\(\int \frac {1}{x (a+b \log (c x^n))} \, dx\) [69]

Optimal result

Integrand size = 16, antiderivative size = 18 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\log \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2339, 29} \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\log \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[In]

Int[1/(x*(a + b*Log[c*x^n])),x]

[Out]

Log[a + b*Log[c*x^n]]/(b*n)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\log \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[In]

Integrate[1/(x*(a + b*Log[c*x^n])),x]

[Out]

Log[a + b*Log[c*x^n]]/(b*n)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \ln \left (c \,x^{n}\right )\right )}{b n}\)	\(19\)
default	\(\frac {\ln \left (a +b \ln \left (c \,x^{n}\right )\right )}{b n}\)	\(19\)
parallelrisch	\(\frac {\ln \left (a +b \ln \left (c \,x^{n}\right )\right )}{b n}\)	\(19\)
norman	\(\frac {\ln \left (a +b \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )\right )}{b n}\)	\(21\)
risch	\(\frac {\ln \left (\ln \left (x^{n}\right )-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-2 b \ln \left (c \right )-2 a}{2 b}\right )}{b n}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\log \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b n} \]

[In]

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

[Out]

log(b*n*log(x) + b*log(c) + a)/(b*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).

Time = 0.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\frac {\log {\left (x \right )}}{a + b \log {\left (c \right )}} & \text {for}\: n = 0 \\\frac {\log {\left (\frac {a}{b} + \log {\left (c x^{n} \right )} \right )}}{b n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(x)/a, Eq(b, 0) & (Eq(b, 0) | Eq(n, 0))), (log(x)/(a + b*log(c)), Eq(n, 0)), (log(a/b + log(c*x*
*n))/(b*n), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\log \left (b \log \left (c x^{n}\right ) + a\right )}{b n} \]

[In]

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

[Out]

log(b*log(c*x^n) + a)/(b*n)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (18) = 36\).

Time = 0.41 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.50 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\log \left (\frac {1}{4} \, {\left (\pi b n {\left (\mathrm {sgn}\left (x\right ) - 1\right )} + \pi b {\left (\mathrm {sgn}\left (c\right ) - 1\right )}\right )}^{2} + {\left (b n \log \left ({\left | x \right |}\right ) + b \log \left ({\left | c \right |}\right ) + a\right )}^{2}\right )}{2 \, b n} \]

[In]

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

[Out]

1/2*log(1/4*(pi*b*n*(sgn(x) - 1) + pi*b*(sgn(c) - 1))^2 + (b*n*log(abs(x)) + b*log(abs(c)) + a)^2)/(b*n)

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {\ln \left (a+b\,\ln \left (c\,x^n\right )\right )}{b\,n} \]

[In]

int(1/(x*(a + b*log(c*x^n))),x)

[Out]

log(a + b*log(c*x^n))/(b*n)