\(\int x^2 (-a-b x)^{-n} (a+b x)^n \, dx\) [821]

Optimal result

Integrand size = 23, antiderivative size = 26 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {1}{3} x^3 (-a-b x)^{-n} (a+b x)^n \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {23, 30} \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {1}{3} x^3 (-a-b x)^{-n} (a+b x)^n \]

[In]

Int[(x^2*(a + b*x)^n)/(-a - b*x)^n,x]

[Out]

(x^3*(a + b*x)^n)/(3*(-a - b*x)^n)

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \left ((-a-b x)^{-n} (a+b x)^n\right ) \int x^2 \, dx \\ & = \frac {1}{3} x^3 (-a-b x)^{-n} (a+b x)^n \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {1}{3} x^3 (-a-b x)^{-n} (a+b x)^n \]

[In]

Integrate[(x^2*(a + b*x)^n)/(-a - b*x)^n,x]

[Out]

(x^3*(a + b*x)^n)/(3*(-a - b*x)^n)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
gosper	\(\frac {x^{3} \left (b x +a \right )^{n} \left (-b x -a \right )^{-n}}{3}\)	\(25\)
parallelrisch	\(\frac {x^{3} \left (b x +a \right )^{n} \left (-b x -a \right )^{-n}}{3}\)	\(25\)
risch	\(\frac {x^{3} {\mathrm e}^{-i n \pi \left (\operatorname {csgn}\left (i \left (b x +a \right )\right )^{3}-\operatorname {csgn}\left (i \left (b x +a \right )\right )^{2}+1\right )}}{3}\)	\(38\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.38 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {1}{3} \, x^{3} e^{\left (i \, \pi n\right )} \]

[In]

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

[Out]

1/3*x^3*e^(I*pi*n)

Sympy [A] (verification not implemented)

Time = 3.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {x^{3} \left (- a - b x\right )^{- n} \left (a + b x\right )^{n}}{3} \]

[In]

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

[Out]

x**3*(a + b*x)**n/(3*(-a - b*x)**n)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.31 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {1}{3} \, \left (-1\right )^{n} x^{3} \]

[In]

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

[Out]

1/3*(-1)^n*x^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.19 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {1}{3} \, x^{3} \]

[In]

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

[Out]

1/3*x^3

Mupad [B] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int x^2 (-a-b x)^{-n} (a+b x)^n \, dx=\frac {x^3\,{\left (a+b\,x\right )}^n}{3\,{\left (-a-b\,x\right )}^n} \]

[In]

int((x^2*(a + b*x)^n)/(- a - b*x)^n,x)

[Out]

(x^3*(a + b*x)^n)/(3*(- a - b*x)^n)