3.3023 \(\int x^2 (a+b (c x^n)^{\frac{1}{n}})^p \, dx\)

Optimal. Leaf size=126 \[ \frac{a^2 x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^3 (p+1)}-\frac{2 a x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^3 (p+2)}+\frac{x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+3}}{b^3 (p+3)} \]

[Out]

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

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Rubi [A]  time = 0.0453388, antiderivative size = 126, 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} \[ \frac{a^2 x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^3 (p+1)}-\frac{2 a x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^3 (p+2)}+\frac{x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+3}}{b^3 (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

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

Mathematica [A]  time = 0.0494296, size = 95, normalized size = 0.75 \[ \frac{x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1} \left (2 a^2-2 a b (p+1) \left (c x^n\right )^{\frac{1}{n}}+b^2 \left (p^2+3 p+2\right ) \left (c x^n\right )^{2/n}\right )}{b^3 (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

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

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Maple [C]  time = 0.539, size = 2256, normalized size = 17.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*(c*x^n)^(1/n))^p,x)

[Out]

x^3/(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^2/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)-2
*x^2/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)+4*(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/(3+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)+4*(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+p)/(2+p)*a/(c^(1/n))^2/b^2/(1+p)^2*exp((I*Pi*csgn(I*c*x^n)*csg
n(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)+4*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/(3+p)/(2+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*Pi*cs
gn(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)-4*(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)/(3+p)*a^2/(c^(1/n))^3/(2+p)/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)-2*a/(1+p)^2/(c^(1/n))/b*((x^n)^(1/n)*c^(1/n)*b*exp(-1/2*I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x
^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+a)^p/((x^n)^(1/n))*x^3*exp(1/2*I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n)
)*(csgn(I*c*x^n)-csgn(I*c))/n)-2*a^2/(1+p)^2/(c^(1/n))^2/b^2*((x^n)^(1/n)*c^(1/n)*b*exp(-1/2*I*Pi*csgn(I*c*x^n
)*(csgn(I*c*x^n)-csgn(I*x^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+a)^p/((x^n)^(1/n))^2*x^3*exp(I*Pi*csgn(I*c*x^n)*(cs
gn(I*c*x^n)-csgn(I*x^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+2*a/(1+p)^2/(c^(1/n))/b*((x^n)^(1/n)*c^(1/n)*b*exp(-1/2*
I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+a)^p/(2+p)/((x^n)^(1/n))*x^3*exp(1
/2*I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+4*a^2/(1+p)^2/(c^(1/n))^2/b^2*(
(x^n)^(1/n)*c^(1/n)*b*exp(-1/2*I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+a)^
p/(2+p)/((x^n)^(1/n))^2*x^3*exp(I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n))*(csgn(I*c*x^n)-csgn(I*c))/n)+2*
a^3/(1+p)^2/(c^(1/n))^3/b^3*((x^n)^(1/n)*c^(1/n)*b*exp(-1/2*I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n))*(cs
gn(I*c*x^n)-csgn(I*c))/n)+a)^p/(2+p)/((x^n)^(1/n))^3*x^3*exp(3/2*I*Pi*csgn(I*c*x^n)*(csgn(I*c*x^n)-csgn(I*x^n)
)*(csgn(I*c*x^n)-csgn(I*c))/n)

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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^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*(c*x^n)^(1/n))^p,x, algorithm="maxima")

[Out]

integrate(((c*x^n)^(1/n)*b + a)^p*x^2, x)

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Fricas [A]  time = 1.57907, size = 244, normalized size = 1.94 \begin{align*} -\frac{{\left (2 \, a^{2} b c^{\left (\frac{1}{n}\right )} p x -{\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} c^{\frac{3}{n}} x^{3} -{\left (a b^{2} p^{2} + a b^{2} p\right )} c^{\frac{2}{n}} x^{2} - 2 \, a^{3}\right )}{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p}}{{\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )} c^{\frac{3}{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*(c*x^n)^(1/n))^p,x, algorithm="fricas")

[Out]

-(2*a^2*b*c^(1/n)*p*x - (b^3*p^2 + 3*b^3*p + 2*b^3)*c^(3/n)*x^3 - (a*b^2*p^2 + a*b^2*p)*c^(2/n)*x^2 - 2*a^3)*(
b*c^(1/n)*x + a)^p/((b^3*p^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)*c^(3/n))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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**2*(a+b*(c*x**n)**(1/n))**p,x)

[Out]

Integral(x**2*(a + b*(c*x**n)**(1/n))**p, x)

________________________________________________________________________________________

Giac [A]  time = 1.33438, size = 328, normalized size = 2.6 \begin{align*} \frac{{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{3} c^{\frac{3}{n}} p^{2} x^{3} +{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a b^{2} c^{\frac{2}{n}} p^{2} x^{2} + 3 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{3} c^{\frac{3}{n}} p x^{3} +{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a b^{2} c^{\frac{2}{n}} p x^{2} + 2 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{3} c^{\frac{3}{n}} x^{3} - 2 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{2} b c^{\left (\frac{1}{n}\right )} p x + 2 \,{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{3}}{b^{3} c^{\frac{3}{n}} p^{3} + 6 \, b^{3} c^{\frac{3}{n}} p^{2} + 11 \, b^{3} c^{\frac{3}{n}} p + 6 \, b^{3} c^{\frac{3}{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*(c*x^n)^(1/n))^p,x, algorithm="giac")

[Out]

((b*c^(1/n)*x + a)^p*b^3*c^(3/n)*p^2*x^3 + (b*c^(1/n)*x + a)^p*a*b^2*c^(2/n)*p^2*x^2 + 3*(b*c^(1/n)*x + a)^p*b
^3*c^(3/n)*p*x^3 + (b*c^(1/n)*x + a)^p*a*b^2*c^(2/n)*p*x^2 + 2*(b*c^(1/n)*x + a)^p*b^3*c^(3/n)*x^3 - 2*(b*c^(1
/n)*x + a)^p*a^2*b*c^(1/n)*p*x + 2*(b*c^(1/n)*x + a)^p*a^3)/(b^3*c^(3/n)*p^3 + 6*b^3*c^(3/n)*p^2 + 11*b^3*c^(3
/n)*p + 6*b^3*c^(3/n))