3.2.47 \(\int \frac {\sqrt {1+\log (x)}}{x \log (x)} \, dx\) [147]

Optimal. Leaf size=22 \[ -2 \tanh ^{-1}\left (\sqrt {1+\log (x)}\right )+2 \sqrt {1+\log (x)} \]

[Out]

-2*arctanh((1+ln(x))^(1/2))+2*(1+ln(x))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2412, 52, 65, 213} \begin {gather*} 2 \sqrt {\log (x)+1}-2 \tanh ^{-1}\left (\sqrt {\log (x)+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Log[x]]/(x*Log[x]),x]

[Out]

-2*ArcTanh[Sqrt[1 + Log[x]]] + 2*Sqrt[1 + Log[x]]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2412

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

Rubi steps

\begin {align*} \int \frac {\sqrt {1+\log (x)}}{x \log (x)} \, dx &=\text {Subst}\left (\int \frac {\sqrt {1+x}}{x} \, dx,x,\log (x)\right )\\ &=2 \sqrt {1+\log (x)}+\text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,\log (x)\right )\\ &=2 \sqrt {1+\log (x)}+2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\log (x)}\right )\\ &=-2 \tanh ^{-1}\left (\sqrt {1+\log (x)}\right )+2 \sqrt {1+\log (x)}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 22, normalized size = 1.00 \begin {gather*} -2 \tanh ^{-1}\left (\sqrt {1+\log (x)}\right )+2 \sqrt {1+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Log[x]]/(x*Log[x]),x]

[Out]

-2*ArcTanh[Sqrt[1 + Log[x]]] + 2*Sqrt[1 + Log[x]]

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Maple [A]
time = 0.01, size = 30, normalized size = 1.36

method	result	size
derivativedivides	\(2 \sqrt {1+\ln \left (x \right )}+\ln \left (\sqrt {1+\ln \left (x \right )}-1\right )-\ln \left (\sqrt {1+\ln \left (x \right )}+1\right )\)	\(30\)
default	\(2 \sqrt {1+\ln \left (x \right )}+\ln \left (\sqrt {1+\ln \left (x \right )}-1\right )-\ln \left (\sqrt {1+\ln \left (x \right )}+1\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+ln(x))^(1/2)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

2*(1+ln(x))^(1/2)+ln((1+ln(x))^(1/2)-1)-ln((1+ln(x))^(1/2)+1)

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Maxima [A]
time = 0.28, size = 29, normalized size = 1.32 \begin {gather*} 2 \, \sqrt {\log \left (x\right ) + 1} - \log \left (\sqrt {\log \left (x\right ) + 1} + 1\right ) + \log \left (\sqrt {\log \left (x\right ) + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))^(1/2)/x/log(x),x, algorithm="maxima")

[Out]

2*sqrt(log(x) + 1) - log(sqrt(log(x) + 1) + 1) + log(sqrt(log(x) + 1) - 1)

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Fricas [A]
time = 0.37, size = 29, normalized size = 1.32 \begin {gather*} 2 \, \sqrt {\log \left (x\right ) + 1} - \log \left (\sqrt {\log \left (x\right ) + 1} + 1\right ) + \log \left (\sqrt {\log \left (x\right ) + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))^(1/2)/x/log(x),x, algorithm="fricas")

[Out]

2*sqrt(log(x) + 1) - log(sqrt(log(x) + 1) + 1) + log(sqrt(log(x) + 1) - 1)

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Sympy [A]
time = 2.07, size = 32, normalized size = 1.45 \begin {gather*} 2 \sqrt {\log {\left (x \right )} + 1} + \log {\left (\sqrt {\log {\left (x \right )} + 1} - 1 \right )} - \log {\left (\sqrt {\log {\left (x \right )} + 1} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+ln(x))**(1/2)/x/ln(x),x)

[Out]

2*sqrt(log(x) + 1) + log(sqrt(log(x) + 1) - 1) - log(sqrt(log(x) + 1) + 1)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))^(1/2)/x/log(x),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 0.41, size = 18, normalized size = 0.82 \begin {gather*} 2\,\sqrt {\ln \left (x\right )+1}-2\,\mathrm {atanh}\left (\sqrt {\ln \left (x\right )+1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) + 1)^(1/2)/(x*log(x)),x)

[Out]

2*(log(x) + 1)^(1/2) - 2*atanh((log(x) + 1)^(1/2))

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