3.5.9 \(\int \frac {1}{x (\sqrt {a+b x}+\sqrt {c+b x})^2} \, dx\) [409]

Optimal. Leaf size=133 \[ \frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{(a-c)^2}-\frac {2 (a+c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {c+b x}}\right )}{(a-c)^2}+\frac {4 \sqrt {a} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+b x}}\right )}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6821, 103, 163, 65, 223, 212, 95, 214} \begin {gather*} \frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {b x+c}}{(a-c)^2}-\frac {2 (a+c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {b x+c}}\right )}{(a-c)^2}+\frac {4 \sqrt {a} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {b x+c}}\right )}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b
*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + p + 1))), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

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 6821

Int[(u_.)*((e_.)*Sqrt[(a_.) + (b_.)*(x_)^(n_.)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)^(n_.)])^(m_), x_Symbol] :> Dis
t[(a*e^2 - c*f^2)^m, Int[ExpandIntegrand[u/(e*Sqrt[a + b*x^n] - f*Sqrt[c + d*x^n])^m, x], x], x] /; FreeQ[{a,
b, c, d, e, f, n}, x] && ILtQ[m, 0] && EqQ[b*e^2 - d*f^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{x \left (\sqrt {a+b x}+\sqrt {c+b x}\right )^2} \, dx &=\frac {\int \left (2 b+\frac {a \left (1+\frac {c}{a}\right )}{x}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{x}\right ) \, dx}{(a-c)^2}\\ &=\frac {2 b x}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2}-\frac {2 \int \frac {\sqrt {a+b x} \sqrt {c+b x}}{x} \, dx}{(a-c)^2}\\ &=\frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2}+\frac {2 \int \frac {-a c-\frac {1}{2} b (a+c) x}{x \sqrt {a+b x} \sqrt {c+b x}} \, dx}{(a-c)^2}\\ &=\frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2}-\frac {(2 a c) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+b x}} \, dx}{(a-c)^2}-\frac {(b (a+c)) \int \frac {1}{\sqrt {a+b x} \sqrt {c+b x}} \, dx}{(a-c)^2}\\ &=\frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2}-\frac {(4 a c) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+b x}}\right )}{(a-c)^2}-\frac {(2 (a+c)) \text {Subst}\left (\int \frac {1}{\sqrt {-a+c+x^2}} \, dx,x,\sqrt {a+b x}\right )}{(a-c)^2}\\ &=\frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{(a-c)^2}+\frac {4 \sqrt {a} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+b x}}\right )}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2}-\frac {(2 (a+c)) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+b x}}\right )}{(a-c)^2}\\ &=\frac {2 b x}{(a-c)^2}-\frac {2 \sqrt {a+b x} \sqrt {c+b x}}{(a-c)^2}-\frac {2 (a+c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {c+b x}}\right )}{(a-c)^2}+\frac {4 \sqrt {a} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+b x}}\right )}{(a-c)^2}+\frac {(a+c) \log (x)}{(a-c)^2}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 137, normalized size = 1.03 \begin {gather*} \frac {\log \left (\sqrt {a} \sqrt {c}+b x-\sqrt {a+b x} \sqrt {c+b x}\right )}{\left (\sqrt {a}+\sqrt {c}\right )^2}+\frac {2 \left (c+b x-\sqrt {a+b x} \sqrt {c+b x}\right )+\left (\sqrt {a}+\sqrt {c}\right )^2 \log \left (\sqrt {a} \sqrt {c}-b x+\sqrt {a+b x} \sqrt {c+b x}\right )}{(a-c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.02, size = 258, normalized size = 1.94

method	result	size
default	\(\frac {a \ln \left (x \right )}{\left (a -c \right )^{2}}+\frac {c \ln \left (x \right )}{\left (a -c \right )^{2}}+\frac {2 b x}{\left (a -c \right )^{2}}+\frac {\sqrt {b x +a}\, \sqrt {b x +c}\, \left (2 \,\mathrm {csgn}\left (b \right ) \ln \left (\frac {a b x +b c x +2 \sqrt {a c}\, \sqrt {b^{2} x^{2}+a b x +b c x +a c}+2 a c}{x}\right ) a c -2 \,\mathrm {csgn}\left (b \right ) \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \sqrt {a c}-\ln \left (\frac {\left (2 \,\mathrm {csgn}\left (b \right ) \sqrt {b^{2} x^{2}+a b x +b c x +a c}+2 b x +a +c \right ) \mathrm {csgn}\left (b \right )}{2}\right ) \sqrt {a c}\, a -\ln \left (\frac {\left (2 \,\mathrm {csgn}\left (b \right ) \sqrt {b^{2} x^{2}+a b x +b c x +a c}+2 b x +a +c \right ) \mathrm {csgn}\left (b \right )}{2}\right ) \sqrt {a c}\, c \right ) \mathrm {csgn}\left (b \right )}{\left (a -c \right )^{2} \sqrt {a c}\, \sqrt {b^{2} x^{2}+a b x +b c x +a c}}\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 290, normalized size = 2.18 \begin {gather*} \left [\frac {2 \, b x + {\left (a + c\right )} \log \left (-2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b x + c} - a - c\right ) + {\left (a + c\right )} \log \left (x\right ) + 2 \, \sqrt {a c} \log \left (\frac {2 \, a^{2} c + 2 \, a c^{2} + 2 \, {\left (2 \, a c + \sqrt {a c} {\left (a + c\right )}\right )} \sqrt {b x + a} \sqrt {b x + c} + {\left (a^{2} b + 2 \, a b c + b c^{2}\right )} x + 2 \, {\left (2 \, a c + {\left (a b + b c\right )} x\right )} \sqrt {a c}}{x}\right ) - 2 \, \sqrt {b x + a} \sqrt {b x + c}}{a^{2} - 2 \, a c + c^{2}}, \frac {2 \, b x + {\left (a + c\right )} \log \left (-2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b x + c} - a - c\right ) + {\left (a + c\right )} \log \left (x\right ) - 4 \, \sqrt {-a c} \arctan \left (-\frac {\sqrt {-a c} b x - \sqrt {-a c} \sqrt {b x + a} \sqrt {b x + c}}{a c}\right ) - 2 \, \sqrt {b x + a} \sqrt {b x + c}}{a^{2} - 2 \, a c + c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (\sqrt {a + b x} + \sqrt {b x + c}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*(sqrt(a + b*x) + sqrt(b*x + c))**2), x)

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Giac [A]
time = 4.35, size = 194, normalized size = 1.46 \begin {gather*} \frac {4 \, a c \arctan \left (\frac {{\left (\sqrt {b x + a} - \sqrt {b x + c}\right )}^{2} - a - c}{2 \, \sqrt {-a c}}\right )}{{\left (a^{2} - 2 \, a c + c^{2}\right )} \sqrt {-a c}} - \frac {2 \, {\left (a^{2} - 2 \, a c + c^{2}\right )} \sqrt {b x + a} \sqrt {b x + c}}{a^{4} - 4 \, a^{3} c + 6 \, a^{2} c^{2} - 4 \, a c^{3} + c^{4}} + \frac {{\left (a + c\right )} \log \left ({\left (\sqrt {b x + a} - \sqrt {b x + c}\right )}^{2}\right )}{a^{2} - 2 \, a c + c^{2}} + \frac {{\left (a + c\right )} \log \left ({\left | b x \right |}\right )}{a^{2} - 2 \, a c + c^{2}} + \frac {2 \, {\left (b x + a\right )}}{a^{2} - 2 \, a c + c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a*c*arctan(1/2*((sqrt(b*x + a) - sqrt(b*x + c))^2 - a - c)/sqrt(-a*c))/((a^2 - 2*a*c + c^2)*sqrt(-a*c)) - 2*
(a^2 - 2*a*c + c^2)*sqrt(b*x + a)*sqrt(b*x + c)/(a^4 - 4*a^3*c + 6*a^2*c^2 - 4*a*c^3 + c^4) + (a + c)*log((sqr
t(b*x + a) - sqrt(b*x + c))^2)/(a^2 - 2*a*c + c^2) + (a + c)*log(abs(b*x))/(a^2 - 2*a*c + c^2) + 2*(b*x + a)/(
a^2 - 2*a*c + c^2)

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Mupad [B]
time = 11.14, size = 524, normalized size = 3.94 \begin {gather*} \frac {2\,b\,x}{{\left (a-c\right )}^2}-\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+b\,x}-\sqrt {c}}+1\right )\,\left (\frac {4\,c}{{\left (a-c\right )}^2}+\frac {2}{a-c}\right )-\frac {\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (4\,a+4\,c\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^3\,\left (a^2-2\,a\,c+c^2\right )}+\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (4\,a+4\,c\right )}{\left (\sqrt {c+b\,x}-\sqrt {c}\right )\,\left (a^2-2\,a\,c+c^2\right )}-\frac {16\,\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^2\,\left (a^2-2\,a\,c+c^2\right )}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^4}-\frac {2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^2}+1}+\frac {2\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+b\,x}-\sqrt {c}}-1\right )\,\left (a+c\right )}{{\left (a-c\right )}^2}+\frac {\ln \left (x\right )\,\left (a+c\right )}{a^2-2\,a\,c+c^2}+\frac {2\,\sqrt {a}\,\sqrt {c}\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+b\,x}-\sqrt {c}}\right )}{{\left (a-c\right )}^2}-\frac {2\,\sqrt {a}\,\sqrt {c}\,\ln \left (\frac {a\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {c+b\,x}-\sqrt {c}}-\sqrt {a}\,\sqrt {c}+\frac {c\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {c+b\,x}-\sqrt {c}}-\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^2}\right )}{a^2-2\,a\,c+c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*x)/(a - c)^2 - log(((a + b*x)^(1/2) - a^(1/2))/((c + b*x)^(1/2) - c^(1/2)) + 1)*((4*c)/(a - c)^2 + 2/(a -
c)) - ((((a + b*x)^(1/2) - a^(1/2))^3*(4*a + 4*c))/(((c + b*x)^(1/2) - c^(1/2))^3*(a^2 - 2*a*c + c^2)) + (((a
+ b*x)^(1/2) - a^(1/2))*(4*a + 4*c))/(((c + b*x)^(1/2) - c^(1/2))*(a^2 - 2*a*c + c^2)) - (16*a^(1/2)*c^(1/2)*
((a + b*x)^(1/2) - a^(1/2))^2)/(((c + b*x)^(1/2) - c^(1/2))^2*(a^2 - 2*a*c + c^2)))/(((a + b*x)^(1/2) - a^(1/2
))^4/((c + b*x)^(1/2) - c^(1/2))^4 - (2*((a + b*x)^(1/2) - a^(1/2))^2)/((c + b*x)^(1/2) - c^(1/2))^2 + 1) + (2
*log(((a + b*x)^(1/2) - a^(1/2))/((c + b*x)^(1/2) - c^(1/2)) - 1)*(a + c))/(a - c)^2 + (log(x)*(a + c))/(a^2 -
2*a*c + c^2) + (2*a^(1/2)*c^(1/2)*log(((a + b*x)^(1/2) - a^(1/2))/((c + b*x)^(1/2) - c^(1/2))))/(a - c)^2 - (
2*a^(1/2)*c^(1/2)*log((a*((a + b*x)^(1/2) - a^(1/2)))/((c + b*x)^(1/2) - c^(1/2)) - a^(1/2)*c^(1/2) + (c*((a +
b*x)^(1/2) - a^(1/2)))/((c + b*x)^(1/2) - c^(1/2)) - (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^2)/((c + b*
x)^(1/2) - c^(1/2))^2))/(a^2 - 2*a*c + c^2)

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