\(\int \frac {1}{\sqrt {d+e x^2} (a d+(b d+a e) x^2+b e x^4)^{3/2}} \, dx\) [13]

Optimal result

Integrand size = 37, antiderivative size = 209 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx=-\frac {e x}{2 d (b d-a e) \sqrt {d+e x^2} \sqrt {a d+(b d+a e) x^2+b e x^4}}+\frac {b (2 b d+a e) x \sqrt {d+e x^2}}{2 a d (b d-a e)^2 \sqrt {a d+(b d+a e) x^2+b e x^4}}-\frac {e (4 b d-a e) \text {arctanh}\left (\frac {\sqrt {b d-a e} x \sqrt {d+e x^2}}{\sqrt {d} \sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{2 d^{3/2} (b d-a e)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx=\frac {\sqrt {d+e x^2} \left (\frac {\sqrt {d} x \left (a+b x^2\right ) \left (a^2 e^2+a b e^2 x^2+2 b^2 d \left (d+e x^2\right )\right )}{a (b d-a e)^2}+\frac {e (4 b d-a e) \left (a+b x^2\right )^{3/2} \left (d+e x^2\right ) \arctan \left (\frac {-e x \sqrt {a+b x^2}+\sqrt {b} \left (d+e x^2\right )}{\sqrt {d} \sqrt {-b d+a e}}\right )}{(-b d+a e)^{5/2}}\right )}{2 d^{3/2} \left (\left (a+b x^2\right ) \left (d+e x^2\right )\right )^{3/2}} \] Input:

Integrate[1/(Sqrt[d + e*x^2]*(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(3/2)),x]
 

Output:

(Sqrt[d + e*x^2]*((Sqrt[d]*x*(a + b*x^2)*(a^2*e^2 + a*b*e^2*x^2 + 2*b^2*d* 
(d + e*x^2)))/(a*(b*d - a*e)^2) + (e*(4*b*d - a*e)*(a + b*x^2)^(3/2)*(d + 
e*x^2)*ArcTan[(-(e*x*Sqrt[a + b*x^2]) + Sqrt[b]*(d + e*x^2))/(Sqrt[d]*Sqrt 
[-(b*d) + a*e])])/(-(b*d) + a*e)^(5/2)))/(2*d^(3/2)*((a + b*x^2)*(d + e*x^ 
2))^(3/2))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1395, 316, 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.

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

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[d + e*x^2]*(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(3/2)),x]
 

Output:

(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*(-1/2*(e*x)/(d*(b*d - a*e)*Sqrt[a + b*x^2 
]*(d + e*x^2)) + ((b*(2*b*d + a*e)*x)/(a*(b*d - a*e)*Sqrt[a + b*x^2]) - (e 
*(4*b*d - a*e)*ArcTanh[(Sqrt[b*d - a*e]*x)/(Sqrt[d]*Sqrt[a + b*x^2])])/(Sq 
rt[d]*(b*d - a*e)^(3/2)))/(2*d*(b*d - a*e))))/Sqrt[a*d + (b*d + a*e)*x^2 + 
 b*e*x^4]
 

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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

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]
 

rule 1395

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( 
x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d 
+ e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p 
+ q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E 
qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 
1] && EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3489\) vs. \(2(183)=366\).

Time = 0.48 (sec) , antiderivative size = 3490, normalized size of antiderivative = 16.70

method	result	size
default	\(\text {Expression too large to display}\)	\(3490\)

Input:

int(1/(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2),x,method=_RETURNVE 
RBOSE)
 

Output:

1/4*(-4*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x+a*e)/ 
(e*x-(-d*e)^(1/2)))*a*b^(9/2)*d^3*e*x^4-3*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d) 
/e)^(1/2)*e-(-d*e)^(1/2)*b*x+a*e)/(e*x+(-d*e)^(1/2)))*a^4*b^(3/2)*d*e^3*x^ 
2-6*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e-(-d*e)^(1/2)*b*x+a*e)/(e*x 
+(-d*e)^(1/2)))*a^3*b^(5/2)*d^2*e^2*x^2+8*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d) 
/e)^(1/2)*e-(-d*e)^(1/2)*b*x+a*e)/(e*x+(-d*e)^(1/2)))*a^2*b^(7/2)*d^3*e*x^ 
2+3*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x+a*e)/(e*x 
-(-d*e)^(1/2)))*a^4*b^(3/2)*d*e^3*x^2+6*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e 
)^(1/2)*e+(-d*e)^(1/2)*b*x+a*e)/(e*x-(-d*e)^(1/2)))*a^3*b^(5/2)*d^2*e^2*x^ 
2-8*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x+a*e)/(e*x 
-(-d*e)^(1/2)))*a^2*b^(7/2)*d^3*e*x^2+ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^ 
(1/2)*e-(-d*e)^(1/2)*b*x+a*e)/(e*x+(-d*e)^(1/2)))*a^5*e^4*x^2*b^(1/2)-ln(2 
*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x+a*e)/(e*x-(-d*e)^ 
(1/2)))*a^5*e^4*x^2*b^(1/2)+ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e-(- 
d*e)^(1/2)*b*x+a*e)/(e*x+(-d*e)^(1/2)))*a^5*d*e^3*b^(1/2)-ln(2*((b*x^2+a)^ 
(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x+a*e)/(e*x-(-d*e)^(1/2)))*a^5* 
d*e^3*b^(1/2)-ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x 
+a*e)/(e*x-(-d*e)^(1/2)))*a^3*b^(5/2)*e^4*x^6+2*ln(2*((b*x^2+a)^(1/2)*((a* 
e-b*d)/e)^(1/2)*e-(-d*e)^(1/2)*b*x+a*e)/(e*x+(-d*e)^(1/2)))*a^4*b^(3/2)*e^ 
4*x^4-2*ln(2*((b*x^2+a)^(1/2)*((a*e-b*d)/e)^(1/2)*e+(-d*e)^(1/2)*b*x+a*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 553 vs. \(2 (183) = 366\).

Time = 0.10 (sec) , antiderivative size = 1132, normalized size of antiderivative = 5.42 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2),x, algorithm 
="fricas")
 

Output:

[-1/4*((4*a^2*b*d^3*e - a^3*d^2*e^2 + (4*a*b^2*d*e^3 - a^2*b*e^4)*x^6 + (8 
*a*b^2*d^2*e^2 + 2*a^2*b*d*e^3 - a^3*e^4)*x^4 + (4*a*b^2*d^3*e + 7*a^2*b*d 
^2*e^2 - 2*a^3*d*e^3)*x^2)*sqrt(b*d^2 - a*d*e)*log((2*b*d^2*x^2 + (2*b*d*e 
 - a*e^2)*x^4 + a*d^2 + 2*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(b*d^2 
 - a*d*e)*sqrt(e*x^2 + d)*x)/(e^2*x^4 + 2*d*e*x^2 + d^2)) - 2*sqrt(b*e*x^4 
 + (b*d + a*e)*x^2 + a*d)*((2*b^3*d^3*e - a*b^2*d^2*e^2 - a^2*b*d*e^3)*x^3 
 + (2*b^3*d^4 - 2*a*b^2*d^3*e + a^2*b*d^2*e^2 - a^3*d*e^3)*x)*sqrt(e*x^2 + 
 d))/(a^2*b^3*d^7 - 3*a^3*b^2*d^6*e + 3*a^4*b*d^5*e^2 - a^5*d^4*e^3 + (a*b 
^4*d^5*e^2 - 3*a^2*b^3*d^4*e^3 + 3*a^3*b^2*d^3*e^4 - a^4*b*d^2*e^5)*x^6 + 
(2*a*b^4*d^6*e - 5*a^2*b^3*d^5*e^2 + 3*a^3*b^2*d^4*e^3 + a^4*b*d^3*e^4 - a 
^5*d^2*e^5)*x^4 + (a*b^4*d^7 - a^2*b^3*d^6*e - 3*a^3*b^2*d^5*e^2 + 5*a^4*b 
*d^4*e^3 - 2*a^5*d^3*e^4)*x^2), 1/2*((4*a^2*b*d^3*e - a^3*d^2*e^2 + (4*a*b 
^2*d*e^3 - a^2*b*e^4)*x^6 + (8*a*b^2*d^2*e^2 + 2*a^2*b*d*e^3 - a^3*e^4)*x^ 
4 + (4*a*b^2*d^3*e + 7*a^2*b*d^2*e^2 - 2*a^3*d*e^3)*x^2)*sqrt(-b*d^2 + a*d 
*e)*arctan(sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(-b*d^2 + a*d*e)*sqrt 
(e*x^2 + d)*x/(b*d*e*x^4 + a*d^2 + (b*d^2 + a*d*e)*x^2)) + sqrt(b*e*x^4 + 
(b*d + a*e)*x^2 + a*d)*((2*b^3*d^3*e - a*b^2*d^2*e^2 - a^2*b*d*e^3)*x^3 + 
(2*b^3*d^4 - 2*a*b^2*d^3*e + a^2*b*d^2*e^2 - a^3*d*e^3)*x)*sqrt(e*x^2 + d) 
)/(a^2*b^3*d^7 - 3*a^3*b^2*d^6*e + 3*a^4*b*d^5*e^2 - a^5*d^4*e^3 + (a*b^4* 
d^5*e^2 - 3*a^2*b^3*d^4*e^3 + 3*a^3*b^2*d^3*e^4 - a^4*b*d^2*e^5)*x^6 + ...
 

Sympy [F]

\[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (\left (a + b x^{2}\right ) \left (d + e x^{2}\right )\right )^{\frac {3}{2}} \sqrt {d + e x^{2}}}\, dx \] Input:

integrate(1/(e*x**2+d)**(1/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(3/2),x)
 

Output:

Integral(1/(((a + b*x**2)*(d + e*x**2))**(3/2)*sqrt(d + e*x**2)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {3}{2}} \sqrt {e x^{2} + d}} \,d x } \] Input:

integrate(1/(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2),x, algorithm 
="maxima")
 

Output:

integrate(1/((b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(3/2)*sqrt(e*x^2 + d)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {3}{2}} \sqrt {e x^{2} + d}} \,d x } \] Input:

integrate(1/(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2),x, algorithm 
="giac")
 

Output:

integrate(1/((b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(3/2)*sqrt(e*x^2 + d)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {e\,x^2+d}\,{\left (b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d\right )}^{3/2}} \,d x \] Input:

int(1/((d + e*x^2)^(1/2)*(a*d + x^2*(a*e + b*d) + b*e*x^4)^(3/2)),x)
 

Output:

int(1/((d + e*x^2)^(1/2)*(a*d + x^2*(a*e + b*d) + b*e*x^4)^(3/2)), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1275, normalized size of antiderivative = 6.10 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(1/(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2),x)
 

Output:

( - sqrt(d)*sqrt(a*e - b*d)*atan((sqrt(a*e - b*d) - sqrt(e)*sqrt(a + b*x** 
2) - sqrt(e)*sqrt(b)*x)/(sqrt(d)*sqrt(b)))*a**3*d*e**2 - sqrt(d)*sqrt(a*e 
- b*d)*atan((sqrt(a*e - b*d) - sqrt(e)*sqrt(a + b*x**2) - sqrt(e)*sqrt(b)* 
x)/(sqrt(d)*sqrt(b)))*a**3*e**3*x**2 + 4*sqrt(d)*sqrt(a*e - b*d)*atan((sqr 
t(a*e - b*d) - sqrt(e)*sqrt(a + b*x**2) - sqrt(e)*sqrt(b)*x)/(sqrt(d)*sqrt 
(b)))*a**2*b*d**2*e + 3*sqrt(d)*sqrt(a*e - b*d)*atan((sqrt(a*e - b*d) - sq 
rt(e)*sqrt(a + b*x**2) - sqrt(e)*sqrt(b)*x)/(sqrt(d)*sqrt(b)))*a**2*b*d*e* 
*2*x**2 - sqrt(d)*sqrt(a*e - b*d)*atan((sqrt(a*e - b*d) - sqrt(e)*sqrt(a + 
 b*x**2) - sqrt(e)*sqrt(b)*x)/(sqrt(d)*sqrt(b)))*a**2*b*e**3*x**4 + 4*sqrt 
(d)*sqrt(a*e - b*d)*atan((sqrt(a*e - b*d) - sqrt(e)*sqrt(a + b*x**2) - sqr 
t(e)*sqrt(b)*x)/(sqrt(d)*sqrt(b)))*a*b**2*d**2*e*x**2 + 4*sqrt(d)*sqrt(a*e 
 - b*d)*atan((sqrt(a*e - b*d) - sqrt(e)*sqrt(a + b*x**2) - sqrt(e)*sqrt(b) 
*x)/(sqrt(d)*sqrt(b)))*a*b**2*d*e**2*x**4 - sqrt(d)*sqrt(a*e - b*d)*atan(( 
sqrt(a*e - b*d) + sqrt(e)*sqrt(a + b*x**2) + sqrt(e)*sqrt(b)*x)/(sqrt(d)*s 
qrt(b)))*a**3*d*e**2 - sqrt(d)*sqrt(a*e - b*d)*atan((sqrt(a*e - b*d) + sqr 
t(e)*sqrt(a + b*x**2) + sqrt(e)*sqrt(b)*x)/(sqrt(d)*sqrt(b)))*a**3*e**3*x* 
*2 + 4*sqrt(d)*sqrt(a*e - b*d)*atan((sqrt(a*e - b*d) + sqrt(e)*sqrt(a + b* 
x**2) + sqrt(e)*sqrt(b)*x)/(sqrt(d)*sqrt(b)))*a**2*b*d**2*e + 3*sqrt(d)*sq 
rt(a*e - b*d)*atan((sqrt(a*e - b*d) + sqrt(e)*sqrt(a + b*x**2) + sqrt(e)*s 
qrt(b)*x)/(sqrt(d)*sqrt(b)))*a**2*b*d*e**2*x**2 - sqrt(d)*sqrt(a*e - b*...