\(\int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx\) [826]

Optimal result

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

[Out]

-1/2*(b*x+a)^n/x^2/((-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} \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx=-\frac {(-a-b x)^{-n} (a+b x)^n}{2 x^2} \]

[In]

Int[(a + b*x)^n/(x^3*(-a - b*x)^n),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx=-\frac {(-a-b x)^{-n} (a+b x)^n}{2 x^2} \]

[In]

Integrate[(a + b*x)^n/(x^3*(-a - b*x)^n),x]

[Out]

-1/2*(a + b*x)^n/(x^2*(-a - b*x)^n)

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

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

[In]

int((b*x+a)^n/x^3/((-b*x-a)^n),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.38 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx=-\frac {e^{\left (i \, \pi n\right )}}{2 \, x^{2}} \]

[In]

integrate((b*x+a)^n/x^3/((-b*x-a)^n),x, algorithm="fricas")

[Out]

-1/2*e^(I*pi*n)/x^2

Sympy [A] (verification not implemented)

Time = 2.71 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx=- \frac {\left (- a - b x\right )^{- n} \left (a + b x\right )^{n}}{2 x^{2}} \]

[In]

integrate((b*x+a)**n/x**3/((-b*x-a)**n),x)

[Out]

-(a + b*x)**n/(2*x**2*(-a - b*x)**n)

Maxima [A] (verification not implemented)

none

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

[In]

integrate((b*x+a)^n/x^3/((-b*x-a)^n),x, algorithm="maxima")

[Out]

-1/2*(-1)^n/x^2

Giac [F]

\[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (-b x - a\right )}^{n} x^{3}} \,d x } \]

[In]

integrate((b*x+a)^n/x^3/((-b*x-a)^n),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x^3} \, dx=-\frac {{\left (a+b\,x\right )}^n}{2\,x^2\,{\left (-a-b\,x\right )}^n} \]

[In]

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

[Out]

-(a + b*x)^n/(2*x^2*(- a - b*x)^n)