\(\int x^m (a+b x) \, dx\) [703]

Optimal result

Integrand size = 9, antiderivative size = 25 \[ \int x^m (a+b x) \, dx=\frac {a x^{1+m}}{1+m}+\frac {b x^{2+m}}{2+m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, 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 = {45} \[ \int x^m (a+b x) \, dx=\frac {a x^{m+1}}{m+1}+\frac {b x^{m+2}}{m+2} \]

[In]

Int[x^m*(a + b*x),x]

[Out]

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

Rule 45

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int x^m (a+b x) \, dx=x^{1+m} \left (\frac {a}{1+m}+\frac {b x}{2+m}\right ) \]

[In]

Integrate[x^m*(a + b*x),x]

[Out]

x^(1 + m)*(a/(1 + m) + (b*x)/(2 + m))

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
norman	\(\frac {a x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}+\frac {b \,x^{2} {\mathrm e}^{m \ln \left (x \right )}}{2+m}\)	\(30\)
risch	\(\frac {x \left (b x m +a m +b x +2 a \right ) x^{m}}{\left (2+m \right ) \left (1+m \right )}\)	\(30\)
gosper	\(\frac {x^{1+m} \left (b x m +a m +b x +2 a \right )}{\left (1+m \right ) \left (2+m \right )}\)	\(31\)
parallelrisch	\(\frac {x^{2} x^{m} b m +x^{2} x^{m} b +x \,x^{m} a m +2 x \,x^{m} a}{\left (2+m \right ) \left (1+m \right )}\)	\(44\)

[In]

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

[Out]

a/(1+m)*x*exp(m*ln(x))+b/(2+m)*x^2*exp(m*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int x^m (a+b x) \, dx=\frac {{\left ({\left (b m + b\right )} x^{2} + {\left (a m + 2 \, a\right )} x\right )} x^{m}}{m^{2} + 3 \, m + 2} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.48 \[ \int x^m (a+b x) \, dx=\begin {cases} - \frac {a}{x} + b \log {\left (x \right )} & \text {for}\: m = -2 \\a \log {\left (x \right )} + b x & \text {for}\: m = -1 \\\frac {a m x x^{m}}{m^{2} + 3 m + 2} + \frac {2 a x x^{m}}{m^{2} + 3 m + 2} + \frac {b m x^{2} x^{m}}{m^{2} + 3 m + 2} + \frac {b x^{2} x^{m}}{m^{2} + 3 m + 2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a/x + b*log(x), Eq(m, -2)), (a*log(x) + b*x, Eq(m, -1)), (a*m*x*x**m/(m**2 + 3*m + 2) + 2*a*x*x**m
/(m**2 + 3*m + 2) + b*m*x**2*x**m/(m**2 + 3*m + 2) + b*x**2*x**m/(m**2 + 3*m + 2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int x^m (a+b x) \, dx=\frac {b x^{m + 2}}{m + 2} + \frac {a x^{m + 1}}{m + 1} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int x^m (a+b x) \, dx=\frac {b m x^{2} x^{m} + a m x x^{m} + b x^{2} x^{m} + 2 \, a x x^{m}}{m^{2} + 3 \, m + 2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int x^m (a+b x) \, dx=\frac {x^{m+1}\,\left (2\,a+a\,m+b\,x+b\,m\,x\right )}{m^2+3\,m+2} \]

[In]

int(x^m*(a + b*x),x)

[Out]

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