3.5.28 \(\int x^m (a+b x) \, dx\) [428]

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

[Out]

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

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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 = {45} \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 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*} \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^{1+m} \left (\frac {a}{1+m}+\frac {b x}{2+m}\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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.70, size = 101, normalized size = 4.04 \begin {gather*} \text {Piecewise}\left [\left \{\left \{-\frac {a}{x}+b \text {Log}\left [x\right ],m\text {==}-2\right \},\left \{a \text {Log}\left [x\right ]+b x,m\text {==}-1\right \}\right \},\frac {2 a x x^m}{2+3 m+m^2}+\frac {a m x x^m}{2+3 m+m^2}+\frac {b x^2 x^m}{2+3 m+m^2}+\frac {b m x^2 x^m}{2+3 m+m^2}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x^m*(a + b*x),x]')

[Out]

Piecewise[{{-a / x + b Log[x], m == -2}, {a Log[x] + b x, m == -1}}, 2 a x x ^ m / (2 + 3 m + m ^ 2) + a m x x
^ m / (2 + 3 m + m ^ 2) + b x ^ 2 x ^ m / (2 + 3 m + m ^ 2) + b m x ^ 2 x ^ m / (2 + 3 m + m ^ 2)]

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Maple [A]
time = 0.01, size = 30, normalized size = 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	\(\frac {x \left (b m x +a m +b x +2 a \right ) x^{m}}{\left (2+m \right ) \left (1+m \right )}\)	\(30\)
gosper	\(\frac {x^{1+m} \left (b m x +a m +b x +2 a \right )}{\left (2+m \right ) \left (1+m \right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Maxima [A]
time = 0.26, 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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Fricas [A]
time = 0.31, 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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Sympy [A]
time = 0.13, size = 87, normalized size = 3.48 \begin {gather*} \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 {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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Giac [A]
time = 0.00, size = 50, normalized size = 2.00 \begin {gather*} \frac {a m x \mathrm {e}^{m \ln x}+2 a x \mathrm {e}^{m \ln x}+b m x^{2} \mathrm {e}^{m \ln x}+b x^{2} \mathrm {e}^{m \ln x}}{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)

[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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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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