3.7.3 \(\int \frac {a+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx\) [603]

Optimal. Leaf size=147 \[ -\frac {c (3 e f+5 d g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (8 a e^2 g^2+c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{4 e^{5/2} g^{5/2}} \]

[Out]

1/4*(8*a*e^2*g^2+c*(3*d^2*g^2+2*d*e*f*g+3*e^2*f^2))*arctanh(g^(1/2)*(e*x+d)^(1/2)/e^(1/2)/(g*x+f)^(1/2))/e^(5/
2)/g^(5/2)+1/2*c*(e*x+d)^(3/2)*(g*x+f)^(1/2)/e^2/g-1/4*c*(5*d*g+3*e*f)*(e*x+d)^(1/2)*(g*x+f)^(1/2)/e^2/g^2

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Rubi [A]
time = 0.09, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {966, 81, 65, 223, 212} \begin {gather*} \frac {\left (8 a e^2 g^2+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{4 e^{5/2} g^{5/2}}-\frac {c \sqrt {d+e x} \sqrt {f+g x} (5 d g+3 e f)}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(c*(3*e*f + 5*d*g)*Sqrt[d + e*x]*Sqrt[f + g*x])/(e^2*g^2) + (c*(d + e*x)^(3/2)*Sqrt[f + g*x])/(2*e^2*g) +
((8*a*e^2*g^2 + c*(3*e^2*f^2 + 2*d*e*f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[g]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[f + g*x]
)])/(4*e^(5/2)*g^(5/2))

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 81

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

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 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 966

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d
+ e*x)^(m + 2*p)*((f + g*x)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0]
&& NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {a+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx &=\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\int \frac {\frac {1}{2} \left (4 a e^2 g-c d (3 e f+d g)\right )-\frac {1}{2} c e (3 e f+5 d g) x}{\sqrt {d+e x} \sqrt {f+g x}} \, dx}{2 e^2 g}\\ &=-\frac {c (3 e f+5 d g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {1}{8} \left (8 a+\frac {c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )}{e^2 g^2}\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {f+g x}} \, dx\\ &=-\frac {c (3 e f+5 d g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (8 a+\frac {c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )}{e^2 g^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {f-\frac {d g}{e}+\frac {g x^2}{e}}} \, dx,x,\sqrt {d+e x}\right )}{4 e}\\ &=-\frac {c (3 e f+5 d g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (8 a+\frac {c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )}{e^2 g^2}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {g x^2}{e}} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{4 e}\\ &=-\frac {c (3 e f+5 d g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (8 a e^2 g^2+c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{4 e^{5/2} g^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 123, normalized size = 0.84 \begin {gather*} \frac {c \sqrt {d+e x} \sqrt {f+g x} (-3 e f-3 d g+2 e g x)}{4 e^2 g^2}+\frac {\left (8 a e^2 g^2+c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{4 e^{5/2} g^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[d + e*x]*Sqrt[f + g*x]*(-3*e*f - 3*d*g + 2*e*g*x))/(4*e^2*g^2) + ((8*a*e^2*g^2 + c*(3*e^2*f^2 + 2*d*e*
f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/(Sqrt[g]*Sqrt[d + e*x])])/(4*e^(5/2)*g^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(305\) vs. \(2(121)=242\).
time = 0.08, size = 306, normalized size = 2.08

method	result	size
default	\(\frac {\left (4 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c e g x +3 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,d^{2} g^{2}+2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c d e f g +3 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,e^{2} f^{2}+8 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) a \,e^{2} g^{2}-6 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, c d g -6 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, c e f \right ) \sqrt {e x +d}\, \sqrt {g x +f}}{8 \sqrt {e g}\, g^{2} e^{2} \sqrt {\left (e x +d \right ) \left (g x +f \right )}}\)	\(306\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)*c*e*g*x+3*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g
+e*f)/(e*g)^(1/2))*c*d^2*g^2+2*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d
*e*f*g+3*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*e^2*f^2+8*ln(1/2*(2*e*g
*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*a*e^2*g^2-6*(e*g)^(1/2)*((e*x+d)*(g*x+f))^(1/2)
*c*d*g-6*(e*g)^(1/2)*((e*x+d)*(g*x+f))^(1/2)*c*e*f)*(e*x+d)^(1/2)*(g*x+f)^(1/2)/(e*g)^(1/2)/g^2/e^2/((e*x+d)*(
g*x+f))^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-%e*f>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.77, size = 338, normalized size = 2.30 \begin {gather*} \left [\frac {{\left ({\left (3 \, c d^{2} g^{2} + 2 \, c d f g e + {\left (3 \, c f^{2} + 8 \, a g^{2}\right )} e^{2}\right )} \sqrt {g} e^{\frac {1}{2}} \log \left (d^{2} g^{2} + 4 \, {\left (d g + {\left (2 \, g x + f\right )} e\right )} \sqrt {g x + f} \sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} + {\left (8 \, g^{2} x^{2} + 8 \, f g x + f^{2}\right )} e^{2} + 2 \, {\left (4 \, d g^{2} x + 3 \, d f g\right )} e\right ) - 4 \, {\left (3 \, c d g^{2} e - {\left (2 \, c g^{2} x - 3 \, c f g\right )} e^{2}\right )} \sqrt {g x + f} \sqrt {x e + d}\right )} e^{\left (-3\right )}}{16 \, g^{3}}, -\frac {{\left ({\left (3 \, c d^{2} g^{2} + 2 \, c d f g e + {\left (3 \, c f^{2} + 8 \, a g^{2}\right )} e^{2}\right )} \sqrt {-g e} \arctan \left (\frac {{\left (d g + {\left (2 \, g x + f\right )} e\right )} \sqrt {g x + f} \sqrt {-g e} \sqrt {x e + d}}{2 \, {\left ({\left (g^{2} x^{2} + f g x\right )} e^{2} + {\left (d g^{2} x + d f g\right )} e\right )}}\right ) + 2 \, {\left (3 \, c d g^{2} e - {\left (2 \, c g^{2} x - 3 \, c f g\right )} e^{2}\right )} \sqrt {g x + f} \sqrt {x e + d}\right )} e^{\left (-3\right )}}{8 \, g^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((3*c*d^2*g^2 + 2*c*d*f*g*e + (3*c*f^2 + 8*a*g^2)*e^2)*sqrt(g)*e^(1/2)*log(d^2*g^2 + 4*(d*g + (2*g*x + f
)*e)*sqrt(g*x + f)*sqrt(x*e + d)*sqrt(g)*e^(1/2) + (8*g^2*x^2 + 8*f*g*x + f^2)*e^2 + 2*(4*d*g^2*x + 3*d*f*g)*e
) - 4*(3*c*d*g^2*e - (2*c*g^2*x - 3*c*f*g)*e^2)*sqrt(g*x + f)*sqrt(x*e + d))*e^(-3)/g^3, -1/8*((3*c*d^2*g^2 +
2*c*d*f*g*e + (3*c*f^2 + 8*a*g^2)*e^2)*sqrt(-g*e)*arctan(1/2*(d*g + (2*g*x + f)*e)*sqrt(g*x + f)*sqrt(-g*e)*sq
rt(x*e + d)/((g^2*x^2 + f*g*x)*e^2 + (d*g^2*x + d*f*g)*e)) + 2*(3*c*d*g^2*e - (2*c*g^2*x - 3*c*f*g)*e^2)*sqrt(
g*x + f)*sqrt(x*e + d))*e^(-3)/g^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + c x^{2}}{\sqrt {d + e x} \sqrt {f + g x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.16, size = 160, normalized size = 1.09 \begin {gather*} \frac {{\left (\sqrt {d g^{2} + {\left (g x + f\right )} g e - f g e} \sqrt {g x + f} {\left (\frac {2 \, {\left (g x + f\right )} c e^{\left (-1\right )}}{g^{3}} - \frac {{\left (3 \, c d g^{6} e + 5 \, c f g^{5} e^{2}\right )} e^{\left (-3\right )}}{g^{8}}\right )} - \frac {{\left (3 \, c d^{2} g^{2} + 2 \, c d f g e + 3 \, c f^{2} e^{2} + 8 \, a g^{2} e^{2}\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -\sqrt {g x + f} \sqrt {g} e^{\frac {1}{2}} + \sqrt {d g^{2} + {\left (g x + f\right )} g e - f g e} \right |}\right )}{g^{\frac {5}{2}}}\right )} g}{4 \, {\left | g \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(d*g^2 + (g*x + f)*g*e - f*g*e)*sqrt(g*x + f)*(2*(g*x + f)*c*e^(-1)/g^3 - (3*c*d*g^6*e + 5*c*f*g^5*e^
2)*e^(-3)/g^8) - (3*c*d^2*g^2 + 2*c*d*f*g*e + 3*c*f^2*e^2 + 8*a*g^2*e^2)*e^(-5/2)*log(abs(-sqrt(g*x + f)*sqrt(
g)*e^(1/2) + sqrt(d*g^2 + (g*x + f)*g*e - f*g*e)))/g^(5/2))*g/abs(g)

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Mupad [B]
time = 20.13, size = 569, normalized size = 3.87 \begin {gather*} \frac {c\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{2\,e^{5/2}\,g^{5/2}}-\frac {4\,a\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}{\sqrt {-e\,g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {-e\,g}}-\frac {\frac {\left (\sqrt {d+e\,x}-\sqrt {d}\right )\,\left (\frac {3\,c\,d^2\,e\,g^2}{2}+c\,d\,e^2\,f\,g+\frac {3\,c\,e^3\,f^2}{2}\right )}{g^6\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{g^5\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}+\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7\,\left (\frac {3\,c\,d^2\,g^2}{2}+c\,d\,e\,f\,g+\frac {3\,c\,e^2\,f^2}{2}\right )}{e^2\,g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^7}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{e\,g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^5}+\frac {\sqrt {d}\,\sqrt {f}\,\left (32\,c\,d\,g+32\,c\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^8}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^8}+\frac {e^4}{g^4}-\frac {4\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^6}-\frac {4\,e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}+\frac {6\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*atanh((g^(1/2)*((d + e*x)^(1/2) - d^(1/2)))/(e^(1/2)*((f + g*x)^(1/2) - f^(1/2))))*(3*d^2*g^2 + 3*e^2*f^2 +
2*d*e*f*g))/(2*e^(5/2)*g^(5/2)) - (4*a*atan((e*((f + g*x)^(1/2) - f^(1/2)))/((-e*g)^(1/2)*((d + e*x)^(1/2) -
d^(1/2)))))/(-e*g)^(1/2) - ((((d + e*x)^(1/2) - d^(1/2))*((3*c*e^3*f^2)/2 + (3*c*d^2*e*g^2)/2 + c*d*e^2*f*g))/
(g^6*((f + g*x)^(1/2) - f^(1/2))) - (((d + e*x)^(1/2) - d^(1/2))^3*((11*c*d^2*g^2)/2 + (11*c*e^2*f^2)/2 + 25*c
*d*e*f*g))/(g^5*((f + g*x)^(1/2) - f^(1/2))^3) + (((d + e*x)^(1/2) - d^(1/2))^7*((3*c*d^2*g^2)/2 + (3*c*e^2*f^
2)/2 + c*d*e*f*g))/(e^2*g^3*((f + g*x)^(1/2) - f^(1/2))^7) - (((d + e*x)^(1/2) - d^(1/2))^5*((11*c*d^2*g^2)/2
+ (11*c*e^2*f^2)/2 + 25*c*d*e*f*g))/(e*g^4*((f + g*x)^(1/2) - f^(1/2))^5) + (d^(1/2)*f^(1/2)*(32*c*d*g + 32*c*
e*f)*((d + e*x)^(1/2) - d^(1/2))^4)/(g^4*((f + g*x)^(1/2) - f^(1/2))^4))/(((d + e*x)^(1/2) - d^(1/2))^8/((f +
g*x)^(1/2) - f^(1/2))^8 + e^4/g^4 - (4*e*((d + e*x)^(1/2) - d^(1/2))^6)/(g*((f + g*x)^(1/2) - f^(1/2))^6) - (4
*e^3*((d + e*x)^(1/2) - d^(1/2))^2)/(g^3*((f + g*x)^(1/2) - f^(1/2))^2) + (6*e^2*((d + e*x)^(1/2) - d^(1/2))^4
)/(g^2*((f + g*x)^(1/2) - f^(1/2))^4))

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