\(\int \frac {(d+c^2 d x^2)^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx\) [18]

Optimal result

Integrand size = 24, antiderivative size = 126 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=-b c^3 d^2 \sqrt {1+c^2 x^2}-\frac {b c d^2 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {11}{6} b c^3 d^2 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right ) \] Output:

-b*c^3*d^2*(c^2*x^2+1)^(1/2)-1/6*b*c*d^2*(c^2*x^2+1)^(1/2)/x^2-1/3*d^2*(a+ 
b*arcsinh(c*x))/x^3-2*c^2*d^2*(a+b*arcsinh(c*x))/x+c^4*d^2*x*(a+b*arcsinh( 
c*x))-11/6*b*c^3*d^2*arctanh((c^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\frac {d^2 \left (-2 a-12 a c^2 x^2+6 a c^4 x^4-b c x \sqrt {1+c^2 x^2}-6 b c^3 x^3 \sqrt {1+c^2 x^2}+2 b \left (-1-6 c^2 x^2+3 c^4 x^4\right ) \text {arcsinh}(c x)+11 b c^3 x^3 \log (x)-11 b c^3 x^3 \log \left (1+\sqrt {1+c^2 x^2}\right )\right )}{6 x^3} \] Input:

Integrate[((d + c^2*d*x^2)^2*(a + b*ArcSinh[c*x]))/x^4,x]
 

Output:

(d^2*(-2*a - 12*a*c^2*x^2 + 6*a*c^4*x^4 - b*c*x*Sqrt[1 + c^2*x^2] - 6*b*c^ 
3*x^3*Sqrt[1 + c^2*x^2] + 2*b*(-1 - 6*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] + 
11*b*c^3*x^3*Log[x] - 11*b*c^3*x^3*Log[1 + Sqrt[1 + c^2*x^2]]))/(6*x^3)
 

Rubi [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6218, 27, 1578, 1192, 25, 1471, 25, 27, 299, 219}

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.

\(\displaystyle \int \frac {\left (c^2 d x^2+d\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx\)

\(\Big \downarrow \) 6218

\(\displaystyle -b c \int -\frac {d^2 \left (-3 c^4 x^4+6 c^2 x^2+1\right )}{3 x^3 \sqrt {c^2 x^2+1}}dx+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} b c d^2 \int \frac {-3 c^4 x^4+6 c^2 x^2+1}{x^3 \sqrt {c^2 x^2+1}}dx+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {1}{6} b c d^2 \int \frac {-3 c^4 x^4+6 c^2 x^2+1}{x^4 \sqrt {c^2 x^2+1}}dx^2+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 1192

\(\displaystyle \frac {b d^2 \int -\frac {3 c^4 x^8-12 c^4 x^4+8 c^4}{\left (1-x^4\right )^2}d\sqrt {c^2 x^2+1}}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b d^2 \int \frac {3 c^4 x^8-12 c^4 x^4+8 c^4}{\left (1-x^4\right )^2}d\sqrt {c^2 x^2+1}}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {b d^2 \left (\frac {1}{2} \int -\frac {c^4 \left (17-6 x^4\right )}{1-x^4}d\sqrt {c^2 x^2+1}+\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} \int \frac {c^4 \left (17-6 x^4\right )}{1-x^4}d\sqrt {c^2 x^2+1}\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \int \frac {17-6 x^4}{1-x^4}d\sqrt {c^2 x^2+1}\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \left (11 \int \frac {1}{1-x^4}d\sqrt {c^2 x^2+1}+6 \sqrt {c^2 x^2+1}\right )\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \left (11 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )+6 \sqrt {c^2 x^2+1}\right )\right )}{3 c}\)

Input:

Int[((d + c^2*d*x^2)^2*(a + b*ArcSinh[c*x]))/x^4,x]
 

Output:

-1/3*(d^2*(a + b*ArcSinh[c*x]))/x^3 - (2*c^2*d^2*(a + b*ArcSinh[c*x]))/x + 
 c^4*d^2*x*(a + b*ArcSinh[c*x]) + (b*d^2*((c^4*Sqrt[1 + c^2*x^2])/(2*(1 - 
x^4)) - (c^4*(6*Sqrt[1 + c^2*x^2] + 11*ArcTanh[Sqrt[1 + c^2*x^2]]))/2))/(3 
*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 1192

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^( 
2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + 
 c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

rule 6218

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_ 
)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp 
[(a + b*ArcSinh[c*x])   u, x] - Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 + 
 c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] 
&& IGtQ[p, 0]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89

method	result	size
parts	\(a \,d^{2} \left (c^{4} x -\frac {1}{3 x^{3}}-\frac {2 c^{2}}{x}\right )+d^{2} b \,c^{3} \left (x c \,\operatorname {arcsinh}\left (x c \right )-\frac {\operatorname {arcsinh}\left (x c \right )}{3 x^{3} c^{3}}-\frac {2 \,\operatorname {arcsinh}\left (x c \right )}{x c}-\frac {\sqrt {c^{2} x^{2}+1}}{6 x^{2} c^{2}}-\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\sqrt {c^{2} x^{2}+1}\right )\)	\(112\)
derivativedivides	\(c^{3} \left (a \,d^{2} \left (x c -\frac {1}{3 x^{3} c^{3}}-\frac {2}{x c}\right )+d^{2} b \left (x c \,\operatorname {arcsinh}\left (x c \right )-\frac {\operatorname {arcsinh}\left (x c \right )}{3 x^{3} c^{3}}-\frac {2 \,\operatorname {arcsinh}\left (x c \right )}{x c}-\frac {\sqrt {c^{2} x^{2}+1}}{6 x^{2} c^{2}}-\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(114\)
default	\(c^{3} \left (a \,d^{2} \left (x c -\frac {1}{3 x^{3} c^{3}}-\frac {2}{x c}\right )+d^{2} b \left (x c \,\operatorname {arcsinh}\left (x c \right )-\frac {\operatorname {arcsinh}\left (x c \right )}{3 x^{3} c^{3}}-\frac {2 \,\operatorname {arcsinh}\left (x c \right )}{x c}-\frac {\sqrt {c^{2} x^{2}+1}}{6 x^{2} c^{2}}-\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(114\)

Input:

int((c^2*d*x^2+d)^2*(a+b*arcsinh(x*c))/x^4,x,method=_RETURNVERBOSE)
 

Output:

a*d^2*(c^4*x-1/3/x^3-2*c^2/x)+d^2*b*c^3*(x*c*arcsinh(x*c)-1/3*arcsinh(x*c) 
/x^3/c^3-2*arcsinh(x*c)/x/c-1/6/x^2/c^2*(c^2*x^2+1)^(1/2)-11/6*arctanh(1/( 
c^2*x^2+1)^(1/2))-(c^2*x^2+1)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (114) = 228\).

Time = 0.12 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.93 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\frac {6 \, a c^{4} d^{2} x^{4} - 11 \, b c^{3} d^{2} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} + 1\right ) + 11 \, b c^{3} d^{2} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} - 1\right ) - 12 \, a c^{2} d^{2} x^{2} - 2 \, {\left (3 \, b c^{4} - 6 \, b c^{2} - b\right )} d^{2} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, a d^{2} + 2 \, {\left (3 \, b c^{4} d^{2} x^{4} - 6 \, b c^{2} d^{2} x^{2} - {\left (3 \, b c^{4} - 6 \, b c^{2} - b\right )} d^{2} x^{3} - b d^{2}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (6 \, b c^{3} d^{2} x^{3} + b c d^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}{6 \, x^{3}} \] Input:

integrate((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")
 

Output:

1/6*(6*a*c^4*d^2*x^4 - 11*b*c^3*d^2*x^3*log(-c*x + sqrt(c^2*x^2 + 1) + 1) 
+ 11*b*c^3*d^2*x^3*log(-c*x + sqrt(c^2*x^2 + 1) - 1) - 12*a*c^2*d^2*x^2 - 
2*(3*b*c^4 - 6*b*c^2 - b)*d^2*x^3*log(-c*x + sqrt(c^2*x^2 + 1)) - 2*a*d^2 
+ 2*(3*b*c^4*d^2*x^4 - 6*b*c^2*d^2*x^2 - (3*b*c^4 - 6*b*c^2 - b)*d^2*x^3 - 
 b*d^2)*log(c*x + sqrt(c^2*x^2 + 1)) - (6*b*c^3*d^2*x^3 + b*c*d^2*x)*sqrt( 
c^2*x^2 + 1))/x^3
 

Sympy [F]

\[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=d^{2} \left (\int a c^{4}\, dx + \int \frac {a}{x^{4}}\, dx + \int \frac {2 a c^{2}}{x^{2}}\, dx + \int b c^{4} \operatorname {asinh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {2 b c^{2} \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx\right ) \] Input:

integrate((c**2*d*x**2+d)**2*(a+b*asinh(c*x))/x**4,x)
 

Output:

d**2*(Integral(a*c**4, x) + Integral(a/x**4, x) + Integral(2*a*c**2/x**2, 
x) + Integral(b*c**4*asinh(c*x), x) + Integral(b*asinh(c*x)/x**4, x) + Int 
egral(2*b*c**2*asinh(c*x)/x**2, x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.09 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=a c^{4} d^{2} x + {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b c^{3} d^{2} - 2 \, {\left (c \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arsinh}\left (c x\right )}{x}\right )} b c^{2} d^{2} + \frac {1}{6} \, {\left ({\left (c^{2} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac {2 \, \operatorname {arsinh}\left (c x\right )}{x^{3}}\right )} b d^{2} - \frac {2 \, a c^{2} d^{2}}{x} - \frac {a d^{2}}{3 \, x^{3}} \] Input:

integrate((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x))/x^4,x, algorithm="maxima")
 

Output:

a*c^4*d^2*x + (c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c^3*d^2 - 2*(c*arcs 
inh(1/(c*abs(x))) + arcsinh(c*x)/x)*b*c^2*d^2 + 1/6*((c^2*arcsinh(1/(c*abs 
(x))) - sqrt(c^2*x^2 + 1)/x^2)*c - 2*arcsinh(c*x)/x^3)*b*d^2 - 2*a*c^2*d^2 
/x - 1/3*a*d^2/x^3
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x))/x^4,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^2}{x^4} \,d x \] Input:

int(((a + b*asinh(c*x))*(d + c^2*d*x^2)^2)/x^4,x)
 

Output:

int(((a + b*asinh(c*x))*(d + c^2*d*x^2)^2)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\frac {d^{2} \left (6 \mathit {asinh} \left (c x \right ) b \,c^{4} x^{4}-12 \mathit {asinh} \left (c x \right ) b \,c^{2} x^{2}-2 \mathit {asinh} \left (c x \right ) b -6 \sqrt {c^{2} x^{2}+1}\, b \,c^{3} x^{3}-\sqrt {c^{2} x^{2}+1}\, b c x +11 \,\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x -1\right ) b \,c^{3} x^{3}-11 \,\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x +1\right ) b \,c^{3} x^{3}+6 a \,c^{4} x^{4}-12 a \,c^{2} x^{2}-2 a \right )}{6 x^{3}} \] Input:

int((c^2*d*x^2+d)^2*(a+b*asinh(c*x))/x^4,x)
 

Output:

(d**2*(6*asinh(c*x)*b*c**4*x**4 - 12*asinh(c*x)*b*c**2*x**2 - 2*asinh(c*x) 
*b - 6*sqrt(c**2*x**2 + 1)*b*c**3*x**3 - sqrt(c**2*x**2 + 1)*b*c*x + 11*lo 
g(sqrt(c**2*x**2 + 1) + c*x - 1)*b*c**3*x**3 - 11*log(sqrt(c**2*x**2 + 1) 
+ c*x + 1)*b*c**3*x**3 + 6*a*c**4*x**4 - 12*a*c**2*x**2 - 2*a))/(6*x**3)