\(\int \frac {1}{a x+b x \log (c x^n)} \, dx\) [250]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\log \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {31} \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\log \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[In]

Int[(a*x + b*x*Log[c*x^n])^(-1),x]

[Out]

Log[a + b*Log[c*x^n]]/(b*n)

Rule 31

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\log \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[In]

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

[Out]

Log[a + b*Log[c*x^n]]/(b*n)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method result size
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 \(\frac {\ln \left (b \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )+a \right )}{b n}\) \(21\)
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/(a*x+b*x*ln(c*x^n)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\log \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b n} \]

[In]

integrate(1/(a*x+b*x*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 = 0.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\frac {\log {\left (x \right )}}{a + b \log {\left (c \right )}} & \text {for}\: n = 0 \\\frac {\log {\left (\frac {a}{b} + \log {\left (c x^{n} \right )} \right )}}{b n} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a*x+b*x*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 = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\log \left (\frac {b \log \left (c\right ) + b \log \left (x^{n}\right ) + a}{b}\right )}{b n} \]

[In]

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

[Out]

log((b*log(c) + b*log(x^n) + a)/b)/(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 = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.50 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\log \left (\frac {1}{4} \, {\left (\pi b n {\left (\mathrm {sgn}\left (x\right ) - 1\right )} + \pi b {\left (\mathrm {sgn}\left (c\right ) - 1\right )}\right )}^{2} + {\left (b n \log \left ({\left | x \right |}\right ) + b \log \left ({\left | c \right |}\right ) + a\right )}^{2}\right )}{2 \, b n} \]

[In]

integrate(1/(a*x+b*x*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 = 1.53 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a x+b x \log \left (c x^n\right )} \, dx=\frac {\ln \left (a+b\,\ln \left (c\,x^n\right )\right )}{b\,n} \]

[In]

int(1/(a*x + b*x*log(c*x^n)),x)

[Out]

log(a + b*log(c*x^n))/(b*n)