\(\int \frac {x (A+B x+C x^2+D x^3)}{(a+b x^2)^{5/2}} \, dx\) [112]

Optimal result

Integrand size = 28, antiderivative size = 100 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {A b-a C+(b B-a D) x}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {3 a C-(b B-4 a D) x}{3 a b^2 \sqrt {a+b x^2}}+\frac {D \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:

-1/3*(A*b-C*a+(B*b-D*a)*x)/b^2/(b*x^2+a)^(3/2)-1/3*(3*C*a-(B*b-4*D*a)*x)/a 
/b^2/(b*x^2+a)^(1/2)+D*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.91 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {b^2 B x^3-a^2 (2 C+3 D x)-a b \left (A+3 C x^2+4 D x^3\right )}{3 a b^2 \left (a+b x^2\right )^{3/2}}-\frac {D \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{5/2}} \] Input:

Integrate[(x*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(5/2),x]
 

Output:

(b^2*B*x^3 - a^2*(2*C + 3*D*x) - a*b*(A + 3*C*x^2 + 4*D*x^3))/(3*a*b^2*(a 
+ b*x^2)^(3/2)) - (D*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/b^(5/2)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2335, 25, 2345, 27, 224, 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.

\(\displaystyle \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2335

\(\displaystyle -\frac {\int -\frac {3 a D x^2+(A b+2 a C) x+\frac {a (b B-a D)}{b}}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {3 a D x^2+(A b+2 a C) x+\frac {a (b B-a D)}{b}}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {-\frac {\int -\frac {3 a^2 D}{b \sqrt {b x^2+a}}dx}{a}-\frac {-x (b B-4 a D)+2 a C+A b}{b \sqrt {a+b x^2}}}{3 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 a D \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}-\frac {-x (b B-4 a D)+2 a C+A b}{b \sqrt {a+b x^2}}}{3 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

Input:

Int[(x*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(5/2),x]
 

Output:

-1/3*(x*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^(3/2)) + (-((A 
*b + 2*a*C - (b*B - 4*a*D)*x)/(b*Sqrt[a + b*x^2])) + (3*a*D*ArcTanh[(Sqrt[ 
b]*x)/Sqrt[a + b*x^2]])/b^(3/2))/(3*a*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 2335

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq 
, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] 
+ Simp[c/(2*a*b*(p + 1))   Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu 
m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, 
 b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67

method	result	size
default	\(C \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+D \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )-\frac {A}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+B \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )\)	\(167\)

Input:

int(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

C*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))+D*(-1/3*x^3/b/(b*x^2+ 
a)^(3/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)) 
))-1/3*A/b/(b*x^2+a)^(3/2)+B*(-1/2*x/b/(b*x^2+a)^(3/2)+1/2*a/b*(1/3*x/a/(b 
*x^2+a)^(3/2)+2/3*x/a^2/(b*x^2+a)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.91 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (D a b^{2} x^{4} + 2 \, D a^{2} b x^{2} + D a^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, C a b^{2} x^{2} + 3 \, D a^{2} b x + 2 \, C a^{2} b + A a b^{2} + {\left (4 \, D a b^{2} - B b^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}, -\frac {3 \, {\left (D a b^{2} x^{4} + 2 \, D a^{2} b x^{2} + D a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, C a b^{2} x^{2} + 3 \, D a^{2} b x + 2 \, C a^{2} b + A a b^{2} + {\left (4 \, D a b^{2} - B b^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}\right ] \] Input:

integrate(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[1/6*(3*(D*a*b^2*x^4 + 2*D*a^2*b*x^2 + D*a^3)*sqrt(b)*log(-2*b*x^2 - 2*sqr 
t(b*x^2 + a)*sqrt(b)*x - a) - 2*(3*C*a*b^2*x^2 + 3*D*a^2*b*x + 2*C*a^2*b + 
 A*a*b^2 + (4*D*a*b^2 - B*b^3)*x^3)*sqrt(b*x^2 + a))/(a*b^5*x^4 + 2*a^2*b^ 
4*x^2 + a^3*b^3), -1/3*(3*(D*a*b^2*x^4 + 2*D*a^2*b*x^2 + D*a^3)*sqrt(-b)*a 
rctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*C*a*b^2*x^2 + 3*D*a^2*b*x + 2*C*a^2 
*b + A*a*b^2 + (4*D*a*b^2 - B*b^3)*x^3)*sqrt(b*x^2 + a))/(a*b^5*x^4 + 2*a^ 
2*b^4*x^2 + a^3*b^3)]
 

Sympy [A] (verification not implemented)

Time = 10.41 (sec) , antiderivative size = 496, normalized size of antiderivative = 4.96 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=A \left (\begin {cases} - \frac {1}{3 a b \sqrt {a + b x^{2}} + 3 b^{2} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {B x^{3}}{3 a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {3}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + C \left (\begin {cases} - \frac {2 a}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} - \frac {3 b x^{2}}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + D \left (\frac {3 a^{\frac {39}{2}} b^{11} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{\frac {37}{2}} b^{12} x^{2} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a^{19} b^{\frac {23}{2}} x}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {4 a^{18} b^{\frac {25}{2}} x^{3}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:

integrate(x*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(5/2),x)
 

Output:

A*Piecewise((-1/(3*a*b*sqrt(a + b*x**2) + 3*b**2*x**2*sqrt(a + b*x**2)), N 
e(b, 0)), (x**2/(2*a**(5/2)), True)) + B*x**3/(3*a**(5/2)*sqrt(1 + b*x**2/ 
a) + 3*a**(3/2)*b*x**2*sqrt(1 + b*x**2/a)) + C*Piecewise((-2*a/(3*a*b**2*s 
qrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)) - 3*b*x**2/(3*a*b**2*sqrt( 
a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(5/2)) 
, True)) + D*(3*a**(39/2)*b**11*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a) 
)/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*s 
qrt(1 + b*x**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt( 
b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**( 
29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39/2)*b**(27/2 
)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4* 
a**18*b**(25/2)*x**3/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/ 
2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.52 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, D x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} - \frac {C x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {D x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {B x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {B x}{3 \, \sqrt {b x^{2} + a} a b} + \frac {D \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, C a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {A}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \] Input:

integrate(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

-1/3*D*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) - C*x 
^2/((b*x^2 + a)^(3/2)*b) - 1/3*D*x/(sqrt(b*x^2 + a)*b^2) - 1/3*B*x/((b*x^2 
 + a)^(3/2)*b) + 1/3*B*x/(sqrt(b*x^2 + a)*a*b) + D*arcsinh(b*x/sqrt(a*b))/ 
b^(5/2) - 2/3*C*a/((b*x^2 + a)^(3/2)*b^2) - 1/3*A/((b*x^2 + a)^(3/2)*b)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left (x {\left (\frac {3 \, C}{b} + \frac {{\left (4 \, D a b^{3} - B b^{4}\right )} x}{a b^{4}}\right )} + \frac {3 \, D a}{b^{2}}\right )} x + \frac {2 \, C a^{2} b^{2} + A a b^{3}}{a b^{4}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {D \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:

integrate(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

-1/3*((x*(3*C/b + (4*D*a*b^3 - B*b^4)*x/(a*b^4)) + 3*D*a/b^2)*x + (2*C*a^2 
*b^2 + A*a*b^3)/(a*b^4))/(b*x^2 + a)^(3/2) - D*log(abs(-sqrt(b)*x + sqrt(b 
*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

int((x*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(5/2),x)
 

Output:

int((x*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.45 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-\sqrt {b \,x^{2}+a}\, a^{2} b^{2}-2 \sqrt {b \,x^{2}+a}\, a^{2} b c -3 \sqrt {b \,x^{2}+a}\, a^{2} b d x -3 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}-4 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}+\sqrt {b \,x^{2}+a}\, b^{4} x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d +6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{4}+\sqrt {b}\, a^{2} b^{2}+2 \sqrt {b}\, a \,b^{3} x^{2}+\sqrt {b}\, b^{4} x^{4}}{3 a \,b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int(x*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x)
 

Output:

( - sqrt(a + b*x**2)*a**2*b**2 - 2*sqrt(a + b*x**2)*a**2*b*c - 3*sqrt(a + 
b*x**2)*a**2*b*d*x - 3*sqrt(a + b*x**2)*a*b**2*c*x**2 - 4*sqrt(a + b*x**2) 
*a*b**2*d*x**3 + sqrt(a + b*x**2)*b**4*x**3 + 3*sqrt(b)*log((sqrt(a + b*x* 
*2) + sqrt(b)*x)/sqrt(a))*a**3*d + 6*sqrt(b)*log((sqrt(a + b*x**2) + sqrt( 
b)*x)/sqrt(a))*a**2*b*d*x**2 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x 
)/sqrt(a))*a*b**2*d*x**4 + sqrt(b)*a**2*b**2 + 2*sqrt(b)*a*b**3*x**2 + sqr 
t(b)*b**4*x**4)/(3*a*b**3*(a**2 + 2*a*b*x**2 + b**2*x**4))