### 3.144 $$\int \frac{1+\log (x)}{x (3+2 \log (x))^2} \, dx$$

Optimal. Leaf size=24 $\frac{1}{4} \log (2 \log (x)+3)+\frac{1}{4 (2 \log (x)+3)}$

[Out]

1/(4*(3 + 2*Log[x])) + Log[3 + 2*Log[x]]/4

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Rubi [A]  time = 0.0405776, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {2365, 43} $\frac{1}{4} \log (2 \log (x)+3)+\frac{1}{4 (2 \log (x)+3)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 + Log[x])/(x*(3 + 2*Log[x])^2),x]

[Out]

1/(4*(3 + 2*Log[x])) + Log[3 + 2*Log[x]]/4

Rule 2365

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]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1+\log (x)}{x (3+2 \log (x))^2} \, dx &=\operatorname{Subst}\left (\int \frac{1+x}{(3+2 x)^2} \, dx,x,\log (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{1}{2 (3+2 x)^2}+\frac{1}{2 (3+2 x)}\right ) \, dx,x,\log (x)\right )\\ &=\frac{1}{4 (3+2 \log (x))}+\frac{1}{4} \log (3+2 \log (x))\\ \end{align*}

Mathematica [A]  time = 0.0266525, size = 20, normalized size = 0.83 $\frac{1}{4} \left (\log (2 \log (x)+3)+\frac{1}{2 \log (x)+3}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(1 + Log[x])/(x*(3 + 2*Log[x])^2),x]

[Out]

((3 + 2*Log[x])^(-1) + Log[3 + 2*Log[x]])/4

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Maple [A]  time = 0.01, size = 21, normalized size = 0.9 \begin{align*}{\frac{1}{12+8\,\ln \left ( x \right ) }}+{\frac{\ln \left ( 3+2\,\ln \left ( x \right ) \right ) }{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1+ln(x))/x/(3+2*ln(x))^2,x)

[Out]

1/4/(3+2*ln(x))+1/4*ln(3+2*ln(x))

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Maxima [A]  time = 1.10067, size = 27, normalized size = 1.12 \begin{align*} \frac{1}{4 \,{\left (2 \, \log \left (x\right ) + 3\right )}} + \frac{1}{4} \, \log \left (2 \, \log \left (x\right ) + 3\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))/x/(3+2*log(x))^2,x, algorithm="maxima")

[Out]

1/4/(2*log(x) + 3) + 1/4*log(2*log(x) + 3)

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Fricas [A]  time = 1.95505, size = 80, normalized size = 3.33 \begin{align*} \frac{{\left (2 \, \log \left (x\right ) + 3\right )} \log \left (2 \, \log \left (x\right ) + 3\right ) + 1}{4 \,{\left (2 \, \log \left (x\right ) + 3\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))/x/(3+2*log(x))^2,x, algorithm="fricas")

[Out]

1/4*((2*log(x) + 3)*log(2*log(x) + 3) + 1)/(2*log(x) + 3)

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Sympy [A]  time = 0.120399, size = 17, normalized size = 0.71 \begin{align*} \frac{\log{\left (\log{\left (x \right )} + \frac{3}{2} \right )}}{4} + \frac{1}{8 \log{\left (x \right )} + 12} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+ln(x))/x/(3+2*ln(x))**2,x)

[Out]

log(log(x) + 3/2)/4 + 1/(8*log(x) + 12)

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Giac [A]  time = 1.21718, size = 46, normalized size = 1.92 \begin{align*} \frac{1}{4 \,{\left (2 \, \log \left (x\right ) + 3\right )}} + \frac{1}{8} \, \log \left (\pi ^{2}{\left (\mathrm{sgn}\left (x\right ) - 1\right )}^{2} +{\left (2 \, \log \left ({\left | x \right |}\right ) + 3\right )}^{2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))/x/(3+2*log(x))^2,x, algorithm="giac")

[Out]

1/4/(2*log(x) + 3) + 1/8*log(pi^2*(sgn(x) - 1)^2 + (2*log(abs(x)) + 3)^2)