\(\int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx\) [37]

Optimal result

Integrand size = 22, antiderivative size = 224 \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx=\frac {\left (e^2+\frac {2 c d^2-b e^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (e^2-\frac {2 c d^2-b e^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b+\sqrt {b^2-4 a c}}}-\frac {2 d e \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx=\frac {\frac {\sqrt {2} \left (2 c d^2+\left (-b+\sqrt {b^2-4 a c}\right ) e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \left (-2 c d^2+\left (b+\sqrt {b^2-4 a c}\right ) e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {c} \sqrt {b+\sqrt {b^2-4 a c}}}+2 d e \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )-2 d e \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{2 \sqrt {b^2-4 a c}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2202, 27, 1432, 1083, 219, 1480, 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.

\(\displaystyle \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 2202

\(\displaystyle \int \frac {d^2+e^2 x^2}{c x^4+b x^2+a}dx+\int \frac {2 d e x}{c x^4+b x^2+a}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {d^2+e^2 x^2}{c x^4+b x^2+a}dx+2 d e \int \frac {x}{c x^4+b x^2+a}dx\)

\(\Big \downarrow \) 1432

\(\displaystyle \int \frac {d^2+e^2 x^2}{c x^4+b x^2+a}dx+d e \int \frac {1}{c x^4+b x^2+a}dx^2\)

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 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 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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1432

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 2202

Int[(Pn_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{n 
 = Expon[Pn, x], k}, Int[Sum[Coeff[Pn, x, 2*k]*x^(2*k), {k, 0, n/2}]*(a + b 
*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pn, x, 2*k + 1]*x^(2*k), {k, 0, (n - 
1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pn, x] 
 &&  !PolyQ[Pn, x^2]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.24

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{2} e^{2}+2 \textit {\_R} d e +d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} c +\textit {\_R} b}\right )}{2}\)	\(54\)
default	\(4 c \left (-\frac {\sqrt {-4 a c +b^{2}}\, \left (-e d \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )+\frac {\left (\sqrt {-4 a c +b^{2}}\, e^{2}+b \,e^{2}-2 c \,d^{2}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 c \left (4 a c -b^{2}\right )}-\frac {\sqrt {-4 a c +b^{2}}\, \left (e d \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )+\frac {\left (-\sqrt {-4 a c +b^{2}}\, e^{2}+b \,e^{2}-2 c \,d^{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 c \left (4 a c -b^{2}\right )}\right )\)	\(253\)

Input:

int((e*x+d)^2/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

1/2*sum((_R^2*e^2+2*_R*d*e+d^2)/(2*_R^3*c+_R*b)*ln(x-_R),_R=RootOf(_Z^4*c+ 
_Z^2*b+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 76.28 (sec) , antiderivative size = 540080, normalized size of antiderivative = 2411.07 \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((e*x + d)^2/(c*x^4 + b*x^2 + a), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1723 vs. \(2 (186) = 372\).

Time = 0.79 (sec) , antiderivative size = 1723, normalized size of antiderivative = 7.69 \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

(b^2*c^2 - 4*a*c^3 - 2*b*c^3 + c^4)*sqrt(b^2 - 4*a*c)*d*e*log(x^2 + 1/2*(b 
 - sqrt(b^2 - 4*a*c))/c)/((b^4 - 8*a*b^2*c - 2*b^3*c + 16*a^2*c^2 + 8*a*b* 
c^2 + b^2*c^2 - 4*a*c^3)*c^2) + 1/2*(b^5*c - 8*a*b^3*c^2 - 2*b^4*c^2 + 16* 
a^2*b*c^3 + 8*a*b^2*c^3 + b^3*c^3 - 4*a*b*c^4 + (b^4*c - 6*a*b^2*c^2 - 2*b 
^3*c^2 + 8*a^2*c^3 + 4*a*b*c^3 + b^2*c^3 - 2*a*c^4)*sqrt(b^2 - 4*a*c))*d*e 
*log(x^2 + 1/2*(b + sqrt(b^2 - 4*a*c))/c)/((a*b^4 - 8*a^2*b^2*c - 2*a*b^3* 
c + 16*a^3*c^2 + 8*a^2*b*c^2 + a*b^2*c^2 - 4*a^2*c^3)*c^2) + 1/4*((sqrt(2) 
*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4 - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a 
*c)*c)*a*b^2*c - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c - 2*b^4*c 
 + 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^2 + 8*sqrt(2)*sqrt(b*c 
 + sqrt(b^2 - 4*a*c)*c)*a*b*c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)* 
b^2*c^2 + 16*a*b^2*c^2 + 2*b^3*c^2 - 4*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c 
)*c)*a*c^3 - 32*a^2*c^3 - 8*a*b*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + 
 sqrt(b^2 - 4*a*c)*c)*b^3 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^ 
2 - 4*a*c)*c)*a*b*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4* 
a*c)*c)*b^2*c - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)* 
b*c^2 + 2*(b^2 - 4*a*c)*b^2*c - 8*(b^2 - 4*a*c)*a*c^2 - 2*(b^2 - 4*a*c)*b* 
c^2)*d^2 - 2*(2*a*b^2*c^2 - 8*a^2*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c 
 + sqrt(b^2 - 4*a*c)*c)*a*b^2 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqr 
t(b^2 - 4*a*c)*c)*a^2*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b...
 

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 3046, normalized size of antiderivative = 13.60 \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

symsum(log(3*c^2*d^4*e^2 - a*c*e^6 - 8*root(16*a*b^4*c*z^4 - 128*a^2*b^2*c 
^2*z^4 + 256*a^3*c^3*z^4 - 48*a*b^2*c*d^2*e^2*z^2 - 16*a^2*b*c*e^4*z^2 - 1 
6*a*b*c^2*d^4*z^2 + 192*a^2*c^2*d^2*e^2*z^2 + 4*b^3*c*d^4*z^2 + 4*a*b^3*e^ 
4*z^2 + 8*b^2*c*d^5*e*z + 32*a^2*c*d*e^5*z - 32*a*c^2*d^5*e*z - 8*a*b^2*d* 
e^5*z + 2*b*c*d^6*e^2 + 2*a*c*d^4*e^4 + 2*a*b*d^2*e^6 + b^2*d^4*e^4 + c^2* 
d^8 + a^2*e^8, z, k)^3*b^3*c^2*x + 4*c^2*d^3*e^3*x + 4*root(16*a*b^4*c*z^4 
 - 128*a^2*b^2*c^2*z^4 + 256*a^3*c^3*z^4 - 48*a*b^2*c*d^2*e^2*z^2 - 16*a^2 
*b*c*e^4*z^2 - 16*a*b*c^2*d^4*z^2 + 192*a^2*c^2*d^2*e^2*z^2 + 4*b^3*c*d^4* 
z^2 + 4*a*b^3*e^4*z^2 + 8*b^2*c*d^5*e*z + 32*a^2*c*d*e^5*z - 32*a*c^2*d^5* 
e*z - 8*a*b^2*d*e^5*z + 2*b*c*d^6*e^2 + 2*a*c*d^4*e^4 + 2*a*b*d^2*e^6 + b^ 
2*d^4*e^4 + c^2*d^8 + a^2*e^8, z, k)^2*b^2*c^2*d^2 + b*c*d^2*e^4 - 4*root( 
16*a*b^4*c*z^4 - 128*a^2*b^2*c^2*z^4 + 256*a^3*c^3*z^4 - 48*a*b^2*c*d^2*e^ 
2*z^2 - 16*a^2*b*c*e^4*z^2 - 16*a*b*c^2*d^4*z^2 + 192*a^2*c^2*d^2*e^2*z^2 
+ 4*b^3*c*d^4*z^2 + 4*a*b^3*e^4*z^2 + 8*b^2*c*d^5*e*z + 32*a^2*c*d*e^5*z - 
 32*a*c^2*d^5*e*z - 8*a*b^2*d*e^5*z + 2*b*c*d^6*e^2 + 2*a*c*d^4*e^4 + 2*a* 
b*d^2*e^6 + b^2*d^4*e^4 + c^2*d^8 + a^2*e^8, z, k)*c^3*d^4*x - 16*root(16* 
a*b^4*c*z^4 - 128*a^2*b^2*c^2*z^4 + 256*a^3*c^3*z^4 - 48*a*b^2*c*d^2*e^2*z 
^2 - 16*a^2*b*c*e^4*z^2 - 16*a*b*c^2*d^4*z^2 + 192*a^2*c^2*d^2*e^2*z^2 + 4 
*b^3*c*d^4*z^2 + 4*a*b^3*e^4*z^2 + 8*b^2*c*d^5*e*z + 32*a^2*c*d*e^5*z - 32 
*a*c^2*d^5*e*z - 8*a*b^2*d*e^5*z + 2*b*c*d^6*e^2 + 2*a*c*d^4*e^4 + 2*a*...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 863, normalized size of antiderivative = 3.85 \[ \int \frac {(d+e x)^2}{a+b x^2+c x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2 
*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d*e 
- 4*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) 
- 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*e**2 + 2*sqrt(a)*sqrt(2*sq 
rt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2 
*sqrt(c)*sqrt(a) + b))*b*c*d**2 + 2*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*at 
an((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b) 
)*a*b*e**2 - 4*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sq 
rt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d**2 - 8*sqrt(2 
*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqr 
t(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d*e + 4*sqrt(a)* 
sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)* 
x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*e**2 - 2*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a 
) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqr 
t(a) + b))*b*c*d**2 - 2*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*s 
qrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*e**2 + 
 4*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 
 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d**2 + 2*sqrt(a)*sqrt(2*sqr 
t(c)*sqrt(a) - b)*log( - sqrt(2*sqrt(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c) 
*x**2)*a*c*e**2 - sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) - b)*log( - sqrt(2*sqr...