3.1.64 \(\int \frac {\log ^p(x)}{x} \, dx\) [64]

Optimal. Leaf size=12 \[ \frac {\log ^{1+p}(x)}{1+p} \]

[Out]

ln(x)^(1+p)/(1+p)

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Rubi [A]
time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2339, 30} \begin {gather*} \frac {\log ^{p+1}(x)}{p+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[x]^p/x,x]

[Out]

Log[x]^(1 + p)/(1 + p)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps

\begin {align*} \int \frac {\log ^p(x)}{x} \, dx &=\text {Subst}\left (\int x^p \, dx,x,\log (x)\right )\\ &=\frac {\log ^{1+p}(x)}{1+p}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} \frac {\log ^{1+p}(x)}{1+p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[x]^p/x,x]

[Out]

Log[x]^(1 + p)/(1 + p)

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.96, size = 21, normalized size = 1.75 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {\text {Log}\left [x\right ]^{1+p}}{1+p},p\text {!=}-1\right \}\right \},\text {Log}\left [\text {Log}\left [x\right ]\right ]\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(Log[x])^p/x,x]')

[Out]

Piecewise[{{Log[x] ^ (1 + p) / (1 + p), p != -1}}, Log[Log[x]]]

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Maple [A]
time = 0.01, size = 13, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {\ln \left (x \right )^{1+p}}{1+p}\)	\(13\)
default	\(\frac {\ln \left (x \right )^{1+p}}{1+p}\)	\(13\)
risch	\(\frac {\ln \left (x \right ) \ln \left (x \right )^{p}}{1+p}\)	\(13\)
norman	\(\frac {\ln \left (x \right ) {\mathrm e}^{p \ln \left (\ln \left (x \right )\right )}}{1+p}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)^p/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)^(1+p)/(1+p)

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Maxima [A]
time = 0.26, size = 12, normalized size = 1.00 \begin {gather*} \frac {\log \left (x\right )^{p + 1}}{p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^p/x,x, algorithm="maxima")

[Out]

log(x)^(p + 1)/(p + 1)

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Fricas [A]
time = 0.34, size = 12, normalized size = 1.00 \begin {gather*} \frac {\log \left (x\right )^{p} \log \left (x\right )}{p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^p/x,x, algorithm="fricas")

[Out]

log(x)^p*log(x)/(p + 1)

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Sympy [A]
time = 0.44, size = 15, normalized size = 1.25 \begin {gather*} \begin {cases} \frac {\log {\left (x \right )}^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (\log {\left (x \right )} \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)**p/x,x)

[Out]

Piecewise((log(x)**(p + 1)/(p + 1), Ne(p, -1)), (log(log(x)), True))

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Giac [A]
time = 0.00, size = 11, normalized size = 0.92 \begin {gather*} \frac {\left (\ln x\right )^{p+1}}{p+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^p/x,x)

[Out]

log(x)^(p + 1)/(p + 1)

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Mupad [B]
time = 0.15, size = 22, normalized size = 1.83 \begin {gather*} \left \{\begin {array}{cl} \ln \left (\ln \left (x\right )\right ) & \text {\ if\ \ }p=-1\\ \frac {{\ln \left (x\right )}^{p+1}}{p+1} & \text {\ if\ \ }p\neq -1 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)^p/x,x)

[Out]

piecewise(p == -1, log(log(x)), p ~= -1, log(x)^(p + 1)/(p + 1))

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