3.805 \(\int \frac {(d^2-e^2 x^2)^{7/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=142 \[ \frac {7 d^2 \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {7 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}+\frac {7}{8} d^3 x \sqrt {d^2-e^2 x^2} \]

[Out]

7/12*d^2*(-e^2*x^2+d^2)^(3/2)/e+7/20*d*(-e*x+d)*(-e^2*x^2+d^2)^(3/2)/e+1/5*(-e*x+d)^2*(-e^2*x^2+d^2)^(3/2)/e+7
/8*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e+7/8*d^3*x*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {655, 671, 641, 195, 217, 203} \[ \frac {7}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {7 d^2 \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {7 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^3,x]

[Out]

(7*d^3*x*Sqrt[d^2 - e^2*x^2])/8 + (7*d^2*(d^2 - e^2*x^2)^(3/2))/(12*e) + (7*d*(d - e*x)*(d^2 - e^2*x^2)^(3/2))
/(20*e) + ((d - e*x)^2*(d^2 - e^2*x^2)^(3/2))/(5*e) + (7*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

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 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 655

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

Rule 671

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

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^3} \, dx &=\int (d-e x)^3 \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {1}{5} (7 d) \int (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {1}{4} \left (7 d^2\right ) \int (d-e x) \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {7 d^2 \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {1}{4} \left (7 d^3\right ) \int \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {7}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {7 d^2 \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {1}{8} \left (7 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {7}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {7 d^2 \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {1}{8} \left (7 d^5\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 {7}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {7 d^2 \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {7 d (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{20 e}+\frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {7 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 91, normalized size = 0.64 \[ \frac {105 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (136 d^4+15 d^3 e x-112 d^2 e^2 x^2+90 d e^3 x^3-24 e^4 x^4\right )}{120 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^3,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(136*d^4 + 15*d^3*e*x - 112*d^2*e^2*x^2 + 90*d*e^3*x^3 - 24*e^4*x^4) + 105*d^5*ArcTan[(e*
x)/Sqrt[d^2 - e^2*x^2]])/(120*e)

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fricas [A]  time = 0.86, size = 94, normalized size = 0.66 \[ -\frac {210 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (24 \, e^{4} x^{4} - 90 \, d e^{3} x^{3} + 112 \, d^{2} e^{2} x^{2} - 15 \, d^{3} e x - 136 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

-1/120*(210*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (24*e^4*x^4 - 90*d*e^3*x^3 + 112*d^2*e^2*x^2 - 15*
d^3*e*x - 136*d^4)*sqrt(-e^2*x^2 + d^2))/e

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-2*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^10*exp(2)^2-28*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^2*exp(1)^10*exp(2)^2-12*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2
)^3+37*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)^3+19*d^5*(-1/2*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^4+2*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(1)^6*exp(2)^4+6*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^
3*exp(1)^4*exp(2)^5-d^5*exp(1)^8*exp(2)^3-5*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*
exp(1)^4*exp(2)^5-14*d^5*exp(1)^6*exp(2)^4-11*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^
3*exp(2)^7+19*d^5*exp(1)^4*exp(2)^5-4*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)
^7-4*d^5*exp(2)^7-2*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^12*exp(2)+37/2*d^
5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^7/x/exp(2)-13*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp
(1))*exp(1)^4*exp(2)^5/x/exp(2)-57/2*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^4/x/exp(2
)+22*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^3/x/exp(2)+d^5*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))*exp(1)^10*exp(2)^2/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*
exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^2/(-exp(1)^11-exp(1)^7*exp(2)^2-2*exp(1)^9*exp(2)
)+1/2*(16*d^5*exp(1)^10*exp(2)^2-114*d^5*exp(1)^8*exp(2)^3+4*d^5*exp(1)^6*exp(2)^4+186*d^5*exp(1)^4*exp(2)^5-9
2*d^5*exp(2)^7)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqr
t(-exp(1)^4+exp(2)^2)/(exp(1)^13+exp(1)^9*exp(2)^2+2*exp(1)^11*exp(2))+7/8*d^5*sign(d)*asin(x*exp(2)/d/exp(1))
/exp(1)+2*((((-192*exp(1)^9*1/1920/exp(1)^6*x+720*exp(1)^8*d*1/1920/exp(1)^6)*x-896*exp(1)^7*d^2*1/1920/exp(1)
^6)*x+120*exp(1)^6*d^3*1/1920/exp(1)^6)*x+1088*exp(1)^5*d^4*1/1920/exp(1)^6)*sqrt(d^2-x^2*exp(2))

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maple [B]  time = 0.05, size = 274, normalized size = 1.93 \[ \frac {7 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}+\frac {7 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3} x}{8}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d x}{12}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} x}{15 d}+\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{5 d^{2} e}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{3 \left (x +\frac {d}{e}\right )^{3} d \,e^{4}}+\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{5 \left (x +\frac {d}{e}\right )^{2} d^{2} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(7/2)/(e*x+d)^3,x)

[Out]

1/3/e^4/d/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)+2/5/e^3/d^2/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9
/2)+2/5/e/d^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+7/15/d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)*x+7/12*d*(2*(x+d/
e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+7/8*d^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+7/8*d^5/(e^2)^(1/2)*arctan((e^2)^(
1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)

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maxima [C]  time = 3.07, size = 160, normalized size = 1.13 \[ -\frac {7 i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e} + \frac {7}{8} \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x + \frac {7 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{4 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{5 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{20 \, {\left (e^{2} x + d e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{12 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^3,x, algorithm="maxima")

[Out]

-7/8*I*d^5*arcsin(e*x/d + 2)/e + 7/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^3*x + 7/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^
2)*d^4/e + 1/5*(-e^2*x^2 + d^2)^(7/2)/(e^3*x^2 + 2*d*e^2*x + d^2*e) + 7/20*(-e^2*x^2 + d^2)^(5/2)*d/(e^2*x + d
*e) + 7/12*(-e^2*x^2 + d^2)^(3/2)*d^2/e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{7/2}}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(7/2)/(d + e*x)^3,x)

[Out]

int((d^2 - e^2*x^2)^(7/2)/(d + e*x)^3, x)

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sympy [C]  time = 18.06, size = 439, normalized size = 3.09 \[ d^{3} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) - 3 d^{2} e \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - e^{3} \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(7/2)/(e*x+d)**3,x)

[Out]

d**3*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 +
e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) -
3*d**2*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + 3*d*e**2*
Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1
+ e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**4*asin(e*x/d)
/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*
sqrt(1 - e**2*x**2/d**2)), True)) - e**3*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*sqrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True))

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