\(\int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx\) [21]

Optimal result

Integrand size = 24, antiderivative size = 298 \[ \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx=\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f}-\frac {\sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f}}+\frac {\sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.45 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx=\frac {-\sqrt {c} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )+\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a c e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 c^{3/2} d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a \sqrt {c} f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{f} \] Input:

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

Output:

(-(Sqrt[c]*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]]) + RootSum[a^2*f + 2*a*Sqrt 
[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (a*c*e* 
Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(3/2)*d*Log[-(Sqrt[c]*x) + 
Sqrt[a + c*x^2] - #1]*#1 - 2*a*Sqrt[c]*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2 
] - #1]*#1 - c*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c] 
*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ])/f
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1322, 224, 219, 1367, 488, 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.

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

\(\Big \downarrow \) 1322

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1367

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 488

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

rule 1322

Int[Sqrt[(a_) + (c_.)*(x_)^2]/((d_) + (e_.)*(x_) + (f_.)*(x_)^2), x_Symbol] 
 :> Simp[c/f   Int[1/Sqrt[a + c*x^2], x], x] - Simp[1/f   Int[(c*d - a*f + 
c*e*x)/(Sqrt[a + c*x^2]*(d + e*x + f*x^2)), x], x] /; FreeQ[{a, c, d, e, f} 
, x] && NeQ[e^2 - 4*d*f, 0]
 

rule 1367

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1211\) vs. \(2(259)=518\).

Time = 2.13 (sec) , antiderivative size = 1212, normalized size of antiderivative = 4.07

method	result	size
default	\(\text {Expression too large to display}\)	\(1212\)

Input:

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

Output:

1/(-4*d*f+e^2)^(1/2)*(1/2*(4*c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2-4*c*(e- 
(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+2*(-(-4*d*f+e^2)^( 
1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2)-1/2*c^(1/2)*(e-(-4*d*f+e^2)^(1/ 
2))/f*ln((-1/2*c*(e-(-4*d*f+e^2)^(1/2))/f+c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2 
))))/c^(1/2)+(c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2-c*(e-(-4*d*f+e^2)^(1/2 
))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^ 
2-2*d*f*c+c*e^2)/f^2)^(1/2))-1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+ 
c*e^2)/f^2*2^(1/2)/((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^( 
1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2-c*(e-(-4*d*f+ 
e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*2^(1/2)*((-(-4*d*f+e^2 
)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x-1/2/f*(-e+(-4*d*f+e^ 
2)^(1/2)))^2-4*c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)) 
)+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2))/(x-1/2/f*( 
-e+(-4*d*f+e^2)^(1/2)))))-1/(-4*d*f+e^2)^(1/2)*(1/2*(4*c*(x+1/2*(e+(-4*d*f 
+e^2)^(1/2))/f)^2-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2 
))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2)-1/2*c^(1 
/2)*(e+(-4*d*f+e^2)^(1/2))/f*ln((-1/2*c*(e+(-4*d*f+e^2)^(1/2))/f+c*(x+1/2* 
(e+(-4*d*f+e^2)^(1/2))/f))/c^(1/2)+(c*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2-c 
*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*((-4*d*f+e^ 
2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2))-1/2*((-4*d*f+e^2)^(1/2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1188 vs. \(2 (257) = 514\).

Time = 118.11 (sec) , antiderivative size = 2384, normalized size of antiderivative = 8.00 \[ \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(sqrt(2)*f*sqrt((c*e^2 - 2*c*d*f + 2*a*f^2 + (e^2*f^2 - 4*d*f^3)*sqrt 
(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/(e^2*f^2 - 4*d*f^3))*log((4*c^2*d*e*x - 2*a 
*c*e^2 + sqrt(2)*(c*e^3 - 4*c*d*e*f - (e^3*f^2 - 4*d*e*f^3)*sqrt(c^2*e^2/( 
e^2*f^4 - 4*d*f^5)))*sqrt(c*x^2 + a)*sqrt((c*e^2 - 2*c*d*f + 2*a*f^2 + (e^ 
2*f^2 - 4*d*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/(e^2*f^2 - 4*d*f^3)) + 
 2*(a*e^2*f^2 - 4*a*d*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/x) - sqrt(2) 
*f*sqrt((c*e^2 - 2*c*d*f + 2*a*f^2 + (e^2*f^2 - 4*d*f^3)*sqrt(c^2*e^2/(e^2 
*f^4 - 4*d*f^5)))/(e^2*f^2 - 4*d*f^3))*log((4*c^2*d*e*x - 2*a*c*e^2 - sqrt 
(2)*(c*e^3 - 4*c*d*e*f - (e^3*f^2 - 4*d*e*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d 
*f^5)))*sqrt(c*x^2 + a)*sqrt((c*e^2 - 2*c*d*f + 2*a*f^2 + (e^2*f^2 - 4*d*f 
^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/(e^2*f^2 - 4*d*f^3)) + 2*(a*e^2*f^2 
 - 4*a*d*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/x) + sqrt(2)*f*sqrt((c*e^ 
2 - 2*c*d*f + 2*a*f^2 - (e^2*f^2 - 4*d*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^ 
5)))/(e^2*f^2 - 4*d*f^3))*log((4*c^2*d*e*x - 2*a*c*e^2 + sqrt(2)*(c*e^3 - 
4*c*d*e*f + (e^3*f^2 - 4*d*e*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))*sqrt( 
c*x^2 + a)*sqrt((c*e^2 - 2*c*d*f + 2*a*f^2 - (e^2*f^2 - 4*d*f^3)*sqrt(c^2* 
e^2/(e^2*f^4 - 4*d*f^5)))/(e^2*f^2 - 4*d*f^3)) - 2*(a*e^2*f^2 - 4*a*d*f^3) 
*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/x) - sqrt(2)*f*sqrt((c*e^2 - 2*c*d*f + 
 2*a*f^2 - (e^2*f^2 - 4*d*f^3)*sqrt(c^2*e^2/(e^2*f^4 - 4*d*f^5)))/(e^2*f^2 
 - 4*d*f^3))*log((4*c^2*d*e*x - 2*a*c*e^2 - sqrt(2)*(c*e^3 - 4*c*d*e*f ...
 

Sympy [F]

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

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

Output:

Integral(sqrt(a + c*x**2)/(d + e*x + f*x**2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` for 
 more deta
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx=\frac {-\sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{2}+a}-\sqrt {c}\, x \right )+\sqrt {c}\, \mathrm {log}\left (\sqrt {c \,x^{2}+a}+\sqrt {c}\, x \right )+2 \left (\int \frac {\sqrt {c \,x^{2}+a}}{c f \,x^{4}+c e \,x^{3}+a f \,x^{2}+c d \,x^{2}+a e x +a d}d x \right ) a f -2 \left (\int \frac {\sqrt {c \,x^{2}+a}}{c f \,x^{4}+c e \,x^{3}+a f \,x^{2}+c d \,x^{2}+a e x +a d}d x \right ) c d -2 \left (\int \frac {\sqrt {c \,x^{2}+a}\, x}{c f \,x^{4}+c e \,x^{3}+a f \,x^{2}+c d \,x^{2}+a e x +a d}d x \right ) c e}{2 f} \] Input:

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

Output:

( - sqrt(c)*log(sqrt(a + c*x**2) - sqrt(c)*x) + sqrt(c)*log(sqrt(a + c*x** 
2) + sqrt(c)*x) + 2*int(sqrt(a + c*x**2)/(a*d + a*e*x + a*f*x**2 + c*d*x** 
2 + c*e*x**3 + c*f*x**4),x)*a*f - 2*int(sqrt(a + c*x**2)/(a*d + a*e*x + a* 
f*x**2 + c*d*x**2 + c*e*x**3 + c*f*x**4),x)*c*d - 2*int((sqrt(a + c*x**2)* 
x)/(a*d + a*e*x + a*f*x**2 + c*d*x**2 + c*e*x**3 + c*f*x**4),x)*c*e)/(2*f)