\(\int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx\) [36]

Optimal result

Integrand size = 25, antiderivative size = 209 \[ \int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx=\frac {\left (c-\frac {c e-2 a f}{\sqrt {e^2-4 d f}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {e-\sqrt {e^2-4 d f}}}\right )}{\sqrt {2} \sqrt {f} \sqrt {e-\sqrt {e^2-4 d f}}}+\frac {\left (c+\frac {c e-2 a f}{\sqrt {e^2-4 d f}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {e+\sqrt {e^2-4 d f}}}\right )}{\sqrt {2} \sqrt {f} \sqrt {e+\sqrt {e^2-4 d f}}}-\frac {b \text {arctanh}\left (\frac {e+2 f x^2}{\sqrt {e^2-4 d f}}\right )}{\sqrt {e^2-4 d f}} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.12 \[ \int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx=\frac {\frac {\sqrt {2} \left (2 a f+c \left (-e+\sqrt {e^2-4 d f}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {e-\sqrt {e^2-4 d f}}}\right )}{\sqrt {f} \sqrt {e-\sqrt {e^2-4 d f}}}+\frac {\sqrt {2} \left (-2 a f+c \left (e+\sqrt {e^2-4 d f}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {e+\sqrt {e^2-4 d f}}}\right )}{\sqrt {f} \sqrt {e+\sqrt {e^2-4 d f}}}+b \log \left (-e+\sqrt {e^2-4 d f}-2 f x^2\right )-b \log \left (e+\sqrt {e^2-4 d f}+2 f x^2\right )}{2 \sqrt {e^2-4 d f}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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.

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

\(\Big \downarrow \) 2202

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1432

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((c - (c*e - 2*a*f)/Sqrt[e^2 - 4*d*f])*ArcTan[(Sqrt[2]*Sqrt[f]*x)/Sqrt[e - 
 Sqrt[e^2 - 4*d*f]]])/(Sqrt[2]*Sqrt[f]*Sqrt[e - Sqrt[e^2 - 4*d*f]]) + ((c 
+ (c*e - 2*a*f)/Sqrt[e^2 - 4*d*f])*ArcTan[(Sqrt[2]*Sqrt[f]*x)/Sqrt[e + Sqr 
t[e^2 - 4*d*f]]])/(Sqrt[2]*Sqrt[f]*Sqrt[e + Sqrt[e^2 - 4*d*f]]) - (b*ArcTa 
nh[(e + 2*f*x^2)/Sqrt[e^2 - 4*d*f]])/Sqrt[e^2 - 4*d*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 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 = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.23

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{4}+\textit {\_Z}^{2} e +d \right )}{\sum }\frac {\left (c \,\textit {\_R}^{2}+\textit {\_R} b +a \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} f +\textit {\_R} e}\right )}{2}\)	\(48\)
default	\(4 f \left (-\frac {\sqrt {-4 d f +e^{2}}\, \left (-\frac {b \ln \left (2 f \,x^{2}+\sqrt {-4 d f +e^{2}}+e \right )}{2}+\frac {\left (\sqrt {-4 d f +e^{2}}\, c -2 a f +c e \right ) \sqrt {2}\, \arctan \left (\frac {f x \sqrt {2}}{\sqrt {\left (e +\sqrt {-4 d f +e^{2}}\right ) f}}\right )}{2 \sqrt {\left (e +\sqrt {-4 d f +e^{2}}\right ) f}}\right )}{4 f \left (4 d f -e^{2}\right )}-\frac {\sqrt {-4 d f +e^{2}}\, \left (\frac {b \ln \left (-2 f \,x^{2}+\sqrt {-4 d f +e^{2}}-e \right )}{2}+\frac {\left (-\sqrt {-4 d f +e^{2}}\, c -2 a f +c e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {f x \sqrt {2}}{\sqrt {\left (\sqrt {-4 d f +e^{2}}-e \right ) f}}\right )}{2 \sqrt {\left (\sqrt {-4 d f +e^{2}}-e \right ) f}}\right )}{4 f \left (4 d f -e^{2}\right )}\right )\)	\(240\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 26.82 (sec) , antiderivative size = 578003, normalized size of antiderivative = 2765.56 \[ \int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1712 vs. \(2 (171) = 342\).

Time = 0.88 (sec) , antiderivative size = 1712, normalized size of antiderivative = 8.19 \[ \int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(e^2*f^2 - 4*d*f^3 - 2*e*f^3 + f^4)*sqrt(e^2 - 4*d*f)*b*log(x^2 + 1/2* 
(e - sqrt(e^2 - 4*d*f))/f)/((e^4 - 8*d*e^2*f - 2*e^3*f + 16*d^2*f^2 + 8*d* 
e*f^2 + e^2*f^2 - 4*d*f^3)*f^2) + 1/4*((sqrt(2)*sqrt(e*f + sqrt(e^2 - 4*d* 
f)*f)*e^4 - 8*sqrt(2)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*e^2*f - 2*sqrt(2)* 
sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*e^3*f - 2*e^4*f + 16*sqrt(2)*sqrt(e*f + sq 
rt(e^2 - 4*d*f)*f)*d^2*f^2 + 8*sqrt(2)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*e 
*f^2 + sqrt(2)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*e^2*f^2 + 16*d*e^2*f^2 + 2* 
e^3*f^2 - 4*sqrt(2)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*f^3 - 32*d^2*f^3 - 8 
*d*e*f^3 - sqrt(2)*sqrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*e^3 + 
 4*sqrt(2)*sqrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*e*f + 2*sqr 
t(2)*sqrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*e^2*f - sqrt(2)*sqr 
t(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*e*f^2 + 2*(e^2 - 4*d*f)*e^2 
*f - 8*(e^2 - 4*d*f)*d*f^2 - 2*(e^2 - 4*d*f)*e*f^2)*a - 2*(2*d*e^2*f^2 - 8 
*d^2*f^3 - sqrt(2)*sqrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*e^2 
 + 4*sqrt(2)*sqrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d^2*f + 2*s 
qrt(2)*sqrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*e*f - sqrt(2)*s 
qrt(e^2 - 4*d*f)*sqrt(e*f + sqrt(e^2 - 4*d*f)*f)*d*f^2 - 2*(e^2 - 4*d*f)*d 
*f^2)*c)*arctan(2*sqrt(1/2)*x/sqrt((e + sqrt(e^2 - 4*d*f))/f))/((d*e^4 - 8 
*d^2*e^2*f - 2*d*e^3*f + 16*d^3*f^2 + 8*d^2*e*f^2 + d*e^2*f^2 - 4*d^2*f^3) 
*abs(f)) + 1/4*((sqrt(2)*sqrt(e*f - sqrt(e^2 - 4*d*f)*f)*e^4 - 8*sqrt(2...
 

Mupad [B] (verification not implemented)

Time = 19.01 (sec) , antiderivative size = 3942, normalized size of antiderivative = 18.86 \[ \int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

symsum(log(a*b^2*f^2 - a^2*c*f^2 + b^3*f^2*x - c^3*d*f - 8*root(16*d*e^4*f 
*z^4 - 128*d^2*e^2*f^2*z^4 + 256*d^3*f^3*z^4 - 16*a*c*d*e^2*f*z^2 - 16*c^2 
*d^2*e*f*z^2 - 8*b^2*d*e^2*f*z^2 - 16*a^2*d*e*f^2*z^2 + 64*a*c*d^2*f^2*z^2 
 + 32*b^2*d^2*f^2*z^2 + 4*c^2*d*e^3*z^2 + 4*a^2*e^3*f*z^2 + 16*b*c^2*d^2*f 
*z + 4*a^2*b*e^2*f*z - 4*b*c^2*d*e^2*z - 16*a^2*b*d*f^2*z - 4*a*b^2*c*d*f 
+ 2*a^2*c^2*d*f - 2*a^3*c*e*f - 2*a*c^3*d*e + b^2*c^2*d*e + a^2*b^2*e*f + 
b^4*d*f + a^2*c^2*e^2 + c^4*d^2 + a^4*f^2, z, k)^3*e^3*f^2*x + a*c^2*e*f - 
 16*root(16*d*e^4*f*z^4 - 128*d^2*e^2*f^2*z^4 + 256*d^3*f^3*z^4 - 16*a*c*d 
*e^2*f*z^2 - 16*c^2*d^2*e*f*z^2 - 8*b^2*d*e^2*f*z^2 - 16*a^2*d*e*f^2*z^2 + 
 64*a*c*d^2*f^2*z^2 + 32*b^2*d^2*f^2*z^2 + 4*c^2*d*e^3*z^2 + 4*a^2*e^3*f*z 
^2 + 16*b*c^2*d^2*f*z + 4*a^2*b*e^2*f*z - 4*b*c^2*d*e^2*z - 16*a^2*b*d*f^2 
*z - 4*a*b^2*c*d*f + 2*a^2*c^2*d*f - 2*a^3*c*e*f - 2*a*c^3*d*e + b^2*c^2*d 
*e + a^2*b^2*e*f + b^4*d*f + a^2*c^2*e^2 + c^4*d^2 + a^4*f^2, z, k)^2*a*d* 
f^3 - 4*root(16*d*e^4*f*z^4 - 128*d^2*e^2*f^2*z^4 + 256*d^3*f^3*z^4 - 16*a 
*c*d*e^2*f*z^2 - 16*c^2*d^2*e*f*z^2 - 8*b^2*d*e^2*f*z^2 - 16*a^2*d*e*f^2*z 
^2 + 64*a*c*d^2*f^2*z^2 + 32*b^2*d^2*f^2*z^2 + 4*c^2*d*e^3*z^2 + 4*a^2*e^3 
*f*z^2 + 16*b*c^2*d^2*f*z + 4*a^2*b*e^2*f*z - 4*b*c^2*d*e^2*z - 16*a^2*b*d 
*f^2*z - 4*a*b^2*c*d*f + 2*a^2*c^2*d*f - 2*a^3*c*e*f - 2*a*c^3*d*e + b^2*c 
^2*d*e + a^2*b^2*e*f + b^4*d*f + a^2*c^2*e^2 + c^4*d^2 + a^4*f^2, z, k)*a^ 
2*f^3*x + 4*root(16*d*e^4*f*z^4 - 128*d^2*e^2*f^2*z^4 + 256*d^3*f^3*z^4...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 829, normalized size of antiderivative = 3.97 \[ \int \frac {a+b x+c x^2}{d+e x^2+f x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 4*sqrt(2*sqrt(f)*sqrt(d) + e)*sqrt(2*sqrt(f)*sqrt(d) - e)*atan((sqrt(2 
*sqrt(f)*sqrt(d) - e) - 2*sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*b*d*f + 
2*sqrt(d)*sqrt(2*sqrt(f)*sqrt(d) + e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) - 
2*sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*a*e*f - 4*sqrt(d)*sqrt(2*sqrt(f) 
*sqrt(d) + e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) - 2*sqrt(f)*x)/sqrt(2*sqrt 
(f)*sqrt(d) + e))*c*d*f - 4*sqrt(f)*sqrt(2*sqrt(f)*sqrt(d) + e)*atan((sqrt 
(2*sqrt(f)*sqrt(d) - e) - 2*sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*a*d*f 
+ 2*sqrt(f)*sqrt(2*sqrt(f)*sqrt(d) + e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) 
- 2*sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*c*d*e - 4*sqrt(2*sqrt(f)*sqrt( 
d) + e)*sqrt(2*sqrt(f)*sqrt(d) - e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) + 2* 
sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*b*d*f - 2*sqrt(d)*sqrt(2*sqrt(f)*s 
qrt(d) + e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) + 2*sqrt(f)*x)/sqrt(2*sqrt(f 
)*sqrt(d) + e))*a*e*f + 4*sqrt(d)*sqrt(2*sqrt(f)*sqrt(d) + e)*atan((sqrt(2 
*sqrt(f)*sqrt(d) - e) + 2*sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*c*d*f + 
4*sqrt(f)*sqrt(2*sqrt(f)*sqrt(d) + e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) + 
2*sqrt(f)*x)/sqrt(2*sqrt(f)*sqrt(d) + e))*a*d*f - 2*sqrt(f)*sqrt(2*sqrt(f) 
*sqrt(d) + e)*atan((sqrt(2*sqrt(f)*sqrt(d) - e) + 2*sqrt(f)*x)/sqrt(2*sqrt 
(f)*sqrt(d) + e))*c*d*e - sqrt(d)*sqrt(2*sqrt(f)*sqrt(d) - e)*log( - sqrt( 
2*sqrt(f)*sqrt(d) - e)*x + sqrt(d) + sqrt(f)*x**2)*a*e*f + 2*sqrt(d)*sqrt( 
2*sqrt(f)*sqrt(d) - e)*log( - sqrt(2*sqrt(f)*sqrt(d) - e)*x + sqrt(d) +...