\(\int \frac {A+B x}{\sqrt {x} (a^2+2 a b x+b^2 x^2)^2} \, dx\) [399]

Optimal result

Integrand size = 29, antiderivative size = 130 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(A b-a B) \sqrt {x}}{3 a b (a+b x)^3}+\frac {(5 A b+a B) \sqrt {x}}{12 a^2 b (a+b x)^2}+\frac {(5 A b+a B) \sqrt {x}}{8 a^3 b (a+b x)}+\frac {(5 A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{7/2} b^{3/2}} \] Output:

1/3*(A*b-B*a)*x^(1/2)/a/b/(b*x+a)^3+1/12*(5*A*b+B*a)*x^(1/2)/a^2/b/(b*x+a) 
^2+1/8*(5*A*b+B*a)*x^(1/2)/a^3/b/(b*x+a)+1/8*(5*A*b+B*a)*arctan(b^(1/2)*x^ 
(1/2)/a^(1/2))/a^(7/2)/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {x} \left (-3 a^3 B+15 A b^3 x^2+a b^2 x (40 A+3 B x)+a^2 b (33 A+8 B x)\right )}{24 a^3 b (a+b x)^3}+\frac {(5 A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{7/2} b^{3/2}} \] Input:

Integrate[(A + B*x)/(Sqrt[x]*(a^2 + 2*a*b*x + b^2*x^2)^2),x]
 

Output:

(Sqrt[x]*(-3*a^3*B + 15*A*b^3*x^2 + a*b^2*x*(40*A + 3*B*x) + a^2*b*(33*A + 
 8*B*x)))/(24*a^3*b*(a + b*x)^3) + ((5*A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x]) 
/Sqrt[a]])/(8*a^(7/2)*b^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1184, 27, 87, 52, 52, 73, 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.

\(\displaystyle \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1184

\(\displaystyle b^4 \int \frac {A+B x}{b^4 \sqrt {x} (a+b x)^4}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {A+B x}{\sqrt {x} (a+b x)^4}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(a B+5 A b) \int \frac {1}{\sqrt {x} (a+b x)^3}dx}{6 a b}+\frac {\sqrt {x} (A b-a B)}{3 a b (a+b x)^3}\)

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a B+5 A b) \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}+\frac {\sqrt {x}}{a (a+b x)}\right )}{4 a}+\frac {\sqrt {x}}{2 a (a+b x)^2}\right )}{6 a b}+\frac {\sqrt {x} (A b-a B)}{3 a b (a+b x)^3}\)

Input:

Int[(A + B*x)/(Sqrt[x]*(a^2 + 2*a*b*x + b^2*x^2)^2),x]
 

Output:

((A*b - a*B)*Sqrt[x])/(3*a*b*(a + b*x)^3) + ((5*A*b + a*B)*(Sqrt[x]/(2*a*( 
a + b*x)^2) + (3*(Sqrt[x]/(a*(a + b*x)) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a] 
]/(a^(3/2)*Sqrt[b])))/(4*a)))/(6*a*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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {\left (5 A b +B a \right ) b \,x^{\frac {5}{2}}}{8 a^{3}}+\frac {\left (5 A b +B a \right ) x^{\frac {3}{2}}}{3 a^{2}}+\frac {\left (11 A b -B a \right ) \sqrt {x}}{8 a b}}{\left (b x +a \right )^{3}}+\frac {\left (5 A b +B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 a^{3} b \sqrt {a b}}\)	\(97\)
default	\(\frac {\frac {\left (5 A b +B a \right ) b \,x^{\frac {5}{2}}}{8 a^{3}}+\frac {\left (5 A b +B a \right ) x^{\frac {3}{2}}}{3 a^{2}}+\frac {\left (11 A b -B a \right ) \sqrt {x}}{8 a b}}{\left (b x +a \right )^{3}}+\frac {\left (5 A b +B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 a^{3} b \sqrt {a b}}\)	\(97\)

Input:

int((B*x+A)/x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
 

Output:

2*(1/16*(5*A*b+B*a)/a^3*b*x^(5/2)+1/6/a^2*(5*A*b+B*a)*x^(3/2)+1/16*(11*A*b 
-B*a)/a/b*x^(1/2))/(b*x+a)^3+1/8*(5*A*b+B*a)/a^3/b/(a*b)^(1/2)*arctan(b*x^ 
(1/2)/(a*b)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.15 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [-\frac {3 \, {\left (B a^{4} + 5 \, A a^{3} b + {\left (B a b^{3} + 5 \, A b^{4}\right )} x^{3} + 3 \, {\left (B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{2} + 3 \, {\left (B a^{3} b + 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (3 \, B a^{4} b - 33 \, A a^{3} b^{2} - 3 \, {\left (B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{2} - 8 \, {\left (B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{48 \, {\left (a^{4} b^{5} x^{3} + 3 \, a^{5} b^{4} x^{2} + 3 \, a^{6} b^{3} x + a^{7} b^{2}\right )}}, -\frac {3 \, {\left (B a^{4} + 5 \, A a^{3} b + {\left (B a b^{3} + 5 \, A b^{4}\right )} x^{3} + 3 \, {\left (B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{2} + 3 \, {\left (B a^{3} b + 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (3 \, B a^{4} b - 33 \, A a^{3} b^{2} - 3 \, {\left (B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{2} - 8 \, {\left (B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{5} x^{3} + 3 \, a^{5} b^{4} x^{2} + 3 \, a^{6} b^{3} x + a^{7} b^{2}\right )}}\right ] \] Input:

integrate((B*x+A)/x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
 

Output:

[-1/48*(3*(B*a^4 + 5*A*a^3*b + (B*a*b^3 + 5*A*b^4)*x^3 + 3*(B*a^2*b^2 + 5* 
A*a*b^3)*x^2 + 3*(B*a^3*b + 5*A*a^2*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sq 
rt(-a*b)*sqrt(x))/(b*x + a)) + 2*(3*B*a^4*b - 33*A*a^3*b^2 - 3*(B*a^2*b^3 
+ 5*A*a*b^4)*x^2 - 8*(B*a^3*b^2 + 5*A*a^2*b^3)*x)*sqrt(x))/(a^4*b^5*x^3 + 
3*a^5*b^4*x^2 + 3*a^6*b^3*x + a^7*b^2), -1/24*(3*(B*a^4 + 5*A*a^3*b + (B*a 
*b^3 + 5*A*b^4)*x^3 + 3*(B*a^2*b^2 + 5*A*a*b^3)*x^2 + 3*(B*a^3*b + 5*A*a^2 
*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (3*B*a^4*b - 33*A*a^3*b 
^2 - 3*(B*a^2*b^3 + 5*A*a*b^4)*x^2 - 8*(B*a^3*b^2 + 5*A*a^2*b^3)*x)*sqrt(x 
))/(a^4*b^5*x^3 + 3*a^5*b^4*x^2 + 3*a^6*b^3*x + a^7*b^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2300 vs. \(2 (119) = 238\).

Time = 32.98 (sec) , antiderivative size = 2300, normalized size of antiderivative = 17.69 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)/x**(1/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)
 

Output:

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0) 
), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a**4, Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 
2*B/(5*x**(5/2)))/b**4, Eq(a, 0)), (15*A*a**3*b*log(sqrt(x) - sqrt(-a/b))/ 
(48*a**6*b**2*sqrt(-a/b) + 144*a**5*b**3*x*sqrt(-a/b) + 144*a**4*b**4*x**2 
*sqrt(-a/b) + 48*a**3*b**5*x**3*sqrt(-a/b)) - 15*A*a**3*b*log(sqrt(x) + sq 
rt(-a/b))/(48*a**6*b**2*sqrt(-a/b) + 144*a**5*b**3*x*sqrt(-a/b) + 144*a**4 
*b**4*x**2*sqrt(-a/b) + 48*a**3*b**5*x**3*sqrt(-a/b)) + 66*A*a**2*b**2*sqr 
t(x)*sqrt(-a/b)/(48*a**6*b**2*sqrt(-a/b) + 144*a**5*b**3*x*sqrt(-a/b) + 14 
4*a**4*b**4*x**2*sqrt(-a/b) + 48*a**3*b**5*x**3*sqrt(-a/b)) + 45*A*a**2*b* 
*2*x*log(sqrt(x) - sqrt(-a/b))/(48*a**6*b**2*sqrt(-a/b) + 144*a**5*b**3*x* 
sqrt(-a/b) + 144*a**4*b**4*x**2*sqrt(-a/b) + 48*a**3*b**5*x**3*sqrt(-a/b)) 
 - 45*A*a**2*b**2*x*log(sqrt(x) + sqrt(-a/b))/(48*a**6*b**2*sqrt(-a/b) + 1 
44*a**5*b**3*x*sqrt(-a/b) + 144*a**4*b**4*x**2*sqrt(-a/b) + 48*a**3*b**5*x 
**3*sqrt(-a/b)) + 80*A*a*b**3*x**(3/2)*sqrt(-a/b)/(48*a**6*b**2*sqrt(-a/b) 
 + 144*a**5*b**3*x*sqrt(-a/b) + 144*a**4*b**4*x**2*sqrt(-a/b) + 48*a**3*b* 
*5*x**3*sqrt(-a/b)) + 45*A*a*b**3*x**2*log(sqrt(x) - sqrt(-a/b))/(48*a**6* 
b**2*sqrt(-a/b) + 144*a**5*b**3*x*sqrt(-a/b) + 144*a**4*b**4*x**2*sqrt(-a/ 
b) + 48*a**3*b**5*x**3*sqrt(-a/b)) - 45*A*a*b**3*x**2*log(sqrt(x) + sqrt(- 
a/b))/(48*a**6*b**2*sqrt(-a/b) + 144*a**5*b**3*x*sqrt(-a/b) + 144*a**4*b** 
4*x**2*sqrt(-a/b) + 48*a**3*b**5*x**3*sqrt(-a/b)) + 30*A*b**4*x**(5/2)*...
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {3 \, {\left (B a b^{2} + 5 \, A b^{3}\right )} x^{\frac {5}{2}} + 8 \, {\left (B a^{2} b + 5 \, A a b^{2}\right )} x^{\frac {3}{2}} - 3 \, {\left (B a^{3} - 11 \, A a^{2} b\right )} \sqrt {x}}{24 \, {\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} + \frac {{\left (B a + 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} b} \] Input:

integrate((B*x+A)/x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/24*(3*(B*a*b^2 + 5*A*b^3)*x^(5/2) + 8*(B*a^2*b + 5*A*a*b^2)*x^(3/2) - 3* 
(B*a^3 - 11*A*a^2*b)*sqrt(x))/(a^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + 
 a^6*b) + 1/8*(B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {{\left (B a + 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} b} + \frac {3 \, B a b^{2} x^{\frac {5}{2}} + 15 \, A b^{3} x^{\frac {5}{2}} + 8 \, B a^{2} b x^{\frac {3}{2}} + 40 \, A a b^{2} x^{\frac {3}{2}} - 3 \, B a^{3} \sqrt {x} + 33 \, A a^{2} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} a^{3} b} \] Input:

integrate((B*x+A)/x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
 

Output:

1/8*(B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b) + 1/24*(3* 
B*a*b^2*x^(5/2) + 15*A*b^3*x^(5/2) + 8*B*a^2*b*x^(3/2) + 40*A*a*b^2*x^(3/2 
) - 3*B*a^3*sqrt(x) + 33*A*a^2*b*sqrt(x))/((b*x + a)^3*a^3*b)
 

Mupad [B] (verification not implemented)

Time = 10.86 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {x^{3/2}\,\left (5\,A\,b+B\,a\right )}{3\,a^2}+\frac {\sqrt {x}\,\left (11\,A\,b-B\,a\right )}{8\,a\,b}+\frac {b\,x^{5/2}\,\left (5\,A\,b+B\,a\right )}{8\,a^3}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (5\,A\,b+B\,a\right )}{8\,a^{7/2}\,b^{3/2}} \] Input:

int((A + B*x)/(x^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^2),x)
 

Output:

((x^(3/2)*(5*A*b + B*a))/(3*a^2) + (x^(1/2)*(11*A*b - B*a))/(8*a*b) + (b*x 
^(5/2)*(5*A*b + B*a))/(8*a^3))/(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x) + 
 (atan((b^(1/2)*x^(1/2))/a^(1/2))*(5*A*b + B*a))/(8*a^(7/2)*b^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a b x +3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{2}+5 \sqrt {x}\, a^{2} b +3 \sqrt {x}\, a \,b^{2} x}{4 a^{3} b \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:

int((B*x+A)/x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)
 

Output:

(3*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2 + 6*sqrt(b)*sq 
rt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b*x + 3*sqrt(b)*sqrt(a)*atan(( 
sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**2*x**2 + 5*sqrt(x)*a**2*b + 3*sqrt(x)*a*b 
**2*x)/(4*a**3*b*(a**2 + 2*a*b*x + b**2*x**2))