\(\int x^3 \log ^3(c (a+b x^2)^p) \, dx\) [92]

Optimal result

Integrand size = 18, antiderivative size = 211 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {3 a p^3 x^2}{b}-\frac {3 p^3 \left (a+b x^2\right )^2}{16 b^2}-\frac {3 a p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {3 p^2 \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{8 b^2}+\frac {3 a p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {3 p \left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{8 b^2}-\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{4 b^2} \] Output:

3*a*p^3*x^2/b-3/16*p^3*(b*x^2+a)^2/b^2-3*a*p^2*(b*x^2+a)*ln(c*(b*x^2+a)^p) 
/b^2+3/8*p^2*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)/b^2+3/2*a*p*(b*x^2+a)*ln(c*(b*x 
^2+a)^p)^2/b^2-3/8*p*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)^2/b^2-1/2*a*(b*x^2+a)*l 
n(c*(b*x^2+a)^p)^3/b^2+1/4*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)^3/b^2
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.69 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {3 b p^3 x^2 \left (-14 a+b x^2\right )+6 a^2 p^3 \log \left (a+b x^2\right )+6 p^2 \left (6 a^2+6 a b x^2-b^2 x^4\right ) \log \left (c \left (a+b x^2\right )^p\right )-6 p \left (3 a^2+2 a b x^2-b^2 x^4\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+4 \left (a^2-b^2 x^4\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{16 b^2} \] Input:

Integrate[x^3*Log[c*(a + b*x^2)^p]^3,x]
 

Output:

-1/16*(3*b*p^3*x^2*(-14*a + b*x^2) + 6*a^2*p^3*Log[a + b*x^2] + 6*p^2*(6*a 
^2 + 6*a*b*x^2 - b^2*x^4)*Log[c*(a + b*x^2)^p] - 6*p*(3*a^2 + 2*a*b*x^2 - 
b^2*x^4)*Log[c*(a + b*x^2)^p]^2 + 4*(a^2 - b^2*x^4)*Log[c*(a + b*x^2)^p]^3 
)/b^2
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2904, 2848, 2009}

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.

\(\displaystyle \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2904

\(\displaystyle \frac {1}{2} \int x^2 \log ^3\left (c \left (b x^2+a\right )^p\right )dx^2\)

\(\Big \downarrow \) 2848

\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b x^2+a\right ) \log ^3\left (c \left (b x^2+a\right )^p\right )}{b}-\frac {a \log ^3\left (c \left (b x^2+a\right )^p\right )}{b}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {3 p^2 \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {6 a p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{b^2}-\frac {3 p \left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{4 b^2}+\frac {3 a p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{b^2}-\frac {3 p^3 \left (a+b x^2\right )^2}{8 b^2}+\frac {6 a p^3 x^2}{b}\right )\)

Input:

Int[x^3*Log[c*(a + b*x^2)^p]^3,x]
 

Output:

((6*a*p^3*x^2)/b - (3*p^3*(a + b*x^2)^2)/(8*b^2) - (6*a*p^2*(a + b*x^2)*Lo 
g[c*(a + b*x^2)^p])/b^2 + (3*p^2*(a + b*x^2)^2*Log[c*(a + b*x^2)^p])/(4*b^ 
2) + (3*a*p*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/b^2 - (3*p*(a + b*x^2)^2*L 
og[c*(a + b*x^2)^p]^2)/(4*b^2) - (a*(a + b*x^2)*Log[c*(a + b*x^2)^p]^3)/b^ 
2 + ((a + b*x^2)^2*Log[c*(a + b*x^2)^p]^3)/(2*b^2))/2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2848

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d 
 + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - 
 d*g, 0] && IGtQ[q, 0]
 

rule 2904

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L 
og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, 
 x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & 
&  !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
 

Maple [A] (verified)

Time = 15.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(-\frac {-4 x^{4} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3} b^{2}+6 x^{4} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} b^{2} p -6 x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{2} p^{2}+3 x^{4} b^{2} p^{3}-12 x^{2} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a b p +36 x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a b \,p^{2}-42 a b \,p^{3} x^{2}+78 \ln \left (b \,x^{2}+a \right ) a^{2} p^{3}+4 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3} a^{2}-18 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a^{2} p -36 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} p^{2}+42 a^{2} p^{3}}{16 b^{2}}\)	\(223\)
risch	\(\text {Expression too large to display}\)	\(241126\)

Input:

int(x^3*ln(c*(b*x^2+a)^p)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/16*(-4*x^4*ln(c*(b*x^2+a)^p)^3*b^2+6*x^4*ln(c*(b*x^2+a)^p)^2*b^2*p-6*x^ 
4*ln(c*(b*x^2+a)^p)*b^2*p^2+3*x^4*b^2*p^3-12*x^2*ln(c*(b*x^2+a)^p)^2*a*b*p 
+36*x^2*ln(c*(b*x^2+a)^p)*a*b*p^2-42*a*b*p^3*x^2+78*ln(b*x^2+a)*a^2*p^3+4* 
ln(c*(b*x^2+a)^p)^3*a^2-18*ln(c*(b*x^2+a)^p)^2*a^2*p-36*ln(c*(b*x^2+a)^p)* 
a^2*p^2+42*a^2*p^3)/b^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.30 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {3 \, b^{2} p^{3} x^{4} - 4 \, b^{2} x^{4} \log \left (c\right )^{3} - 42 \, a b p^{3} x^{2} - 4 \, {\left (b^{2} p^{3} x^{4} - a^{2} p^{3}\right )} \log \left (b x^{2} + a\right )^{3} + 6 \, {\left (b^{2} p^{3} x^{4} - 2 \, a b p^{3} x^{2} - 3 \, a^{2} p^{3} - 2 \, {\left (b^{2} p^{2} x^{4} - a^{2} p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )^{2} + 6 \, {\left (b^{2} p x^{4} - 2 \, a b p x^{2}\right )} \log \left (c\right )^{2} - 6 \, {\left (b^{2} p^{3} x^{4} - 6 \, a b p^{3} x^{2} - 7 \, a^{2} p^{3} + 2 \, {\left (b^{2} p x^{4} - a^{2} p\right )} \log \left (c\right )^{2} - 2 \, {\left (b^{2} p^{2} x^{4} - 2 \, a b p^{2} x^{2} - 3 \, a^{2} p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} p^{2} x^{4} - 6 \, a b p^{2} x^{2}\right )} \log \left (c\right )}{16 \, b^{2}} \] Input:

integrate(x^3*log(c*(b*x^2+a)^p)^3,x, algorithm="fricas")
 

Output:

-1/16*(3*b^2*p^3*x^4 - 4*b^2*x^4*log(c)^3 - 42*a*b*p^3*x^2 - 4*(b^2*p^3*x^ 
4 - a^2*p^3)*log(b*x^2 + a)^3 + 6*(b^2*p^3*x^4 - 2*a*b*p^3*x^2 - 3*a^2*p^3 
 - 2*(b^2*p^2*x^4 - a^2*p^2)*log(c))*log(b*x^2 + a)^2 + 6*(b^2*p*x^4 - 2*a 
*b*p*x^2)*log(c)^2 - 6*(b^2*p^3*x^4 - 6*a*b*p^3*x^2 - 7*a^2*p^3 + 2*(b^2*p 
*x^4 - a^2*p)*log(c)^2 - 2*(b^2*p^2*x^4 - 2*a*b*p^2*x^2 - 3*a^2*p^2)*log(c 
))*log(b*x^2 + a) - 6*(b^2*p^2*x^4 - 6*a*b*p^2*x^2)*log(c))/b^2
 

Sympy [A] (verification not implemented)

Time = 2.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.06 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} - \frac {21 a^{2} p^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{8 b^{2}} + \frac {9 a^{2} p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{8 b^{2}} - \frac {a^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{4 b^{2}} + \frac {21 a p^{3} x^{2}}{8 b} - \frac {9 a p^{2} x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 b} + \frac {3 a p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{4 b} - \frac {3 p^{3} x^{4}}{16} + \frac {3 p^{2} x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{8} - \frac {3 p x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{8} + \frac {x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{4} & \text {for}\: b \neq 0 \\\frac {x^{4} \log {\left (a^{p} c \right )}^{3}}{4} & \text {otherwise} \end {cases} \] Input:

integrate(x**3*ln(c*(b*x**2+a)**p)**3,x)
 

Output:

Piecewise((-21*a**2*p**2*log(c*(a + b*x**2)**p)/(8*b**2) + 9*a**2*p*log(c* 
(a + b*x**2)**p)**2/(8*b**2) - a**2*log(c*(a + b*x**2)**p)**3/(4*b**2) + 2 
1*a*p**3*x**2/(8*b) - 9*a*p**2*x**2*log(c*(a + b*x**2)**p)/(4*b) + 3*a*p*x 
**2*log(c*(a + b*x**2)**p)**2/(4*b) - 3*p**3*x**4/16 + 3*p**2*x**4*log(c*( 
a + b*x**2)**p)/8 - 3*p*x**4*log(c*(a + b*x**2)**p)**2/8 + x**4*log(c*(a + 
 b*x**2)**p)**3/4, Ne(b, 0)), (x**4*log(a**p*c)**3/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.96 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3} - \frac {3}{8} \, b p {\left (\frac {2 \, a^{2} \log \left (b x^{2} + a\right )}{b^{3}} + \frac {b x^{4} - 2 \, a x^{2}}{b^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} - \frac {1}{16} \, b p {\left (\frac {{\left (3 \, b^{2} x^{4} + 4 \, a^{2} \log \left (b x^{2} + a\right )^{3} - 42 \, a b x^{2} + 18 \, a^{2} \log \left (b x^{2} + a\right )^{2} + 42 \, a^{2} \log \left (b x^{2} + a\right )\right )} p^{2}}{b^{3}} - \frac {6 \, {\left (b^{2} x^{4} - 6 \, a b x^{2} + 2 \, a^{2} \log \left (b x^{2} + a\right )^{2} + 6 \, a^{2} \log \left (b x^{2} + a\right )\right )} p \log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{b^{3}}\right )} \] Input:

integrate(x^3*log(c*(b*x^2+a)^p)^3,x, algorithm="maxima")
 

Output:

1/4*x^4*log((b*x^2 + a)^p*c)^3 - 3/8*b*p*(2*a^2*log(b*x^2 + a)/b^3 + (b*x^ 
4 - 2*a*x^2)/b^2)*log((b*x^2 + a)^p*c)^2 - 1/16*b*p*((3*b^2*x^4 + 4*a^2*lo 
g(b*x^2 + a)^3 - 42*a*b*x^2 + 18*a^2*log(b*x^2 + a)^2 + 42*a^2*log(b*x^2 + 
 a))*p^2/b^3 - 6*(b^2*x^4 - 6*a*b*x^2 + 2*a^2*log(b*x^2 + a)^2 + 6*a^2*log 
(b*x^2 + a))*p*log((b*x^2 + a)^p*c)/b^3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.82 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {4 \, {\left (b x^{2} + a\right )}^{2} p^{3} \log \left (b x^{2} + a\right )^{3} - 6 \, {\left (b x^{2} + a\right )}^{2} p^{3} \log \left (b x^{2} + a\right )^{2} + 12 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (b x^{2} + a\right )^{2} \log \left (c\right ) + 6 \, {\left (b x^{2} + a\right )}^{2} p^{3} \log \left (b x^{2} + a\right ) - 12 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (b x^{2} + a\right ) \log \left (c\right ) + 12 \, {\left (b x^{2} + a\right )}^{2} p \log \left (b x^{2} + a\right ) \log \left (c\right )^{2} - 3 \, {\left (b x^{2} + a\right )}^{2} p^{3} + 6 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (c\right ) - 6 \, {\left (b x^{2} + a\right )}^{2} p \log \left (c\right )^{2} + 4 \, {\left (b x^{2} + a\right )}^{2} \log \left (c\right )^{3}}{16 \, b^{2}} - \frac {{\left ({\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{3} - 6 \, b x^{2} - 3 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} + 6 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) - 6 \, a\right )} a p^{3} + 3 \, {\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} a p^{2} \log \left (c\right ) - 3 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} a p \log \left (c\right )^{2} + {\left (b x^{2} + a\right )} a \log \left (c\right )^{3}}{2 \, b^{2}} \] Input:

integrate(x^3*log(c*(b*x^2+a)^p)^3,x, algorithm="giac")
 

Output:

1/16*(4*(b*x^2 + a)^2*p^3*log(b*x^2 + a)^3 - 6*(b*x^2 + a)^2*p^3*log(b*x^2 
 + a)^2 + 12*(b*x^2 + a)^2*p^2*log(b*x^2 + a)^2*log(c) + 6*(b*x^2 + a)^2*p 
^3*log(b*x^2 + a) - 12*(b*x^2 + a)^2*p^2*log(b*x^2 + a)*log(c) + 12*(b*x^2 
 + a)^2*p*log(b*x^2 + a)*log(c)^2 - 3*(b*x^2 + a)^2*p^3 + 6*(b*x^2 + a)^2* 
p^2*log(c) - 6*(b*x^2 + a)^2*p*log(c)^2 + 4*(b*x^2 + a)^2*log(c)^3)/b^2 - 
1/2*(((b*x^2 + a)*log(b*x^2 + a)^3 - 6*b*x^2 - 3*(b*x^2 + a)*log(b*x^2 + a 
)^2 + 6*(b*x^2 + a)*log(b*x^2 + a) - 6*a)*a*p^3 + 3*(2*b*x^2 + (b*x^2 + a) 
*log(b*x^2 + a)^2 - 2*(b*x^2 + a)*log(b*x^2 + a) + 2*a)*a*p^2*log(c) - 3*( 
b*x^2 - (b*x^2 + a)*log(b*x^2 + a) + a)*a*p*log(c)^2 + (b*x^2 + a)*a*log(c 
)^3)/b^2
 

Mupad [B] (verification not implemented)

Time = 14.76 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.68 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx={\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {9\,a^2\,p}{8\,b^2}-\frac {3\,p\,x^4}{8}+\frac {3\,a\,p\,x^2}{4\,b}\right )-\frac {3\,p^3\,x^4}{16}+\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,\left (\frac {3\,p^2\,x^4}{8}-\frac {9\,a\,p^2\,x^2}{4\,b}\right )+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^3\,\left (\frac {x^4}{4}-\frac {a^2}{4\,b^2}\right )+\frac {21\,a\,p^3\,x^2}{8\,b}-\frac {21\,a^2\,p^3\,\ln \left (b\,x^2+a\right )}{8\,b^2} \] Input:

int(x^3*log(c*(a + b*x^2)^p)^3,x)
 

Output:

log(c*(a + b*x^2)^p)^2*((9*a^2*p)/(8*b^2) - (3*p*x^4)/8 + (3*a*p*x^2)/(4*b 
)) - (3*p^3*x^4)/16 + log(c*(a + b*x^2)^p)*((3*p^2*x^4)/8 - (9*a*p^2*x^2)/ 
(4*b)) + log(c*(a + b*x^2)^p)^3*(x^4/4 - a^2/(4*b^2)) + (21*a*p^3*x^2)/(8* 
b) - (21*a^2*p^3*log(a + b*x^2))/(8*b^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.94 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-4 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{3} a^{2}+4 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{3} b^{2} x^{4}+18 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a^{2} p +12 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a b p \,x^{2}-6 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} b^{2} p \,x^{4}-42 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{2} p^{2}-36 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a b \,p^{2} x^{2}+6 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} p^{2} x^{4}+42 a b \,p^{3} x^{2}-3 b^{2} p^{3} x^{4}}{16 b^{2}} \] Input:

int(x^3*log(c*(b*x^2+a)^p)^3,x)
 

Output:

( - 4*log((a + b*x**2)**p*c)**3*a**2 + 4*log((a + b*x**2)**p*c)**3*b**2*x* 
*4 + 18*log((a + b*x**2)**p*c)**2*a**2*p + 12*log((a + b*x**2)**p*c)**2*a* 
b*p*x**2 - 6*log((a + b*x**2)**p*c)**2*b**2*p*x**4 - 42*log((a + b*x**2)** 
p*c)*a**2*p**2 - 36*log((a + b*x**2)**p*c)*a*b*p**2*x**2 + 6*log((a + b*x* 
*2)**p*c)*b**2*p**2*x**4 + 42*a*b*p**3*x**2 - 3*b**2*p**3*x**4)/(16*b**2)