\(\int x^n \sqrt {a^2+x^{1+n}} \, dx\) [2754]

Optimal result

Integrand size = 17, antiderivative size = 22 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2 \left (a^2+x^{1+n}\right )^{3/2}}{3 (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {267} \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2 \left (a^2+x^{n+1}\right )^{3/2}}{3 (n+1)} \]

[In]

Int[x^n*Sqrt[a^2 + x^(1 + n)],x]

[Out]

(2*(a^2 + x^(1 + n))^(3/2))/(3*(1 + 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 {2 \left (a^2+x^{1+n}\right )^{3/2}}{3 (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2 \left (a^2+x^{1+n}\right )^{3/2}}{3 (1+n)} \]

[In]

Integrate[x^n*Sqrt[a^2 + x^(1 + n)],x]

[Out]

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

Maple [A] (verified)

Time = 3.86 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {2 \left (a^{2}+x \,x^{n}\right )^{\frac {3}{2}}}{3 \left (1+n \right )}\)	\(19\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2 \, {\left (a^{2} + x^{n + 1}\right )}^{\frac {3}{2}}}{3 \, {\left (n + 1\right )}} \]

[In]

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

[Out]

2/3*(a^2 + x^(n + 1))^(3/2)/(n + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).

Time = 0.45 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.64 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\begin {cases} \frac {2 a^{2} \sqrt {a^{2} + x^{n + 1}}}{3 n + 3} + \frac {2 x^{n + 1} \sqrt {a^{2} + x^{n + 1}}}{3 n + 3} & \text {for}\: n \neq -1 \\\sqrt {a^{2} + 1} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**n*(a**2+x**(1+n))**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2 \, {\left (a^{2} + x^{n + 1}\right )}^{\frac {3}{2}}}{3 \, {\left (n + 1\right )}} \]

[In]

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

[Out]

2/3*(a^2 + x^(n + 1))^(3/2)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2 \, {\left (a^{2} + x^{n + 1}\right )}^{\frac {3}{2}}}{3 \, {\left (n + 1\right )}} \]

[In]

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

[Out]

2/3*(a^2 + x^(n + 1))^(3/2)/(n + 1)

Mupad [B] (verification not implemented)

Time = 5.88 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int x^n \sqrt {a^2+x^{1+n}} \, dx=\frac {2\,{\left (x^{n+1}+a^2\right )}^{3/2}}{3\,\left (n+1\right )} \]

[In]

int(x^n*(x^(n + 1) + a^2)^(1/2),x)

[Out]

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