\(\int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx\) [144]

Optimal result

Integrand size = 16, antiderivative size = 24 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {1}{4 (3+2 \log (x))}+\frac {1}{4} \log (3+2 \log (x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2412, 45} \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {1}{4} \log (2 \log (x)+3)+\frac {1}{4 (2 \log (x)+3)} \]

[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 45

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])

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*} \text {integral}& = \text {Subst}\left (\int \frac {1+x}{(3+2 x)^2} \, dx,x,\log (x)\right ) \\ & = \text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {1}{4} \left (\frac {1}{3+2 \log (x)}+\log (3+2 \log (x))\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {1}{12+8 \ln \left (x \right )}+\frac {\ln \left (\frac {3}{2}+\ln \left (x \right )\right )}{4}\)	\(19\)
derivativedivides	\(\frac {1}{12+8 \ln \left (x \right )}+\frac {\ln \left (3+2 \ln \left (x \right )\right )}{4}\)	\(21\)
default	\(\frac {1}{12+8 \ln \left (x \right )}+\frac {\ln \left (3+2 \ln \left (x \right )\right )}{4}\)	\(21\)
norman	\(\frac {1}{12+8 \ln \left (x \right )}+\frac {\ln \left (3+2 \ln \left (x \right )\right )}{4}\)	\(21\)
parallelrisch	\(\frac {1+2 \ln \left (\frac {3}{2}+\ln \left (x \right )\right ) \ln \left (x \right )+3 \ln \left (\frac {3}{2}+\ln \left (x \right )\right )}{12+8 \ln \left (x \right )}\)	\(29\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {{\left (2 \, \log \left (x\right ) + 3\right )} \log \left (2 \, \log \left (x\right ) + 3\right ) + 1}{4 \, {\left (2 \, \log \left (x\right ) + 3\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {\log {\left (\log {\left (x \right )} + \frac {3}{2} \right )}}{4} + \frac {1}{8 \log {\left (x \right )} + 12} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {1}{4 \, {\left (2 \, \log \left (x\right ) + 3\right )}} + \frac {1}{4} \, \log \left (2 \, \log \left (x\right ) + 3\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {1}{4 \, {\left (2 \, \log \left (x\right ) + 3\right )}} + \frac {1}{8} \, \log \left (\pi ^{2} {\left (\mathrm {sgn}\left (x\right ) - 1\right )}^{2} + {\left (2 \, \log \left ({\left | x \right |}\right ) + 3\right )}^{2}\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {1+\log (x)}{x (3+2 \log (x))^2} \, dx=\frac {\ln \left (2\,\ln \left (x\right )+3\right )}{4}+\frac {1}{4\,\left (2\,\ln \left (x\right )+3\right )} \]

[In]

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

[Out]

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