\(\int \frac {(c x)^{-1+n}}{(a+b x^n)^2} \, dx\) [2780]

Optimal result

Integrand size = 17, antiderivative size = 24 \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=\frac {(c x)^n}{a c n \left (a+b x^n\right )} \]

[Out]

(c*x)^n/a/c/n/(a+b*x^n)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {270} \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=\frac {(c x)^n}{a c n \left (a+b x^n\right )} \]

[In]

Int[(c*x)^(-1 + n)/(a + b*x^n)^2,x]

[Out]

(c*x)^n/(a*c*n*(a + b*x^n))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {(c x)^n}{a c n \left (a+b x^n\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=-\frac {x^{1-n} (c x)^{-1+n}}{b n \left (a+b x^n\right )} \]

[In]

Integrate[(c*x)^(-1 + n)/(a + b*x^n)^2,x]

[Out]

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

Maple [A] (verified)

Time = 3.79 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {x \left (c x \right )^{-1+n}}{a n \left (a +b \,x^{n}\right )}\)	\(25\)
risch	\(\frac {x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \operatorname {csgn}\left (i c x \right )^{3} \pi +i \operatorname {csgn}\left (i c x \right )^{2} \operatorname {csgn}\left (i c \right ) \pi +i \operatorname {csgn}\left (i c x \right )^{2} \operatorname {csgn}\left (i x \right ) \pi -i \operatorname {csgn}\left (i c x \right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x \right ) \pi +2 \ln \left (x \right )+2 \ln \left (c \right )\right )}{2}}}{a n \left (a +b \,x^{n}\right )}\)	\(99\)

[In]

int((c*x)^(-1+n)/(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=-\frac {c^{n - 1}}{b^{2} n x^{n} + a b n} \]

[In]

integrate((c*x)^(-1+n)/(a+b*x^n)^2,x, algorithm="fricas")

[Out]

-c^(n - 1)/(b^2*n*x^n + a*b*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (15) = 30\).

Time = 0.72 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.04 \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=\begin {cases} \frac {\tilde {\infty } \log {\left (x \right )}}{c} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\- \frac {x x^{- 2 n} \left (c x\right )^{n - 1}}{b^{2} n} & \text {for}\: a = 0 \\\frac {\tilde {\infty } x \left (c x\right )^{n - 1}}{n} & \text {for}\: b = - a x^{- n} \\\frac {\log {\left (x \right )}}{c \left (a + b\right )^{2}} & \text {for}\: n = 0 \\\frac {x \left (c x\right )^{n - 1}}{a^{2} n + a b n x^{n}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x)**(-1+n)/(a+b*x**n)**2,x)

[Out]

Piecewise((zoo*log(x)/c, Eq(a, 0) & Eq(b, 0) & Eq(n, 0)), (-x*(c*x)**(n - 1)/(b**2*n*x**(2*n)), Eq(a, 0)), (zo
o*x*(c*x)**(n - 1)/n, Eq(b, -a/x**n)), (log(x)/(c*(a + b)**2), Eq(n, 0)), (x*(c*x)**(n - 1)/(a**2*n + a*b*n*x*
*n), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=-\frac {c^{n}}{b^{2} c n x^{n} + a b c n} \]

[In]

integrate((c*x)^(-1+n)/(a+b*x^n)^2,x, algorithm="maxima")

[Out]

-c^n/(b^2*c*n*x^n + a*b*c*n)

Giac [F]

\[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=\int { \frac {\left (c x\right )^{n - 1}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]

[In]

integrate((c*x)^(-1+n)/(a+b*x^n)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {(c x)^{-1+n}}{\left (a+b x^n\right )^2} \, dx=\frac {x\,{\left (c\,x\right )}^{n-1}}{a\,b\,n\,\left (x^n+\frac {a}{b}\right )} \]

[In]

int((c*x)^(n - 1)/(a + b*x^n)^2,x)

[Out]

(x*(c*x)^(n - 1))/(a*b*n*(x^n + a/b))