3.109 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^2 (d+e x)} \, dx\)

Optimal. Leaf size=115 \[ -\frac {1}{2} d e (2 d+3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac {3}{2} d^3 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+d^3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[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 = 0.12, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {850, 813, 815, 844, 217, 203, 266, 63, 208} \[ -\frac {1}{2} d e (2 d+3 e x) \sqrt {d^2-e^2 x^2}-\frac {(3 d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac {3}{2} d^3 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+d^3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^2*(d + e*x)),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 63

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 203

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

Rule 208

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

Rule 217

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 266

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 813

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 815

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 844

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]

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 \operatorname {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 = 0.13, size = 114, normalized size = 0.99 \[ -d^3 e \log (x)+d^3 e \log \left (\sqrt {d^2-e^2 x^2}+d\right )-\frac {3}{2} d^3 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (-\frac {d^3}{x}-\frac {4 d^2 e}{3}-\frac {1}{2} d e^2 x+\frac {e^3 x^2}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^2*(d + e*x)),x]

[Out]

Sqrt[d^2 - e^2*x^2]*((-4*d^2*e)/3 - d^3/x - (d*e^2*x)/2 + (e^3*x^2)/3) - (3*d^3*e*ArcTan[(e*x)/Sqrt[d^2 - e^2*
x^2]])/2 - d^3*e*Log[x] + d^3*e*Log[d + Sqrt[d^2 - e^2*x^2]]

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fricas [A]  time = 0.67, size = 123, normalized size = 1.07 \[ \frac {18 \, d^{3} e x \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 6 \, d^{3} e x \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - 8 \, d^{3} e x + {\left (2 \, e^{3} x^{3} - 3 \, d e^{2} x^{2} - 8 \, d^{2} e x - 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(12*d^3*exp(1)^4*exp(2)^2-8*d^3*exp(
2)^4-4*d^3*exp(1)^6*exp(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+ex
p(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^4/exp(1)-d^3*x*exp(2)^3/(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/e
xp(1)/exp(2)+1/4*d^3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^4/exp(1)^4/x/exp(1)/exp(2)^2+d^3*exp(2
)*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/exp(1)-3/2*d^3*sign(d)*asin(x*exp(2)/d/
exp(1))*exp(2)/exp(1)+2*((4*exp(1)^5*1/24/exp(1)^2*x-6*exp(1)^4*d*1/24/exp(1)^2)*x-16*exp(1)^3*d^2*1/24/exp(1)
^2)*sqrt(d^2-x^2*exp(2))

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maple [B]  time = 0.01, size = 380, normalized size = 3.30 \[ \frac {d^{4} e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {3 d^{3} e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {15 d^{3} e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2} x}{8}+\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d \,e^{2} x}{8}-\sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e -\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2} x}{4 d}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{2} x}{4 d}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}{3}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2} x}{d^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{5 d^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e}{5 d^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{d^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d),x)

[Out]

-1/d^3/x*(-e^2*x^2+d^2)^(7/2)-1/d^3*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/4*(-e^2*x^2+d^2)^(3/2)/d*e^2*x-15/8*(-e^2*x^2
+d^2)^(1/2)*d*e^2*x-15/8/(e^2)^(1/2)*d^3*e^2*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/5*e/d^2*(-e^2*x^2+d^
2)^(5/2)-1/3*(-e^2*x^2+d^2)^(3/2)*e-(-e^2*x^2+d^2)^(1/2)*d^2*e+1/(d^2)^(1/2)*d^4*e*ln((2*d^2+2*(d^2)^(1/2)*(-e
^2*x^2+d^2)^(1/2))/x)+1/5*e/d^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)+1/4*e^2/d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3
/2)*x+3/8*e^2*d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+3/8*e^2*d^3/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*
e-(x+d/e)^2*e^2)^(1/2)*x)

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maxima [A]  time = 0.99, size = 131, normalized size = 1.14 \[ -\frac {3}{2} \, d^{3} e \arcsin \left (\frac {e x}{d}\right ) + d^{3} e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{2} x - \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e - \frac {1}{3} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^2\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(5/2)/(x^2*(d + e*x)),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x^2*(d + e*x)), x)

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sympy [C]  time = 10.20, size = 386, normalized size = 3.36 \[ d^{3} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**2/(e*x+d),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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