Integrand size = 19, antiderivative size = 140 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=\frac {c^2 x \sqrt {1-a^2 x^2}}{16 a^3}+\frac {c^2 x^3 \sqrt {1-a^2 x^2}}{24 a}-\frac {1}{6} a c^2 x^5 \sqrt {1-a^2 x^2}-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}+\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^4}-\frac {c^2 \arcsin (a x)}{16 a^4} \] Output:
1/16*c^2*x*(-a^2*x^2+1)^(1/2)/a^3+1/24*c^2*x^3*(-a^2*x^2+1)^(1/2)/a-1/6*a* c^2*x^5*(-a^2*x^2+1)^(1/2)-1/3*c^2*(-a^2*x^2+1)^(3/2)/a^4+1/5*c^2*(-a^2*x^ 2+1)^(5/2)/a^4-1/16*c^2*arcsin(a*x)/a^4
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.64 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=-\frac {c^2 \left (\sqrt {1-a^2 x^2} \left (32-15 a x+16 a^2 x^2-10 a^3 x^3-48 a^4 x^4+40 a^5 x^5\right )-60 \arcsin (a x)-150 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{240 a^4} \] Input:
Integrate[E^ArcTanh[a*x]*x^3*(c - a*c*x)^2,x]
Output:
-1/240*(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) - 60*ArcSin[a*x] - 150*ArcSin[Sqrt[1 - a*x]/Sqrt[2] ]))/a^4
Time = 0.61 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {6678, 27, 533, 27, 533, 25, 27, 533, 25, 27, 455, 211, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 e^{\text {arctanh}(a x)} (c-a c x)^2 \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int c x^3 (1-a x) \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \int x^3 (1-a x) \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle c^2 \left (\frac {\int -3 a x^2 (1-2 a x) \sqrt {1-a^2 x^2}dx}{6 a^2}+\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\int x^2 (1-2 a x) \sqrt {1-a^2 x^2}dx}{2 a}\right )\) |
\(\Big \downarrow \) 533 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {\int -a x (4-5 a x) \sqrt {1-a^2 x^2}dx}{5 a^2}+\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\int a x (4-5 a x) \sqrt {1-a^2 x^2}dx}{5 a^2}}{2 a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\int x (4-5 a x) \sqrt {1-a^2 x^2}dx}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 533 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {\int -a (5-16 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}+\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{4 a}}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {\int a (5-16 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {\int (5-16 a x) \sqrt {1-a^2 x^2}dx}{4 a}}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {5 \int \sqrt {1-a^2 x^2}dx+\frac {16 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {5 \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {16 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}}{2 a}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^2 \left (\frac {x^3 \left (1-a^2 x^2\right )^{3/2}}{6 a}-\frac {\frac {2 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {5 \left (\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {16 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}}{2 a}\right )\) |
Input:
Int[E^ArcTanh[a*x]*x^3*(c - a*c*x)^2,x]
Output:
c^2*((x^3*(1 - a^2*x^2)^(3/2))/(6*a) - ((2*x^2*(1 - a^2*x^2)^(3/2))/(5*a) - ((5*x*(1 - a^2*x^2)^(3/2))/(4*a) - ((16*(1 - a^2*x^2)^(3/2))/(3*a) + 5*( (x*Sqrt[1 - a^2*x^2])/2 + ArcSin[a*x]/(2*a)))/(4*a))/(5*a))/(2*a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[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]
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73
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 | \(c^{2} \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}+a^{3} \left (-\frac {x^{5} \sqrt {-a^{2} x^{2}+1}}{6 a^{2}}+\frac {-\frac {5 x^{3} \sqrt {-a^{2} x^{2}+1}}{24 a^{2}}+\frac {5 \left (-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}\right )}{6 a^{2}}}{a^{2}}\right )-a \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )-a^{2} \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )\right )\) | \(295\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/240*(40*a^5*x^5-48*a^4*x^4-10*a^3*x^3+16*a^2*x^2-15*a*x+32)*(a^2*x^2-1)/ a^4/(-a^2*x^2+1)^(1/2)*c^2-1/16/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2 *x^2+1)^(1/2))*c^2
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.74 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=\frac {30 \, c^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (40 \, a^{5} c^{2} x^{5} - 48 \, a^{4} c^{2} x^{4} - 10 \, a^{3} c^{2} x^{3} + 16 \, a^{2} c^{2} x^{2} - 15 \, a c^{2} x + 32 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{240 \, a^{4}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x, algorithm="fricas ")
Output:
1/240*(30*c^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (40*a^5*c^2*x^5 - 4 8*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)*sqr t(-a^2*x^2 + 1))/a^4
Time = 0.90 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.16 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (- \frac {a c^{2} x^{5}}{6} + \frac {c^{2} x^{4}}{5} + \frac {c^{2} x^{3}}{24 a} - \frac {c^{2} x^{2}}{15 a^{2}} + \frac {c^{2} x}{16 a^{3}} - \frac {2 c^{2}}{15 a^{4}}\right ) - \frac {c^{2} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{16 a^{3} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {a^{3} c^{2} x^{7}}{7} - \frac {a^{2} c^{2} x^{6}}{6} - \frac {a c^{2} x^{5}}{5} + \frac {c^{2} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**3*(-a*c*x+c)**2,x)
Output:
Piecewise((sqrt(-a**2*x**2 + 1)*(-a*c**2*x**5/6 + c**2*x**4/5 + c**2*x**3/ (24*a) - c**2*x**2/(15*a**2) + c**2*x/(16*a**3) - 2*c**2/(15*a**4)) - c**2 *log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(16*a**3*sqrt(-a**2)) , Ne(a**2, 0)), (a**3*c**2*x**7/7 - a**2*c**2*x**6/6 - a*c**2*x**5/5 + c** 2*x**4/4, True))
Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=-\frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} a c^{2} x^{5} + \frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} c^{2} x^{4} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x^{3}}{24 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x^{2}}{15 \, a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x}{16 \, a^{3}} - \frac {c^{2} \arcsin \left (a x\right )}{16 \, a^{4}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{2}}{15 \, a^{4}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x, algorithm="maxima ")
Output:
-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
Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.66 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=-\frac {1}{240} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, a c^{2} x - 6 \, c^{2}\right )} x - \frac {5 \, c^{2}}{a}\right )} x + \frac {8 \, c^{2}}{a^{2}}\right )} x - \frac {15 \, c^{2}}{a^{3}}\right )} x + \frac {32 \, c^{2}}{a^{4}}\right )} - \frac {c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{16 \, a^{3} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x, algorithm="giac")
Output:
-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))
Time = 0.04 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.10 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=\frac {c^2\,x^4\,\sqrt {1-a^2\,x^2}}{5}-\frac {2\,c^2\,\sqrt {1-a^2\,x^2}}{15\,a^4}+\frac {c^2\,x\,\sqrt {1-a^2\,x^2}}{16\,a^3}-\frac {a\,c^2\,x^5\,\sqrt {1-a^2\,x^2}}{6}-\frac {c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{16\,a^3\,\sqrt {-a^2}}+\frac {c^2\,x^3\,\sqrt {1-a^2\,x^2}}{24\,a}-\frac {c^2\,x^2\,\sqrt {1-a^2\,x^2}}{15\,a^2} \] Input:
int((x^3*(c - a*c*x)^2*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
(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)
Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.86 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^2 \, dx=\frac {c^{2} \left (-15 \mathit {asin} \left (a x \right )-40 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}+48 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+10 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+15 \sqrt {-a^{2} x^{2}+1}\, a x -32 \sqrt {-a^{2} x^{2}+1}+32\right )}{240 a^{4}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a*c*x+c)^2,x)
Output:
(c**2*( - 15*asin(a*x) - 40*sqrt( - a**2*x**2 + 1)*a**5*x**5 + 48*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 10*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 16*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 15*sqrt( - a**2*x**2 + 1)*a*x - 32*sqrt( - a* *2*x**2 + 1) + 32))/(240*a**4)