\(\int x (a+b x^n) \, dx\) [2453]

Optimal result

Integrand size = 9, antiderivative size = 21 \[ \int x \left (a+b x^n\right ) \, dx=\frac {a x^2}{2}+\frac {b x^{2+n}}{2+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {14} \[ \int x \left (a+b x^n\right ) \, dx=\frac {a x^2}{2}+\frac {b x^{n+2}}{n+2} \]

[In]

Int[x*(a + b*x^n),x]

[Out]

(a*x^2)/2 + (b*x^(2 + n))/(2 + n)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x+b x^{1+n}\right ) \, dx \\ & = \frac {a x^2}{2}+\frac {b x^{2+n}}{2+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int x \left (a+b x^n\right ) \, dx=\frac {a x^2}{2}+\frac {b x^{2+n}}{2+n} \]

[In]

Integrate[x*(a + b*x^n),x]

[Out]

(a*x^2)/2 + (b*x^(2 + n))/(2 + n)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {b \,x^{2} x^{n}}{2+n}+\frac {a \,x^{2}}{2}\)	\(21\)
norman	\(\frac {b \,x^{2} {\mathrm e}^{n \ln \left (x \right )}}{2+n}+\frac {a \,x^{2}}{2}\)	\(23\)
parallelrisch	\(\frac {2 x^{2} x^{n} b +x^{2} a n +2 a \,x^{2}}{4+2 n}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int x \left (a+b x^n\right ) \, dx=\frac {2 \, b x^{2} x^{n} + {\left (a n + 2 \, a\right )} x^{2}}{2 \, {\left (n + 2\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (15) = 30\).

Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43 \[ \int x \left (a+b x^n\right ) \, dx=\begin {cases} \frac {a n x^{2}}{2 n + 4} + \frac {2 a x^{2}}{2 n + 4} + \frac {2 b x^{2} x^{n}}{2 n + 4} & \text {for}\: n \neq -2 \\\frac {a x^{2}}{2} + b \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x*(a+b*x**n),x)

[Out]

Piecewise((a*n*x**2/(2*n + 4) + 2*a*x**2/(2*n + 4) + 2*b*x**2*x**n/(2*n + 4), Ne(n, -2)), (a*x**2/2 + b*log(x)
, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int x \left (a+b x^n\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {b x^{n + 2}}{n + 2} \]

[In]

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

[Out]

1/2*a*x^2 + b*x^(n + 2)/(n + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int x \left (a+b x^n\right ) \, dx=\frac {2 \, b x^{2} x^{n} + a n x^{2} + 2 \, a x^{2}}{2 \, {\left (n + 2\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int x \left (a+b x^n\right ) \, dx=\frac {a\,x^2}{2}+\frac {b\,x^n\,x^2}{n+2} \]

[In]

int(x*(a + b*x^n),x)

[Out]

(a*x^2)/2 + (b*x^n*x^2)/(n + 2)