\(\int \frac {\log (c (d+e x^n)^p)}{x} \, dx\) [73]

Optimal result

Integrand size = 16, antiderivative size = 44 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]

[Out]

ln(-e*x^n/d)*ln(c*(d+e*x^n)^p)/n+p*polylog(2,1+e*x^n/d)/n

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2441, 2352} \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n} \]

[In]

Int[Log[c*(d + e*x^n)^p]/x,x]

[Out]

(Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/n + (p*PolyLog[2, 1 + (e*x^n)/d])/n

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n} \\ & = \frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {(e p) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n} \\ & = \frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \operatorname {PolyLog}\left (2,\frac {d+e x^n}{d}\right )}{n} \]

[In]

Integrate[Log[c*(d + e*x^n)^p]/x,x]

[Out]

(Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + p*PolyLog[2, (d + e*x^n)/d])/n

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 3.86

method	result	size
risch	\(\ln \left (x \right ) \ln \left (\left (d +e \,x^{n}\right )^{p}\right )+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (x \right )-\frac {p \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\)	\(170\)

[In]

int(ln(c*(d+e*x^n)^p)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*ln((d+e*x^n)^p)+(1/2*I*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)^2-1/2*I*Pi*csgn(I*(d+e*x^n)^p)*csgn(
I*c*(d+e*x^n)^p)*csgn(I*c)-1/2*I*Pi*csgn(I*c*(d+e*x^n)^p)^3+1/2*I*Pi*csgn(I*c*(d+e*x^n)^p)^2*csgn(I*c)+ln(c))*
ln(x)-p/n*dilog((d+e*x^n)/d)-p*ln(x)*ln((d+e*x^n)/d)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {n p \log \left (e x^{n} + d\right ) \log \left (x\right ) - n p \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) + n \log \left (c\right ) \log \left (x\right ) - p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right )}{n} \]

[In]

integrate(log(c*(d+e*x^n)^p)/x,x, algorithm="fricas")

[Out]

(n*p*log(e*x^n + d)*log(x) - n*p*log(x)*log((e*x^n + d)/d) + n*log(c)*log(x) - p*dilog(-(e*x^n + d)/d + 1))/n

Sympy [F]

\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]

[In]

integrate(ln(c*(d+e*x**n)**p)/x,x)

[Out]

Integral(log(c*(d + e*x**n)**p)/x, x)

Maxima [F]

\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]

[In]

integrate(log(c*(d+e*x^n)^p)/x,x, algorithm="maxima")

[Out]

d*n*p*integrate(log(x)/(e*x*x^n + d*x), x) - 1/2*n*p*log(x)^2 + log((e*x^n + d)^p)*log(x) + log(c)*log(x)

Giac [F]

\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]

[In]

integrate(log(c*(d+e*x^n)^p)/x,x, algorithm="giac")

[Out]

integrate(log((e*x^n + d)^p*c)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}{x} \,d x \]

[In]

int(log(c*(d + e*x^n)^p)/x,x)

[Out]

int(log(c*(d + e*x^n)^p)/x, x)