\(\int \frac {x^3 (c+d x)^2}{(a+b x^2)^{5/2}} \, dx\) [1267]

Optimal result

Integrand size = 22, antiderivative size = 110 \[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {a (c+d x)^2}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {3 b c^2-5 a d^2+8 b c d x}{3 b^3 \sqrt {a+b x^2}}+\frac {d^2 \sqrt {a+b x^2}}{b^3}+\frac {2 c d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:

1/3*a*(d*x+c)^2/b^2/(b*x^2+a)^(3/2)-1/3*(8*b*c*d*x-5*a*d^2+3*b*c^2)/b^3/(b 
*x^2+a)^(1/2)+d^2*(b*x^2+a)^(1/2)/b^3+2*c*d*arctanh(b^(1/2)*x/(b*x^2+a)^(1 
/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-2 a b c^2+8 a^2 d^2-6 a b c d x-3 b^2 c^2 x^2+12 a b d^2 x^2-8 b^2 c d x^3+3 b^2 d^2 x^4}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {2 c d \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{5/2}} \] Input:

Integrate[(x^3*(c + d*x)^2)/(a + b*x^2)^(5/2),x]
 

Output:

(-2*a*b*c^2 + 8*a^2*d^2 - 6*a*b*c*d*x - 3*b^2*c^2*x^2 + 12*a*b*d^2*x^2 - 8 
*b^2*c*d*x^3 + 3*b^2*d^2*x^4)/(3*b^3*(a + b*x^2)^(3/2)) - (2*c*d*Log[-(Sqr 
t[b]*x) + Sqrt[a + b*x^2]])/b^(5/2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {531, 2176, 27, 455, 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.

\(\displaystyle \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[(x^3*(c + d*x)^2)/(a + b*x^2)^(5/2),x]
 

Output:

(a*(c + d*x)^2)/(3*b^2*(a + b*x^2)^(3/2)) - ((a*(c + d*x)*(3*c + 5*d*x))/( 
b*Sqrt[a + b*x^2]) - (2*a*d*((4*d*Sqrt[a + b*x^2])/b + (3*c*ArcTanh[(Sqrt[ 
b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/b)/(3*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 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 455

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

rule 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

rule 2176

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema 
inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 
, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p 
 + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p 
 + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + 
b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && R 
ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.46

method	result	size
default	\(c^{2} \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+d^{2} \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )+2 c d \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )\)	\(161\)
risch	\(\frac {d^{2} \sqrt {b \,x^{2}+a}}{b^{3}}+\frac {\frac {2 c d \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {\left (-\frac {\sqrt {-a b}\, a \,d^{2}}{4}+\frac {\sqrt {-a b}\, b \,c^{2}}{4}-\frac {a b c d}{2}\right ) \left (-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x -\frac {\sqrt {-a b}}{b}\right )}\right )}{b^{2}}+\frac {\left (\frac {\sqrt {-a b}\, a \,d^{2}}{4}-\frac {\sqrt {-a b}\, b \,c^{2}}{4}-\frac {a b c d}{2}\right ) \left (\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x +\frac {\sqrt {-a b}}{b}\right )}\right )}{b^{2}}+\frac {\left (-3 a b c d +2 \sqrt {-a b}\, a \,d^{2}-\sqrt {-a b}\, b \,c^{2}\right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 a \,b^{2} \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {\left (2 \sqrt {-a b}\, a \,d^{2}-\sqrt {-a b}\, b \,c^{2}+3 a b c d \right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{2 a \,b^{2} \left (x -\frac {\sqrt {-a b}}{b}\right )}}{b^{2}}\)	\(549\)

Input:

int(x^3*(d*x+c)^2/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

c^2*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))+d^2*(x^4/b/(b*x^2+a 
)^(3/2)-4*a/b*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2)))+2*c*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))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.85 \[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} c d x^{4} + 2 \, a b c d x^{2} + a^{2} c d\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + {\left (3 \, b^{2} d^{2} x^{4} - 8 \, b^{2} c d x^{3} - 6 \, a b c d x - 2 \, a b c^{2} + 8 \, a^{2} d^{2} - 3 \, {\left (b^{2} c^{2} - 4 \, a b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac {6 \, {\left (b^{2} c d x^{4} + 2 \, a b c d x^{2} + a^{2} c d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, b^{2} d^{2} x^{4} - 8 \, b^{2} c d x^{3} - 6 \, a b c d x - 2 \, a b c^{2} + 8 \, a^{2} d^{2} - 3 \, {\left (b^{2} c^{2} - 4 \, a b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:

integrate(x^3*(d*x+c)^2/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[1/3*(3*(b^2*c*d*x^4 + 2*a*b*c*d*x^2 + a^2*c*d)*sqrt(b)*log(-2*b*x^2 - 2*s 
qrt(b*x^2 + a)*sqrt(b)*x - a) + (3*b^2*d^2*x^4 - 8*b^2*c*d*x^3 - 6*a*b*c*d 
*x - 2*a*b*c^2 + 8*a^2*d^2 - 3*(b^2*c^2 - 4*a*b*d^2)*x^2)*sqrt(b*x^2 + a)) 
/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3), -1/3*(6*(b^2*c*d*x^4 + 2*a*b*c*d*x^2 + 
 a^2*c*d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (3*b^2*d^2*x^4 - 8 
*b^2*c*d*x^3 - 6*a*b*c*d*x - 2*a*b*c^2 + 8*a^2*d^2 - 3*(b^2*c^2 - 4*a*b*d^ 
2)*x^2)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)]
 

Sympy [F]

\[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (c + d x\right )^{2}}{\left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**3*(d*x+c)**2/(b*x**2+a)**(5/2),x)
 

Output:

Integral(x**3*(c + d*x)**2/(a + b*x**2)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.55 \[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {2}{3} \, c d x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {d^{2} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {c^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {4 \, a d^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, c d x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {2 \, c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, a c^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2} d^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \] Input:

integrate(x^3*(d*x+c)^2/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

-2/3*c*d*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) + d 
^2*x^4/((b*x^2 + a)^(3/2)*b) - c^2*x^2/((b*x^2 + a)^(3/2)*b) + 4*a*d^2*x^2 
/((b*x^2 + a)^(3/2)*b^2) - 2/3*c*d*x/(sqrt(b*x^2 + a)*b^2) + 2*c*d*arcsinh 
(b*x/sqrt(a*b))/b^(5/2) - 2/3*a*c^2/((b*x^2 + a)^(3/2)*b^2) + 8/3*a^2*d^2/ 
((b*x^2 + a)^(3/2)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.17 \[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {2 \, c d \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} + \frac {{\left ({\left ({\left (\frac {3 \, d^{2} x}{b} - \frac {8 \, c d}{b}\right )} x - \frac {3 \, {\left (a b^{4} c^{2} - 4 \, a^{2} b^{3} d^{2}\right )}}{a b^{5}}\right )} x - \frac {6 \, a c d}{b^{2}}\right )} x - \frac {2 \, {\left (a^{2} b^{3} c^{2} - 4 \, a^{3} b^{2} d^{2}\right )}}{a b^{5}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \] Input:

integrate(x^3*(d*x+c)^2/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

-2*c*d*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) + 1/3*((((3*d^2*x/b 
- 8*c*d/b)*x - 3*(a*b^4*c^2 - 4*a^2*b^3*d^2)/(a*b^5))*x - 6*a*c*d/b^2)*x - 
 2*(a^2*b^3*c^2 - 4*a^3*b^2*d^2)/(a*b^5))/(b*x^2 + a)^(3/2)
 

Mupad [F(-1)]

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

int((x^3*(c + d*x)^2)/(a + b*x^2)^(5/2),x)
 

Output:

int((x^3*(c + d*x)^2)/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.14 \[ \int \frac {x^3 (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {8 \sqrt {b \,x^{2}+a}\, a^{2} d^{2}-2 \sqrt {b \,x^{2}+a}\, a b \,c^{2}-6 \sqrt {b \,x^{2}+a}\, a b c d x +12 \sqrt {b \,x^{2}+a}\, a b \,d^{2} x^{2}-3 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} x^{2}-8 \sqrt {b \,x^{2}+a}\, b^{2} c d \,x^{3}+3 \sqrt {b \,x^{2}+a}\, b^{2} d^{2} x^{4}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c d +12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b c d \,x^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} c d \,x^{4}}{3 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int(x^3*(d*x+c)^2/(b*x^2+a)^(5/2),x)
 

Output:

(8*sqrt(a + b*x**2)*a**2*d**2 - 2*sqrt(a + b*x**2)*a*b*c**2 - 6*sqrt(a + b 
*x**2)*a*b*c*d*x + 12*sqrt(a + b*x**2)*a*b*d**2*x**2 - 3*sqrt(a + b*x**2)* 
b**2*c**2*x**2 - 8*sqrt(a + b*x**2)*b**2*c*d*x**3 + 3*sqrt(a + b*x**2)*b** 
2*d**2*x**4 + 6*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c 
*d + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*c*d*x**2 + 
 6*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**2*c*d*x**4)/(3*b 
**3*(a**2 + 2*a*b*x**2 + b**2*x**4))