3.2.59 \(\int \frac {x^3 (d^2-e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=171 \[ -\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^4}-\frac {d^5 x \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {d^3 (88 d-105 e x) \left (d^2-e^2 x^2\right )^{3/2}}{420 e^4} \]

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Rubi [A]  time = 0.22, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {852, 1809, 833, 780, 195, 217, 203} \begin {gather*} -\frac {d^5 x \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {d^3 (88 d-105 e x) \left (d^2-e^2 x^2\right )^{3/2}}{420 e^4}-\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

-(d^5*x*Sqrt[d^2 - e^2*x^2])/(8*e^3) - (11*d^2*x^2*(d^2 - e^2*x^2)^(3/2))/(35*e^2) + (d*x^3*(d^2 - e^2*x^2)^(3
/2))/(3*e) - (x^4*(d^2 - e^2*x^2)^(3/2))/7 - (d^3*(88*d - 105*e*x)*(d^2 - e^2*x^2)^(3/2))/(420*e^4) - (d^7*Arc
Tan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^4)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 780

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

Rule 833

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

Rule 852

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int x^3 (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx\\ &=-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {\int x^3 \left (-11 d^2 e^2+14 d e^3 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{7 e^2}\\ &=\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}+\frac {\int x^2 \left (-42 d^3 e^3+66 d^2 e^4 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{42 e^4}\\ &=-\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {\int x \left (-132 d^4 e^4+210 d^3 e^5 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{210 e^6}\\ &=-\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {d^3 (88 d-105 e x) \left (d^2-e^2 x^2\right )^{3/2}}{420 e^4}-\frac {d^5 \int \sqrt {d^2-e^2 x^2} \, dx}{4 e^3}\\ &=-\frac {d^5 x \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {d^3 (88 d-105 e x) \left (d^2-e^2 x^2\right )^{3/2}}{420 e^4}-\frac {d^7 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^3}\\ &=-\frac {d^5 x \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {d^3 (88 d-105 e x) \left (d^2-e^2 x^2\right )^{3/2}}{420 e^4}-\frac {d^7 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}\\ &=-\frac {d^5 x \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {11 d^2 x^2 \left (d^2-e^2 x^2\right )^{3/2}}{35 e^2}+\frac {d x^3 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {1}{7} x^4 \left (d^2-e^2 x^2\right )^{3/2}-\frac {d^3 (88 d-105 e x) \left (d^2-e^2 x^2\right )^{3/2}}{420 e^4}-\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^4}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 113, normalized size = 0.66 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-176 d^6+105 d^5 e x-88 d^4 e^2 x^2+70 d^3 e^3 x^3+144 d^2 e^4 x^4-280 d e^5 x^5+120 e^6 x^6\right )-105 d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{840 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-176*d^6 + 105*d^5*e*x - 88*d^4*e^2*x^2 + 70*d^3*e^3*x^3 + 144*d^2*e^4*x^4 - 280*d*e^5*x
^5 + 120*e^6*x^6) - 105*d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(840*e^4)

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IntegrateAlgebraic [A]  time = 0.65, size = 136, normalized size = 0.80 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-176 d^6+105 d^5 e x-88 d^4 e^2 x^2+70 d^3 e^3 x^3+144 d^2 e^4 x^4-280 d e^5 x^5+120 e^6 x^6\right )}{840 e^4}-\frac {d^7 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-176*d^6 + 105*d^5*e*x - 88*d^4*e^2*x^2 + 70*d^3*e^3*x^3 + 144*d^2*e^4*x^4 - 280*d*e^5*x
^5 + 120*e^6*x^6))/(840*e^4) - (d^7*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*e^5)

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fricas [A]  time = 0.41, size = 116, normalized size = 0.68 \begin {gather*} \frac {210 \, d^{7} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (120 \, e^{6} x^{6} - 280 \, d e^{5} x^{5} + 144 \, d^{2} e^{4} x^{4} + 70 \, d^{3} e^{3} x^{3} - 88 \, d^{4} e^{2} x^{2} + 105 \, d^{5} e x - 176 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{840 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

1/840*(210*d^7*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (120*e^6*x^6 - 280*d*e^5*x^5 + 144*d^2*e^4*x^4 + 70
*d^3*e^3*x^3 - 88*d^4*e^2*x^2 + 105*d^5*e*x - 176*d^6)*sqrt(-e^2*x^2 + d^2))/e^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.02, size = 327, normalized size = 1.91 \begin {gather*} \frac {d^{7} \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 )}{2 \sqrt {e^{2}}\, e^{3}}-\frac {5 d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{3}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} x}{8 e^{3}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{5} x}{2 e^{3}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} x}{12 e^{3}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{3} x}{3 e^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d x}{3 e^{3}}+\frac {4 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} d^{2}}{15 e^{4}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} d^{2}}{3 \left (x +\frac {d}{e}\right )^{2} e^{6}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{7 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x)

[Out]

-1/7/e^4*(-e^2*x^2+d^2)^(7/2)-1/3*d/e^3*x*(-e^2*x^2+d^2)^(5/2)-5/12/e^3*d^3*x*(-e^2*x^2+d^2)^(3/2)-5/8*d^5*x*(
-e^2*x^2+d^2)^(1/2)/e^3-5/8/e^3*d^7/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+4/15/e^4*d^2*(2*(x+
d/e)*d*e-(x+d/e)^2*e^2)^(5/2)+1/3/e^3*d^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+1/2/e^3*d^5*(2*(x+d/e)*d*e-(x+
d/e)^2*e^2)^(1/2)*x+1/2/e^3*d^7/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)-1/3*d^2/
e^6/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)

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maxima [C]  time = 1.04, size = 251, normalized size = 1.47 \begin {gather*} -\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{4 \, {\left (e^{5} x + d e^{4}\right )}} - \frac {i \, d^{7} \arcsin \left (\frac {e x}{d} + 2\right )}{2 \, e^{4}} - \frac {5 \, d^{7} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{4}} + \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5} x}{2 \, e^{3}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5} x}{8 \, e^{3}} + \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{6}}{e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} x}{3 \, e^{3}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}}{12 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x}{3 \, e^{3}} + \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{5 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{7 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="maxima")

[Out]

-1/4*(-e^2*x^2 + d^2)^(5/2)*d^3/(e^5*x + d*e^4) - 1/2*I*d^7*arcsin(e*x/d + 2)/e^4 - 5/8*d^7*arcsin(e*x/d)/e^4
+ 1/2*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^5*x/e^3 - 5/8*sqrt(-e^2*x^2 + d^2)*d^5*x/e^3 + sqrt(e^2*x^2 + 4*d*e*x
+ 3*d^2)*d^6/e^4 + 1/3*(-e^2*x^2 + d^2)^(3/2)*d^3*x/e^3 - 5/12*(-e^2*x^2 + d^2)^(3/2)*d^4/e^4 - 1/3*(-e^2*x^2
+ d^2)^(5/2)*d*x/e^3 + 3/5*(-e^2*x^2 + d^2)^(5/2)*d^2/e^4 - 1/7*(-e^2*x^2 + d^2)^(7/2)/e^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x)

[Out]

int((x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2, x)

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sympy [A]  time = 12.03, size = 450, normalized size = 2.63 \begin {gather*} d^{2} \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{5}} + \frac {i d^{5} x}{16 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{3}}{48 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d x^{5}}{24 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{5}} - \frac {d^{5} x}{16 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{3}}{48 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d x^{5}}{24 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {8 d^{6} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{6}} - \frac {4 d^{4} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{4}} - \frac {d^{2} x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{35 e^{2}} + \frac {x^{6} \sqrt {d^{2} - e^{2} x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {x^{6} \sqrt {d^{2}}}{6} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**2,x)

[Out]

d**2*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*s
qrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) - 2*d*e*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**
5) + I*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x*
*5/(24*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d*
*6*asin(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/d
**2)) + 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x**2/d**2)), True)) + e**2*Piece
wise((-8*d**6*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4*x**2*sqrt(d**2 - e**2*x**2)/(105*e**4) - d**2*x**4*sq
rt(d**2 - e**2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x**2)/7, Ne(e, 0)), (x**6*sqrt(d**2)/6, True))

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