3.3.95 \(\int e^{\tanh ^{-1}(a x)} x^3 (c-a c x)^2 \, dx\) [295]

Optimal. Leaf size=124 \[ -\frac {c^2 x \sqrt {1-a^2 x^2}}{16 a^3}-\frac {c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a^2}+\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {c^2 (16-15 a x) \left (1-a^2 x^2\right )^{3/2}}{120 a^4}-\frac {c^2 \text {ArcSin}(a x)}{16 a^4} \]

[Out]

-1/5*c^2*x^2*(-a^2*x^2+1)^(3/2)/a^2+1/6*c^2*x^3*(-a^2*x^2+1)^(3/2)/a-1/120*c^2*(-15*a*x+16)*(-a^2*x^2+1)^(3/2)
/a^4-1/16*c^2*arcsin(a*x)/a^4-1/16*c^2*x*(-a^2*x^2+1)^(1/2)/a^3

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Rubi [A]
time = 0.10, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6263, 847, 794, 201, 222} \begin {gather*} -\frac {c^2 \text {ArcSin}(a x)}{16 a^4}-\frac {c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a^2}+\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {c^2 (16-15 a x) \left (1-a^2 x^2\right )^{3/2}}{120 a^4}-\frac {c^2 x \sqrt {1-a^2 x^2}}{16 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^3*(c - a*c*x)^2,x]

[Out]

-1/16*(c^2*x*Sqrt[1 - a^2*x^2])/a^3 - (c^2*x^2*(1 - a^2*x^2)^(3/2))/(5*a^2) + (c^2*x^3*(1 - a^2*x^2)^(3/2))/(6
*a) - (c^2*(16 - 15*a*x)*(1 - a^2*x^2)^(3/2))/(120*a^4) - (c^2*ArcSin[a*x])/(16*a^4)

Rule 201

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 794

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 847

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 6263

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} x^3 (c-a c x)^2 \, dx &=c \int x^3 (c-a c x) \sqrt {1-a^2 x^2} \, dx\\ &=\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {c \int x^2 \left (3 a c-6 a^2 c x\right ) \sqrt {1-a^2 x^2} \, dx}{6 a^2}\\ &=-\frac {c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a^2}+\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}+\frac {c \int x \left (12 a^2 c-15 a^3 c x\right ) \sqrt {1-a^2 x^2} \, dx}{30 a^4}\\ &=-\frac {c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a^2}+\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {c^2 (16-15 a x) \left (1-a^2 x^2\right )^{3/2}}{120 a^4}-\frac {c^2 \int \sqrt {1-a^2 x^2} \, dx}{8 a^3}\\ &=-\frac {c^2 x \sqrt {1-a^2 x^2}}{16 a^3}-\frac {c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a^2}+\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {c^2 (16-15 a x) \left (1-a^2 x^2\right )^{3/2}}{120 a^4}-\frac {c^2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{16 a^3}\\ &=-\frac {c^2 x \sqrt {1-a^2 x^2}}{16 a^3}-\frac {c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a^2}+\frac {c^2 x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {c^2 (16-15 a x) \left (1-a^2 x^2\right )^{3/2}}{120 a^4}-\frac {c^2 \sin ^{-1}(a x)}{16 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 70, normalized size = 0.56 \begin {gather*} \frac {c^2 \left (\sqrt {1-a^2 x^2} \left (-32+15 a x-16 a^2 x^2+10 a^3 x^3+48 a^4 x^4-40 a^5 x^5\right )-15 \text {ArcSin}(a x)\right )}{240 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^3*(c - a*c*x)^2,x]

[Out]

(c^2*(Sqrt[1 - a^2*x^2]*(-32 + 15*a*x - 16*a^2*x^2 + 10*a^3*x^3 + 48*a^4*x^4 - 40*a^5*x^5) - 15*ArcSin[a*x]))/
(240*a^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(106)=212\).
time = 0.92, size = 295, normalized size = 2.38

method	result	size
risch	\(\frac {\left (40 a^{5} x^{5}-48 a^{4} x^{4}-10 a^{3} x^{3}+16 a^{2} x^{2}-15 a x +32\right ) \left (a^{2} x^{2}-1\right ) c^{2}}{240 a^{4} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{2}}{16 a^{3} \sqrt {a^{2}}}\)	\(102\)
meijerg	\(-\frac {c^{2} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {7}{2}} \left (56 a^{4} x^{4}+70 a^{2} x^{2}+105\right ) \sqrt {-a^{2} x^{2}+1}}{168 a^{6}}+\frac {5 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {7}{2}} \arcsin \left (a x \right )}{8 a^{7}}\right )}{2 a^{3} \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{2} \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}+8 a^{2} x^{2}+16\right ) \sqrt {-a^{2} x^{2}+1}}{15}\right )}{2 a^{4} \sqrt {\pi }}-\frac {c^{2} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{2 a^{3} \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{2} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{2 a^{4} \sqrt {\pi }}\)	\(258\)
default	\(c^{2} \left (a^{3} \left (-\frac {x^{5} \sqrt {-a^{2} x^{2}+1}}{6 a^{2}}+\frac {-\frac {5 x^{3} \sqrt {-a^{2} x^{2}+1}}{24 a^{2}}+\frac {5 \left (-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}\right )}{6 a^{2}}}{a^{2}}\right )-a^{2} \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )-a \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )\)	\(295\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

c^2*(a^3*(-1/6*x^5/a^2*(-a^2*x^2+1)^(1/2)+5/6/a^2*(-1/4*x^3*(-a^2*x^2+1)^(1/2)/a^2+3/4/a^2*(-1/2*x*(-a^2*x^2+1
)^(1/2)/a^2+1/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2)))))-a^2*(-1/5*x^4/a^2*(-a^2*x^2+1)^(1/
2)+4/5/a^2*(-1/3*x^2*(-a^2*x^2+1)^(1/2)/a^2-2/3*(-a^2*x^2+1)^(1/2)/a^4))-a*(-1/4*x^3*(-a^2*x^2+1)^(1/2)/a^2+3/
4/a^2*(-1/2*x*(-a^2*x^2+1)^(1/2)/a^2+1/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))))-1/3*x^2*(-
a^2*x^2+1)^(1/2)/a^2-2/3*(-a^2*x^2+1)^(1/2)/a^4)

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Maxima [A]
time = 0.46, size = 141, normalized size = 1.14 \begin {gather*} -\frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} a c^{2} x^{5} + \frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} c^{2} x^{4} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x^{3}}{24 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x^{2}}{15 \, a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x}{16 \, a^{3}} - \frac {c^{2} \arcsin \left (a x\right )}{16 \, a^{4}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{2}}{15 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x, algorithm="maxima")

[Out]

-1/6*sqrt(-a^2*x^2 + 1)*a*c^2*x^5 + 1/5*sqrt(-a^2*x^2 + 1)*c^2*x^4 + 1/24*sqrt(-a^2*x^2 + 1)*c^2*x^3/a - 1/15*
sqrt(-a^2*x^2 + 1)*c^2*x^2/a^2 + 1/16*sqrt(-a^2*x^2 + 1)*c^2*x/a^3 - 1/16*c^2*arcsin(a*x)/a^4 - 2/15*sqrt(-a^2
*x^2 + 1)*c^2/a^4

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Fricas [A]
time = 0.35, size = 104, normalized size = 0.84 \begin {gather*} \frac {30 \, c^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (40 \, a^{5} c^{2} x^{5} - 48 \, a^{4} c^{2} x^{4} - 10 \, a^{3} c^{2} x^{3} + 16 \, a^{2} c^{2} x^{2} - 15 \, a c^{2} x + 32 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{240 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x, algorithm="fricas")

[Out]

1/240*(30*c^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (40*a^5*c^2*x^5 - 48*a^4*c^2*x^4 - 10*a^3*c^2*x^3 + 16*
a^2*c^2*x^2 - 15*a*c^2*x + 32*c^2)*sqrt(-a^2*x^2 + 1))/a^4

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Sympy [A]
time = 12.73, size = 486, normalized size = 3.92 \begin {gather*} a^{3} c^{2} \left (\begin {cases} - \frac {i x^{7}}{6 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{5}}{24 a^{2} \sqrt {a^{2} x^{2} - 1}} - \frac {5 i x^{3}}{48 a^{4} \sqrt {a^{2} x^{2} - 1}} + \frac {5 i x}{16 a^{6} \sqrt {a^{2} x^{2} - 1}} - \frac {5 i \operatorname {acosh}{\left (a x \right )}}{16 a^{7}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{7}}{6 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{5}}{24 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {5 x^{3}}{48 a^{4} \sqrt {- a^{2} x^{2} + 1}} - \frac {5 x}{16 a^{6} \sqrt {- a^{2} x^{2} + 1}} + \frac {5 \operatorname {asin}{\left (a x \right )}}{16 a^{7}} & \text {otherwise} \end {cases}\right ) - a^{2} c^{2} \left (\begin {cases} - \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{5 a^{2}} - \frac {4 x^{2} \sqrt {- a^{2} x^{2} + 1}}{15 a^{4}} - \frac {8 \sqrt {- a^{2} x^{2} + 1}}{15 a^{6}} & \text {for}\: a \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right ) - a c^{2} \left (\begin {cases} - \frac {i x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{3}}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {3 i x}{8 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \operatorname {acosh}{\left (a x \right )}}{8 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{3}}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {3 x}{8 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \operatorname {asin}{\left (a x \right )}}{8 a^{5}} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**3*(-a*c*x+c)**2,x)

[Out]

a**3*c**2*Piecewise((-I*x**7/(6*sqrt(a**2*x**2 - 1)) - I*x**5/(24*a**2*sqrt(a**2*x**2 - 1)) - 5*I*x**3/(48*a**
4*sqrt(a**2*x**2 - 1)) + 5*I*x/(16*a**6*sqrt(a**2*x**2 - 1)) - 5*I*acosh(a*x)/(16*a**7), Abs(a**2*x**2) > 1),
(x**7/(6*sqrt(-a**2*x**2 + 1)) + x**5/(24*a**2*sqrt(-a**2*x**2 + 1)) + 5*x**3/(48*a**4*sqrt(-a**2*x**2 + 1)) -
5*x/(16*a**6*sqrt(-a**2*x**2 + 1)) + 5*asin(a*x)/(16*a**7), True)) - a**2*c**2*Piecewise((-x**4*sqrt(-a**2*x*
*2 + 1)/(5*a**2) - 4*x**2*sqrt(-a**2*x**2 + 1)/(15*a**4) - 8*sqrt(-a**2*x**2 + 1)/(15*a**6), Ne(a, 0)), (x**6/
6, True)) - a*c**2*Piecewise((-I*x**5/(4*sqrt(a**2*x**2 - 1)) - I*x**3/(8*a**2*sqrt(a**2*x**2 - 1)) + 3*I*x/(8
*a**4*sqrt(a**2*x**2 - 1)) - 3*I*acosh(a*x)/(8*a**5), Abs(a**2*x**2) > 1), (x**5/(4*sqrt(-a**2*x**2 + 1)) + x*
*3/(8*a**2*sqrt(-a**2*x**2 + 1)) - 3*x/(8*a**4*sqrt(-a**2*x**2 + 1)) + 3*asin(a*x)/(8*a**5), True)) + c**2*Pie
cewise((-x**2*sqrt(-a**2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x**2 + 1)/(3*a**4), Ne(a, 0)), (x**4/4, True))

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Giac [A]
time = 0.41, size = 92, normalized size = 0.74 \begin {gather*} -\frac {1}{240} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, a c^{2} x - 6 \, c^{2}\right )} x - \frac {5 \, c^{2}}{a}\right )} x + \frac {8 \, c^{2}}{a^{2}}\right )} x - \frac {15 \, c^{2}}{a^{3}}\right )} x + \frac {32 \, c^{2}}{a^{4}}\right )} - \frac {c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{16 \, a^{3} {\left | a \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x, algorithm="giac")

[Out]

-1/240*sqrt(-a^2*x^2 + 1)*((2*((4*(5*a*c^2*x - 6*c^2)*x - 5*c^2/a)*x + 8*c^2/a^2)*x - 15*c^2/a^3)*x + 32*c^2/a
^4) - 1/16*c^2*arcsin(a*x)*sgn(a)/(a^3*abs(a))

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Mupad [B]
time = 0.79, size = 154, normalized size = 1.24 \begin {gather*} \frac {c^2\,x^4\,\sqrt {1-a^2\,x^2}}{5}-\frac {2\,c^2\,\sqrt {1-a^2\,x^2}}{15\,a^4}+\frac {c^2\,x\,\sqrt {1-a^2\,x^2}}{16\,a^3}-\frac {a\,c^2\,x^5\,\sqrt {1-a^2\,x^2}}{6}-\frac {c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{16\,a^3\,\sqrt {-a^2}}+\frac {c^2\,x^3\,\sqrt {1-a^2\,x^2}}{24\,a}-\frac {c^2\,x^2\,\sqrt {1-a^2\,x^2}}{15\,a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(c - a*c*x)^2*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)

[Out]

(c^2*x^4*(1 - a^2*x^2)^(1/2))/5 - (2*c^2*(1 - a^2*x^2)^(1/2))/(15*a^4) + (c^2*x*(1 - a^2*x^2)^(1/2))/(16*a^3)
- (a*c^2*x^5*(1 - a^2*x^2)^(1/2))/6 - (c^2*asinh(x*(-a^2)^(1/2)))/(16*a^3*(-a^2)^(1/2)) + (c^2*x^3*(1 - a^2*x^
2)^(1/2))/(24*a) - (c^2*x^2*(1 - a^2*x^2)^(1/2))/(15*a^2)

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