\(\int \frac {c+d x^2}{\sqrt {a+b x^2} (e+f x^2)^3} \, dx\) [321]

Optimal result

Integrand size = 28, antiderivative size = 197 \[ \int \frac {c+d x^2}{\sqrt {a+b x^2} \left (e+f x^2\right )^3} \, dx=\frac {(d e-c f) x \sqrt {a+b x^2}}{4 e (b e-a f) \left (e+f x^2\right )^2}+\frac {(2 b e (d e-3 c f)+a f (d e+3 c f)) x \sqrt {a+b x^2}}{8 e^2 (b e-a f)^2 \left (e+f x^2\right )}+\frac {\left (8 b^2 c e^2-4 a b e (d e+2 c f)+a^2 f (d e+3 c f)\right ) \text {arctanh}\left (\frac {\sqrt {b e-a f} x}{\sqrt {e} \sqrt {a+b x^2}}\right )}{8 e^{5/2} (b e-a f)^{5/2}} \] Output:

1/4*(-c*f+d*e)*x*(b*x^2+a)^(1/2)/e/(-a*f+b*e)/(f*x^2+e)^2+1/8*(2*b*e*(-3*c 
*f+d*e)+a*f*(3*c*f+d*e))*x*(b*x^2+a)^(1/2)/e^2/(-a*f+b*e)^2/(f*x^2+e)+1/8* 
(8*b^2*c*e^2-4*a*b*e*(2*c*f+d*e)+a^2*f*(3*c*f+d*e))*arctanh((-a*f+b*e)^(1/ 
2)*x/e^(1/2)/(b*x^2+a)^(1/2))/e^(5/2)/(-a*f+b*e)^(5/2)
 

Mathematica [A] (verified)

Time = 11.21 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.48 \[ \int \frac {c+d x^2}{\sqrt {a+b x^2} \left (e+f x^2\right )^3} \, dx=\frac {x \left (-4 d e^2 (b e-a f) \left (e+f x^2\right ) \left (f \left (a+b x^2\right )-\frac {(2 b e-a f) \left (e+f x^2\right ) \text {arctanh}\left (\sqrt {\frac {(b e-a f) x^2}{e \left (a+b x^2\right )}}\right )}{e \sqrt {\frac {(b e-a f) x^2}{e \left (a+b x^2\right )}}}\right )+(d e-c f) \left (e f \left (a+b x^2\right ) \left (2 b e \left (4 e+3 f x^2\right )-a f \left (5 e+3 f x^2\right )\right )-\frac {\left (8 b^2 e^2-8 a b e f+3 a^2 f^2\right ) \left (e+f x^2\right )^2 \text {arctanh}\left (\sqrt {\frac {(b e-a f) x^2}{e \left (a+b x^2\right )}}\right )}{\sqrt {\frac {(b e-a f) x^2}{e \left (a+b x^2\right )}}}\right )\right )}{8 e^3 f (b e-a f)^2 \sqrt {a+b x^2} \left (e+f x^2\right )^2} \] Input:

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

Output:

(x*(-4*d*e^2*(b*e - a*f)*(e + f*x^2)*(f*(a + b*x^2) - ((2*b*e - a*f)*(e + 
f*x^2)*ArcTanh[Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^2))]])/(e*Sqrt[((b*e - a 
*f)*x^2)/(e*(a + b*x^2))])) + (d*e - c*f)*(e*f*(a + b*x^2)*(2*b*e*(4*e + 3 
*f*x^2) - a*f*(5*e + 3*f*x^2)) - ((8*b^2*e^2 - 8*a*b*e*f + 3*a^2*f^2)*(e + 
 f*x^2)^2*ArcTanh[Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^2))]])/Sqrt[((b*e - a 
*f)*x^2)/(e*(a + b*x^2))])))/(8*e^3*f*(b*e - a*f)^2*Sqrt[a + b*x^2]*(e + f 
*x^2)^2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {402, 402, 27, 291, 221}

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.

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\int \frac {f (d e+3 c f) a^2-4 b e (d e+2 c f) a+8 b^2 c e^2}{\sqrt {b x^2+a} \left (f x^2+e\right )}dx}{2 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (a f (3 c f+d e)+2 b e (d e-3 c f))}{2 e \left (e+f x^2\right ) (b e-a f)}}{4 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (d e-c f)}{4 e \left (e+f x^2\right )^2 (b e-a f)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\left (a^2 f (3 c f+d e)-4 a b e (2 c f+d e)+8 b^2 c e^2\right ) \int \frac {1}{\sqrt {b x^2+a} \left (f x^2+e\right )}dx}{2 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (a f (3 c f+d e)+2 b e (d e-3 c f))}{2 e \left (e+f x^2\right ) (b e-a f)}}{4 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (d e-c f)}{4 e \left (e+f x^2\right )^2 (b e-a f)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\left (a^2 f (3 c f+d e)-4 a b e (2 c f+d e)+8 b^2 c e^2\right ) \int \frac {1}{e-\frac {(b e-a f) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (a f (3 c f+d e)+2 b e (d e-3 c f))}{2 e \left (e+f x^2\right ) (b e-a f)}}{4 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (d e-c f)}{4 e \left (e+f x^2\right )^2 (b e-a f)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {x \sqrt {b e-a f}}{\sqrt {e} \sqrt {a+b x^2}}\right ) \left (a^2 f (3 c f+d e)-4 a b e (2 c f+d e)+8 b^2 c e^2\right )}{2 e^{3/2} (b e-a f)^{3/2}}+\frac {x \sqrt {a+b x^2} (a f (3 c f+d e)+2 b e (d e-3 c f))}{2 e \left (e+f x^2\right ) (b e-a f)}}{4 e (b e-a f)}+\frac {x \sqrt {a+b x^2} (d e-c f)}{4 e \left (e+f x^2\right )^2 (b e-a f)}\)

Input:

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

Output:

((d*e - c*f)*x*Sqrt[a + b*x^2])/(4*e*(b*e - a*f)*(e + f*x^2)^2) + (((2*b*e 
*(d*e - 3*c*f) + a*f*(d*e + 3*c*f))*x*Sqrt[a + b*x^2])/(2*e*(b*e - a*f)*(e 
 + f*x^2)) + ((8*b^2*c*e^2 - 4*a*b*e*(d*e + 2*c*f) + a^2*f*(d*e + 3*c*f))* 
ArcTanh[(Sqrt[b*e - a*f]*x)/(Sqrt[e]*Sqrt[a + b*x^2])])/(2*e^(3/2)*(b*e - 
a*f)^(3/2)))/(4*e*(b*e - a*f))
 

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 221

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

rule 291

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

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {-\frac {3 \left (-\frac {4 b \left (a d -2 b c \right ) e^{2}}{3}+\frac {f a \left (a d -8 b c \right ) e}{3}+a^{2} c \,f^{2}\right ) \left (f \,x^{2}+e \right )^{2} \arctan \left (\frac {e \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a f -b e \right ) e}}\right )}{8}+\frac {5 \sqrt {\left (a f -b e \right ) e}\, \left (\frac {4 b d \,e^{3}}{5}-\frac {\left (a d +8 b \left (-\frac {x^{2} d}{4}+c \right )\right ) f \,e^{2}}{5}+\left (\left (\frac {x^{2} d}{5}+c \right ) a -\frac {6 x^{2} b c}{5}\right ) f^{2} e +\frac {3 a c \,f^{3} x^{2}}{5}\right ) \sqrt {b \,x^{2}+a}\, x}{8}}{\sqrt {\left (a f -b e \right ) e}\, \left (a f -b e \right )^{2} e^{2} \left (f \,x^{2}+e \right )^{2}}\)	\(197\)
default	\(\text {Expression too large to display}\)	\(1897\)

Input:

int((d*x^2+c)/(b*x^2+a)^(1/2)/(f*x^2+e)^3,x,method=_RETURNVERBOSE)
 

Output:

5/8/((a*f-b*e)*e)^(1/2)*(-3/5*(-4/3*b*(a*d-2*b*c)*e^2+1/3*f*a*(a*d-8*b*c)* 
e+a^2*c*f^2)*(f*x^2+e)^2*arctan(e*(b*x^2+a)^(1/2)/x/((a*f-b*e)*e)^(1/2))+( 
(a*f-b*e)*e)^(1/2)*(4/5*b*d*e^3-1/5*(a*d+8*b*(-1/4*x^2*d+c))*f*e^2+((1/5*x 
^2*d+c)*a-6/5*x^2*b*c)*f^2*e+3/5*a*c*f^3*x^2)*(b*x^2+a)^(1/2)*x)/(a*f-b*e) 
^2/e^2/(f*x^2+e)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (177) = 354\).

Time = 2.75 (sec) , antiderivative size = 1094, normalized size of antiderivative = 5.55 \[ \int \frac {c+d x^2}{\sqrt {a+b x^2} \left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate((d*x^2+c)/(b*x^2+a)^(1/2)/(f*x^2+e)^3,x, algorithm="fricas")
 

Output:

[1/32*((3*a^2*c*e^2*f^2 + 4*(2*b^2*c - a*b*d)*e^4 - (8*a*b*c - a^2*d)*e^3* 
f + (3*a^2*c*f^4 + 4*(2*b^2*c - a*b*d)*e^2*f^2 - (8*a*b*c - a^2*d)*e*f^3)* 
x^4 + 2*(3*a^2*c*e*f^3 + 4*(2*b^2*c - a*b*d)*e^3*f - (8*a*b*c - a^2*d)*e^2 
*f^2)*x^2)*sqrt(b*e^2 - a*e*f)*log(((8*b^2*e^2 - 8*a*b*e*f + a^2*f^2)*x^4 
+ a^2*e^2 + 2*(4*a*b*e^2 - 3*a^2*e*f)*x^2 + 4*((2*b*e - a*f)*x^3 + a*e*x)* 
sqrt(b*e^2 - a*e*f)*sqrt(b*x^2 + a))/(f^2*x^4 + 2*e*f*x^2 + e^2)) + 4*((2* 
b^2*d*e^4*f - 3*a^2*c*e*f^4 - (6*b^2*c + a*b*d)*e^3*f^2 + (9*a*b*c - a^2*d 
)*e^2*f^3)*x^3 + (4*b^2*d*e^5 - 5*a^2*c*e^2*f^3 - (8*b^2*c + 5*a*b*d)*e^4* 
f + (13*a*b*c + a^2*d)*e^3*f^2)*x)*sqrt(b*x^2 + a))/(b^3*e^8 - 3*a*b^2*e^7 
*f + 3*a^2*b*e^6*f^2 - a^3*e^5*f^3 + (b^3*e^6*f^2 - 3*a*b^2*e^5*f^3 + 3*a^ 
2*b*e^4*f^4 - a^3*e^3*f^5)*x^4 + 2*(b^3*e^7*f - 3*a*b^2*e^6*f^2 + 3*a^2*b* 
e^5*f^3 - a^3*e^4*f^4)*x^2), -1/16*((3*a^2*c*e^2*f^2 + 4*(2*b^2*c - a*b*d) 
*e^4 - (8*a*b*c - a^2*d)*e^3*f + (3*a^2*c*f^4 + 4*(2*b^2*c - a*b*d)*e^2*f^ 
2 - (8*a*b*c - a^2*d)*e*f^3)*x^4 + 2*(3*a^2*c*e*f^3 + 4*(2*b^2*c - a*b*d)* 
e^3*f - (8*a*b*c - a^2*d)*e^2*f^2)*x^2)*sqrt(-b*e^2 + a*e*f)*arctan(1/2*sq 
rt(-b*e^2 + a*e*f)*((2*b*e - a*f)*x^2 + a*e)*sqrt(b*x^2 + a)/((b^2*e^2 - a 
*b*e*f)*x^3 + (a*b*e^2 - a^2*e*f)*x)) - 2*((2*b^2*d*e^4*f - 3*a^2*c*e*f^4 
- (6*b^2*c + a*b*d)*e^3*f^2 + (9*a*b*c - a^2*d)*e^2*f^3)*x^3 + (4*b^2*d*e^ 
5 - 5*a^2*c*e^2*f^3 - (8*b^2*c + 5*a*b*d)*e^4*f + (13*a*b*c + a^2*d)*e^3*f 
^2)*x)*sqrt(b*x^2 + a))/(b^3*e^8 - 3*a*b^2*e^7*f + 3*a^2*b*e^6*f^2 - a^...
 

Sympy [F(-1)]

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

integrate((d*x**2+c)/(b*x**2+a)**(1/2)/(f*x**2+e)**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*x^2+c)/(b*x^2+a)^(1/2)/(f*x^2+e)^3,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)/(sqrt(b*x^2 + a)*(f*x^2 + e)^3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 907 vs. \(2 (177) = 354\).

Time = 0.37 (sec) , antiderivative size = 907, normalized size of antiderivative = 4.60 \[ \int \frac {c+d x^2}{\sqrt {a+b x^2} \left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate((d*x^2+c)/(b*x^2+a)^(1/2)/(f*x^2+e)^3,x, algorithm="giac")
 

Output:

-1/8*(8*b^(5/2)*c*e^2 - 4*a*b^(3/2)*d*e^2 - 8*a*b^(3/2)*c*e*f + a^2*sqrt(b 
)*d*e*f + 3*a^2*sqrt(b)*c*f^2)*arctan(1/2*((sqrt(b)*x - sqrt(b*x^2 + a))^2 
*f + 2*b*e - a*f)/sqrt(-b^2*e^2 + a*b*e*f))/((b^2*e^4 - 2*a*b*e^3*f + a^2* 
e^2*f^2)*sqrt(-b^2*e^2 + a*b*e*f)) - 1/4*(8*(sqrt(b)*x - sqrt(b*x^2 + a))^ 
6*b^(5/2)*c*e^2*f^2 - 4*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(3/2)*d*e^2*f^ 
2 - 8*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(3/2)*c*e*f^3 + (sqrt(b)*x - sqr 
t(b*x^2 + a))^6*a^2*sqrt(b)*d*e*f^3 + 3*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^ 
2*sqrt(b)*c*f^4 - 16*(sqrt(b)*x - sqrt(b*x^2 + a))^4*b^(7/2)*d*e^4 + 48*(s 
qrt(b)*x - sqrt(b*x^2 + a))^4*b^(7/2)*c*e^3*f + 8*(sqrt(b)*x - sqrt(b*x^2 
+ a))^4*a*b^(5/2)*d*e^3*f - 72*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b^(5/2)*c 
*e^2*f^2 + 2*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(3/2)*d*e^2*f^2 + 42*(s 
qrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(3/2)*c*e*f^3 - 3*(sqrt(b)*x - sqrt(b* 
x^2 + a))^4*a^3*sqrt(b)*d*e*f^3 - 9*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*sq 
rt(b)*c*f^4 - 16*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^(5/2)*d*e^3*f + 40* 
(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^(5/2)*c*e^2*f^2 + 4*(sqrt(b)*x - sqr 
t(b*x^2 + a))^2*a^3*b^(3/2)*d*e^2*f^2 - 40*(sqrt(b)*x - sqrt(b*x^2 + a))^2 
*a^3*b^(3/2)*c*e*f^3 + 3*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*sqrt(b)*d*e*f 
^3 + 9*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*sqrt(b)*c*f^4 - 2*a^4*b^(3/2)*d 
*e^2*f^2 + 6*a^4*b^(3/2)*c*e*f^3 - a^5*sqrt(b)*d*e*f^3 - 3*a^5*sqrt(b)*c*f 
^4)/((b^2*e^4*f - 2*a*b*e^3*f^2 + a^2*e^2*f^3)*((sqrt(b)*x - sqrt(b*x^2...
 

Mupad [F(-1)]

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

int((c + d*x^2)/((a + b*x^2)^(1/2)*(e + f*x^2)^3),x)
 

Output:

int((c + d*x^2)/((a + b*x^2)^(1/2)*(e + f*x^2)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 3591, normalized size of antiderivative = 18.23 \[ \int \frac {c+d x^2}{\sqrt {a+b x^2} \left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((d*x^2+c)/(b*x^2+a)^(1/2)/(f*x^2+e)^3,x)
 

Output:

( - 6*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x 
**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*c*e**2*f**4 - 12*sqrt(e) 
*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f 
)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*c*e*f**5*x**2 - 6*sqrt(e)*sqrt(a*f - 
b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x) 
/(sqrt(e)*sqrt(b)))*a**3*c*f**6*x**4 - 2*sqrt(e)*sqrt(a*f - b*e)*atan((sqr 
t(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt 
(b)))*a**3*d*e**3*f**3 - 4*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - 
 sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*d*e 
**2*f**4*x**2 - 2*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)* 
sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*d*e*f**5*x** 
4 + 28*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b* 
x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**2*b*c*e**3*f**3 + 56*sqrt 
(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqr 
t(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**2*b*c*e**2*f**4*x**2 + 28*sqrt(e)*sq 
rt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*s 
qrt(b)*x)/(sqrt(e)*sqrt(b)))*a**2*b*c*e*f**5*x**4 + 12*sqrt(e)*sqrt(a*f - 
b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x) 
/(sqrt(e)*sqrt(b)))*a**2*b*d*e**4*f**2 + 24*sqrt(e)*sqrt(a*f - b*e)*atan(( 
sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e...