\(\int x^{-1+n} (a+b x^n)^2 \, dx\) [2531]

Optimal result

Integrand size = 15, antiderivative size = 19 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {\left (a+b x^n\right )^3}{3 b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {\left (a+b x^n\right )^3}{3 b n} \]

[In]

Int[x^(-1 + n)*(a + b*x^n)^2,x]

[Out]

(a + b*x^n)^3/(3*b*n)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x^n\right )^3}{3 b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {\left (a+b x^n\right )^3}{3 b n} \]

[In]

Integrate[x^(-1 + n)*(a + b*x^n)^2,x]

[Out]

(a + b*x^n)^3/(3*b*n)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(35\) vs. \(2(17)=34\).

Time = 3.75 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89

method	result	size
risch	\(\frac {a^{2} x^{n}}{n}+\frac {a b \,x^{2 n}}{n}+\frac {b^{2} x^{3 n}}{3 n}\)	\(36\)
norman	\(\frac {a^{2} {\mathrm e}^{n \ln \left (x \right )}}{n}+\frac {a b \,{\mathrm e}^{2 n \ln \left (x \right )}}{n}+\frac {b^{2} {\mathrm e}^{3 n \ln \left (x \right )}}{3 n}\)	\(42\)
parallelrisch	\(\frac {x \,x^{2 n} x^{-1+n} b^{2}+3 x \,x^{n} x^{-1+n} a b +3 x \,x^{-1+n} a^{2}}{3 n}\)	\(46\)

[In]

int(x^(-1+n)*(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {b^{2} x^{3 \, n} + 3 \, a b x^{2 \, n} + 3 \, a^{2} x^{n}}{3 \, n} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (12) = 24\).

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.79 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\begin {cases} \frac {a^{2} x x^{n - 1}}{n} + \frac {a b x x^{n} x^{n - 1}}{n} + \frac {b^{2} x x^{2 n} x^{n - 1}}{3 n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{2} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*x*x**(n - 1)/n + a*b*x*x**n*x**(n - 1)/n + b**2*x*x**(2*n)*x**(n - 1)/(3*n), Ne(n, 0)), ((a +
b)**2*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {{\left (b x^{n} + a\right )}^{3}}{3 \, b n} \]

[In]

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

[Out]

1/3*(b*x^n + a)^3/(b*n)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {b^{2} x^{3 \, n} + 3 \, a b x^{2 \, n} + 3 \, a^{2} x^{n}}{3 \, n} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int x^{-1+n} \left (a+b x^n\right )^2 \, dx=\frac {x^n\,\left (a^2+\frac {b^2\,x^{2\,n}}{3}+a\,b\,x^n\right )}{n} \]

[In]

int(x^(n - 1)*(a + b*x^n)^2,x)

[Out]

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