\(\int \frac {1}{a x+b x^{1-n}} \, dx\) [357]

Optimal result

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

[Out]

ln(b+a*x^n)/a/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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1607, 266} \[ \int \frac {1}{a x+b x^{1-n}} \, dx=\frac {\log \left (a x^n+b\right )}{a n} \]

[In]

Int[(a*x + b*x^(1 - n))^(-1),x]

[Out]

Log[b + a*x^n]/(a*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 1607

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a x+b x^{1-n}} \, dx=\frac {\log \left (b+a x^n\right )}{a n} \]

[In]

Integrate[(a*x + b*x^(1 - n))^(-1),x]

[Out]

Log[b + a*x^n]/(a*n)

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07

method	result	size
parallelrisch	\(\frac {n \ln \left (x \right )-\ln \left (x \right )+\ln \left (a x +b \,x^{1-n}\right )}{a n}\)	\(31\)
norman	\(\frac {\left (-1+n \right ) \ln \left (x \right )}{a n}+\frac {\ln \left (a x +b \,{\mathrm e}^{\left (1-n \right ) \ln \left (x \right )}\right )}{a n}\)	\(37\)
risch	\(-\frac {\ln \left (x \right )}{a n}+\frac {\ln \left (x \right )}{a}+\frac {\ln \left (x^{1-n}+\frac {a x}{b}\right )}{a n}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.87 \[ \int \frac {1}{a x+b x^{1-n}} \, dx=\frac {{\left (n - 1\right )} \log \left (x\right ) + \log \left (a x + b x^{-n + 1}\right )}{a n} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (10) = 20\).

Time = 0.35 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.40 \[ \int \frac {1}{a x+b x^{1-n}} \, dx=\begin {cases} \tilde {\infty } \log {\left (x \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\\frac {x x^{n - 1}}{b n} & \text {for}\: a = 0 \\\frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{a + b} & \text {for}\: n = 0 \\\frac {\log {\left (x \right )}}{a} - \frac {\log {\left (x \right )}}{a n} + \frac {\log {\left (\frac {a x}{b} + x^{1 - n} \right )}}{a n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {1}{a x+b x^{1-n}} \, dx=\frac {\log \left (\frac {a x^{n} + b}{a}\right )}{a n} \]

[In]

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

[Out]

log((a*x^n + b)/a)/(a*n)

Giac [F]

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

[In]

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

[Out]

integrate(1/(a*x + b*x^(-n + 1)), x)

Mupad [B] (verification not implemented)

Time = 8.95 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.27 \[ \int \frac {1}{a x+b x^{1-n}} \, dx=\frac {\ln \left (a\,x+b\,x^{1-n}\right )}{a\,n}+\frac {\ln \left (x\right )\,\left (n-1\right )}{a\,n} \]

[In]

int(1/(a*x + b*x^(1 - n)),x)

[Out]

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