3.1.10 \(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=120 \[ \frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 120, 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 = {811, 813, 844, 217, 203, 266, 63, 208} \begin {gather*} \frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d e^3 \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^4,x]

[Out]

(e^2*(2*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/(2*x) - ((2*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/(6*x^3) + d*e^3*ArcTan[(
e*x)/Sqrt[d^2 - e^2*x^2]] + (3*d*e^3*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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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^4} \, dx &=-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}-\frac {\int \frac {\left (4 d^3 e^2+6 d^2 e^3 x\right ) \sqrt {d^2-e^2 x^2}}{x^2} \, dx}{4 d^2}\\ &=\frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+\frac {\int \frac {-12 d^4 e^3+8 d^3 e^4 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{8 d^2}\\ &=\frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}-\frac {1}{2} \left (3 d^2 e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx+\left (d e^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}-\frac {1}{4} \left (3 d^2 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )+\left (d e^4\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 {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{2} \left (3 d^2 e\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 {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 111, normalized size = 0.92 \begin {gather*} -\frac {e^3 \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^4}-\frac {d^3 \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{3 x^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^4,x]

[Out]

-1/3*(d^3*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (e^2*x^2)/d^2])/(x^3*Sqrt[1 - (e^2*x^2)/d^2]
) - (e^3*(d^2 - e^2*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 - (e^2*x^2)/d^2])/(5*d^4)

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IntegrateAlgebraic [A]  time = 0.59, size = 141, normalized size = 1.18 \begin {gather*} d \sqrt {-e^2} e^2 \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )-3 d e^3 \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 (-2 d^3-3 d^2 e x+8 d e^2 x^2-6 e^3 x^3\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^4,x]

[Out]

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

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fricas [A]  time = 0.42, size = 129, normalized size = 1.08 \begin {gather*} -\frac {12 \, d e^{3} x^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 9 \, d e^{3} x^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + 6 \, d e^{3} x^{3} + {\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 3 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

-1/6*(12*d*e^3*x^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 9*d*e^3*x^3*log(-(d - sqrt(-e^2*x^2 + d^2))/x)
+ 6*d*e^3*x^3 + (6*e^3*x^3 - 8*d*e^2*x^2 + 3*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 + d^2))/x^3

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giac [B]  time = 0.23, size = 261, normalized size = 2.18 \begin {gather*} d \arcsin \left (\frac {x e}{d}\right ) e^{3} \mathrm {sgn}\relax (d) + \frac {3}{2} \, d e^{3} \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 {{\left (d e^{8} + \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{6}}{x} - \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{4}}{x^{2}}\right )} x^{3} e}{24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3}} + \frac {1}{24} \, {\left (\frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{16}}{x} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{14}}{x^{2}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{12}}{x^{3}}\right )} e^{\left (-15\right )} - \sqrt {-x^{2} e^{2} + d^{2}} e^{3} \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^4,x, algorithm="giac")

[Out]

d*arcsin(x*e/d)*e^3*sgn(d) + 3/2*d*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/24*(d
*e^8 + 3*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^6/x - 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d*e^4/x^2)*x^3*e/(d*e +
sqrt(-x^2*e^2 + d^2)*e)^3 + 1/24*(15*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^16/x - 3*(d*e + sqrt(-x^2*e^2 + d^2)*e
)^2*d*e^14/x^2 - (d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^12/x^3)*e^(-15) - sqrt(-x^2*e^2 + d^2)*e^3

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maple [B]  time = 0.02, size = 235, normalized size = 1.96 \begin {gather*} \frac {3 d^{2} e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}+\frac {d \,e^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4} x}{d}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}{2}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4} x}{3 d^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}{2 d^{2}}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}{3 d^{3} x}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{2 d^{2} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{3 d \,x^{3}} \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^4,x)

[Out]

-1/3/d/x^3*(-e^2*x^2+d^2)^(5/2)+2/3*e^2/d^3/x*(-e^2*x^2+d^2)^(5/2)+2/3*e^4/d^3*x*(-e^2*x^2+d^2)^(3/2)+e^4/d*x*
(-e^2*x^2+d^2)^(1/2)+d*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/2*e/d^2/x^2*(-e^2*x^2+d^2)
^(5/2)-1/2*e^3/d^2*(-e^2*x^2+d^2)^(3/2)-3/2*e^3*(-e^2*x^2+d^2)^(1/2)+3/2*e^3*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 = 0.98, size = 184, normalized size = 1.53 \begin {gather*} d e^{3} \arcsin \left (\frac {e x}{d}\right ) + \frac {3}{2} \, d e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{4} x}{d} - \frac {3}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{2 \, d^{2}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{3 \, d x} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{2 \, d^{2} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{3 \, d x^{3}} \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^4,x, algorithm="maxima")

[Out]

d*e^3*arcsin(e*x/d) + 3/2*d*e^3*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) + sqrt(-e^2*x^2 + d^2)*e^4
*x/d - 3/2*sqrt(-e^2*x^2 + d^2)*e^3 - 1/2*(-e^2*x^2 + d^2)^(3/2)*e^3/d^2 + 2/3*(-e^2*x^2 + d^2)^(3/2)*e^2/(d*x
) - 1/2*(-e^2*x^2 + d^2)^(5/2)*e/(d^2*x^2) - 1/3*(-e^2*x^2 + d^2)^(5/2)/(d*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )}{x^4} \,d x \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^4,x)

[Out]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^4, x)

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sympy [C]  time = 8.93, size = 457, normalized size = 3.81 \begin {gather*} d^{3} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + d^{2} e \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 e^{2} \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 ) - e^{3} \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 ) \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**4,x)

[Out]

d**3*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e
**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), Tru
e)) + d**2*e*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*e**2*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)) - e**3*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))

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