\(\int \log (c (a+\frac {b}{x})^p) \, dx\) [30]

Optimal result

Integrand size = 12, antiderivative size = 27 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=x \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p \log (b+a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2498, 269, 31} \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=x \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p \log (a x+b)}{a} \]

[In]

Int[Log[c*(a + b/x)^p],x]

[Out]

x*Log[c*(a + b/x)^p] + (b*p*Log[b + a*x])/a

Rule 31

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

Rule 269

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

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b p \log \left (a+\frac {b}{x}\right )}{a}+x \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p \log (x)}{a} \]

[In]

Integrate[Log[c*(a + b/x)^p],x]

[Out]

(b*p*Log[a + b/x])/a + x*Log[c*(a + b/x)^p] + (b*p*Log[x])/a

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
parts	\(x \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )+\frac {b p \ln \left (a x +b \right )}{a}\)	\(28\)
default	\(x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right )+\frac {b p \ln \left (a x +b \right )}{a}\)	\(30\)
parallelrisch	\(-\frac {-\ln \left (x \right ) b^{2} p^{2}-x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a b p -\ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{2} p}{a b p}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a p x \log \left (\frac {a x + b}{x}\right ) + b p \log \left (a x + b\right ) + a x \log \left (c\right )}{a} \]

[In]

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

[Out]

(a*p*x*log((a*x + b)/x) + b*p*log(a*x + b) + a*x*log(c))/a

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )} + \frac {b p \log {\left (a x + b \right )}}{a} & \text {for}\: a \neq 0 \\p x + x \log {\left (c \left (\frac {b}{x}\right )^{p} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*log(c*(a + b/x)**p) + b*p*log(a*x + b)/a, Ne(a, 0)), (p*x + x*log(c*(b/x)**p), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=x \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) + \frac {b p \log \left (a x + b\right )}{a} \]

[In]

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

[Out]

x*log((a + b/x)^p*c) + b*p*log(a*x + b)/a

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (27) = 54\).

Time = 0.34 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.56 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {\frac {b^{2} p \log \left (-a + \frac {a x + b}{x}\right )}{a} + \frac {b^{2} p \log \left (\frac {a x + b}{x}\right )}{a - \frac {a x + b}{x}} - \frac {b^{2} p \log \left (\frac {a x + b}{x}\right )}{a} + \frac {b^{2} \log \left (c\right )}{a - \frac {a x + b}{x}}}{b} \]

[In]

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

[Out]

-(b^2*p*log(-a + (a*x + b)/x)/a + b^2*p*log((a*x + b)/x)/(a - (a*x + b)/x) - b^2*p*log((a*x + b)/x)/a + b^2*lo
g(c)/(a - (a*x + b)/x))/b

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=x\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )+\frac {b\,p\,\ln \left (b+a\,x\right )}{a} \]

[In]

int(log(c*(a + b/x)^p),x)

[Out]

x*log(c*(a + b/x)^p) + (b*p*log(b + a*x))/a