\(\int \frac {1}{x (a c+b c x^2+d \sqrt {a+b x^2})} \, dx\) [51]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a} \left (a c^2-d^2\right )}+\frac {c \log (x)}{a c^2-d^2}-\frac {c \log \left (d+c \sqrt {a+b x^2}\right )}{a c^2-d^2} \] Output:

d*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(1/2)/(a*c^2-d^2)+c*ln(x)/(a*c^2-d^2) 
-c*ln(d+c*(b*x^2+a)^(1/2))/(a*c^2-d^2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {\frac {2 d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+c \log \left (b x^2\right )-2 c \log \left (d+c \sqrt {a+b x^2}\right )}{2 a c^2-2 d^2} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.72 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2586, 7267, 25, 479, 452, 219, 240}

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.

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

\(\Big \downarrow \) 2586

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 479

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

\(\Big \downarrow \) 452

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 240

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 452

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

rule 479

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

rule 2586

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] 
), x_Symbol] :> Simp[1/n   Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a 
+ b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* 
d, 0] && IntegerQ[(m + 1)/n]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1696\) vs. \(2(80)=160\).

Time = 0.04 (sec) , antiderivative size = 1697, normalized size of antiderivative = 19.28

method	result	size
default	\(\text {Expression too large to display}\)	\(1697\)

Input:

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

Output:

c*ln(x)/(a*c^2-d^2)-1/2*a*c^3/(a*c^2-d^2)/d^2*ln(b*c^2*x^2+a*c^2-d^2)+1/2* 
c/d^2*ln(b*c^2*x^2+a*c^2-d^2)-d*(1/a/(a*c^2-d^2)*((b*x^2+a)^(1/2)-a^(1/2)* 
ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))-1/2*b*c^2/a/((-a*b)^(1/2)*c^2+(-(a* 
c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*(((x- 
(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b))^(1/2)+(-a*b)^(1/2)* 
ln(((x-(-a*b)^(1/2)/b)*b+(-a*b)^(1/2))/b^(1/2)+((x-(-a*b)^(1/2)/b)^2*b+2*( 
-a*b)^(1/2)*(x-(-a*b)^(1/2)/b))^(1/2))/b^(1/2))-1/2*b*c^2/a/((-a*b)^(1/2)* 
c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/ 
2))*(((x+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2)-(-a* 
b)^(1/2)*ln(((x+(-a*b)^(1/2)/b)*b-(-a*b)^(1/2))/b^(1/2)+((x+(-a*b)^(1/2)/b 
)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2))/b^(1/2))+1/2*b*c^4/(a*c^2- 
d^2)/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a* 
c^2-d^2)*b*c^2)^(1/2))*(((x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2*b+2/c^2*(- 
(a*c^2-d^2)*b*c^2)^(1/2)*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/ 
2)+1/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*ln((1/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)+( 
x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)*b)/b^(1/2)+((x-(-(a*c^2-d^2)*b*c^2)^(1 
/2)/b/c^2)^2*b+2/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(x-(-(a*c^2-d^2)*b*c^2)^(1 
/2)/b/c^2)+d^2/c^2)^(1/2))/b^(1/2)-d^2/c^2/(d^2/c^2)^(1/2)*ln((2*d^2/c^2+2 
/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+2*(d^ 
2/c^2)^(1/2)*((x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2*b+2/c^2*(-(a*c^2-d...
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.62 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\left [-\frac {2 \, a c \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a c \log \left (x\right ) + a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) + 2 \, \sqrt {a} d \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right )}{4 \, {\left (a^{2} c^{2} - a d^{2}\right )}}, -\frac {2 \, a c \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a c \log \left (x\right ) + a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) + 4 \, \sqrt {-a} d \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right )}{4 \, {\left (a^{2} c^{2} - a d^{2}\right )}}\right ] \] Input:

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

Output:

[-1/4*(2*a*c*log(b*c^2*x^2 + a*c^2 - d^2) - 4*a*c*log(x) + a*c*log(-(b*c^2 
*x^2 + a*c^2 + 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) - a*c*log(-(b*c^2*x^2 + a 
*c^2 - 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) + 2*sqrt(a)*d*log(-(b*x^2 - 2*sqr 
t(b*x^2 + a)*sqrt(a) + 2*a)/x^2))/(a^2*c^2 - a*d^2), -1/4*(2*a*c*log(b*c^2 
*x^2 + a*c^2 - d^2) - 4*a*c*log(x) + a*c*log(-(b*c^2*x^2 + a*c^2 + 2*sqrt( 
b*x^2 + a)*c*d + d^2)/x^2) - a*c*log(-(b*c^2*x^2 + a*c^2 - 2*sqrt(b*x^2 + 
a)*c*d + d^2)/x^2) + 4*sqrt(-a)*d*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a))/(a^2 
*c^2 - a*d^2)]
 

Sympy [A] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {b c^{2} \left (\begin {cases} \frac {\sqrt {a + b x^{2}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{2}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{2 \left (a c^{2} - d^{2}\right )} - \frac {b \left (- \frac {c \log {\left (- b x^{2} \right )}}{2} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}}\right )}{2 \left (a c^{2} - d^{2}\right )}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \frac {x^{2} \log {\left (x^{2} \right )}}{2 \sqrt {a} d x^{2} + 2 a c x^{2}} & \text {for}\: 2 \sqrt {a} d + 2 a c \neq 0 \\\tilde {\infty } x^{2} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(-b*c**2*Piecewise((sqrt(a + b*x**2)/d, Eq(c, 0)), (log(c*sqr 
t(a + b*x**2) + d)/c, True))/(2*(a*c**2 - d**2)) - b*(-c*log(-b*x**2)/2 + 
d*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a))/(2*(a*c**2 - d**2)))/b, Ne(b, 
0)), (Piecewise((x**2*log(x**2)/(2*sqrt(a)*d*x**2 + 2*a*c*x**2), Ne(2*sqrt 
(a)*d + 2*a*c, 0)), (zoo*x**2, True)), True))
 

Maxima [F]

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

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

Output:

integrate(1/((b*c*x^2 + a*c + sqrt(b*x^2 + a)*d)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {c^{2} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{a c^{3} - c d^{2}} + \frac {c \log \left (b x^{2}\right )}{2 \, {\left (a c^{2} - d^{2}\right )}} - \frac {d \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{{\left (a c^{2} - d^{2}\right )} \sqrt {-a}} \] Input:

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

Output:

-c^2*log(abs(sqrt(b*x^2 + a)*c + d))/(a*c^3 - c*d^2) + 1/2*c*log(b*x^2)/(a 
*c^2 - d^2) - d*arctan(sqrt(b*x^2 + a)/sqrt(-a))/((a*c^2 - d^2)*sqrt(-a))
 

Mupad [B] (verification not implemented)

Time = 23.98 (sec) , antiderivative size = 1270, normalized size of antiderivative = 14.43 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

(c*log(x))/(a*c^2 - d^2) - (c*atan(((c*(4*c^6*d^2*(a + b*x^2)^(1/2) + (c*( 
4*c^4*d^5 - 8*a*c^6*d^3 + 4*a^2*c^8*d - (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 
 8*c^4*d^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - 
 d^2)))*1i)/(2*(a*c^2 - d^2)) + (c*(4*c^6*d^2*(a + b*x^2)^(1/2) - (c*(4*c^ 
4*d^5 - 8*a*c^6*d^3 + 4*a^2*c^8*d + (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c 
^4*d^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2 
)))*1i)/(2*(a*c^2 - d^2)))/((c*(4*c^6*d^2*(a + b*x^2)^(1/2) + (c*(4*c^4*d^ 
5 - 8*a*c^6*d^3 + 4*a^2*c^8*d - (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d 
^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2)))) 
/(2*(a*c^2 - d^2)) - (c*(4*c^6*d^2*(a + b*x^2)^(1/2) - (c*(4*c^4*d^5 - 8*a 
*c^6*d^3 + 4*a^2*c^8*d + (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d^6 - 8* 
a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2))))/(2*(a* 
c^2 - d^2))))*1i)/(a*c^2 - d^2) - (c*log(a*c^2 - d^2 + b*c^2*x^2))/(2*a*c^ 
2 - 2*d^2) - (d*atan(((d*(4*c^6*d^2*(a + b*x^2)^(1/2) + (d*(4*c^4*d^5 - 8* 
a*c^6*d^3 + 4*a^2*c^8*d - (d*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d^6 - 8 
*a*c^6*d^4 - 8*a^2*c^8*d^2))/(a^(1/2)*(2*a*c^2 - 2*d^2))))/(a^(1/2)*(2*a*c 
^2 - 2*d^2)))*1i)/(a^(1/2)*(2*a*c^2 - 2*d^2)) + (d*(4*c^6*d^2*(a + b*x^2)^ 
(1/2) - (d*(4*c^4*d^5 - 8*a*c^6*d^3 + 4*a^2*c^8*d + (d*(a + b*x^2)^(1/2)*( 
8*a^3*c^10 + 8*c^4*d^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(a^(1/2)*(2*a*c^2 - 
 2*d^2))))/(a^(1/2)*(2*a*c^2 - 2*d^2)))*1i)/(a^(1/2)*(2*a*c^2 - 2*d^2))...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 364, normalized size of antiderivative = 4.14 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {-2 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) d +2 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) d -\mathrm {log}\left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, a c x -\sqrt {b \,x^{2}+a}\, a d -\sqrt {b}\, a d x +a^{2} c +a b c \,x^{2}}{\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x}\right ) a c -\mathrm {log}\left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, a c x +\sqrt {b \,x^{2}+a}\, a d +\sqrt {b}\, a d x +a^{2} c +a b c \,x^{2}}{\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x}\right ) a c +2 \,\mathrm {log}\left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +b \,x^{2}}{\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x}\right ) a c +\mathrm {log}\left (\frac {2 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, c x -2 \sqrt {b \,x^{2}+a}\, d -2 \sqrt {b}\, d x +2 a c +2 b c \,x^{2}}{\sqrt {a}}\right ) a c -\mathrm {log}\left (\frac {2 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, c x +2 \sqrt {b \,x^{2}+a}\, d +2 \sqrt {b}\, d x +2 a c +2 b c \,x^{2}}{\sqrt {a}}\right ) a c}{2 a \left (a \,c^{2}-d^{2}\right )} \] Input:

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

Output:

( - 2*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*d + 2* 
sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*d - log((sqr 
t(b)*sqrt(a + b*x**2)*a*c*x - sqrt(a + b*x**2)*a*d - sqrt(b)*a*d*x + a**2* 
c + a*b*c*x**2)/(sqrt(a)*sqrt(a + b*x**2) + sqrt(b)*sqrt(a)*x))*a*c - log( 
(sqrt(b)*sqrt(a + b*x**2)*a*c*x + sqrt(a + b*x**2)*a*d + sqrt(b)*a*d*x + a 
**2*c + a*b*c*x**2)/(sqrt(a)*sqrt(a + b*x**2) + sqrt(b)*sqrt(a)*x))*a*c + 
2*log((sqrt(b)*sqrt(a + b*x**2)*x + b*x**2)/(sqrt(a)*sqrt(a + b*x**2) + sq 
rt(b)*sqrt(a)*x))*a*c + log((2*sqrt(b)*sqrt(a + b*x**2)*c*x - 2*sqrt(a + b 
*x**2)*d - 2*sqrt(b)*d*x + 2*a*c + 2*b*c*x**2)/sqrt(a))*a*c - log((2*sqrt( 
b)*sqrt(a + b*x**2)*c*x + 2*sqrt(a + b*x**2)*d + 2*sqrt(b)*d*x + 2*a*c + 2 
*b*c*x**2)/sqrt(a))*a*c)/(2*a*(a*c**2 - d**2))