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

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {a x^{m+1}}{m+1}+\frac {b x^{m+2}}{m+2} \end {gather*}

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*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.88 \begin {gather*} x^{m+1} \left (\frac {a}{m+1}+\frac {b x}{m+2}\right ) \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.01, size = 0, normalized size = 0.00 \begin {gather*} \int x^m (a+b x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^m*(a + b*x),x]

[Out]

Defer[IntegrateAlgebraic][x^m*(a + b*x), x]

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fricas [A]  time = 0.96, size = 33, normalized size = 1.32 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.01, size = 43, normalized size = 1.72 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.00, size = 31, normalized size = 1.24 \begin {gather*} \frac {\left (b m x +a m +b x +2 a \right ) x^{m +1}}{\left (m +2\right ) \left (m +1\right )} \end {gather*}

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 [A]  time = 1.34, size = 25, normalized size = 1.00 \begin {gather*} \frac {b x^{m + 2}}{m + 2} + \frac {a x^{m + 1}}{m + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.31, size = 30, normalized size = 1.20 \begin {gather*} \frac {x^{m+1}\,\left (2\,a+a\,m+b\,x+b\,m\,x\right )}{m^2+3\,m+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.30, size = 87, normalized size = 3.48 \begin {gather*} \begin {cases} - \frac {a}{x} + b \log {\relax (x )} & \text {for}\: m = -2 \\a \log {\relax (x )} + 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 {gather*}

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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