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

Optimal. Leaf size=44 \[ \frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2441, 2352} \begin {gather*} \frac {p \text {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}+\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.00, size = 43, normalized size = 0.98 \begin {gather*} \frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \text {Li}_2\left (\frac {d+e x^n}{d}\right )}{n} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.23, size = 177, normalized size = 4.02

method	result	size
risch	\(\ln \left (x \right ) \ln \left (\left (d +e \,x^{n}\right )^{p}\right )+\frac {i \ln \left (x \right ) \pi \,\mathrm {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \mathrm {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )^{2}}{2}-\frac {i \ln \left (x \right ) \pi \,\mathrm {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \mathrm {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{2}-\frac {i \ln \left (x \right ) \pi \mathrm {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )^{3}}{2}+\frac {i \ln \left (x \right ) \pi \mathrm {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}+\ln \left (c \right ) \ln \left (x \right )-\frac {p \dilog \left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.38, size = 63, normalized size = 1.43 \begin {gather*} \frac {n p \log \left (x^{n} e + d\right ) \log \left (x\right ) - n p \log \left (x\right ) \log \left (\frac {x^{n} e + d}{d}\right ) + n \log \left (c\right ) \log \left (x\right ) - p {\rm Li}_2\left (-\frac {x^{n} e + d}{d} + 1\right )}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(d + e*x**n)**p)/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((x^n*e + d)^p*c)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^n)^p)/x,x)

[Out]

int(log(c*(d + e*x^n)^p)/x, x)

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