3.701 \(\int x^m (a+b x)^3 \, dx\)

Optimal. Leaf size=61 \[ \frac{3 a^2 b x^{m+2}}{m+2}+\frac{a^3 x^{m+1}}{m+1}+\frac{3 a b^2 x^{m+3}}{m+3}+\frac{b^3 x^{m+4}}{m+4} \]

[Out]

(a^3*x^(1 + m))/(1 + m) + (3*a^2*b*x^(2 + m))/(2 + m) + (3*a*b^2*x^(3 + m))/(3 + m) + (b^3*x^(4 + m))/(4 + m)

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Rubi [A]  time = 0.0182403, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{3 a^2 b x^{m+2}}{m+2}+\frac{a^3 x^{m+1}}{m+1}+\frac{3 a b^2 x^{m+3}}{m+3}+\frac{b^3 x^{m+4}}{m+4} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x)^3,x]

[Out]

(a^3*x^(1 + m))/(1 + m) + (3*a^2*b*x^(2 + m))/(2 + m) + (3*a*b^2*x^(3 + m))/(3 + m) + (b^3*x^(4 + m))/(4 + 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)^3 \, dx &=\int \left (a^3 x^m+3 a^2 b x^{1+m}+3 a b^2 x^{2+m}+b^3 x^{3+m}\right ) \, dx\\ &=\frac{a^3 x^{1+m}}{1+m}+\frac{3 a^2 b x^{2+m}}{2+m}+\frac{3 a b^2 x^{3+m}}{3+m}+\frac{b^3 x^{4+m}}{4+m}\\ \end{align*}

Mathematica [A]  time = 0.0300431, size = 54, normalized size = 0.89 \[ x^{m+1} \left (\frac{3 a^2 b x}{m+2}+\frac{a^3}{m+1}+\frac{3 a b^2 x^2}{m+3}+\frac{b^3 x^3}{m+4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x)^3,x]

[Out]

x^(1 + m)*(a^3/(1 + m) + (3*a^2*b*x)/(2 + m) + (3*a*b^2*x^2)/(3 + m) + (b^3*x^3)/(4 + m))

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Maple [B]  time = 0., size = 170, normalized size = 2.8 \begin{align*}{\frac{{x}^{1+m} \left ({b}^{3}{m}^{3}{x}^{3}+3\,a{b}^{2}{m}^{3}{x}^{2}+6\,{b}^{3}{m}^{2}{x}^{3}+3\,{a}^{2}b{m}^{3}x+21\,a{b}^{2}{m}^{2}{x}^{2}+11\,{b}^{3}m{x}^{3}+{a}^{3}{m}^{3}+24\,{a}^{2}b{m}^{2}x+42\,a{b}^{2}m{x}^{2}+6\,{b}^{3}{x}^{3}+9\,{a}^{3}{m}^{2}+57\,{a}^{2}bmx+24\,a{b}^{2}{x}^{2}+26\,{a}^{3}m+36\,{a}^{2}bx+24\,{a}^{3} \right ) }{ \left ( 4+m \right ) \left ( 3+m \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)^3,x)

[Out]

x^(1+m)*(b^3*m^3*x^3+3*a*b^2*m^3*x^2+6*b^3*m^2*x^3+3*a^2*b*m^3*x+21*a*b^2*m^2*x^2+11*b^3*m*x^3+a^3*m^3+24*a^2*
b*m^2*x+42*a*b^2*m*x^2+6*b^3*x^3+9*a^3*m^2+57*a^2*b*m*x+24*a*b^2*x^2+26*a^3*m+36*a^2*b*x+24*a^3)/(4+m)/(3+m)/(
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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.58413, size = 336, normalized size = 5.51 \begin{align*} \frac{{\left ({\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 3 \,{\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} + 3 \,{\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} +{\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(b*x+a)^3,x, algorithm="fricas")

[Out]

((b^3*m^3 + 6*b^3*m^2 + 11*b^3*m + 6*b^3)*x^4 + 3*(a*b^2*m^3 + 7*a*b^2*m^2 + 14*a*b^2*m + 8*a*b^2)*x^3 + 3*(a^
2*b*m^3 + 8*a^2*b*m^2 + 19*a^2*b*m + 12*a^2*b)*x^2 + (a^3*m^3 + 9*a^3*m^2 + 26*a^3*m + 24*a^3)*x)*x^m/(m^4 + 1
0*m^3 + 35*m^2 + 50*m + 24)

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Sympy [A]  time = 0.769065, size = 663, normalized size = 10.87 \begin{align*} \begin{cases} - \frac{a^{3}}{3 x^{3}} - \frac{3 a^{2} b}{2 x^{2}} - \frac{3 a b^{2}}{x} + b^{3} \log{\left (x \right )} & \text{for}\: m = -4 \\- \frac{a^{3}}{2 x^{2}} - \frac{3 a^{2} b}{x} + 3 a b^{2} \log{\left (x \right )} + b^{3} x & \text{for}\: m = -3 \\- \frac{a^{3}}{x} + 3 a^{2} b \log{\left (x \right )} + 3 a b^{2} x + \frac{b^{3} x^{2}}{2} & \text{for}\: m = -2 \\a^{3} \log{\left (x \right )} + 3 a^{2} b x + \frac{3 a b^{2} x^{2}}{2} + \frac{b^{3} x^{3}}{3} & \text{for}\: m = -1 \\\frac{a^{3} m^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{9 a^{3} m^{2} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{26 a^{3} m x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{24 a^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{3 a^{2} b m^{3} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{24 a^{2} b m^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{57 a^{2} b m x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{36 a^{2} b x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{3 a b^{2} m^{3} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{21 a b^{2} m^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{42 a b^{2} m x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{24 a b^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{b^{3} m^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{6 b^{3} m^{2} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{11 b^{3} m x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{6 b^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(b*x+a)**3,x)

[Out]

Piecewise((-a**3/(3*x**3) - 3*a**2*b/(2*x**2) - 3*a*b**2/x + b**3*log(x), Eq(m, -4)), (-a**3/(2*x**2) - 3*a**2
*b/x + 3*a*b**2*log(x) + b**3*x, Eq(m, -3)), (-a**3/x + 3*a**2*b*log(x) + 3*a*b**2*x + b**3*x**2/2, Eq(m, -2))
, (a**3*log(x) + 3*a**2*b*x + 3*a*b**2*x**2/2 + b**3*x**3/3, Eq(m, -1)), (a**3*m**3*x*x**m/(m**4 + 10*m**3 + 3
5*m**2 + 50*m + 24) + 9*a**3*m**2*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*a**3*m*x*x**m/(m**4 + 10*
m**3 + 35*m**2 + 50*m + 24) + 24*a**3*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 3*a**2*b*m**3*x**2*x**m/
(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 24*a**2*b*m**2*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 57*
a**2*b*m*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 36*a**2*b*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50
*m + 24) + 3*a*b**2*m**3*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 21*a*b**2*m**2*x**3*x**m/(m**4 + 1
0*m**3 + 35*m**2 + 50*m + 24) + 42*a*b**2*m*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 24*a*b**2*x**3*
x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + b**3*m**3*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 6*b
**3*m**2*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 11*b**3*m*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50
*m + 24) + 6*b**3*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))

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Giac [B]  time = 1.07291, size = 302, normalized size = 4.95 \begin{align*} \frac{b^{3} m^{3} x^{4} x^{m} + 3 \, a b^{2} m^{3} x^{3} x^{m} + 6 \, b^{3} m^{2} x^{4} x^{m} + 3 \, a^{2} b m^{3} x^{2} x^{m} + 21 \, a b^{2} m^{2} x^{3} x^{m} + 11 \, b^{3} m x^{4} x^{m} + a^{3} m^{3} x x^{m} + 24 \, a^{2} b m^{2} x^{2} x^{m} + 42 \, a b^{2} m x^{3} x^{m} + 6 \, b^{3} x^{4} x^{m} + 9 \, a^{3} m^{2} x x^{m} + 57 \, a^{2} b m x^{2} x^{m} + 24 \, a b^{2} x^{3} x^{m} + 26 \, a^{3} m x x^{m} + 36 \, a^{2} b x^{2} x^{m} + 24 \, a^{3} x x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(b*x+a)^3,x, algorithm="giac")

[Out]

(b^3*m^3*x^4*x^m + 3*a*b^2*m^3*x^3*x^m + 6*b^3*m^2*x^4*x^m + 3*a^2*b*m^3*x^2*x^m + 21*a*b^2*m^2*x^3*x^m + 11*b
^3*m*x^4*x^m + a^3*m^3*x*x^m + 24*a^2*b*m^2*x^2*x^m + 42*a*b^2*m*x^3*x^m + 6*b^3*x^4*x^m + 9*a^3*m^2*x*x^m + 5
7*a^2*b*m*x^2*x^m + 24*a*b^2*x^3*x^m + 26*a^3*m*x*x^m + 36*a^2*b*x^2*x^m + 24*a^3*x*x^m)/(m^4 + 10*m^3 + 35*m^
2 + 50*m + 24)