\(\int \log (c \log ^p(d x)) \, dx\) [55]

Optimal result

Integrand size = 9, antiderivative size = 22 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=x \log \left (c \log ^p(d x)\right )-\frac {p \operatorname {LogIntegral}(d x)}{d} \]

[Out]

-p*Li(d*x)/d+x*ln(c*ln(d*x)^p)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2600, 2335} \[ \int \log \left (c \log ^p(d x)\right ) \, dx=x \log \left (c \log ^p(d x)\right )-\frac {p \operatorname {LogIntegral}(d x)}{d} \]

[In]

Int[Log[c*Log[d*x]^p],x]

[Out]

x*Log[c*Log[d*x]^p] - (p*LogIntegral[d*x])/d

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2600

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=x \log \left (c \log ^p(d x)\right )-\frac {p \operatorname {LogIntegral}(d x)}{d} \]

[In]

Integrate[Log[c*Log[d*x]^p],x]

[Out]

x*Log[c*Log[d*x]^p] - (p*LogIntegral[d*x])/d

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
default	\(x \ln \left (c \ln \left (d x \right )^{p}\right )+\frac {p \,\operatorname {Ei}_{1}\left (-\ln \left (d x \right )\right )}{d}\)	\(26\)

[In]

int(ln(c*ln(d*x)^p),x,method=_RETURNVERBOSE)

[Out]

x*ln(c*ln(d*x)^p)+p/d*Ei(1,-ln(d*x))

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=\frac {d p x \log \left (\log \left (d x\right )\right ) + d x \log \left (c\right ) - p \operatorname {log\_integral}\left (d x\right )}{d} \]

[In]

integrate(log(c*log(d*x)^p),x, algorithm="fricas")

[Out]

(d*p*x*log(log(d*x)) + d*x*log(c) - p*log_integral(d*x))/d

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=x \log {\left (c \log {\left (d x \right )}^{p} \right )} - \frac {p \operatorname {li}{\left (d x \right )}}{d} \]

[In]

integrate(ln(c*ln(d*x)**p),x)

[Out]

x*log(c*log(d*x)**p) - p*li(d*x)/d

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=x \log \left (c \log \left (d x\right )^{p}\right ) - \frac {p {\rm Ei}\left (\log \left (d x\right )\right )}{d} \]

[In]

integrate(log(c*log(d*x)^p),x, algorithm="maxima")

[Out]

x*log(c*log(d*x)^p) - p*Ei(log(d*x))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=p x \log \left (\log \left (d\right ) + \log \left (x\right )\right ) + x \log \left (c\right ) - \frac {p {\rm Ei}\left (\log \left (d\right ) + \log \left (x\right )\right )}{d} \]

[In]

integrate(log(c*log(d*x)^p),x, algorithm="giac")

[Out]

p*x*log(log(d) + log(x)) + x*log(c) - p*Ei(log(d) + log(x))/d

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \log \left (c \log ^p(d x)\right ) \, dx=x\,\ln \left (c\,{\ln \left (d\,x\right )}^p\right )-\frac {p\,\mathrm {logint}\left (d\,x\right )}{d} \]

[In]

int(log(c*log(d*x)^p),x)

[Out]

x*log(c*log(d*x)^p) - (p*logint(d*x))/d