\(\int x^{1+2 n} \, dx\) [463]

Optimal result

Integrand size = 7, antiderivative size = 16 \[ \int x^{1+2 n} \, dx=\frac {x^{2 (1+n)}}{2 (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {30} \[ \int x^{1+2 n} \, dx=\frac {x^{2 (n+1)}}{2 (n+1)} \]

[In]

Int[x^(1 + 2*n),x]

[Out]

x^(2*(1 + n))/(2*(1 + n))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{2 (1+n)}}{2 (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int x^{1+2 n} \, dx=\frac {x^{2+2 n}}{2+2 n} \]

[In]

Integrate[x^(1 + 2*n),x]

[Out]

x^(2 + 2*n)/(2 + 2*n)

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
gosper	\(\frac {x^{2+2 n}}{2+2 n}\)	\(15\)
default	\(\frac {x^{2+2 n}}{2+2 n}\)	\(16\)
risch	\(\frac {x \,x^{1+2 n}}{2+2 n}\)	\(16\)
parallelrisch	\(\frac {x \,x^{1+2 n}}{2+2 n}\)	\(16\)
norman	\(\frac {x \,{\mathrm e}^{\left (1+2 n \right ) \ln \left (x \right )}}{2+2 n}\)	\(18\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int x^{1+2 n} \, dx=\frac {x x^{2 \, n + 1}}{2 \, {\left (n + 1\right )}} \]

[In]

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

[Out]

1/2*x*x^(2*n + 1)/(n + 1)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int x^{1+2 n} \, dx=\begin {cases} \frac {x^{2 n + 2}}{2 n + 2} & \text {for}\: n \neq -1 \\\log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int x^{1+2 n} \, dx=\frac {x^{2 \, n + 2}}{2 \, {\left (n + 1\right )}} \]

[In]

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

[Out]

1/2*x^(2*n + 2)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int x^{1+2 n} \, dx=\frac {x^{2 \, n + 2}}{2 \, {\left (n + 1\right )}} \]

[In]

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

[Out]

1/2*x^(2*n + 2)/(n + 1)

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int x^{1+2 n} \, dx=\left \{\begin {array}{cl} \ln \left (x\right ) & \text {\ if\ \ }n=-1\\ \frac {x^{2\,n+2}}{2\,\left (n+1\right )} & \text {\ if\ \ }n\neq -1 \end {array}\right . \]

[In]

int(x^(2*n + 1),x)

[Out]

piecewise(n == -1, log(x), n ~= -1, x^(2*n + 2)/(2*(n + 1)))