\(\int \frac {i \pi +4 x+\log (\frac {1}{10} (50-\log ^2(\frac {5}{3})))}{x (i \pi +\log (\frac {1}{10} (50-\log ^2(\frac {5}{3}))))} \, dx\) [2891]

Optimal result

Integrand size = 51, antiderivative size = 27 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\log (x)+\frac {4 x}{i \pi +\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )} \]

[Out]

ln(x)+4/ln(1/10*ln(5/3)^2-5)*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 45} \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\log (x)+\frac {4 x}{\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )+i \pi } \]

[In]

Int[(I*Pi + 4*x + Log[(50 - Log[5/3]^2)/10])/(x*(I*Pi + Log[(50 - Log[5/3]^2)/10])),x]

[Out]

Log[x] + (4*x)/(I*Pi + Log[5 - Log[5/3]^2/10])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x} \, dx}{i \pi +\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )} \\ & = \frac {\int \left (4+\frac {i \left (\pi -i \log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )}{x}\right ) \, dx}{i \pi +\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )} \\ & = \log (x)+\frac {4 x}{i \pi +\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\frac {-4 i x+\log (x) \left (\pi -i \log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )\right )}{\pi -i \log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )} \]

[In]

Integrate[(I*Pi + 4*x + Log[(50 - Log[5/3]^2)/10])/(x*(I*Pi + Log[(50 - Log[5/3]^2)/10])),x]

[Out]

((-4*I)*x + Log[x]*(Pi - I*Log[5 - Log[5/3]^2/10]))/(Pi - I*Log[5 - Log[5/3]^2/10])

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
norman	\(\frac {4 x}{-\ln \left (10\right )+\ln \left (\ln \left (\frac {5}{3}\right )^{2}-50\right )}+\ln \left (x \right )\)	\(21\)
risch	\(\frac {4 x}{\ln \left (\frac {\left (\ln \left (5\right )-\ln \left (3\right )\right )^{2}}{10}-5\right )}+\ln \left (x \right )\)	\(23\)
default	\(\frac {4 x +\ln \left (\frac {\ln \left (\frac {5}{3}\right )^{2}}{10}-5\right ) \ln \left (x \right )}{\ln \left (\frac {\ln \left (\frac {5}{3}\right )^{2}}{10}-5\right )}\)	\(29\)
parallelrisch	\(\frac {4 x +\ln \left (\frac {\ln \left (\frac {5}{3}\right )^{2}}{10}-5\right ) \ln \left (x \right )}{\ln \left (\frac {\ln \left (\frac {5}{3}\right )^{2}}{10}-5\right )}\)	\(29\)

[In]

int((ln(1/10*ln(5/3)^2-5)+4*x)/x/ln(1/10*ln(5/3)^2-5),x,method=_RETURNVERBOSE)

[Out]

4/(-ln(10)+ln(ln(5/3)^2-50))*x+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\frac {\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right ) \log \left (x\right ) + 4 \, x}{\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right )} \]

[In]

integrate((log(1/10*log(5/3)^2-5)+4*x)/x/log(1/10*log(5/3)^2-5),x, algorithm="fricas")

[Out]

(log(1/10*log(5/3)^2 - 5)*log(x) + 4*x)/log(1/10*log(5/3)^2 - 5)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (20) = 40\).

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\frac {4 x + \left (\log {\left (- \frac {\log {\left (5 \right )}^{2}}{10} - \frac {\log {\left (3 \right )}^{2}}{10} + \frac {\log {\left (3 \right )} \log {\left (5 \right )}}{5} + 5 \right )} + i \pi \right ) \log {\left (x \right )}}{\log {\left (- \frac {\log {\left (5 \right )}^{2}}{10} - \frac {\log {\left (3 \right )}^{2}}{10} + \frac {\log {\left (3 \right )} \log {\left (5 \right )}}{5} + 5 \right )} + i \pi } \]

[In]

integrate((ln(1/10*ln(5/3)**2-5)+4*x)/x/ln(1/10*ln(5/3)**2-5),x)

[Out]

(4*x + (log(-log(5)**2/10 - log(3)**2/10 + log(3)*log(5)/5 + 5) + I*pi)*log(x))/(log(-log(5)**2/10 - log(3)**2
/10 + log(3)*log(5)/5 + 5) + I*pi)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\frac {\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right ) \log \left (x\right ) + 4 \, x}{\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right )} \]

[In]

integrate((log(1/10*log(5/3)^2-5)+4*x)/x/log(1/10*log(5/3)^2-5),x, algorithm="maxima")

[Out]

(log(1/10*log(5/3)^2 - 5)*log(x) + 4*x)/log(1/10*log(5/3)^2 - 5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\frac {\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right ) \log \left ({\left | x \right |}\right ) + 4 \, x}{\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right )} \]

[In]

integrate((log(1/10*log(5/3)^2-5)+4*x)/x/log(1/10*log(5/3)^2-5),x, algorithm="giac")

[Out]

(log(1/10*log(5/3)^2 - 5)*log(abs(x)) + 4*x)/log(1/10*log(5/3)^2 - 5)

Mupad [B] (verification not implemented)

Time = 8.98 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {i \pi +4 x+\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )}{x \left (i \pi +\log \left (\frac {1}{10} \left (50-\log ^2\left (\frac {5}{3}\right )\right )\right )\right )} \, dx=\ln \left (x\right )+\frac {4\,x}{\ln \left (\frac {{\ln \left (\frac {5}{3}\right )}^2}{10}-5\right )} \]

[In]

int((4*x + log(log(5/3)^2/10 - 5))/(x*log(log(5/3)^2/10 - 5)),x)

[Out]

log(x) + (4*x)/log(log(5/3)^2/10 - 5)