3.3022 \(\int x^3 (a+b (c x^n)^{\frac{1}{n}})^p \, dx\)
Optimal. Leaf size=171 \[ -\frac{a^3 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^4 (p+1)}+\frac{3 a^2 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^4 (p+2)}-\frac{3 a x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+3}}{b^4 (p+3)}+\frac{x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+4}}{b^4 (p+4)} \]
[Out]
-((a^3*x^4*(a + b*(c*x^n)^n^(-1))^(1 + p))/(b^4*(1 + p)*(c*x^n)^(4/n))) + (3*a^2*x^4*(a + b*(c*x^n)^n^(-1))^(2
+ p))/(b^4*(2 + p)*(c*x^n)^(4/n)) - (3*a*x^4*(a + b*(c*x^n)^n^(-1))^(3 + p))/(b^4*(3 + p)*(c*x^n)^(4/n)) + (x
^4*(a + b*(c*x^n)^n^(-1))^(4 + p))/(b^4*(4 + p)*(c*x^n)^(4/n))
________________________________________________________________________________________
Rubi [A] time = 0.063625, antiderivative size = 171, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used =
{368, 43} \[ -\frac{a^3 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^4 (p+1)}+\frac{3 a^2 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^4 (p+2)}-\frac{3 a x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+3}}{b^4 (p+3)}+\frac{x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+4}}{b^4 (p+4)} \]
Antiderivative was successfully verified.
[In]
Int[x^3*(a + b*(c*x^n)^n^(-1))^p,x]
[Out]
-((a^3*x^4*(a + b*(c*x^n)^n^(-1))^(1 + p))/(b^4*(1 + p)*(c*x^n)^(4/n))) + (3*a^2*x^4*(a + b*(c*x^n)^n^(-1))^(2
+ p))/(b^4*(2 + p)*(c*x^n)^(4/n)) - (3*a*x^4*(a + b*(c*x^n)^n^(-1))^(3 + p))/(b^4*(3 + p)*(c*x^n)^(4/n)) + (x
^4*(a + b*(c*x^n)^n^(-1))^(4 + p))/(b^4*(4 + p)*(c*x^n)^(4/n))
Rule 368
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
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])
Rubi steps
\begin{align*} \int x^3 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^p \, dx &=\left (x^4 \left (c x^n\right )^{-4/n}\right ) \operatorname{Subst}\left (\int x^3 (a+b x)^p \, dx,x,\left (c x^n\right )^{\frac{1}{n}}\right )\\ &=\left (x^4 \left (c x^n\right )^{-4/n}\right ) \operatorname{Subst}\left (\int \left (-\frac{a^3 (a+b x)^p}{b^3}+\frac{3 a^2 (a+b x)^{1+p}}{b^3}-\frac{3 a (a+b x)^{2+p}}{b^3}+\frac{(a+b x)^{3+p}}{b^3}\right ) \, dx,x,\left (c x^n\right )^{\frac{1}{n}}\right )\\ &=-\frac{a^3 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{1+p}}{b^4 (1+p)}+\frac{3 a^2 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{2+p}}{b^4 (2+p)}-\frac{3 a x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{3+p}}{b^4 (3+p)}+\frac{x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{4+p}}{b^4 (4+p)}\\ \end{align*}
Mathematica [A] time = 0.11799, size = 113, normalized size = 0.66 \[ \frac{x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1} \left (\frac{3 a^2 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{p+2}-\frac{a^3}{p+1}-\frac{3 a \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^2}{p+3}+\frac{\left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^3}{p+4}\right )}{b^4} \]
Antiderivative was successfully verified.
[In]
Integrate[x^3*(a + b*(c*x^n)^n^(-1))^p,x]
[Out]
(x^4*(a + b*(c*x^n)^n^(-1))^(1 + p)*(-(a^3/(1 + p)) + (3*a^2*(a + b*(c*x^n)^n^(-1)))/(2 + p) - (3*a*(a + b*(c*
x^n)^n^(-1))^2)/(3 + p) + (a + b*(c*x^n)^n^(-1))^3/(4 + p)))/(b^4*(c*x^n)^(4/n))
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Maple [C] time = 0.524, size = 2920, normalized size = 17.1 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(a+b*(c*x^n)^(1/n))^p,x)
[Out]
x^4/(1+p)*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c
*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^p+(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*
csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n
*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^p*a/(1+p)/(c^(1/n))*x^3/b*exp(1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-
I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)-3
*x^3/b*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^
n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)/(1+p)^2*exp(-1/2*(-I*Pi*csgn(I*
c*x^n)^3+I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x
^n)+2*ln(c)+2*ln(x^n)-2*n*ln(x))/n)+9*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)
^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)*
x^3/(4+p)/(c^(1/n))/b/(1+p)^2*exp(1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^
n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)+9*(b*exp(-1/2*(I*Pi*csgn(I*c*x^
n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+
2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)*x^2/(4+p)/(3+p)*a/(c^(1/n))^2/b^2/(1+p)^2*exp((I*Pi*csgn(I*c*x^n)*c
sgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*
ln(x)-2*ln(x^n))/n)+9*p*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)
-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)*x^2/(4+p)/(3+p
)*a/(c^(1/n))^2/b^2/(1+p)^2*exp((I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*P
i*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)+18*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*cs
gn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*l
n(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)/(4+p)*a^3/(2+p)/(c^(1/n))^4/(3+p)/b^4/(1+p)^2*exp(2*(I*Pi*csgn(I*c*x^n)*
csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n
*ln(x)-2*ln(x^n))/n)-18*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)
-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)*x/(4+p)/(2+p)/
(3+p)*a^2/(c^(1/n))^3/b^3/(1+p)^2*exp(3/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(
I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)-18*p*(b*exp(-1/2*(I*Pi*csgn
(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*
x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(1+p)*x/(4+p)/(2+p)/(3+p)*a^2/(c^(1/n))^3/b^3/(1+p)^2*exp(3/2*(I*P
i*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csg
n(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)-3*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n
)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^p*x^3
/(3+p)*a/(1+p)/(c^(1/n))/b*exp(1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-
I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)-3*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*
csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n
*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^p*x^2/(3+p)/(2+p)*p*a^2/(1+p)/(c^(1/n))^2/b^2*exp((I*Pi*csgn(I*c*x^n)*csgn(I
*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)
-2*ln(x^n))/n)-6*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*c
sgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^p/(3+p)/(2+p)*a^4/(1+p)^2/(
c^(1/n))^4/b^4*exp(2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*
x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)+6*(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn
(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)
-2*ln(x^n))/n)*x+a)^p*x*p/(3+p)/(2+p)*a^3/(c^(1/n))^3/b^3/(1+p)^2*exp(3/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I
*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n)
)/n)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\left (c x^{n}\right )^{\left (\frac{1}{n}\right )} b + a\right )}^{p} x^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*(c*x^n)^(1/n))^p,x, algorithm="maxima")
[Out]
integrate(((c*x^n)^(1/n)*b + a)^p*x^3, x)
________________________________________________________________________________________
Fricas [A] time = 1.60411, size = 359, normalized size = 2.1 \begin{align*} \frac{{\left (6 \, a^{3} b c^{\left (\frac{1}{n}\right )} p x +{\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} c^{\frac{4}{n}} x^{4} +{\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} c^{\frac{3}{n}} x^{3} - 6 \, a^{4} - 3 \,{\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} c^{\frac{2}{n}} x^{2}\right )}{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p}}{{\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )} c^{\frac{4}{n}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*(c*x^n)^(1/n))^p,x, algorithm="fricas")
[Out]
(6*a^3*b*c^(1/n)*p*x + (b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*c^(4/n)*x^4 + (a*b^3*p^3 + 3*a*b^3*p^2 + 2*a*b
^3*p)*c^(3/n)*x^3 - 6*a^4 - 3*(a^2*b^2*p^2 + a^2*b^2*p)*c^(2/n)*x^2)*(b*c^(1/n)*x + a)^p/((b^4*p^4 + 10*b^4*p^
3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4)*c^(4/n))
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \left (c x^{n}\right )^{\frac{1}{n}}\right )^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(a+b*(c*x**n)**(1/n))**p,x)
[Out]
Integral(x**3*(a + b*(c*x**n)**(1/n))**p, x)
________________________________________________________________________________________
Giac [B] time = 1.33463, size = 518, normalized size = 3.03 \begin{align*} \frac{{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac{4}{n}} p^{3} x^{4} +{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac{3}{n}} p^{3} x^{3} + 6 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac{4}{n}} p^{2} x^{4} + 3 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac{3}{n}} p^{2} x^{3} + 11 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac{4}{n}} p x^{4} - 3 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{2} b^{2} c^{\frac{2}{n}} p^{2} x^{2} + 2 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac{3}{n}} p x^{3} + 6 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac{4}{n}} x^{4} - 3 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{2} b^{2} c^{\frac{2}{n}} p x^{2} + 6 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{3} b c^{\left (\frac{1}{n}\right )} p x - 6 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{4}}{b^{4} c^{\frac{4}{n}} p^{4} + 10 \, b^{4} c^{\frac{4}{n}} p^{3} + 35 \, b^{4} c^{\frac{4}{n}} p^{2} + 50 \, b^{4} c^{\frac{4}{n}} p + 24 \, b^{4} c^{\frac{4}{n}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*(c*x^n)^(1/n))^p,x, algorithm="giac")
[Out]
((b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p^3*x^4 + (b*c^(1/n)*x + a)^p*a*b^3*c^(3/n)*p^3*x^3 + 6*(b*c^(1/n)*x + a)^p*b
^4*c^(4/n)*p^2*x^4 + 3*(b*c^(1/n)*x + a)^p*a*b^3*c^(3/n)*p^2*x^3 + 11*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p*x^4 -
3*(b*c^(1/n)*x + a)^p*a^2*b^2*c^(2/n)*p^2*x^2 + 2*(b*c^(1/n)*x + a)^p*a*b^3*c^(3/n)*p*x^3 + 6*(b*c^(1/n)*x + a
)^p*b^4*c^(4/n)*x^4 - 3*(b*c^(1/n)*x + a)^p*a^2*b^2*c^(2/n)*p*x^2 + 6*(b*c^(1/n)*x + a)^p*a^3*b*c^(1/n)*p*x -
6*(b*c^(1/n)*x + a)^p*a^4)/(b^4*c^(4/n)*p^4 + 10*b^4*c^(4/n)*p^3 + 35*b^4*c^(4/n)*p^2 + 50*b^4*c^(4/n)*p + 24*
b^4*c^(4/n))