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

Optimal. Leaf size=27 \[ \log (x)+\frac {4 x}{i \pi +\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 43} \begin {gather*} \log (x)+\frac {4 x}{\log \left (5-\frac {1}{10} \log ^2\left (\frac {5}{3}\right )\right )+i \pi } \end {gather*}

Antiderivative was successfully verified.

[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 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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 50, normalized size = 1.85 \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.93, size = 28, normalized size = 1.04 \begin {gather*} \frac {\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right ) \log \relax (x) + 4 \, x}{\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.20, size = 29, normalized size = 1.07 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.12, size = 21, normalized size = 0.78

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.42, size = 28, normalized size = 1.04 \begin {gather*} \frac {\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right ) \log \relax (x) + 4 \, x}{\log \left (\frac {1}{10} \, \log \left (\frac {5}{3}\right )^{2} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.66, size = 17, normalized size = 0.63 \begin {gather*} \ln \relax (x)+\frac {4\,x}{\ln \left (\frac {{\ln \left (\frac {5}{3}\right )}^2}{10}-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 0.19, size = 66, normalized size = 2.44 \begin {gather*} \frac {- 4 x - \left (\log {\left (- \frac {\log {\relax (5 )}^{2}}{10} - \frac {\log {\relax (3 )}^{2}}{10} + \frac {\log {\relax (3 )} \log {\relax (5 )}}{5} + 5 \right )} + i \pi \right ) \log {\relax (x )}}{- \log {\left (- \frac {\log {\relax (5 )}^{2}}{10} - \frac {\log {\relax (3 )}^{2}}{10} + \frac {\log {\relax (3 )} \log {\relax (5 )}}{5} + 5 \right )} - i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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