\(\int \frac {x^{12}}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\) [409]

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

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {x}{b^6}+\frac {a^5 x}{10 b^6 \left (a+b x^2\right )^5}-\frac {51 a^4 x}{80 b^6 \left (a+b x^2\right )^4}+\frac {281 a^3 x}{160 b^6 \left (a+b x^2\right )^3}-\frac {359 a^2 x}{128 b^6 \left (a+b x^2\right )^2}+\frac {843 a x}{256 b^6 \left (a+b x^2\right )}-\frac {693 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 b^{13/2}} \] Output:

x/b^6+1/10*a^5*x/b^6/(b*x^2+a)^5-51/80*a^4*x/b^6/(b*x^2+a)^4+281/160*a^3*x 
/b^6/(b*x^2+a)^3-359/128*a^2*x/b^6/(b*x^2+a)^2+843/256*a*x/b^6/(b*x^2+a)-6 
93/256*a^(1/2)*arctan(b^(1/2)*x/a^(1/2))/b^(13/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {\sqrt {b} x \left (3465 a^5+16170 a^4 b x^2+29568 a^3 b^2 x^4+26070 a^2 b^3 x^6+10615 a b^4 x^8+1280 b^5 x^{10}\right )}{\left (a+b x^2\right )^5}-3465 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{1280 b^{13/2}} \] Input:

Integrate[x^12/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
 

Output:

((Sqrt[b]*x*(3465*a^5 + 16170*a^4*b*x^2 + 29568*a^3*b^2*x^4 + 26070*a^2*b^ 
3*x^6 + 10615*a*b^4*x^8 + 1280*b^5*x^10))/(a + b*x^2)^5 - 3465*Sqrt[a]*Arc 
Tan[(Sqrt[b]*x)/Sqrt[a]])/(1280*b^(13/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1380, 27, 252, 252, 252, 252, 252, 262, 218}

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.

\(\displaystyle \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx\)

\(\Big \downarrow \) 1380

\(\displaystyle b^6 \int \frac {x^{12}}{b^6 \left (b x^2+a\right )^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^{12}}{\left (a+b x^2\right )^6}dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {11 \int \frac {x^{10}}{\left (b x^2+a\right )^5}dx}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {11 \left (\frac {9 \int \frac {x^8}{\left (b x^2+a\right )^4}dx}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \int \frac {x^6}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \int \frac {x^4}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {x^2}{b x^2+a}dx}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {a \int \frac {1}{b x^2+a}dx}{b}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\)

Input:

Int[x^12/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
 

Output:

-1/10*x^11/(b*(a + b*x^2)^5) + (11*(-1/8*x^9/(b*(a + b*x^2)^4) + (9*(-1/6* 
x^7/(b*(a + b*x^2)^3) + (7*(-1/4*x^5/(b*(a + b*x^2)^2) + (5*(-1/2*x^3/(b*( 
a + b*x^2)) + (3*(x/b - (Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2)))/(2 
*b)))/(4*b)))/(6*b)))/(8*b)))/(10*b)
 

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 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 252
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 1380
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 
Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64

method result size
default \(\frac {x}{b^{6}}-\frac {a \left (\frac {-\frac {843}{256} b^{4} x^{9}-\frac {1327}{128} a \,b^{3} x^{7}-\frac {131}{10} a^{2} b^{2} x^{5}-\frac {977}{128} a^{3} b \,x^{3}-\frac {437}{256} a^{4} x}{\left (b \,x^{2}+a \right )^{5}}+\frac {693 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \sqrt {a b}}\right )}{b^{6}}\) \(84\)
risch \(\frac {x}{b^{6}}+\frac {\frac {843}{256} a \,b^{4} x^{9}+\frac {1327}{128} a^{2} b^{3} x^{7}+\frac {131}{10} a^{3} b^{2} x^{5}+\frac {977}{128} a^{4} b \,x^{3}+\frac {437}{256} x \,a^{5}}{b^{6} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}+\frac {693 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{512 b^{7}}-\frac {693 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{512 b^{7}}\) \(136\)

Input:

int(x^12/(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

x/b^6-1/b^6*a*((-843/256*b^4*x^9-1327/128*a*b^3*x^7-131/10*a^2*b^2*x^5-977 
/128*a^3*b*x^3-437/256*a^4*x)/(b*x^2+a)^5+693/256/(a*b)^(1/2)*arctan(b/(a* 
b)^(1/2)*x))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.05 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\left [\frac {2560 \, b^{5} x^{11} + 21230 \, a b^{4} x^{9} + 52140 \, a^{2} b^{3} x^{7} + 59136 \, a^{3} b^{2} x^{5} + 32340 \, a^{4} b x^{3} + 6930 \, a^{5} x + 3465 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{2560 \, {\left (b^{11} x^{10} + 5 \, a b^{10} x^{8} + 10 \, a^{2} b^{9} x^{6} + 10 \, a^{3} b^{8} x^{4} + 5 \, a^{4} b^{7} x^{2} + a^{5} b^{6}\right )}}, \frac {1280 \, b^{5} x^{11} + 10615 \, a b^{4} x^{9} + 26070 \, a^{2} b^{3} x^{7} + 29568 \, a^{3} b^{2} x^{5} + 16170 \, a^{4} b x^{3} + 3465 \, a^{5} x - 3465 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{1280 \, {\left (b^{11} x^{10} + 5 \, a b^{10} x^{8} + 10 \, a^{2} b^{9} x^{6} + 10 \, a^{3} b^{8} x^{4} + 5 \, a^{4} b^{7} x^{2} + a^{5} b^{6}\right )}}\right ] \] Input:

integrate(x^12/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")
 

Output:

[1/2560*(2560*b^5*x^11 + 21230*a*b^4*x^9 + 52140*a^2*b^3*x^7 + 59136*a^3*b 
^2*x^5 + 32340*a^4*b*x^3 + 6930*a^5*x + 3465*(b^5*x^10 + 5*a*b^4*x^8 + 10* 
a^2*b^3*x^6 + 10*a^3*b^2*x^4 + 5*a^4*b*x^2 + a^5)*sqrt(-a/b)*log((b*x^2 - 
2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^11*x^10 + 5*a*b^10*x^8 + 10*a^2*b^9 
*x^6 + 10*a^3*b^8*x^4 + 5*a^4*b^7*x^2 + a^5*b^6), 1/1280*(1280*b^5*x^11 + 
10615*a*b^4*x^9 + 26070*a^2*b^3*x^7 + 29568*a^3*b^2*x^5 + 16170*a^4*b*x^3 
+ 3465*a^5*x - 3465*(b^5*x^10 + 5*a*b^4*x^8 + 10*a^2*b^3*x^6 + 10*a^3*b^2* 
x^4 + 5*a^4*b*x^2 + a^5)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^11*x^10 + 5 
*a*b^10*x^8 + 10*a^2*b^9*x^6 + 10*a^3*b^8*x^4 + 5*a^4*b^7*x^2 + a^5*b^6)]
 

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.36 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {693 \sqrt {- \frac {a}{b^{13}}} \log {\left (- b^{6} \sqrt {- \frac {a}{b^{13}}} + x \right )}}{512} - \frac {693 \sqrt {- \frac {a}{b^{13}}} \log {\left (b^{6} \sqrt {- \frac {a}{b^{13}}} + x \right )}}{512} + \frac {2185 a^{5} x + 9770 a^{4} b x^{3} + 16768 a^{3} b^{2} x^{5} + 13270 a^{2} b^{3} x^{7} + 4215 a b^{4} x^{9}}{1280 a^{5} b^{6} + 6400 a^{4} b^{7} x^{2} + 12800 a^{3} b^{8} x^{4} + 12800 a^{2} b^{9} x^{6} + 6400 a b^{10} x^{8} + 1280 b^{11} x^{10}} + \frac {x}{b^{6}} \] Input:

integrate(x**12/(b**2*x**4+2*a*b*x**2+a**2)**3,x)
 

Output:

693*sqrt(-a/b**13)*log(-b**6*sqrt(-a/b**13) + x)/512 - 693*sqrt(-a/b**13)* 
log(b**6*sqrt(-a/b**13) + x)/512 + (2185*a**5*x + 9770*a**4*b*x**3 + 16768 
*a**3*b**2*x**5 + 13270*a**2*b**3*x**7 + 4215*a*b**4*x**9)/(1280*a**5*b**6 
 + 6400*a**4*b**7*x**2 + 12800*a**3*b**8*x**4 + 12800*a**2*b**9*x**6 + 640 
0*a*b**10*x**8 + 1280*b**11*x**10) + x/b**6
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {4215 \, a b^{4} x^{9} + 13270 \, a^{2} b^{3} x^{7} + 16768 \, a^{3} b^{2} x^{5} + 9770 \, a^{4} b x^{3} + 2185 \, a^{5} x}{1280 \, {\left (b^{11} x^{10} + 5 \, a b^{10} x^{8} + 10 \, a^{2} b^{9} x^{6} + 10 \, a^{3} b^{8} x^{4} + 5 \, a^{4} b^{7} x^{2} + a^{5} b^{6}\right )}} - \frac {693 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{6}} + \frac {x}{b^{6}} \] Input:

integrate(x^12/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")
 

Output:

1/1280*(4215*a*b^4*x^9 + 13270*a^2*b^3*x^7 + 16768*a^3*b^2*x^5 + 9770*a^4* 
b*x^3 + 2185*a^5*x)/(b^11*x^10 + 5*a*b^10*x^8 + 10*a^2*b^9*x^6 + 10*a^3*b^ 
8*x^4 + 5*a^4*b^7*x^2 + a^5*b^6) - 693/256*a*arctan(b*x/sqrt(a*b))/(sqrt(a 
*b)*b^6) + x/b^6
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {693 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{6}} + \frac {x}{b^{6}} + \frac {4215 \, a b^{4} x^{9} + 13270 \, a^{2} b^{3} x^{7} + 16768 \, a^{3} b^{2} x^{5} + 9770 \, a^{4} b x^{3} + 2185 \, a^{5} x}{1280 \, {\left (b x^{2} + a\right )}^{5} b^{6}} \] Input:

integrate(x^12/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")
 

Output:

-693/256*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^6) + x/b^6 + 1/1280*(4215*a* 
b^4*x^9 + 13270*a^2*b^3*x^7 + 16768*a^3*b^2*x^5 + 9770*a^4*b*x^3 + 2185*a^ 
5*x)/((b*x^2 + a)^5*b^6)
 

Mupad [B] (verification not implemented)

Time = 17.91 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {437\,a^5\,x}{256}+\frac {977\,a^4\,b\,x^3}{128}+\frac {131\,a^3\,b^2\,x^5}{10}+\frac {1327\,a^2\,b^3\,x^7}{128}+\frac {843\,a\,b^4\,x^9}{256}}{a^5\,b^6+5\,a^4\,b^7\,x^2+10\,a^3\,b^8\,x^4+10\,a^2\,b^9\,x^6+5\,a\,b^{10}\,x^8+b^{11}\,x^{10}}+\frac {x}{b^6}-\frac {693\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{256\,b^{13/2}} \] Input:

int(x^12/(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)
 

Output:

((437*a^5*x)/256 + (977*a^4*b*x^3)/128 + (843*a*b^4*x^9)/256 + (131*a^3*b^ 
2*x^5)/10 + (1327*a^2*b^3*x^7)/128)/(a^5*b^6 + b^11*x^10 + 5*a*b^10*x^8 + 
5*a^4*b^7*x^2 + 10*a^3*b^8*x^4 + 10*a^2*b^9*x^6) + x/b^6 - (693*a^(1/2)*at 
an((b^(1/2)*x)/a^(1/2)))/(256*b^(13/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.02 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {-3465 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{5}-17325 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} b \,x^{2}-34650 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b^{2} x^{4}-34650 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{3} x^{6}-17325 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{4} x^{8}-3465 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{5} x^{10}+3465 a^{5} b x +16170 a^{4} b^{2} x^{3}+29568 a^{3} b^{3} x^{5}+26070 a^{2} b^{4} x^{7}+10615 a \,b^{5} x^{9}+1280 b^{6} x^{11}}{1280 b^{7} \left (b^{5} x^{10}+5 a \,b^{4} x^{8}+10 a^{2} b^{3} x^{6}+10 a^{3} b^{2} x^{4}+5 a^{4} b \,x^{2}+a^{5}\right )} \] Input:

int(x^12/(b^2*x^4+2*a*b*x^2+a^2)^3,x)
 

Output:

( - 3465*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5 - 17325*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b*x**2 - 34650*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**2*x**4 - 34650*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**3*x**6 - 17325*sqrt(b)*sqrt(a)*atan((b* 
x)/(sqrt(b)*sqrt(a)))*a*b**4*x**8 - 3465*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt( 
b)*sqrt(a)))*b**5*x**10 + 3465*a**5*b*x + 16170*a**4*b**2*x**3 + 29568*a** 
3*b**3*x**5 + 26070*a**2*b**4*x**7 + 10615*a*b**5*x**9 + 1280*b**6*x**11)/ 
(1280*b**7*(a**5 + 5*a**4*b*x**2 + 10*a**3*b**2*x**4 + 10*a**2*b**3*x**6 + 
 5*a*b**4*x**8 + b**5*x**10))