\(\int x^m (a+b x)^3 \, dx\) [701]

Optimal result

Integrand size = 11, antiderivative size = 61 \[ \int x^m (a+b x)^3 \, dx=\frac {a^3 x^{1+m}}{1+m}+\frac {3 a^2 b x^{2+m}}{2+m}+\frac {3 a b^2 x^{3+m}}{3+m}+\frac {b^3 x^{4+m}}{4+m} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int x^m (a+b x)^3 \, dx=\frac {a^3 x^{m+1}}{m+1}+\frac {3 a^2 b x^{m+2}}{m+2}+\frac {3 a b^2 x^{m+3}}{m+3}+\frac {b^3 x^{m+4}}{m+4} \]

[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 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^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] (verified)

Time = 0.00 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int x^m (a+b x)^3 \, dx=x^{1+m} \left (\frac {a^3}{1+m}+\frac {3 a^2 b x}{2+m}+\frac {3 a b^2 x^2}{3+m}+\frac {b^3 x^3}{4+m}\right ) \]

[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))

Maple [A] (verified)

Time = 0.00 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18

method	result	size
norman	\(\frac {a^{3} x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}+\frac {b^{3} x^{4} {\mathrm e}^{m \ln \left (x \right )}}{4+m}+\frac {3 a \,b^{2} x^{3} {\mathrm e}^{m \ln \left (x \right )}}{3+m}+\frac {3 a^{2} b \,x^{2} {\mathrm e}^{m \ln \left (x \right )}}{2+m}\)	\(72\)
risch	\(\frac {x \left (b^{3} m^{3} x^{3}+3 a \,b^{2} m^{3} x^{2}+6 b^{3} m^{2} x^{3}+3 a^{2} b \,m^{3} x +21 a \,b^{2} m^{2} x^{2}+11 b^{3} m \,x^{3}+a^{3} m^{3}+24 a^{2} b \,m^{2} x +42 a \,b^{2} m \,x^{2}+6 b^{3} x^{3}+9 a^{3} m^{2}+57 a^{2} b m x +24 a \,b^{2} x^{2}+26 a^{3} m +36 a^{2} b x +24 a^{3}\right ) x^{m}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\)	\(169\)
gosper	\(\frac {x^{1+m} \left (b^{3} m^{3} x^{3}+3 a \,b^{2} m^{3} x^{2}+6 b^{3} m^{2} x^{3}+3 a^{2} b \,m^{3} x +21 a \,b^{2} m^{2} x^{2}+11 b^{3} m \,x^{3}+a^{3} m^{3}+24 a^{2} b \,m^{2} x +42 a \,b^{2} m \,x^{2}+6 b^{3} x^{3}+9 a^{3} m^{2}+57 a^{2} b m x +24 a \,b^{2} x^{2}+26 a^{3} m +36 a^{2} b x +24 a^{3}\right )}{\left (1+m \right ) \left (2+m \right ) \left (3+m \right ) \left (4+m \right )}\)	\(170\)
parallelrisch	\(\frac {x^{4} x^{m} b^{3} m^{3}+6 x^{4} x^{m} b^{3} m^{2}+3 x^{3} x^{m} a \,b^{2} m^{3}+11 x^{4} x^{m} b^{3} m +21 x^{3} x^{m} a \,b^{2} m^{2}+3 x^{2} x^{m} a^{2} b \,m^{3}+6 x^{4} x^{m} b^{3}+42 x^{3} x^{m} a \,b^{2} m +24 x^{2} x^{m} a^{2} b \,m^{2}+x \,x^{m} a^{3} m^{3}+24 x^{3} x^{m} a \,b^{2}+57 x^{2} x^{m} a^{2} b m +9 x \,x^{m} a^{3} m^{2}+36 x^{2} x^{m} a^{2} b +26 x \,x^{m} a^{3} m +24 x \,x^{m} a^{3}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\)	\(225\)

[In]

int(x^m*(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (61) = 122\).

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.57 \[ \int x^m (a+b x)^3 \, dx=\frac {{\left ({\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 3 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} + 3 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 663 vs. \(2 (53) = 106\).

Time = 0.34 (sec) , antiderivative size = 663, normalized size of antiderivative = 10.87 \[ \int x^m (a+b x)^3 \, dx=\begin {cases} - \frac {a^{3}}{3 x^{3}} - \frac {3 a^{2} b}{2 x^{2}} - \frac {3 a b^{2}}{x} + b^{3} \log {\left (x \right )} & \text {for}\: m = -4 \\- \frac {a^{3}}{2 x^{2}} - \frac {3 a^{2} b}{x} + 3 a b^{2} \log {\left (x \right )} + b^{3} x & \text {for}\: m = -3 \\- \frac {a^{3}}{x} + 3 a^{2} b \log {\left (x \right )} + 3 a b^{2} x + \frac {b^{3} x^{2}}{2} & \text {for}\: m = -2 \\a^{3} \log {\left (x \right )} + 3 a^{2} b x + \frac {3 a b^{2} x^{2}}{2} + \frac {b^{3} x^{3}}{3} & \text {for}\: m = -1 \\\frac {a^{3} m^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 a^{3} m^{2} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 a^{3} m x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 a^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {3 a^{2} b m^{3} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 a^{2} b m^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {57 a^{2} b m x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {36 a^{2} b x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {3 a b^{2} m^{3} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {21 a b^{2} m^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {42 a b^{2} m x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 a b^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {b^{3} m^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b^{3} m^{2} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {11 b^{3} m x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int x^m (a+b x)^3 \, dx=\frac {b^{3} x^{m + 4}}{m + 4} + \frac {3 \, a b^{2} x^{m + 3}}{m + 3} + \frac {3 \, a^{2} b x^{m + 2}}{m + 2} + \frac {a^{3} x^{m + 1}}{m + 1} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (61) = 122\).

Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 3.67 \[ \int x^m (a+b x)^3 \, dx=\frac {b^{3} m^{3} x^{4} x^{m} + 3 \, a b^{2} m^{3} x^{3} x^{m} + 6 \, b^{3} m^{2} x^{4} x^{m} + 3 \, a^{2} b m^{3} x^{2} x^{m} + 21 \, a b^{2} m^{2} x^{3} x^{m} + 11 \, b^{3} m x^{4} x^{m} + a^{3} m^{3} x x^{m} + 24 \, a^{2} b m^{2} x^{2} x^{m} + 42 \, a b^{2} m x^{3} x^{m} + 6 \, b^{3} x^{4} x^{m} + 9 \, a^{3} m^{2} x x^{m} + 57 \, a^{2} b m x^{2} x^{m} + 24 \, a b^{2} x^{3} x^{m} + 26 \, a^{3} m x x^{m} + 36 \, a^{2} b x^{2} x^{m} + 24 \, a^{3} x x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.74 \[ \int x^m (a+b x)^3 \, dx=x^m\,\left (\frac {a^3\,x\,\left (m^3+9\,m^2+26\,m+24\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {b^3\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {3\,a\,b^2\,x^3\,\left (m^3+7\,m^2+14\,m+8\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {3\,a^2\,b\,x^2\,\left (m^3+8\,m^2+19\,m+12\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}\right ) \]

[In]

int(x^m*(a + b*x)^3,x)

[Out]

x^m*((a^3*x*(26*m + 9*m^2 + m^3 + 24))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (b^3*x^4*(11*m + 6*m^2 + m^3 + 6)
)/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (3*a*b^2*x^3*(14*m + 7*m^2 + m^3 + 8))/(50*m + 35*m^2 + 10*m^3 + m^4 +
24) + (3*a^2*b*x^2*(19*m + 8*m^2 + m^3 + 12))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24))