\(\int \frac {1}{x (a+b \sqrt {c+d x})^2} \, dx\) [122]

Optimal result

Integrand size = 19, antiderivative size = 129 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}+\frac {4 a b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac {\left (a^2+b^2 c\right ) \log (x)}{\left (a^2-b^2 c\right )^2}-\frac {2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {\frac {2 a \left (a^2-b^2 c\right )}{a+b \sqrt {c+d x}}+4 a b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\left (a^2+b^2 c\right ) \log (-d x)-2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {896, 25, 1732, 594, 25, 657, 2009}

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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 594

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {2 a}{\left (-b^{2} c +a^{2}\right ) \left (a +b \sqrt {d x +c}\right )}-\frac {2 \left (b^{2} c +a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {-\left (-b^{2} c -a^{2}\right ) \ln \left (-d x \right )+4 a b \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}\)	\(118\)
default	\(\frac {2 a}{\left (-b^{2} c +a^{2}\right ) \left (a +b \sqrt {d x +c}\right )}-\frac {2 \left (b^{2} c +a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {-\left (-b^{2} c -a^{2}\right ) \ln \left (-d x \right )+4 a b \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}\)	\(118\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.42 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=\left [\frac {2 \, a^{2} b^{2} c - 2 \, a^{4} + 2 \, {\left (a b^{3} d x + a b^{3} c - a^{3} b\right )} \sqrt {c} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) + {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (x\right ) - 2 \, {\left (a b^{3} c - a^{3} b\right )} \sqrt {d x + c}}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6} + {\left (b^{6} c^{2} - 2 \, a^{2} b^{4} c + a^{4} b^{2}\right )} d x}, \frac {2 \, a^{2} b^{2} c - 2 \, a^{4} - 4 \, {\left (a b^{3} d x + a b^{3} c - a^{3} b\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - 2 \, {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) + {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (x\right ) - 2 \, {\left (a b^{3} c - a^{3} b\right )} \sqrt {d x + c}}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6} + {\left (b^{6} c^{2} - 2 \, a^{2} b^{4} c + a^{4} b^{2}\right )} d x}\right ] \] Input:

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

Output:

[(2*a^2*b^2*c - 2*a^4 + 2*(a*b^3*d*x + a*b^3*c - a^3*b)*sqrt(c)*log((d*x + 
 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(b^4*c^2 - a^4 + (b^4*c + a^2*b^2)* 
d*x)*log(sqrt(d*x + c)*b + a) + (b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*lo 
g(x) - 2*(a*b^3*c - a^3*b)*sqrt(d*x + c))/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4 
*b^2*c - a^6 + (b^6*c^2 - 2*a^2*b^4*c + a^4*b^2)*d*x), (2*a^2*b^2*c - 2*a^ 
4 - 4*(a*b^3*d*x + a*b^3*c - a^3*b)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x + c) 
) - 2*(b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*log(sqrt(d*x + c)*b + a) + ( 
b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*log(x) - 2*(a*b^3*c - a^3*b)*sqrt(d 
*x + c))/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6 + (b^6*c^2 - 2*a^2*b 
^4*c + a^4*b^2)*d*x)]
 

Sympy [A] (verification not implemented)

Time = 5.53 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=\begin {cases} - \frac {2 a b \left (\begin {cases} \frac {\sqrt {c + d x}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \sqrt {c + d x}\right )} & \text {otherwise} \end {cases}\right )}{a^{2} - b^{2} c} - \frac {2 b \left (a^{2} + b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {c + d x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{\left (a^{2} - b^{2} c\right )^{2}} - \frac {2 \cdot \left (\frac {2 a b c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + \left (- \frac {a^{2}}{2} - \frac {b^{2} c}{2}\right ) \log {\left (- d x \right )}\right )}{\left (a^{2} - b^{2} c\right )^{2}} & \text {for}\: d \neq 0 \\\frac {\log {\left (x \right )}}{\left (a + b \sqrt {c}\right )^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-2*a*b*Piecewise((sqrt(c + d*x)/a**2, Eq(b, 0)), (-1/(b*(a + b* 
sqrt(c + d*x))), True))/(a**2 - b**2*c) - 2*b*(a**2 + b**2*c)*Piecewise((s 
qrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/(a**2 - b** 
2*c)**2 - 2*(2*a*b*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + (-a**2/2 - b* 
*2*c/2)*log(-d*x))/(a**2 - b**2*c)**2, Ne(d, 0)), (log(x)/(a + b*sqrt(c))* 
*2, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=-\frac {2 \, a b \sqrt {c} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} + \frac {{\left (b^{2} c + a^{2}\right )} \log \left (d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {2 \, {\left (b^{2} c + a^{2}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {2 \, a}{a b^{2} c - a^{3} + {\left (b^{3} c - a^{2} b\right )} \sqrt {d x + c}} \] Input:

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

Output:

-2*a*b*sqrt(c)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(b 
^4*c^2 - 2*a^2*b^2*c + a^4) + (b^2*c + a^2)*log(d*x)/(b^4*c^2 - 2*a^2*b^2* 
c + a^4) - 2*(b^2*c + a^2)*log(sqrt(d*x + c)*b + a)/(b^4*c^2 - 2*a^2*b^2*c 
 + a^4) - 2*a/(a*b^2*c - a^3 + (b^3*c - a^2*b)*sqrt(d*x + c))
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=-\frac {4 \, a b c \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {-c}} + \frac {{\left (b^{2} c + a^{2}\right )} \log \left (-d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {2 \, {\left (b^{3} c + a^{2} b\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{5} c^{2} - 2 \, a^{2} b^{3} c + a^{4} b} - \frac {2 \, {\left (a b^{2} c - a^{3}\right )}}{{\left (b^{2} c - a^{2}\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}} \] Input:

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

Output:

-4*a*b*c*arctan(sqrt(d*x + c)/sqrt(-c))/((b^4*c^2 - 2*a^2*b^2*c + a^4)*sqr 
t(-c)) + (b^2*c + a^2)*log(-d*x)/(b^4*c^2 - 2*a^2*b^2*c + a^4) - 2*(b^3*c 
+ a^2*b)*log(abs(sqrt(d*x + c)*b + a))/(b^5*c^2 - 2*a^2*b^3*c + a^4*b) - 2 
*(a*b^2*c - a^3)/((b^2*c - a^2)^2*(sqrt(d*x + c)*b + a))
 

Mupad [B] (verification not implemented)

Time = 9.75 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )}{{\left (a+b\,\sqrt {c}\right )}^2}+\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (\frac {2}{b^2\,c-a^2}-\frac {4\,b^2\,c}{{\left (b^2\,c-a^2\right )}^2}\right )+\frac {\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )}{{\left (a-b\,\sqrt {c}\right )}^2}-\frac {2\,a}{\left (b^2\,c-a^2\right )\,\left (a+b\,\sqrt {c+d\,x}\right )} \] Input:

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

Output:

log((c + d*x)^(1/2) - c^(1/2))/(a + b*c^(1/2))^2 + log(a + b*(c + d*x)^(1/ 
2))*(2/(b^2*c - a^2) - (4*b^2*c)/(b^2*c - a^2)^2) + log((c + d*x)^(1/2) + 
c^(1/2))/(a - b*c^(1/2))^2 - (2*a)/((b^2*c - a^2)*(a + b*(c + d*x)^(1/2)))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 408, normalized size of antiderivative = 3.16 \[ \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {-2 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,b^{2}+2 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,b^{2}+\sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} b +\sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b^{3} c +\sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} b +\sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b^{3} c -2 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{2} b -2 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) b^{3} c -2 \sqrt {d x +c}\, a^{2} b +2 \sqrt {d x +c}\, b^{3} c -2 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} b +2 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} b +\mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{3}+\mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,b^{2} c +\mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{3}+\mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,b^{2} c -2 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{3}-2 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a \,b^{2} c}{\sqrt {d x +c}\, a^{4} b -2 \sqrt {d x +c}\, a^{2} b^{3} c +\sqrt {d x +c}\, b^{5} c^{2}+a^{5}-2 a^{3} b^{2} c +a \,b^{4} c^{2}} \] Input:

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

Output:

( - 2*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a*b**2 + 2*sqrt(c 
)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a*b**2 + sqrt(c + d*x)*log(sq 
rt(c + d*x) - sqrt(c))*a**2*b + sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c)) 
*b**3*c + sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**2*b + sqrt(c + d*x 
)*log(sqrt(c + d*x) + sqrt(c))*b**3*c - 2*sqrt(c + d*x)*log(sqrt(c + d*x)* 
b + a)*a**2*b - 2*sqrt(c + d*x)*log(sqrt(c + d*x)*b + a)*b**3*c - 2*sqrt(c 
 + d*x)*a**2*b + 2*sqrt(c + d*x)*b**3*c - 2*sqrt(c)*log(sqrt(c + d*x) - sq 
rt(c))*a**2*b + 2*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*b + log(sqrt(c 
 + d*x) - sqrt(c))*a**3 + log(sqrt(c + d*x) - sqrt(c))*a*b**2*c + log(sqrt 
(c + d*x) + sqrt(c))*a**3 + log(sqrt(c + d*x) + sqrt(c))*a*b**2*c - 2*log( 
sqrt(c + d*x)*b + a)*a**3 - 2*log(sqrt(c + d*x)*b + a)*a*b**2*c)/(sqrt(c + 
 d*x)*a**4*b - 2*sqrt(c + d*x)*a**2*b**3*c + sqrt(c + d*x)*b**5*c**2 + a** 
5 - 2*a**3*b**2*c + a*b**4*c**2)