3.1.20 \(\int \frac {x^6 (d+e x)}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {16 \sqrt {d^2-e^2 x^2}}{5 e^7}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^7}+\frac {x (5 d+8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {819, 641, 217, 203} \begin {gather*} \frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (5 d+8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}+\frac {16 \sqrt {d^2-e^2 x^2}}{5 e^7}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(x^5*(d + e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (x^3*(5*d + 6*e*x))/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (x*(5*d +
8*e*x))/(5*e^6*Sqrt[d^2 - e^2*x^2]) + (16*Sqrt[d^2 - e^2*x^2])/(5*e^7) - (d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]
)/e^7

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 641

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

Rule 819

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 + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

\begin {align*} \int \frac {x^6 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x^4 \left (5 d^3+6 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {x^2 \left (15 d^5+24 d^4 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4 e^4}\\ &=\frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (5 d+8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {15 d^7+48 d^6 e x}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^6 e^6}\\ &=\frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (5 d+8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}+\frac {16 \sqrt {d^2-e^2 x^2}}{5 e^7}-\frac {d \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^6}\\ &=\frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (5 d+8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}+\frac {16 \sqrt {d^2-e^2 x^2}}{5 e^7}-\frac {d \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}\\ &=\frac {x^5 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d+6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (5 d+8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}+\frac {16 \sqrt {d^2-e^2 x^2}}{5 e^7}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^7}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 142, normalized size = 0.97 \begin {gather*} \frac {48 d^5-33 d^4 e x-87 d^3 e^2 x^2+52 d^2 e^3 x^3-15 d (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2} \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+38 d e^4 x^4-15 e^5 x^5}{15 e^7 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(48*d^5 - 33*d^4*e*x - 87*d^3*e^2*x^2 + 52*d^2*e^3*x^3 + 38*d*e^4*x^4 - 15*e^5*x^5 - 15*d*(d - e*x)^2*(d + e*x
)*Sqrt[d^2 - e^2*x^2]*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(15*e^7*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])

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IntegrateAlgebraic [A]  time = 0.73, size = 137, normalized size = 0.93 \begin {gather*} -\frac {d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^8}-\frac {\sqrt {d^2-e^2 x^2} \left (48 d^5-33 d^4 e x-87 d^3 e^2 x^2+52 d^2 e^3 x^3+38 d e^4 x^4-15 e^5 x^5\right )}{15 e^7 (e x-d)^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^6*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

-1/15*(Sqrt[d^2 - e^2*x^2]*(48*d^5 - 33*d^4*e*x - 87*d^3*e^2*x^2 + 52*d^2*e^3*x^3 + 38*d*e^4*x^4 - 15*e^5*x^5)
)/(e^7*(-d + e*x)^3*(d + e*x)^2) - (d*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^8

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fricas [B]  time = 0.42, size = 263, normalized size = 1.79 \begin {gather*} \frac {48 \, d e^{5} x^{5} - 48 \, d^{2} e^{4} x^{4} - 96 \, d^{3} e^{3} x^{3} + 96 \, d^{4} e^{2} x^{2} + 48 \, d^{5} e x - 48 \, d^{6} + 30 \, {\left (d e^{5} x^{5} - d^{2} e^{4} x^{4} - 2 \, d^{3} e^{3} x^{3} + 2 \, d^{4} e^{2} x^{2} + d^{5} e x - d^{6}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (15 \, e^{5} x^{5} - 38 \, d e^{4} x^{4} - 52 \, d^{2} e^{3} x^{3} + 87 \, d^{3} e^{2} x^{2} + 33 \, d^{4} e x - 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{12} x^{5} - d e^{11} x^{4} - 2 \, d^{2} e^{10} x^{3} + 2 \, d^{3} e^{9} x^{2} + d^{4} e^{8} x - d^{5} e^{7}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(48*d*e^5*x^5 - 48*d^2*e^4*x^4 - 96*d^3*e^3*x^3 + 96*d^4*e^2*x^2 + 48*d^5*e*x - 48*d^6 + 30*(d*e^5*x^5 -
d^2*e^4*x^4 - 2*d^3*e^3*x^3 + 2*d^4*e^2*x^2 + d^5*e*x - d^6)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e
^5*x^5 - 38*d*e^4*x^4 - 52*d^2*e^3*x^3 + 87*d^3*e^2*x^2 + 33*d^4*e*x - 48*d^5)*sqrt(-e^2*x^2 + d^2))/(e^12*x^5
- d*e^11*x^4 - 2*d^2*e^10*x^3 + 2*d^3*e^9*x^2 + d^4*e^8*x - d^5*e^7)

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giac [A]  time = 0.27, size = 109, normalized size = 0.74 \begin {gather*} -d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-7\right )} \mathrm {sgn}\relax (d) - \frac {{\left (48 \, d^{6} e^{\left (-7\right )} + {\left (15 \, d^{5} e^{\left (-6\right )} - {\left (120 \, d^{4} e^{\left (-5\right )} + {\left (35 \, d^{3} e^{\left (-4\right )} - {\left (90 \, d^{2} e^{\left (-3\right )} - {\left (15 \, x e^{\left (-1\right )} - 23 \, d e^{\left (-2\right )}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-d*arcsin(x*e/d)*e^(-7)*sgn(d) - 1/15*(48*d^6*e^(-7) + (15*d^5*e^(-6) - (120*d^4*e^(-5) + (35*d^3*e^(-4) - (90
*d^2*e^(-3) - (15*x*e^(-1) - 23*d*e^(-2))*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

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maple [A]  time = 0.02, size = 195, normalized size = 1.33 \begin {gather*} -\frac {x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {d \,x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}+\frac {6 d^{2} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {8 d^{4} x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}-\frac {d \,x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}+\frac {16 d^{6}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{7}}+\frac {d x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{6}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x)

[Out]

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

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maxima [B]  time = 1.03, size = 278, normalized size = 1.89 \begin {gather*} \frac {1}{15} \, d x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {d x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )}}{3 \, e^{2}} + \frac {6 \, d^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {8 \, d^{4} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}} + \frac {16 \, d^{6}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{7}} + \frac {4 \, d^{3} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6}} - \frac {7 \, d x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{6}} - \frac {d \arcsin \left (\frac {e x}{d}\right )}{e^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*d*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 + d
^2)^(5/2)*e^6)) - x^6/((-e^2*x^2 + d^2)^(5/2)*e) - 1/3*d*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*
x^2 + d^2)^(3/2)*e^4))/e^2 + 6*d^2*x^4/((-e^2*x^2 + d^2)^(5/2)*e^3) - 8*d^4*x^2/((-e^2*x^2 + d^2)^(5/2)*e^5) +
16/5*d^6/((-e^2*x^2 + d^2)^(5/2)*e^7) + 4/15*d^3*x/((-e^2*x^2 + d^2)^(3/2)*e^6) - 7/15*d*x/(sqrt(-e^2*x^2 + d
^2)*e^6) - d*arcsin(e*x/d)/e^7

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^6\,\left (d+e\,x\right )}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^6*(d + e*x))/(d^2 - e^2*x^2)^(7/2), x)

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sympy [C]  time = 62.08, size = 1821, normalized size = 12.39

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(e*x+d)/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*Piecewise((-30*I*d**5*sqrt(-1 + e**2*x**2/d**2)*acosh(e*x/d)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d
**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 15*pi*d**5*sqrt(-1 + e*
*2*x**2/d**2)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e*
*11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 30*I*d**4*e*x/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x*
*2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 60*I*d**3*e**2*x**2*sqrt(-1 + e**2
*x**2/d**2)*acosh(e*x/d)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2
) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 30*pi*d**3*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)/(-30*d**5*e**7
*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2
/d**2)) - 70*I*d**2*e**3*x**3/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2
/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 30*I*d*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)*acosh(e*x/d)/
(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(
-1 + e**2*x**2/d**2)) + 15*pi*d*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) +
60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 46*I*e**5*x**5/(-3
0*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1
+ e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (-15*d**5*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)/(15*d**5*e**7*sqr
t(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2))
+ 15*d**4*e*x/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**1
1*x**4*sqrt(1 - e**2*x**2/d**2)) + 30*d**3*e**2*x**2*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)/(15*d**5*e**7*sqrt(1
- e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)) -
35*d**2*e**3*x**3/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e
**11*x**4*sqrt(1 - e**2*x**2/d**2)) - 15*d*e**4*x**4*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)/(15*d**5*e**7*sqrt(1
- e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)) +
23*e**5*x**5/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*
x**4*sqrt(1 - e**2*x**2/d**2)), True)) + e*Piecewise((16*d**6/(5*d**4*e**8*sqrt(d**2 - e**2*x**2) - 10*d**2*e*
*10*x**2*sqrt(d**2 - e**2*x**2) + 5*e**12*x**4*sqrt(d**2 - e**2*x**2)) - 40*d**4*e**2*x**2/(5*d**4*e**8*sqrt(d
**2 - e**2*x**2) - 10*d**2*e**10*x**2*sqrt(d**2 - e**2*x**2) + 5*e**12*x**4*sqrt(d**2 - e**2*x**2)) + 30*d**2*
e**4*x**4/(5*d**4*e**8*sqrt(d**2 - e**2*x**2) - 10*d**2*e**10*x**2*sqrt(d**2 - e**2*x**2) + 5*e**12*x**4*sqrt(
d**2 - e**2*x**2)) - 5*e**6*x**6/(5*d**4*e**8*sqrt(d**2 - e**2*x**2) - 10*d**2*e**10*x**2*sqrt(d**2 - e**2*x**
2) + 5*e**12*x**4*sqrt(d**2 - e**2*x**2)), Ne(e, 0)), (x**8/(8*(d**2)**(7/2)), True))

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