\(\int \frac {(c x)^{-1+n}}{a+b x^n} \, dx\) [2777]

Optimal result

Integrand size = 17, antiderivative size = 28 \[ \int \frac {(c x)^{-1+n}}{a+b x^n} \, dx=\frac {x^{-n} (c x)^n \log \left (a+b x^n\right )}{b c n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {274, 266} \[ \int \frac {(c x)^{-1+n}}{a+b x^n} \, dx=\frac {x^{-n} (c x)^n \log \left (a+b x^n\right )}{b c n} \]

[In]

Int[(c*x)^(-1 + n)/(a + b*x^n),x]

[Out]

((c*x)^n*Log[a + b*x^n])/(b*c*n*x^n)

Rule 266

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

Rule 274

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPa
rt[m]), Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {(c x)^{-1+n}}{a+b x^n} \, dx=\frac {x^{1-n} (c x)^{-1+n} \log \left (a+b x^n\right )}{b n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {(c x)^{-1+n}}{a+b x^n} \, dx=\frac {c^{n - 1} \log \left (b x^{n} + a\right )}{b n} \]

[In]

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

[Out]

c^(n - 1)*log(b*x^n + a)/(b*n)

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {(c x)^{-1+n}}{a+b x^n} \, dx=\frac {c^{n - 1} \log {\left (1 + \frac {b x^{n}}{a} \right )}}{b n} \]

[In]

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

[Out]

c**(n - 1)*log(1 + b*x**n/a)/(b*n)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {(c x)^{-1+n}}{a+b x^n} \, dx=\frac {c^{n - 1} \log \left (\frac {b x^{n} + a}{b}\right )}{b n} \]

[In]

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

[Out]

c^(n - 1)*log((b*x^n + a)/b)/(b*n)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((c*x)^(n - 1)/(a + b*x^n),x)

[Out]

int((c*x)^(n - 1)/(a + b*x^n), x)