\(\int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx\) [495]

Optimal result

Integrand size = 23, antiderivative size = 15 \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\frac {\log \left (b+c x^n\right )}{c n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, 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 = {1598, 266} \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\frac {\log \left (b+c x^n\right )}{c n} \]

[In]

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

[Out]

Log[b + c*x^n]/(c*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 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\frac {\log \left (b+c x^n\right )}{c n} \]

[In]

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

[Out]

Log[b + c*x^n]/(c*n)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20

method	result	size
norman	\(\frac {\ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}+b \right )}{c n}\)	\(18\)
risch	\(\frac {\ln \left (x^{n}+\frac {b}{c}\right )}{c n}\)	\(18\)

[In]

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

[Out]

1/c/n*ln(c*exp(n*ln(x))+b)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\frac {\log \left (c x^{n} + b\right )}{c n} \]

[In]

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

[Out]

log(c*x^n + b)/(c*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (10) = 20\).

Time = 1.75 (sec) , antiderivative size = 46, normalized size of antiderivative = 3.07 \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{b} & \text {for}\: c = 0 \wedge n = 0 \\\frac {x x^{- n} x^{2 n - 1}}{b n} & \text {for}\: c = 0 \\\frac {\log {\left (x \right )}}{b + c} & \text {for}\: n = 0 \\- \frac {\log {\left (x \right )}}{c} + \frac {\log {\left (\frac {b x^{n}}{c} + x^{2 n} \right )}}{c n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\frac {\log \left (\frac {c x^{n} + b}{c}\right )}{c n} \]

[In]

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

[Out]

log((c*x^n + b)/c)/(c*n)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+2 n}}{b x^n+c x^{2 n}} \, dx=\int \frac {x^{2\,n-1}}{b\,x^n+c\,x^{2\,n}} \,d x \]

[In]

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

[Out]

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