3.1.9 \(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=121 \[ -\frac {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {3}{2} d^2 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d^2 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {813, 844, 217, 203, 266, 63, 208} \begin {gather*} -\frac {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {3}{2} d^2 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d^2 e^2 \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^3,x]

[Out]

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

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 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]

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^3} \, dx &=-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {3}{8} \int \frac {\left (-4 d^2 e+4 d e^2 x\right ) \sqrt {d^2-e^2 x^2}}{x^2} \, dx\\ &=-\frac {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac {3}{16} \int \frac {-8 d^3 e^2-8 d^2 e^3 x}{x \sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {1}{2} \left (3 d^3 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-\frac {1}{2} \left (3 d^2 e^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {1}{4} \left (3 d^3 e^2\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^2 e^3\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 {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {3}{2} d^2 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{2} \left (3 d^3\right ) \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 )\\ &=-\frac {3 d e (d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {3}{2} d^2 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d^2 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 110, normalized size = 0.91 \begin {gather*} -\frac {d^2 e \sqrt {d^2-e^2 x^2} \, _2F_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}}}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};1-\frac {e^2 x^2}{d^2}\right )}{5 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^2*e*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, -1/2, 1/2, (e^2*x^2)/d^2])/(x*Sqrt[1 - (e^2*x^2)/d^2])) -
(e^2*(d^2 - e^2*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 - (e^2*x^2)/d^2])/(5*d^3)

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IntegrateAlgebraic [A]  time = 0.69, size = 146, normalized size = 1.21 \begin {gather*} -\frac {3}{2} d^2 e \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )-3 d^2 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-d^3-2 d^2 e x-2 d e^2 x^2-e^3 x^3\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 133, normalized size = 1.10 \begin {gather*} \frac {6 \, d^{2} e^{2} x^{2} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 3 \, d^{2} e^{2} x^{2} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - 2 \, d^{2} e^{2} x^{2} - {\left (e^{3} x^{3} + 2 \, d e^{2} x^{2} + 2 \, d^{2} e x + d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.28, size = 217, normalized size = 1.79 \begin {gather*} -\frac {3}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{2} \mathrm {sgn}\relax (d) + \frac {3}{2} \, d^{2} e^{2} \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 {1}{8} \, {\left (\frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{2} e^{8}}{x} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{2} e^{6}}{x^{2}}\right )} e^{\left (-8\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (x e^{3} + 2 \, d e^{2}\right )} + \frac {{\left (d^{2} e^{6} + \frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{2} e^{4}}{x}\right )} x^{2}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2}} \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^3,x, algorithm="giac")

[Out]

-3/2*d^2*arcsin(x*e/d)*e^2*sgn(d) + 3/2*d^2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))
- 1/8*(4*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^2*e^8/x + (d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^2*e^6/x^2)*e^(-8) - 1/2
*sqrt(-x^2*e^2 + d^2)*(x*e^3 + 2*d*e^2) + 1/8*(d^2*e^6 + 4*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^2*e^4/x)*x^2/(d*e
+ sqrt(-x^2*e^2 + d^2)*e)^2

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maple [B]  time = 0.02, size = 212, normalized size = 1.75 \begin {gather*} \frac {3 d^{3} e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}-\frac {3 d^{2} e^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x}{2}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2}}{2}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3} x}{d^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}{2 d}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{d^{2} x}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{2 d \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e/d^2/x*(-e^2*x^2+d^2)^(5/2)-e^3/d^2*x*(-e^2*x^2+d^2)^(3/2)-3/2*e^3*x*(-e^2*x^2+d^2)^(1/2)-3/2*e^3*d^2/(e^2)^
(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/2/d/x^2*(-e^2*x^2+d^2)^(5/2)-1/2*e^2/d*(-e^2*x^2+d^2)^(3/2)
-3/2*d*e^2*(-e^2*x^2+d^2)^(1/2)+3/2*e^2*d^3/(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 = 0.99, size = 160, normalized size = 1.32 \begin {gather*} -\frac {3}{2} \, d^{2} e^{2} \arcsin \left (\frac {e x}{d}\right ) + \frac {3}{2} \, d^{2} e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {3}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3} x - \frac {3}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{2} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{2 \, d} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{x} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{2 \, d x^{2}} \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^3,x, algorithm="maxima")

[Out]

-3/2*d^2*e^2*arcsin(e*x/d) + 3/2*d^2*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 3/2*sqrt(-e^2*x
^2 + d^2)*e^3*x - 3/2*sqrt(-e^2*x^2 + d^2)*d*e^2 - 1/2*(-e^2*x^2 + d^2)^(3/2)*e^2/d - (-e^2*x^2 + d^2)^(3/2)*e
/x - 1/2*(-e^2*x^2 + d^2)^(5/2)/(d*x^2)

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mupad [B]  time = 3.74, size = 120, normalized size = 0.99 \begin {gather*} \frac {3\,d^2\,e^2\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{2}-\frac {d^3\,\sqrt {d^2-e^2\,x^2}}{2\,x^2}-d\,e^2\,\sqrt {d^2-e^2\,x^2}-\frac {e\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ \frac {e^2\,x^2}{d^2}\right )}{x\,{\left (1-\frac {e^2\,x^2}{d^2}\right )}^{3/2}} \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^3,x)

[Out]

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

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sympy [C]  time = 9.53, size = 461, normalized size = 3.81 \begin {gather*} d^{3} \left (\begin {cases} - \frac {d^{2}}{2 e x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e}{2 x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) + d^{2} e \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 e^{2} \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 ) - e^{3} \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 ) \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**3,x)

[Out]

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

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