3.1.12 \(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=108 \[ -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d}+\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {807, 266, 47, 63, 208} \begin {gather*} \frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^6,x]

[Out]

(3*e^3*Sqrt[d^2 - e^2*x^2])/(8*x^2) - (e*(d^2 - e^2*x^2)^(3/2))/(4*x^4) - (d^2 - e^2*x^2)^(5/2)/(5*d*x^5) - (3
*e^5*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(8*d)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+e \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {1}{2} e \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {1}{8} \left (3 e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )\\ &=\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {1}{16} \left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )\\ &=\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {1}{8} \left (3 e^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 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 133, normalized size = 1.23 \begin {gather*} -\frac {8 d^6+10 d^5 e x-24 d^4 e^2 x^2-35 d^3 e^3 x^3+24 d^2 e^4 x^4+15 d e^5 x^5 \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )+25 d e^5 x^5-8 e^6 x^6}{40 d x^5 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^6,x]

[Out]

-1/40*(8*d^6 + 10*d^5*e*x - 24*d^4*e^2*x^2 - 35*d^3*e^3*x^3 + 24*d^2*e^4*x^4 + 25*d*e^5*x^5 - 8*e^6*x^6 + 15*d
*e^5*x^5*Sqrt[1 - (e^2*x^2)/d^2]*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]])/(d*x^5*Sqrt[d^2 - e^2*x^2])

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IntegrateAlgebraic [A]  time = 0.62, size = 155, normalized size = 1.44 \begin {gather*} -\frac {3 e^5 \log \left (\sqrt {d^2-e^2 x^2}+d-\sqrt {-e^2} x\right )}{8 d}+\frac {3 e^5 \log \left (-d \sqrt {d^2-e^2 x^2}+d^2+d \sqrt {-e^2} x\right )}{8 d}+\frac {\sqrt {d^2-e^2 x^2} \left (-8 d^4-10 d^3 e x+16 d^2 e^2 x^2+25 d e^3 x^3-8 e^4 x^4\right )}{40 d x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^6,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-8*d^4 - 10*d^3*e*x + 16*d^2*e^2*x^2 + 25*d*e^3*x^3 - 8*e^4*x^4))/(40*d*x^5) - (3*e^5*Lo
g[d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2*x^2]])/(8*d) + (3*e^5*Log[d^2 + d*Sqrt[-e^2]*x - d*Sqrt[d^2 - e^2*x^2]])/(
8*d)

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fricas [A]  time = 0.39, size = 98, normalized size = 0.91 \begin {gather*} \frac {15 \, e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (8 \, e^{4} x^{4} - 25 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} + 10 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{40 \, d x^{5}} \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^6,x, algorithm="fricas")

[Out]

1/40*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (8*e^4*x^4 - 25*d*e^3*x^3 - 16*d^2*e^2*x^2 + 10*d^3*e*x
+ 8*d^4)*sqrt(-e^2*x^2 + d^2))/(d*x^5)

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giac [B]  time = 0.25, size = 368, normalized size = 3.41 \begin {gather*} \frac {x^{5} {\left (\frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{10}}{x} - \frac {10 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{8}}{x^{2}} - \frac {40 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{6}}{x^{3}} + \frac {20 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{4}}{x^{4}} + 2 \, e^{12}\right )} e^{3}}{320 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d} - \frac {3 \, e^{5} \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 )}{8 \, d} - \frac {{\left (\frac {20 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{4} e^{38}}{x} - \frac {40 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{36}}{x^{2}} - \frac {10 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{4} e^{34}}{x^{3}} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{4} e^{32}}{x^{4}} + \frac {2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{4} e^{30}}{x^{5}}\right )} e^{\left (-35\right )}}{320 \, d^{5}} \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^6,x, algorithm="giac")

[Out]

1/320*x^5*(5*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^10/x - 10*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^8/x^2 - 40*(d*e + s
qrt(-x^2*e^2 + d^2)*e)^3*e^6/x^3 + 20*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^4/x^4 + 2*e^12)*e^3/((d*e + sqrt(-x^2
*e^2 + d^2)*e)^5*d) - 3/8*e^5*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d - 1/320*(20*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)*d^4*e^38/x - 40*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*e^36/x^2 - 10*(d*e + sqrt(-x^2
*e^2 + d^2)*e)^3*d^4*e^34/x^3 + 5*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^4*e^32/x^4 + 2*(d*e + sqrt(-x^2*e^2 + d^2
)*e)^5*d^4*e^30/x^5)*e^(-35)/d^5

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maple [A]  time = 0.03, size = 158, normalized size = 1.46 \begin {gather*} -\frac {3 e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}}{8 d^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5}}{8 d^{4}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{8 d^{4} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{4 d^{2} x^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 d \,x^{5}} \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^6,x)

[Out]

-1/5*(-e^2*x^2+d^2)^(5/2)/d/x^5-1/4*e/d^2/x^4*(-e^2*x^2+d^2)^(5/2)+1/8*e^3/d^4/x^2*(-e^2*x^2+d^2)^(5/2)+1/8*e^
5/d^4*(-e^2*x^2+d^2)^(3/2)+3/8*e^5/d^2*(-e^2*x^2+d^2)^(1/2)-3/8*e^5/(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 = 155, normalized size = 1.44 \begin {gather*} -\frac {3 \, e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d} + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}}{8 \, d^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}}{8 \, d^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}}{8 \, d^{4} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{4 \, d^{2} x^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, d x^{5}} \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^6,x, algorithm="maxima")

[Out]

-3/8*e^5*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d + 3/8*sqrt(-e^2*x^2 + d^2)*e^5/d^2 + 1/8*(-e^2*
x^2 + d^2)^(3/2)*e^5/d^4 + 1/8*(-e^2*x^2 + d^2)^(5/2)*e^3/(d^4*x^2) - 1/4*(-e^2*x^2 + d^2)^(5/2)*e/(d^2*x^4) -
1/5*(-e^2*x^2 + d^2)^(5/2)/(d*x^5)

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mupad [B]  time = 4.26, size = 93, normalized size = 0.86 \begin {gather*} \frac {3\,d^2\,e\,\sqrt {d^2-e^2\,x^2}}{8\,x^4}-\frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{5\,d\,x^5}-\frac {3\,e^5\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{8\,d}-\frac {5\,e\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{8\,x^4} \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^6,x)

[Out]

(3*d^2*e*(d^2 - e^2*x^2)^(1/2))/(8*x^4) - (d^2 - e^2*x^2)^(5/2)/(5*d*x^5) - (3*e^5*atanh((d^2 - e^2*x^2)^(1/2)
/d))/(8*d) - (5*e*(d^2 - e^2*x^2)^(3/2))/(8*x^4)

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sympy [C]  time = 11.36, size = 774, normalized size = 7.17 \begin {gather*} d^{3} \left (\begin {cases} \frac {3 i d^{3} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 i d e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 i e^{6} x^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {i e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{3} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 d e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 e^{6} x^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {otherwise} \end {cases}\right ) + d^{2} e \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) - d e^{2} \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 ) - e^{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 ) \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**6,x)

[Out]

d**3*Piecewise((3*I*d**3*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 +
e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*I*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d
**3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), Abs(e**2*x**2/d**2) >
1), (3*d**3*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/
(-15*d**2*x**5 + 15*e**2*x**7) + 2*e**6*x**6*sqrt(1 - e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - e*
*4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), True)) + d**2*e*Piecewise((-d**2/(4*e*x**5*
sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(8*d**2*x*sqrt(d**2/(e**2*x**2) -
1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1)
) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d
/(e*x))/(8*d**3), True)) - d*e**2*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), True)) - e**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))

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