(1/(x**7*sqrt(a + b*x**2 + c*x**4)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/x^7/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (123) = 246\).

Time = 0.13 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.31 \[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=-\frac {{\left (5 \, b^{3} - 12 \, a b c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{16 \, \sqrt {-a} a^{3}} + \frac {15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} b^{3} - 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b c - 40 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a b^{3} + 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b c + 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{3} c^{\frac {3}{2}} + 33 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} b^{3} + 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b c + 48 \, a^{3} b^{2} \sqrt {c} - 32 \, a^{4} c^{\frac {3}{2}}}{48 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{3}} \] Input:

integrate(1/x^7/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

-1/16*(5*b^3 - 12*a*b*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/s 
qrt(-a))/(sqrt(-a)*a^3) + 1/48*(15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a)) 
^5*b^3 - 36*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a*b*c - 40*(sqrt(c)* 
x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a*b^3 + 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b* 
x^2 + a))^3*a^2*b*c + 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^3*c^( 
3/2) + 33*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^2*b^3 + 36*(sqrt(c)*x^ 
2 - sqrt(c*x^4 + b*x^2 + a))*a^3*b*c + 48*a^3*b^2*sqrt(c) - 32*a^4*c^(3/2) 
)/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^3*a^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {1}{x^7\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \] Input:

int(1/(x^7*(a + b*x^2 + c*x^4)^(1/2)),x)
                                                                                    
                                                                                    
 

Output:

int(1/(x^7*(a + b*x^2 + c*x^4)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^7 \sqrt {a+b x^2+c x^4}} \, dx=\text {too large to display} \] Input:

int(1/x^7/(c*x^4+b*x^2+a)^(1/2),x)
 

Output:

( - 128*sqrt(a + b*x**2 + c*x**4)*a**5*c**3 - 96*sqrt(a + b*x**2 + c*x**4) 
*a**4*b**2*c**2 - 128*sqrt(a + b*x**2 + c*x**4)*a**4*b*c**3*x**2 + 832*sqr 
t(a + b*x**2 + c*x**4)*a**4*c**4*x**4 + 80*sqrt(a + b*x**2 + c*x**4)*a**3* 
b**4*c + 336*sqrt(a + b*x**2 + c*x**4)*a**3*b**3*c**2*x**2 - 48*sqrt(a + b 
*x**2 + c*x**4)*a**3*b**2*c**3*x**4 - 20*sqrt(a + b*x**2 + c*x**4)*a**2*b* 
*5*c*x**2 - 824*sqrt(a + b*x**2 + c*x**4)*a**2*b**4*c**2*x**4 - 50*sqrt(a 
+ b*x**2 + c*x**4)*a*b**7*x**2 + 130*sqrt(a + b*x**2 + c*x**4)*a*b**6*c*x* 
*4 + 75*sqrt(a + b*x**2 + c*x**4)*b**8*x**4 + 288*sqrt(a)*log(sqrt(a + b*x 
**2 + c*x**4) - sqrt(a))*a**3*b*c**4*x**6 + 96*sqrt(a)*log(sqrt(a + b*x**2 
 + c*x**4) - sqrt(a))*a**2*b**3*c**3*x**6 - 270*sqrt(a)*log(sqrt(a + b*x** 
2 + c*x**4) - sqrt(a))*a*b**5*c**2*x**6 + 75*sqrt(a)*log(sqrt(a + b*x**2 + 
 c*x**4) - sqrt(a))*b**7*c*x**6 - 288*sqrt(a)*log(sqrt(a + b*x**2 + c*x**4 
) + sqrt(a))*a**3*b*c**4*x**6 - 96*sqrt(a)*log(sqrt(a + b*x**2 + c*x**4) + 
 sqrt(a))*a**2*b**3*c**3*x**6 + 270*sqrt(a)*log(sqrt(a + b*x**2 + c*x**4) 
+ sqrt(a))*a*b**5*c**2*x**6 - 75*sqrt(a)*log(sqrt(a + b*x**2 + c*x**4) + s 
qrt(a))*b**7*c*x**6 - 9216*int(sqrt(a + b*x**2 + c*x**4)/(8*a**3*b*c**2*x* 
*5 + 8*a**3*c**3*x**7 + 6*a**2*b**3*c*x**5 + 14*a**2*b**2*c**2*x**7 + 16*a 
**2*b*c**3*x**9 + 8*a**2*c**4*x**11 - 5*a*b**5*x**5 + a*b**4*c*x**7 + 12*a 
*b**3*c**2*x**9 + 6*a*b**2*c**3*x**11 - 5*b**6*x**7 - 10*b**5*c*x**9 - 5*b 
**4*c**2*x**11),x)*a**7*b**2*c**5*x**6 + 13504*int(sqrt(a + b*x**2 + c*...