3.2721 \(\int x^{-1+2 n} (a+b x^n)^p \, dx\)

Optimal. Leaf size=49 \[ \frac{\left (a+b x^n\right )^{p+2}}{b^2 n (p+2)}-\frac{a \left (a+b x^n\right )^{p+1}}{b^2 n (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0316899, antiderivative size = 49, 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 = {266, 43} \[ \frac{\left (a+b x^n\right )^{p+2}}{b^2 n (p+2)}-\frac{a \left (a+b x^n\right )^{p+1}}{b^2 n (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*(a + b*x^n)^p,x]

[Out]

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

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

Mathematica [A]  time = 0.0231275, size = 40, normalized size = 0.82 \[ \frac{\left (a+b x^n\right )^{p+1} \left (b (p+1) x^n-a\right )}{b^2 n (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 61, normalized size = 1.2 \begin{align*} -{\frac{ \left ( -{b}^{2}p \left ({x}^{n} \right ) ^{2}-ap{x}^{n}b-{b}^{2} \left ({x}^{n} \right ) ^{2}+{a}^{2} \right ) \left ( a+b{x}^{n} \right ) ^{p}}{ \left ( 1+p \right ) \left ( 2+p \right ) n{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0158, size = 69, normalized size = 1.41 \begin{align*} \frac{{\left (b^{2}{\left (p + 1\right )} x^{2 \, n} + a b p x^{n} - a^{2}\right )}{\left (b x^{n} + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.3653, size = 123, normalized size = 2.51 \begin{align*} \frac{{\left (a b p x^{n} - a^{2} +{\left (b^{2} p + b^{2}\right )} x^{2 \, n}\right )}{\left (b x^{n} + a\right )}^{p}}{b^{2} n p^{2} + 3 \, b^{2} n p + 2 \, b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)*(a+b*x**n)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{n} + a\right )}^{p} x^{2 \, n - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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