\(\int \frac {1}{x (1+\log ^2(x))} \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 3 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\arctan (\log (x)) \]

[Out]

arctan(ln(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {209} \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\arctan (\log (x)) \]

[In]

Int[1/(x*(1 + Log[x]^2)),x]

[Out]

ArcTan[Log[x]]

Rule 209

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\log (x)\right ) \\ & = \arctan (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\arctan (\log (x)) \]

[In]

Integrate[1/(x*(1 + Log[x]^2)),x]

[Out]

ArcTan[Log[x]]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\arctan \left (\ln \left (x \right )\right )\)	\(4\)
default	\(\arctan \left (\ln \left (x \right )\right )\)	\(4\)
risch	\(\frac {i \ln \left (\ln \left (x \right )+i\right )}{2}-\frac {i \ln \left (\ln \left (x \right )-i\right )}{2}\)	\(20\)
parallelrisch	\(\frac {i \ln \left (\ln \left (x \right )+i\right )}{2}-\frac {i \ln \left (\ln \left (x \right )-i\right )}{2}\)	\(20\)

[In]

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

[Out]

arctan(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\arctan \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

arctan(log(x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (3) = 6\).

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 5.00 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\operatorname {RootSum} {\left (4 z^{2} + 1, \left ( i \mapsto i \log {\left (2 i + \log {\left (x \right )} \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(4*_z**2 + 1, Lambda(_i, _i*log(2*_i + log(x))))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\arctan \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

arctan(log(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\arctan \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

arctan(log(x))

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log ^2(x)\right )} \, dx=\mathrm {atan}\left (\ln \left (x\right )\right ) \]

[In]

int(1/(x*(log(x)^2 + 1)),x)

[Out]

atan(log(x))