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

Optimal. Leaf size=142 \[ \frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}+\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2} \]

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Rubi [A]  time = 0.18, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^4*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) + (2*d*x^2*(d^2 - e^2*x^2)^(3/2))/(5*e) - (x^3*(d^2 - e^2*x^2)^(3/2))/6
+ (d^2*(32*d - 45*e*x)*(d^2 - e^2*x^2)^(3/2))/(120*e^3) + (3*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

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^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int x^2 (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx\\ &=-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}-\frac {\int x^2 \left (-9 d^2 e^2+12 d e^3 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{6 e^2}\\ &=\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {\int x \left (-24 d^3 e^3+45 d^2 e^4 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{30 e^4}\\ &=\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^4\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{8 e^2}\\ &=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^2}\\ &=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^6\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}\\ &=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 102, normalized size = 0.72 \begin {gather*} \frac {45 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (64 d^5-45 d^4 e x+32 d^3 e^2 x^2+50 d^2 e^3 x^3-96 d e^4 x^4+40 e^5 x^5\right )}{240 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(64*d^5 - 45*d^4*e*x + 32*d^3*e^2*x^2 + 50*d^2*e^3*x^3 - 96*d*e^4*x^4 + 40*e^5*x^5) + 45*
d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(240*e^3)

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IntegrateAlgebraic [A]  time = 0.57, size = 125, normalized size = 0.88 \begin {gather*} \frac {3 d^6 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{16 e^4}+\frac {\sqrt {d^2-e^2 x^2} \left (64 d^5-45 d^4 e x+32 d^3 e^2 x^2+50 d^2 e^3 x^3-96 d e^4 x^4+40 e^5 x^5\right )}{240 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(64*d^5 - 45*d^4*e*x + 32*d^3*e^2*x^2 + 50*d^2*e^3*x^3 - 96*d*e^4*x^4 + 40*e^5*x^5))/(240
*e^3) + (3*d^6*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(16*e^4)

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fricas [A]  time = 0.41, size = 106, normalized size = 0.75 \begin {gather*} -\frac {90 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (40 \, e^{5} x^{5} - 96 \, d e^{4} x^{4} + 50 \, d^{2} e^{3} x^{3} + 32 \, d^{3} e^{2} x^{2} - 45 \, d^{4} e x + 64 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(90*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (40*e^5*x^5 - 96*d*e^4*x^4 + 50*d^2*e^3*x^3 + 32*d^
3*e^2*x^2 - 45*d^4*e*x + 64*d^5)*sqrt(-e^2*x^2 + d^2))/e^3

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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^2*(-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 = 0.01, size = 303, normalized size = 2.13 \begin {gather*} -\frac {d^{6} \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}}\, e^{2}}+\frac {5 d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{2}}+\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} x}{16 e^{2}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4} x}{8 e^{2}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2} x}{24 e^{2}}-\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^{2} x}{12 e^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x}{6 e^{2}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} d}{15 e^{3}}+\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}{3 \left (x +\frac {d}{e}\right )^{2} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(-e^2*x^2+d^2)^(5/2)/e^2+5/24/e^2*d^2*x*(-e^2*x^2+d^2)^(3/2)+5/16*d^4*x*(-e^2*x^2+d^2)^(1/2)/e^2+5/16/e^
2*d^6/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/15*d/e^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)-1/
12*d^2/e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-1/8*d^4/e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-1/8*d^6/e^2
/(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/e^5/(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 = 1.03, size = 230, normalized size = 1.62 \begin {gather*} \frac {i \, d^{6} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} + \frac {5 \, d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{4 \, {\left (e^{4} x + d e^{3}\right )}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4} x}{8 \, e^{2}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e^{2}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5}}{4 \, e^{3}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e^{2}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{12 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*I*d^6*arcsin(e*x/d + 2)/e^3 + 5/16*d^6*arcsin(e*x/d)/e^3 + 1/4*(-e^2*x^2 + d^2)^(5/2)*d^2/(e^4*x + d*e^3)
- 1/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4*x/e^2 + 5/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e^2 - 1/4*sqrt(e^2*x^2 + 4*d
*e*x + 3*d^2)*d^5/e^3 - 7/24*(-e^2*x^2 + d^2)^(3/2)*d^2*x/e^2 + 5/12*(-e^2*x^2 + d^2)^(3/2)*d^3/e^3 + 1/6*(-e^
2*x^2 + d^2)^(5/2)*x/e^2 - 2/5*(-e^2*x^2 + d^2)^(5/2)*d/e^3

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

[Out]

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

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sympy [C]  time = 14.45, size = 541, normalized size = 3.81 \begin {gather*} d^{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 ) - 2 d e \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 ) + e^{2} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**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*sq
rt(-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)) - 2*d*e*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)) + e**
2*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))

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