3.1.8 \(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^2} \, dx\) [8]

Optimal. Leaf size=117 \[ \frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac {3}{2} d^3 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

-1/3*(-e*x+3*d)*(-e^2*x^2+d^2)^(3/2)/x-3/2*d^3*e*arctan(e*x/(-e^2*x^2+d^2)^(1/2))-d^3*e*arctanh((-e^2*x^2+d^2)
^(1/2)/d)+1/2*d*e*(-3*e*x+2*d)*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {827, 829, 858, 223, 209, 272, 65, 214} \begin {gather*} -\frac {3}{2} d^3 e \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-d^3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^2,x]

[Out]

(d*e*(2*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/2 - ((3*d - e*x)*(d^2 - e^2*x^2)^(3/2))/(3*x) - (3*d^3*e*ArcTan[(e*x)/
Sqrt[d^2 - e^2*x^2]])/2 - d^3*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 827

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 829

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

Rule 858

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

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^2} \, dx &=-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac {1}{2} \int \frac {\left (-2 d^2 e+6 d e^2 x\right ) \sqrt {d^2-e^2 x^2}}{x} \, dx\\ &=\frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}+\frac {\int \frac {4 d^4 e^3-6 d^3 e^4 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{4 e^2}\\ &=\frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}+\left (d^4 e\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-\frac {1}{2} \left (3 d^3 e^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}+\frac {1}{2} \left (d^4 e\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-\frac {1}{2} \left (3 d^3 e^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=\frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac {3}{2} d^3 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {d^4 \text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{e}\\ &=\frac {1}{2} d e (2 d-3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac {3}{2} d^3 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 143, normalized size = 1.22 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-6 d^3+8 d^2 e x-3 d e^2 x^2-2 e^3 x^3\right )}{6 x}+2 d^3 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-\frac {3}{2} d^3 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^2,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-6*d^3 + 8*d^2*e*x - 3*d*e^2*x^2 - 2*e^3*x^3))/(6*x) + 2*d^3*e*ArcTanh[(Sqrt[-e^2]*x)/d
- Sqrt[d^2 - e^2*x^2]/d] - (3*d^3*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/2

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Maple [A]
time = 0.08, size = 188, normalized size = 1.61

method	result	size
risch	\(-\frac {d^{3} \sqrt {-e^{2} x^{2}+d^{2}}}{x}-\frac {e^{3} x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3}+\frac {4 e \,d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3}-\frac {e^{2} d x \sqrt {-e^{2} x^{2}+d^{2}}}{2}-\frac {3 e^{2} d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {e \,d^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\)	\(165\)
default	\(d \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{d^{2} x}-\frac {4 e^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{d^{2}}\right )+e \left (\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3}+d^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )\right )\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

d*(-1/d^2/x*(-e^2*x^2+d^2)^(5/2)-4*e^2/d^2*(1/4*x*(-e^2*x^2+d^2)^(3/2)+3/4*d^2*(1/2*x*(-e^2*x^2+d^2)^(1/2)+1/2
*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))))+e*(1/3*(-e^2*x^2+d^2)^(3/2)+d^2*((-e^2*x^2+d^2)
^(1/2)-d^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)))

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Maxima [A]
time = 0.49, size = 128, normalized size = 1.09 \begin {gather*} -\frac {3}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e - d^{3} e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {3}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d x e^{2} + \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e + \frac {1}{3} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} e - \frac {{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^2,x, algorithm="maxima")

[Out]

-3/2*d^3*arcsin(x*e/d)*e - d^3*e*log(2*d^2/abs(x) + 2*sqrt(-x^2*e^2 + d^2)*d/abs(x)) - 3/2*sqrt(-x^2*e^2 + d^2
)*d*x*e^2 + sqrt(-x^2*e^2 + d^2)*d^2*e + 1/3*(-x^2*e^2 + d^2)^(3/2)*e - (-x^2*e^2 + d^2)^(3/2)*d/x

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Fricas [A]
time = 1.86, size = 122, normalized size = 1.04 \begin {gather*} \frac {18 \, d^{3} x \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) e + 6 \, d^{3} x e \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + 8 \, d^{3} x e - {\left (2 \, x^{3} e^{3} + 3 \, d x^{2} e^{2} - 8 \, d^{2} x e + 6 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^2,x, algorithm="fricas")

[Out]

1/6*(18*d^3*x*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x)*e + 6*d^3*x*e*log(-(d - sqrt(-x^2*e^2 + d^2))/x) +
8*d^3*x*e - (2*x^3*e^3 + 3*d*x^2*e^2 - 8*d^2*x*e + 6*d^3)*sqrt(-x^2*e^2 + d^2))/x

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Sympy [C] Result contains complex when optimal does not.
time = 3.69, size = 386, normalized size = 3.30 \begin {gather*} d^{3} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + d^{2} e \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) - e^{3} \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x**2,x)

[Out]

d**3*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1 + e**2*x**2/d**2)),
Abs(e**2*x**2/d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**2/d**
2)), True)) + d**2*e*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/sqrt(d**2/(e**2
*x**2) - 1), Abs(d**2/(e**2*x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*d*asin(d/(e*x)) + I*e*
x/sqrt(-d**2/(e**2*x**2) + 1), True)) - d*e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2
*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e)
+ d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) - e**3*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**
2)**(3/2)/(3*e**2), True))

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Giac [A]
time = 1.42, size = 157, normalized size = 1.34 \begin {gather*} -\frac {3}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e \mathrm {sgn}\left (d\right ) - d^{3} e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) + \frac {d^{3} x e^{3}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{3} e^{\left (-1\right )}}{2 \, x} + \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (8 \, d^{2} e - {\left (2 \, x e^{3} + 3 \, d e^{2}\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^2,x, algorithm="giac")

[Out]

-3/2*d^3*arcsin(x*e/d)*e*sgn(d) - d^3*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/2*d^
3*x*e^3/(d*e + sqrt(-x^2*e^2 + d^2)*e) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^3*e^(-1)/x + 1/6*sqrt(-x^2*e^2 +
d^2)*(8*d^2*e - (2*x*e^3 + 3*d*e^2)*x)

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Mupad [B]
time = 3.51, size = 114, normalized size = 0.97 \begin {gather*} \frac {e\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{3}+d^2\,e\,\sqrt {d^2-e^2\,x^2}-d^3\,e\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )-\frac {d^3\,\sqrt {d^2-e^2\,x^2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ \frac {e^2\,x^2}{d^2}\right )}{x\,\sqrt {1-\frac {e^2\,x^2}{d^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^2,x)

[Out]

(e*(d^2 - e^2*x^2)^(3/2))/3 + d^2*e*(d^2 - e^2*x^2)^(1/2) - d^3*e*atanh((d^2 - e^2*x^2)^(1/2)/d) - (d^3*(d^2 -
e^2*x^2)^(1/2)*hypergeom([-3/2, -1/2], 1/2, (e^2*x^2)/d^2))/(x*(1 - (e^2*x^2)/d^2)^(1/2))

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