3.125 \(\int x^{11} (a^2+2 a b x^3+b^2 x^6)^p \, dx\)
Optimal. Leaf size=172 \[ \frac {\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (p+2)}-\frac {a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (2 p+3)}+\frac {a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (p+1)}-\frac {a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (2 p+1)} \]
[Out]
-1/3*a^3*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(1+2*p)+1/2*a^2*(b*x^3+a)^2*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(1+
p)-a*(b*x^3+a)^3*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(3+2*p)+1/6*(b*x^3+a)^4*(b^2*x^6+2*a*b*x^3+a^2)^p/b^4/(2+p)
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Rubi [A] time = 0.11, antiderivative size = 172, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used =
{1356, 266, 43} \[ \frac {\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (p+2)}-\frac {a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (2 p+3)}+\frac {a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (p+1)}-\frac {a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (2 p+1)} \]
Antiderivative was successfully verified.
[In]
Int[x^11*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]
[Out]
-(a^3*(a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(3*b^4*(1 + 2*p)) + (a^2*(a + b*x^3)^2*(a^2 + 2*a*b*x^3 + b^2
*x^6)^p)/(2*b^4*(1 + p)) - (a*(a + b*x^3)^3*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(b^4*(3 + 2*p)) + ((a + b*x^3)^4*(a
^2 + 2*a*b*x^3 + b^2*x^6)^p)/(6*b^4*(2 + p))
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rule 266
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 1356
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a
+ b*x^n + c*x^(2*n))^FracPart[p])/(1 + (2*c*x^n)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^n)/b)^(2*p), x],
x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[2*p]
Rubi steps
\begin {align*} \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx &=\left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int x^{11} \left (1+\frac {b x^3}{a}\right )^{2 p} \, dx\\ &=\frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \operatorname {Subst}\left (\int x^3 \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,x^3\right )\\ &=\frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \operatorname {Subst}\left (\int \left (-\frac {a^3 \left (1+\frac {b x}{a}\right )^{2 p}}{b^3}+\frac {3 a^3 \left (1+\frac {b x}{a}\right )^{1+2 p}}{b^3}-\frac {3 a^3 \left (1+\frac {b x}{a}\right )^{2+2 p}}{b^3}+\frac {a^3 \left (1+\frac {b x}{a}\right )^{3+2 p}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac {a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (1+2 p)}+\frac {a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (1+p)}-\frac {a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (3+2 p)}+\frac {\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (2+p)}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 110, normalized size = 0.64 \[ \frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (-3 a^3+3 a^2 b (2 p+1) x^3-3 a b^2 \left (2 p^2+3 p+1\right ) x^6+b^3 \left (4 p^3+12 p^2+11 p+3\right ) x^9\right )}{6 b^4 (p+1) (p+2) (2 p+1) (2 p+3)} \]
Antiderivative was successfully verified.
[In]
Integrate[x^11*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]
[Out]
((a + b*x^3)*((a + b*x^3)^2)^p*(-3*a^3 + 3*a^2*b*(1 + 2*p)*x^3 - 3*a*b^2*(1 + 3*p + 2*p^2)*x^6 + b^3*(3 + 11*p
+ 12*p^2 + 4*p^3)*x^9))/(6*b^4*(1 + p)*(2 + p)*(1 + 2*p)*(3 + 2*p))
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fricas [A] time = 0.94, size = 163, normalized size = 0.95 \[ \frac {{\left ({\left (4 \, b^{4} p^{3} + 12 \, b^{4} p^{2} + 11 \, b^{4} p + 3 \, b^{4}\right )} x^{12} + 2 \, {\left (2 \, a b^{3} p^{3} + 3 \, a b^{3} p^{2} + a b^{3} p\right )} x^{9} + 6 \, a^{3} b p x^{3} - 3 \, {\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{6} - 3 \, a^{4}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{6 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="fricas")
[Out]
1/6*((4*b^4*p^3 + 12*b^4*p^2 + 11*b^4*p + 3*b^4)*x^12 + 2*(2*a*b^3*p^3 + 3*a*b^3*p^2 + a*b^3*p)*x^9 + 6*a^3*b*
p*x^3 - 3*(2*a^2*b^2*p^2 + a^2*b^2*p)*x^6 - 3*a^4)*(b^2*x^6 + 2*a*b*x^3 + a^2)^p/(4*b^4*p^4 + 20*b^4*p^3 + 35*
b^4*p^2 + 25*b^4*p + 6*b^4)
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giac [B] time = 0.53, size = 375, normalized size = 2.18 \[ \frac {4 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p^{3} x^{12} + 12 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p^{2} x^{12} + 11 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p x^{12} + 4 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p^{3} x^{9} + 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} x^{12} + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p^{2} x^{9} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p x^{9} - 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{6} - 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b^{2} p x^{6} + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{3} b p x^{3} - 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{4}}{6 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="giac")
[Out]
1/6*(4*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^4*p^3*x^12 + 12*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^4*p^2*x^12 + 11*(b^2*x^
6 + 2*a*b*x^3 + a^2)^p*b^4*p*x^12 + 4*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^3*p^3*x^9 + 3*(b^2*x^6 + 2*a*b*x^3 + a
^2)^p*b^4*x^12 + 6*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^3*p^2*x^9 + 2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^3*p*x^9 -
6*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^2*b^2*p^2*x^6 - 3*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^2*b^2*p*x^6 + 6*(b^2*x^6
+ 2*a*b*x^3 + a^2)^p*a^3*b*p*x^3 - 3*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^4)/(4*b^4*p^4 + 20*b^4*p^3 + 35*b^4*p^2 +
25*b^4*p + 6*b^4)
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maple [A] time = 0.01, size = 150, normalized size = 0.87 \[ -\frac {\left (-4 b^{3} p^{3} x^{9}-12 b^{3} p^{2} x^{9}-11 b^{3} p \,x^{9}-3 b^{3} x^{9}+6 a \,b^{2} p^{2} x^{6}+9 a \,b^{2} p \,x^{6}+3 a \,b^{2} x^{6}-6 a^{2} b p \,x^{3}-3 a^{2} b \,x^{3}+3 a^{3}\right ) \left (b \,x^{3}+a \right ) \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{6 \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right ) b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x)
[Out]
-1/6*(b^2*x^6+2*a*b*x^3+a^2)^p*(-4*b^3*p^3*x^9-12*b^3*p^2*x^9-11*b^3*p*x^9-3*b^3*x^9+6*a*b^2*p^2*x^6+9*a*b^2*p
*x^6+3*a*b^2*x^6-6*a^2*b*p*x^3-3*a^2*b*x^3+3*a^3)*(b*x^3+a)/b^4/(4*p^4+20*p^3+35*p^2+25*p+6)
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maxima [A] time = 0.81, size = 115, normalized size = 0.67 \[ \frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{12} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{9} - 3 \, {\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{6} + 6 \, a^{3} b p x^{3} - 3 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{6 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="maxima")
[Out]
1/6*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^12 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^9 - 3*(2*p^2 + p)*a^2*b^2*x^6 + 6*a^
3*b*p*x^3 - 3*a^4)*(b*x^3 + a)^(2*p)/((4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^4)
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mupad [B] time = 1.31, size = 207, normalized size = 1.20 \[ {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^{12}\,\left (4\,p^3+12\,p^2+11\,p+3\right )}{6\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {a^4}{2\,b^4\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a^3\,p\,x^3}{b^3\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a\,p\,x^9\,\left (2\,p^2+3\,p+1\right )}{3\,b\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {a^2\,p\,x^6\,\left (2\,p+1\right )}{2\,b^2\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^11*(a^2 + b^2*x^6 + 2*a*b*x^3)^p,x)
[Out]
(a^2 + b^2*x^6 + 2*a*b*x^3)^p*((x^12*(11*p + 12*p^2 + 4*p^3 + 3))/(6*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) - a
^4/(2*b^4*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) + (a^3*p*x^3)/(b^3*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) + (a*
p*x^9*(3*p + 2*p^2 + 1))/(3*b*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) - (a^2*p*x^6*(2*p + 1))/(2*b^2*(25*p + 35*
p^2 + 20*p^3 + 4*p^4 + 6)))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**11*(b**2*x**6+2*a*b*x**3+a**2)**p,x)
[Out]
Piecewise((x**12*(a**2)**p/12, Eq(b, 0)), (6*a**3*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(18*a**3*b**4 +
54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 6*a**3*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**
(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 1
2*a**3*log(2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 11*a**3/(18*a**3*b**4 + 54*
a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a**2*b*x**3*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(
18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a**2*b*x**3*log(4*(-1)**(2/3)*a**(2/3)*
(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x
**6 + 18*b**7*x**9) - 36*a**2*b*x**3*log(2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9)
+ 27*a**2*b*x**3/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2*x**6*log(-(-1
)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*
b**2*x**6*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(18*a**3*b
**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 36*a*b**2*x**6*log(2)/(18*a**3*b**4 + 54*a**2*b**5*
x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2*x**6/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18
*b**7*x**9) + 6*b**3*x**9*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a
*b**6*x**6 + 18*b**7*x**9) + 6*b**3*x**9*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1
/b)**(1/3) + 4*x**2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 12*b**3*x**9*log(2)/
(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9), Eq(p, -2)), (Integral(x**11/((a + b*x**3)*
*2)**(3/2), x), Eq(p, -3/2)), (6*a**3*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(6*a*b**4 + 6*b**5*x**3) + 6
*a**3*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(6*a*b**4 + 6*
b**5*x**3) - 12*a**3*log(2)/(6*a*b**4 + 6*b**5*x**3) + 6*a**3/(6*a*b**4 + 6*b**5*x**3) + 6*a**2*b*x**3*log(-(-
1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(6*a*b**4 + 6*b**5*x**3) + 6*a**2*b*x**3*log(4*(-1)**(2/3)*a**(2/3)*(1/b)
**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(6*a*b**4 + 6*b**5*x**3) - 12*a**2*b*x**3*log(2)/(6*
a*b**4 + 6*b**5*x**3) - 3*a*b**2*x**6/(6*a*b**4 + 6*b**5*x**3) + b**3*x**9/(6*a*b**4 + 6*b**5*x**3), Eq(p, -1)
), (Integral(x**11/sqrt((a + b*x**3)**2), x), Eq(p, -1/2)), (-3*a**4*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b*
*4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) + 6*a**3*b*p*x**3*(a**2 + 2*a*b*x**3 + b**2*x*
*6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) - 6*a**2*b**2*p**2*x**6*(a**2 + 2
*a*b*x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) - 3*a**2*b**2*
p*x**6*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4
) + 4*a*b**3*p**3*x**9*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*
b**4*p + 36*b**4) + 6*a*b**3*p**2*x**9*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*
b**4*p**2 + 150*b**4*p + 36*b**4) + 2*a*b**3*p*x**9*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**
4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) + 4*b**4*p**3*x**12*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b**4
*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) + 12*b**4*p**2*x**12*(a**2 + 2*a*b*x**3 + b**2*x
**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) + 11*b**4*p*x**12*(a**2 + 2*a*b*
x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4) + 3*b**4*x**12*(a**
2 + 2*a*b*x**3 + b**2*x**6)**p/(24*b**4*p**4 + 120*b**4*p**3 + 210*b**4*p**2 + 150*b**4*p + 36*b**4), True))
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