\(\int x^2 (a+b x)^n \, dx\) [733]

Optimal result

Integrand size = 11, antiderivative size = 60 \[ \int x^2 (a+b x)^n \, dx=\frac {a^2 (a+b x)^{1+n}}{b^3 (1+n)}-\frac {2 a (a+b x)^{2+n}}{b^3 (2+n)}+\frac {(a+b x)^{3+n}}{b^3 (3+n)} \]

[Out]

a^2*(b*x+a)^(1+n)/b^3/(1+n)-2*a*(b*x+a)^(2+n)/b^3/(2+n)+(b*x+a)^(3+n)/b^3/(3+n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, 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} \[ \int x^2 (a+b x)^n \, dx=\frac {a^2 (a+b x)^{n+1}}{b^3 (n+1)}-\frac {2 a (a+b x)^{n+2}}{b^3 (n+2)}+\frac {(a+b x)^{n+3}}{b^3 (n+3)} \]

[In]

Int[x^2*(a + b*x)^n,x]

[Out]

(a^2*(a + b*x)^(1 + n))/(b^3*(1 + n)) - (2*a*(a + b*x)^(2 + n))/(b^3*(2 + n)) + (a + b*x)^(3 + n)/(b^3*(3 + n)
)

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 (\frac {a^2 (a+b x)^n}{b^2}-\frac {2 a (a+b x)^{1+n}}{b^2}+\frac {(a+b x)^{2+n}}{b^2}\right ) \, dx \\ & = \frac {a^2 (a+b x)^{1+n}}{b^3 (1+n)}-\frac {2 a (a+b x)^{2+n}}{b^3 (2+n)}+\frac {(a+b x)^{3+n}}{b^3 (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int x^2 (a+b x)^n \, dx=\frac {(a+b x)^{1+n} \left (2 a^2-2 a b (1+n) x+b^2 \left (2+3 n+n^2\right ) x^2\right )}{b^3 (1+n) (2+n) (3+n)} \]

[In]

Integrate[x^2*(a + b*x)^n,x]

[Out]

((a + b*x)^(1 + n)*(2*a^2 - 2*a*b*(1 + n)*x + b^2*(2 + 3*n + n^2)*x^2))/(b^3*(1 + n)*(2 + n)*(3 + n))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22

method	result	size
gosper	\(\frac {\left (b x +a \right )^{1+n} \left (b^{2} n^{2} x^{2}+3 b^{2} n \,x^{2}-2 a b n x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right )}{b^{3} \left (n^{3}+6 n^{2}+11 n +6\right )}\)	\(73\)
risch	\(\frac {\left (b^{3} n^{2} x^{3}+a \,b^{2} n^{2} x^{2}+3 b^{3} n \,x^{3}+a \,b^{2} n \,x^{2}+2 b^{3} x^{3}-2 a^{2} b n x +2 a^{3}\right ) \left (b x +a \right )^{n}}{\left (2+n \right ) \left (3+n \right ) \left (1+n \right ) b^{3}}\)	\(88\)
norman	\(\frac {x^{3} {\mathrm e}^{n \ln \left (b x +a \right )}}{3+n}+\frac {a n \,x^{2} {\mathrm e}^{n \ln \left (b x +a \right )}}{b \left (n^{2}+5 n +6\right )}+\frac {2 a^{3} {\mathrm e}^{n \ln \left (b x +a \right )}}{b^{3} \left (n^{3}+6 n^{2}+11 n +6\right )}-\frac {2 n \,a^{2} x \,{\mathrm e}^{n \ln \left (b x +a \right )}}{b^{2} \left (n^{3}+6 n^{2}+11 n +6\right )}\)	\(114\)
parallelrisch	\(\frac {x^{3} \left (b x +a \right )^{n} a \,b^{3} n^{2}+3 x^{3} \left (b x +a \right )^{n} a \,b^{3} n +x^{2} \left (b x +a \right )^{n} a^{2} b^{2} n^{2}+2 x^{3} \left (b x +a \right )^{n} a \,b^{3}+x^{2} \left (b x +a \right )^{n} a^{2} b^{2} n -2 x \left (b x +a \right )^{n} a^{3} b n +2 \left (b x +a \right )^{n} a^{4}}{\left (3+n \right ) \left (2+n \right ) a \left (1+n \right ) b^{3}}\)	\(140\)

[In]

int(x^2*(b*x+a)^n,x,method=_RETURNVERBOSE)

[Out]

1/b^3*(b*x+a)^(1+n)/(n^3+6*n^2+11*n+6)*(b^2*n^2*x^2+3*b^2*n*x^2-2*a*b*n*x+2*b^2*x^2-2*a*b*x+2*a^2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int x^2 (a+b x)^n \, dx=-\frac {{\left (2 \, a^{2} b n x - {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3}\right )} x^{3} - 2 \, a^{3} - {\left (a b^{2} n^{2} + a b^{2} n\right )} x^{2}\right )} {\left (b x + a\right )}^{n}}{b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}} \]

[In]

integrate(x^2*(b*x+a)^n,x, algorithm="fricas")

[Out]

-(2*a^2*b*n*x - (b^3*n^2 + 3*b^3*n + 2*b^3)*x^3 - 2*a^3 - (a*b^2*n^2 + a*b^2*n)*x^2)*(b*x + a)^n/(b^3*n^3 + 6*
b^3*n^2 + 11*b^3*n + 6*b^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (51) = 102\).

Time = 1.03 (sec) , antiderivative size = 597, normalized size of antiderivative = 9.95 \[ \int x^2 (a+b x)^n \, dx=\begin {cases} \frac {a^{n} x^{3}}{3} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {3 a^{2}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {4 a b x \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {4 a b x}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {2 b^{2} x^{2} \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} & \text {for}\: n = -3 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + x \right )}}{a b^{3} + b^{4} x} - \frac {2 a^{2}}{a b^{3} + b^{4} x} - \frac {2 a b x \log {\left (\frac {a}{b} + x \right )}}{a b^{3} + b^{4} x} + \frac {b^{2} x^{2}}{a b^{3} + b^{4} x} & \text {for}\: n = -2 \\\frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{b^{3}} - \frac {a x}{b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: n = -1 \\\frac {2 a^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} - \frac {2 a^{2} b n x \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {a b^{2} n^{2} x^{2} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {a b^{2} n x^{2} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {b^{3} n^{2} x^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {3 b^{3} n x^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {2 b^{3} x^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(b*x+a)**n,x)

[Out]

Piecewise((a**n*x**3/3, Eq(b, 0)), (2*a**2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 3*a**2/(2*a
**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x
/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 2*b**2*x**2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2),
Eq(n, -3)), (-2*a**2*log(a/b + x)/(a*b**3 + b**4*x) - 2*a**2/(a*b**3 + b**4*x) - 2*a*b*x*log(a/b + x)/(a*b**3
+ b**4*x) + b**2*x**2/(a*b**3 + b**4*x), Eq(n, -2)), (a**2*log(a/b + x)/b**3 - a*x/b**2 + x**2/(2*b), Eq(n, -
1)), (2*a**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) - 2*a**2*b*n*x*(a + b*x)**n/(b**3*n**
3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + a*b**2*n**2*x**2*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6
*b**3) + a*b**2*n*x**2*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + b**3*n**2*x**3*(a + b*x)*
*n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + 3*b**3*n*x**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b
**3*n + 6*b**3) + 2*b**3*x**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int x^2 (a+b x)^n \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} x^{3} + {\left (n^{2} + n\right )} a b^{2} x^{2} - 2 \, a^{2} b n x + 2 \, a^{3}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} \]

[In]

integrate(x^2*(b*x+a)^n,x, algorithm="maxima")

[Out]

((n^2 + 3*n + 2)*b^3*x^3 + (n^2 + n)*a*b^2*x^2 - 2*a^2*b*n*x + 2*a^3)*(b*x + a)^n/((n^3 + 6*n^2 + 11*n + 6)*b^
3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (60) = 120\).

Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.33 \[ \int x^2 (a+b x)^n \, dx=\frac {{\left (b x + a\right )}^{n} b^{3} n^{2} x^{3} + {\left (b x + a\right )}^{n} a b^{2} n^{2} x^{2} + 3 \, {\left (b x + a\right )}^{n} b^{3} n x^{3} + {\left (b x + a\right )}^{n} a b^{2} n x^{2} + 2 \, {\left (b x + a\right )}^{n} b^{3} x^{3} - 2 \, {\left (b x + a\right )}^{n} a^{2} b n x + 2 \, {\left (b x + a\right )}^{n} a^{3}}{b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}} \]

[In]

integrate(x^2*(b*x+a)^n,x, algorithm="giac")

[Out]

((b*x + a)^n*b^3*n^2*x^3 + (b*x + a)^n*a*b^2*n^2*x^2 + 3*(b*x + a)^n*b^3*n*x^3 + (b*x + a)^n*a*b^2*n*x^2 + 2*(
b*x + a)^n*b^3*x^3 - 2*(b*x + a)^n*a^2*b*n*x + 2*(b*x + a)^n*a^3)/(b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3)

Mupad [B] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.20 \[ \int x^2 (a+b x)^n \, dx=\left \{\begin {array}{cl} \frac {2\,a^2\,\ln \left (a+b\,x\right )+b^2\,x^2-2\,a\,b\,x}{2\,b^3} & \text {\ if\ \ }n=-1\\ \frac {x}{b^2}-\frac {a^2}{b^3\,\left (a+b\,x\right )}-\frac {2\,a\,\ln \left (a+b\,x\right )}{b^3} & \text {\ if\ \ }n=-2\\ \frac {\ln \left (a+b\,x\right )+\frac {2\,a}{a+b\,x}-\frac {a^2}{2\,{\left (a+b\,x\right )}^2}}{b^3} & \text {\ if\ \ }n=-3\\ \frac {2\,{\left (a+b\,x\right )}^{n+1}\,\left (8\,a^2-8\,a\,b\,n\,x-8\,a\,b\,x+4\,b^2\,n^2\,x^2+12\,b^2\,n\,x^2+8\,b^2\,x^2\right )}{b^3\,\left (8\,n^3+48\,n^2+88\,n+48\right )} & \text {\ if\ \ }n\neq -1\wedge n\neq -2\wedge n\neq -3 \end {array}\right . \]

[In]

int(x^2*(a + b*x)^n,x)

[Out]

piecewise(n == -1, (2*a^2*log(a + b*x) + b^2*x^2 - 2*a*b*x)/(2*b^3), n == -2, x/b^2 - a^2/(b^3*(a + b*x)) - (2
*a*log(a + b*x))/b^3, n == -3, (log(a + b*x) + (2*a)/(a + b*x) - a^2/(2*(a + b*x)^2))/b^3, n ~= -1 & n ~= -2 &
n ~= -3, (2*(a + b*x)^(n + 1)*(8*a^2 + 8*b^2*x^2 + 12*b^2*n*x^2 - 8*a*b*x + 4*b^2*n^2*x^2 - 8*a*b*n*x))/(b^3*
(88*n + 48*n^2 + 8*n^3 + 48)))