3.3024 \(\int x (a+b (c x^n)^{\frac{1}{n}})^p \, dx\)

Optimal. Leaf size=83 \[ \frac{x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^2 (p+2)}-\frac{a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0303635, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {368, 43} \[ \frac{x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^2 (p+2)}-\frac{a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

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

Mathematica [A]  time = 0.0284865, size = 63, normalized size = 0.76 \[ \frac{x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1} \left (b (p+1) \left (c x^n\right )^{\frac{1}{n}}-a\right )}{b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.291, size = 1007, normalized size = 12.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(1+p)*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(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/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)-x
/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)+1/(1+p)^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(c)-2*ln(x^n))/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*P
i*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x^n))/n)*x+a)^(2+p)/(2+p)-(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*c
sgn(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)*a/(1+p)^2/(c^(1/n))/b^2*exp((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*n*ln(x)-2*ln(x^n)-ln(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\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.56547, size = 150, normalized size = 1.81 \begin{align*} \frac{{\left (a b c^{\left (\frac{1}{n}\right )} p x +{\left (b^{2} p + b^{2}\right )} c^{\frac{2}{n}} x^{2} - a^{2}\right )}{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p}}{{\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )} c^{\frac{2}{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.5671, size = 184, normalized size = 2.22 \begin{align*} \frac{{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{2} c^{\frac{2}{n}} p x^{2} +{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a b c^{\left (\frac{1}{n}\right )} p x +{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} b^{2} c^{\frac{2}{n}} x^{2} -{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} a^{2}}{b^{2} c^{\frac{2}{n}} p^{2} + 3 \, b^{2} c^{\frac{2}{n}} p + 2 \, b^{2} c^{\frac{2}{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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