3.428 \(\int x^m (a+b x) \, dx\)

Optimal. Leaf size=25 \[ \frac{a x^{m+1}}{m+1}+\frac{b x^{m+2}}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.0083796, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {43} \[ \frac{a x^{m+1}}{m+1}+\frac{b x^{m+2}}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x),x]

[Out]

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

Rule 43

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*} \int x^m (a+b x) \, dx &=\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]  time = 0.0143177, size = 22, normalized size = 0.88 \[ x^{m+1} \left (\frac{a}{m+1}+\frac{b x}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x),x]

[Out]

x^(1 + m)*(a/(1 + m) + (b*x)/(2 + m))

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Maple [A]  time = 0.002, size = 31, normalized size = 1.2 \begin{align*}{\frac{{x}^{1+m} \left ( bmx+am+bx+2\,a \right ) }{ \left ( 2+m \right ) \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(b*x+a),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55342, size = 72, normalized size = 2.88 \begin{align*} \frac{{\left ({\left (b m + b\right )} x^{2} +{\left (a m + 2 \, a\right )} x\right )} x^{m}}{m^{2} + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.374517, size = 87, normalized size = 3.48 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.17033, size = 58, normalized size = 2.32 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)