3.94 \(\int x^5 (a+b x^2) (a^2+2 a b x^2+b^2 x^4)^p \, dx\)

Optimal. Leaf size=128 \[ \frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+2)}-\frac {a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{b^3 (2 p+3)}+\frac {a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+1)} \]

[Out]

1/4*a^2*(b*x^2+a)^2*(b^2*x^4+2*a*b*x^2+a^2)^p/b^3/(1+p)-a*(b*x^2+a)^3*(b^2*x^4+2*a*b*x^2+a^2)^p/b^3/(3+2*p)+1/
4*(b*x^2+a)^4*(b^2*x^4+2*a*b*x^2+a^2)^p/b^3/(2+p)

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Rubi [A]  time = 0.13, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1249, 770, 21, 43} \[ \frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+2)}-\frac {a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{b^3 (2 p+3)}+\frac {a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

(a^2*(a + b*x^2)^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^p)/(4*b^3*(1 + p)) - (a*(a + b*x^2)^3*(a^2 + 2*a*b*x^2 + b^2*x^
4)^p)/(b^3*(3 + 2*p)) + ((a + b*x^2)^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^p)/(4*b^3*(2 + p))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 1249

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, S
ubst[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] &&
EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && IGtQ[(m + 1)/2, 0]

Rubi steps

\begin {align*} \int x^5 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,x^2\right )\\ &=\frac {1}{2} \left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int x^2 (a+b x) \left (a b+b^2 x\right )^{2 p} \, dx,x,x^2\right )\\ &=\frac {\left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int x^2 \left (a b+b^2 x\right )^{1+2 p} \, dx,x,x^2\right )}{2 b}\\ &=\frac {\left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int \left (\frac {a^2 \left (a b+b^2 x\right )^{1+2 p}}{b^2}-\frac {2 a \left (a b+b^2 x\right )^{2+2 p}}{b^3}+\frac {\left (a b+b^2 x\right )^{3+2 p}}{b^4}\right ) \, dx,x,x^2\right )}{2 b}\\ &=\frac {a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (1+p)}-\frac {a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{b^3 (3+2 p)}+\frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (2+p)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 68, normalized size = 0.53 \[ \frac {\left (\left (a+b x^2\right )^2\right )^{p+1} \left (a^2-2 a b (p+1) x^2+b^2 \left (2 p^2+5 p+3\right ) x^4\right )}{4 b^3 (p+1) (p+2) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

(((a + b*x^2)^2)^(1 + p)*(a^2 - 2*a*b*(1 + p)*x^2 + b^2*(3 + 5*p + 2*p^2)*x^4))/(4*b^3*(1 + p)*(2 + p)*(3 + 2*
p))

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fricas [A]  time = 0.69, size = 140, normalized size = 1.09 \[ \frac {{\left ({\left (2 \, b^{4} p^{2} + 5 \, b^{4} p + 3 \, b^{4}\right )} x^{8} - 2 \, a^{3} b p x^{2} + 4 \, {\left (a b^{3} p^{2} + 2 \, a b^{3} p + a b^{3}\right )} x^{6} + {\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} + a^{4}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (2 \, b^{3} p^{3} + 9 \, b^{3} p^{2} + 13 \, b^{3} p + 6 \, b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

1/4*((2*b^4*p^2 + 5*b^4*p + 3*b^4)*x^8 - 2*a^3*b*p*x^2 + 4*(a*b^3*p^2 + 2*a*b^3*p + a*b^3)*x^6 + (2*a^2*b^2*p^
2 + a^2*b^2*p)*x^4 + a^4)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(2*b^3*p^3 + 9*b^3*p^2 + 13*b^3*p + 6*b^3)

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giac [B]  time = 0.38, size = 331, normalized size = 2.59 \[ \frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{2} x^{8} + 5 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p x^{8} + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{2} x^{6} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} x^{8} + 8 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p x^{6} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{4} + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} x^{6} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p x^{4} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3} b p x^{2} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{4}}{4 \, {\left (2 \, b^{3} p^{3} + 9 \, b^{3} p^{2} + 13 \, b^{3} p + 6 \, b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

1/4*(2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p^2*x^8 + 5*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p*x^8 + 4*(b^2*x^4 + 2*
a*b*x^2 + a^2)^p*a*b^3*p^2*x^6 + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*x^8 + 8*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b
^3*p*x^6 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b^2*p^2*x^4 + 4*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b^3*x^6 + (b^2*
x^4 + 2*a*b*x^2 + a^2)^p*a^2*b^2*p*x^4 - 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^3*b*p*x^2 + (b^2*x^4 + 2*a*b*x^2 +
a^2)^p*a^4)/(2*b^3*p^3 + 9*b^3*p^2 + 13*b^3*p + 6*b^3)

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maple [A]  time = 0.01, size = 99, normalized size = 0.77 \[ \frac {\left (b \,x^{2}+a \right )^{2} \left (2 b^{2} p^{2} x^{4}+5 b^{2} p \,x^{4}+3 b^{2} x^{4}-2 a b p \,x^{2}-2 a b \,x^{2}+a^{2}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (2 p^{3}+9 p^{2}+13 p +6\right ) b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

1/4*(b*x^2+a)^2*(2*b^2*p^2*x^4+5*b^2*p*x^4+3*b^2*x^4-2*a*b*p*x^2-2*a*b*x^2+a^2)*(b^2*x^4+2*a*b*x^2+a^2)^p/b^3/
(2*p^3+9*p^2+13*p+6)

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maxima [A]  time = 0.70, size = 196, normalized size = 1.53 \[ \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )} {\left (b x^{2} + a\right )}^{2 \, p} a}{2 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} + \frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{8} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{6} - 3 \, {\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 3 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{4 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

1/2*((2*p^2 + 3*p + 1)*b^3*x^6 + (2*p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + a^3)*(b*x^2 + a)^(2*p)*a/((4*p^3 + 12
*p^2 + 11*p + 3)*b^3) + 1/4*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^8 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^6 - 3*(2*p^2
+ p)*a^2*b^2*x^4 + 6*a^3*b*p*x^2 - 3*a^4)*(b*x^2 + a)^(2*p)/((4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^3)

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mupad [B]  time = 0.20, size = 169, normalized size = 1.32 \[ {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {a^4}{4\,b^3\,\left (2\,p^3+9\,p^2+13\,p+6\right )}+\frac {a\,x^6\,{\left (p+1\right )}^2}{2\,p^3+9\,p^2+13\,p+6}+\frac {b\,x^8\,\left (2\,p^2+5\,p+3\right )}{4\,\left (2\,p^3+9\,p^2+13\,p+6\right )}-\frac {a^3\,p\,x^2}{2\,b^2\,\left (2\,p^3+9\,p^2+13\,p+6\right )}+\frac {a^2\,p\,x^4\,\left (2\,p+1\right )}{4\,b\,\left (2\,p^3+9\,p^2+13\,p+6\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^p,x)

[Out]

(a^2 + b^2*x^4 + 2*a*b*x^2)^p*(a^4/(4*b^3*(13*p + 9*p^2 + 2*p^3 + 6)) + (a*x^6*(p + 1)^2)/(13*p + 9*p^2 + 2*p^
3 + 6) + (b*x^8*(5*p + 2*p^2 + 3))/(4*(13*p + 9*p^2 + 2*p^3 + 6)) - (a^3*p*x^2)/(2*b^2*(13*p + 9*p^2 + 2*p^3 +
6)) + (a^2*p*x^4*(2*p + 1))/(4*b*(13*p + 9*p^2 + 2*p^3 + 6)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {a x^{6} \left (a^{2}\right )^{p}}{6} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {3 a^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} & \text {for}\: p = -2 \\\int \frac {x^{5} \left (a + b x^{2}\right )}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\\frac {a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{3}} + \frac {a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{4}}{4 b} & \text {for}\: p = -1 \\\frac {a^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} - \frac {2 a^{3} b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {2 a^{2} b^{2} p^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {a^{2} b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {4 a b^{3} p^{2} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {8 a b^{3} p x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {4 a b^{3} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {2 b^{4} p^{2} x^{8} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {5 b^{4} p x^{8} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} + \frac {3 b^{4} x^{8} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 36 b^{3} p^{2} + 52 b^{3} p + 24 b^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Piecewise((a*x**6*(a**2)**p/6, Eq(b, 0)), (2*a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 +
4*b**5*x**4) + 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 3*a**2/(4*a*
*2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2
+ 4*b**5*x**4) + 4*a*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*
x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*
a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x*
*4), Eq(p, -2)), (Integral(x**5*(a + b*x**2)/((a + b*x**2)**2)**(3/2), x), Eq(p, -3/2)), (a**2*log(-I*sqrt(a)*
sqrt(1/b) + x)/(2*b**3) + a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)
), (a**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) - 2*a**3*b*p*x*
*2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 2*a**2*b**2*p**2*x*
*4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + a**2*b**2*p*x**4*(a
**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 4*a*b**3*p**2*x**6*(a**2
+ 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 8*a*b**3*p*x**6*(a**2 + 2*a
*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 4*a*b**3*x**6*(a**2 + 2*a*b*x**2
+ b**2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 2*b**4*p**2*x**8*(a**2 + 2*a*b*x**2 + b**
2*x**4)**p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 5*b**4*p*x**8*(a**2 + 2*a*b*x**2 + b**2*x**4)*
*p/(8*b**3*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3) + 3*b**4*x**8*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3
*p**3 + 36*b**3*p**2 + 52*b**3*p + 24*b**3), True))

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