\(\int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx\) [6606]

Optimal result

Integrand size = 20, antiderivative size = 14 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=4-\frac {e^5 \log (5) \log (x)}{x} \]

[Out]

-exp(5)*ln(5)*ln(x)/x+4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2340} \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=-\frac {e^5 \log (5) \log (x)}{x} \]

[In]

Int[(-(E^5*Log[5]) + E^5*Log[5]*Log[x])/x^2,x]

[Out]

-((E^5*Log[5]*Log[x])/x)

Rule 2340

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[b*(d*x)^(m + 1)*(Log[c*x^n]/(d
*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {e^5 \log (5) \log (x)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=-\frac {e^5 \log (5) \log (x)}{x} \]

[In]

Integrate[(-(E^5*Log[5]) + E^5*Log[5]*Log[x])/x^2,x]

[Out]

-((E^5*Log[5]*Log[x])/x)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
norman	\(-\frac {{\mathrm e}^{5} \ln \left (5\right ) \ln \left (x \right )}{x}\)	\(12\)
risch	\(-\frac {{\mathrm e}^{5} \ln \left (5\right ) \ln \left (x \right )}{x}\)	\(12\)
parallelrisch	\(-\frac {{\mathrm e}^{5} \ln \left (5\right ) \ln \left (x \right )}{x}\)	\(12\)
default	\({\mathrm e}^{5} \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+\frac {{\mathrm e}^{5} \ln \left (5\right )}{x}\)	\(28\)
parts	\({\mathrm e}^{5} \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+\frac {{\mathrm e}^{5} \ln \left (5\right )}{x}\)	\(28\)

[In]

int((exp(5)*ln(5)*ln(x)-exp(5)*ln(5))/x^2,x,method=_RETURNVERBOSE)

[Out]

-exp(5)*ln(5)*ln(x)/x

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=-\frac {e^{5} \log \left (5\right ) \log \left (x\right )}{x} \]

[In]

integrate((exp(5)*log(5)*log(x)-exp(5)*log(5))/x^2,x, algorithm="fricas")

[Out]

-e^5*log(5)*log(x)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=- \frac {e^{5} \log {\left (5 \right )} \log {\left (x \right )}}{x} \]

[In]

integrate((exp(5)*ln(5)*ln(x)-exp(5)*ln(5))/x**2,x)

[Out]

-exp(5)*log(5)*log(x)/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.71 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=-\frac {{\left (e^{5} \log \left (x\right ) + e^{5}\right )} \log \left (5\right )}{x} + \frac {e^{5} \log \left (5\right )}{x} \]

[In]

integrate((exp(5)*log(5)*log(x)-exp(5)*log(5))/x^2,x, algorithm="maxima")

[Out]

-(e^5*log(x) + e^5)*log(5)/x + e^5*log(5)/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=-\frac {e^{5} \log \left (5\right ) \log \left (x\right )}{x} \]

[In]

integrate((exp(5)*log(5)*log(x)-exp(5)*log(5))/x^2,x, algorithm="giac")

[Out]

-e^5*log(5)*log(x)/x

Mupad [B] (verification not implemented)

Time = 11.11 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-e^5 \log (5)+e^5 \log (5) \log (x)}{x^2} \, dx=-\frac {{\mathrm {e}}^5\,\ln \left (5\right )\,\ln \left (x\right )}{x} \]

[In]

int(-(exp(5)*log(5) - exp(5)*log(5)*log(x))/x^2,x)

[Out]

-(exp(5)*log(5)*log(x))/x