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

Optimal. Leaf size=172 \[ -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}+\frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^7 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3} \]

[Out]

-1/24*e^3*(-e^2*x^2+d^2)^(3/2)/d^2/x^4-1/7*(-e^2*x^2+d^2)^(5/2)/d/x^7-1/6*e*(-e^2*x^2+d^2)^(5/2)/d^2/x^6-2/35*
e^2*(-e^2*x^2+d^2)^(5/2)/d^3/x^5-1/16*e^7*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^3+1/16*e^5*(-e^2*x^2+d^2)^(1/2)/d^
2/x^2

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Rubi [A]  time = 0.13, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {835, 807, 266, 47, 63, 208} \[ \frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e^7 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^8,x]

[Out]

(e^5*Sqrt[d^2 - e^2*x^2])/(16*d^2*x^2) - (e^3*(d^2 - e^2*x^2)^(3/2))/(24*d^2*x^4) - (d^2 - e^2*x^2)^(5/2)/(7*d
*x^7) - (e*(d^2 - e^2*x^2)^(5/2))/(6*d^2*x^6) - (2*e^2*(d^2 - e^2*x^2)^(5/2))/(35*d^3*x^5) - (e^7*ArcTanh[Sqrt
[d^2 - e^2*x^2]/d])/(16*d^3)

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]

Rule 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {\int \frac {\left (-7 d^2 e-2 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^7} \, dx}{7 d^2}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}+\frac {\int \frac {\left (12 d^3 e^2+7 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx}{42 d^4}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}+\frac {e^3 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx}{6 d^2}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}+\frac {e^3 \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )}{12 d^2}\\ &=-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^5 \operatorname {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{16 d^2}\\ &=\frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}+\frac {e^7 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{32 d^2}\\ &=\frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^5 \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 )}{16 d^2}\\ &=\frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^7 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 72, normalized size = 0.42 \[ -\frac {\left (d^2-e^2 x^2\right )^{5/2} \left (5 d^7+2 d^5 e^2 x^2+7 e^7 x^7 \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};1-\frac {e^2 x^2}{d^2}\right )\right )}{35 d^8 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^8,x]

[Out]

-1/35*((d^2 - e^2*x^2)^(5/2)*(5*d^7 + 2*d^5*e^2*x^2 + 7*e^7*x^7*Hypergeometric2F1[5/2, 4, 7/2, 1 - (e^2*x^2)/d
^2]))/(d^8*x^7)

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fricas [A]  time = 0.97, size = 120, normalized size = 0.70 \[ \frac {105 \, e^{7} x^{7} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (96 \, e^{6} x^{6} + 105 \, d e^{5} x^{5} + 48 \, d^{2} e^{4} x^{4} - 490 \, d^{3} e^{3} x^{3} - 384 \, d^{4} e^{2} x^{2} + 280 \, d^{5} e x + 240 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{1680 \, d^{3} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

1/1680*(105*e^7*x^7*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (96*e^6*x^6 + 105*d*e^5*x^5 + 48*d^2*e^4*x^4 - 490*d^
3*e^3*x^3 - 384*d^4*e^2*x^2 + 280*d^5*e*x + 240*d^6)*sqrt(-e^2*x^2 + d^2))/(d^3*x^7)

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giac [B]  time = 0.26, size = 494, normalized size = 2.87 \[ \frac {x^{7} {\left (\frac {35 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{14}}{x} - \frac {21 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{12}}{x^{2}} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{10}}{x^{3}} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{8}}{x^{4}} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{6}}{x^{5}} + \frac {315 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} e^{4}}{x^{6}} + 15 \, e^{16}\right )} e^{5}}{13440 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} d^{3}} - \frac {e^{7} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{16 \, d^{3}} - \frac {{\left (\frac {315 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{18} e^{68}}{x} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{18} e^{66}}{x^{2}} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{18} e^{64}}{x^{3}} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{18} e^{62}}{x^{4}} - \frac {21 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{18} e^{60}}{x^{5}} + \frac {35 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} d^{18} e^{58}}{x^{6}} + \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} d^{18} e^{56}}{x^{7}}\right )} e^{\left (-63\right )}}{13440 \, d^{21}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

1/13440*x^7*(35*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^14/x - 21*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^12/x^2 - 105*(d*
e + sqrt(-x^2*e^2 + d^2)*e)^3*e^10/x^3 - 105*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^8/x^4 - 105*(d*e + sqrt(-x^2*e
^2 + d^2)*e)^5*e^6/x^5 + 315*(d*e + sqrt(-x^2*e^2 + d^2)*e)^6*e^4/x^6 + 15*e^16)*e^5/((d*e + sqrt(-x^2*e^2 + d
^2)*e)^7*d^3) - 1/16*e^7*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^3 - 1/13440*(315*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)*d^18*e^68/x - 105*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^18*e^66/x^2 - 105*(d*e + sqrt(
-x^2*e^2 + d^2)*e)^3*d^18*e^64/x^3 - 105*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^18*e^62/x^4 - 21*(d*e + sqrt(-x^2*
e^2 + d^2)*e)^5*d^18*e^60/x^5 + 35*(d*e + sqrt(-x^2*e^2 + d^2)*e)^6*d^18*e^58/x^6 + 15*(d*e + sqrt(-x^2*e^2 +
d^2)*e)^7*d^18*e^56/x^7)*e^(-63)/d^21

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maple [A]  time = 0.04, size = 211, normalized size = 1.23 \[ -\frac {e^{7} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{16 \sqrt {d^{2}}\, d^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{7}}{16 d^{4}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{7}}{48 d^{6}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}{48 d^{6} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{24 d^{4} x^{4}}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}{35 d^{3} x^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{6 d^{2} x^{6}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{7 d \,x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x)

[Out]

-1/7*(-e^2*x^2+d^2)^(5/2)/d/x^7-2/35*e^2*(-e^2*x^2+d^2)^(5/2)/d^3/x^5-1/6*e*(-e^2*x^2+d^2)^(5/2)/d^2/x^6-1/24*
e^3/d^4/x^4*(-e^2*x^2+d^2)^(5/2)+1/48*e^5/d^6/x^2*(-e^2*x^2+d^2)^(5/2)+1/48*e^7/d^6*(-e^2*x^2+d^2)^(3/2)+1/16*
e^7/d^4*(-e^2*x^2+d^2)^(1/2)-1/16*e^7/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 = 0.99, size = 205, normalized size = 1.19 \[ -\frac {e^{7} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{16 \, d^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{7}}{16 \, d^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7}}{48 \, d^{6}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}}{48 \, d^{6} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}}{24 \, d^{4} x^{4}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}}{35 \, d^{3} x^{5}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{6 \, d^{2} x^{6}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{7 \, d x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

-1/16*e^7*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^3 + 1/16*sqrt(-e^2*x^2 + d^2)*e^7/d^4 + 1/48*(
-e^2*x^2 + d^2)^(3/2)*e^7/d^6 + 1/48*(-e^2*x^2 + d^2)^(5/2)*e^5/(d^6*x^2) - 1/24*(-e^2*x^2 + d^2)^(5/2)*e^3/(d
^4*x^4) - 2/35*(-e^2*x^2 + d^2)^(5/2)*e^2/(d^3*x^5) - 1/6*(-e^2*x^2 + d^2)^(5/2)*e/(d^2*x^6) - 1/7*(-e^2*x^2 +
d^2)^(5/2)/(d*x^7)

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mupad [B]  time = 5.33, size = 192, normalized size = 1.12 \[ \frac {8\,d\,e^2\,\sqrt {d^2-e^2\,x^2}}{35\,x^5}-\frac {d^3\,\sqrt {d^2-e^2\,x^2}}{7\,x^7}-\frac {e^4\,\sqrt {d^2-e^2\,x^2}}{35\,d\,x^3}-\frac {2\,e^6\,\sqrt {d^2-e^2\,x^2}}{35\,d^3\,x}-\frac {e\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{6\,x^6}+\frac {d^2\,e\,\sqrt {d^2-e^2\,x^2}}{16\,x^6}-\frac {e\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{16\,d^2\,x^6}+\frac {e^7\,\mathrm {atan}\left (\frac {\sqrt {d^2-e^2\,x^2}\,1{}\mathrm {i}}{d}\right )\,1{}\mathrm {i}}{16\,d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^8,x)

[Out]

(e^7*atan(((d^2 - e^2*x^2)^(1/2)*1i)/d)*1i)/(16*d^3) - (d^3*(d^2 - e^2*x^2)^(1/2))/(7*x^7) - (e*(d^2 - e^2*x^2
)^(3/2))/(6*x^6) - (e^4*(d^2 - e^2*x^2)^(1/2))/(35*d*x^3) - (2*e^6*(d^2 - e^2*x^2)^(1/2))/(35*d^3*x) + (8*d*e^
2*(d^2 - e^2*x^2)^(1/2))/(35*x^5) + (d^2*e*(d^2 - e^2*x^2)^(1/2))/(16*x^6) - (e*(d^2 - e^2*x^2)^(5/2))/(16*d^2
*x^6)

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sympy [C]  time = 16.62, size = 1037, normalized size = 6.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x**8,x)

[Out]

d**3*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(7*x**6) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(35*d**2*x**4) + 4*e*
*5*sqrt(d**2/(e**2*x**2) - 1)/(105*d**4*x**2) + 8*e**7*sqrt(d**2/(e**2*x**2) - 1)/(105*d**6), Abs(d**2/(e**2*x
**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(7*x**6) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(35*d**2*x**4) + 4
*I*e**5*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**4*x**2) + 8*I*e**7*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**6), True))
+ d**2*e*Piecewise((-d**2/(6*e*x**7*sqrt(d**2/(e**2*x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e
**3/(48*d**2*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) + e**6*acosh(d/(e*
x))/(16*d**5), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(6*e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x**5*sqr
t(-d**2/(e**2*x**2) + 1)) - I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(16*d**4*x*sqrt(-d**2/(
e**2*x**2) + 1)) - I*e**6*asin(d/(e*x))/(16*d**5), True)) - d*e**2*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*sqr
t(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)) - e**3*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))

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