\(\int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx\) [4813]

Optimal result

Integrand size = 31, antiderivative size = 20 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=\frac {e^e+x+x^2+\frac {e^4}{\log ^2(5)}}{x} \]

[Out]

(x^2+exp(exp(1))+x+exp(4)/ln(5)^2)/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 14} \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=x+\frac {e^e+\frac {e^4}{\log ^2(5)}}{x} \]

[In]

Int[(-E^4 - E^E*Log[5]^2 + x^2*Log[5]^2)/(x^2*Log[5]^2),x]

[Out]

x + (E^E + E^4/Log[5]^2)/x

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=\frac {e^e}{x}+x+\frac {e^4}{x \log ^2(5)} \]

[In]

Integrate[(-E^4 - E^E*Log[5]^2 + x^2*Log[5]^2)/(x^2*Log[5]^2),x]

[Out]

E^E/x + x + E^4/(x*Log[5]^2)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
risch	\(x +\frac {{\mathrm e}^{{\mathrm e}}}{x}+\frac {{\mathrm e}^{4}}{\ln \left (5\right )^{2} x}\)	\(20\)
gosper	\(\frac {x^{2} \ln \left (5\right )^{2}+\ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}}+{\mathrm e}^{4}}{\ln \left (5\right )^{2} x}\)	\(28\)
parallelrisch	\(\frac {x^{2} \ln \left (5\right )^{2}+\ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}}+{\mathrm e}^{4}}{\ln \left (5\right )^{2} x}\)	\(28\)
default	\(\frac {x \ln \left (5\right )^{2}-\frac {-\ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}}-{\mathrm e}^{4}}{x}}{\ln \left (5\right )^{2}}\)	\(32\)
norman	\(\frac {x^{2} \ln \left (5\right )+\frac {\ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}}+{\mathrm e}^{4}}{\ln \left (5\right )}}{x \ln \left (5\right )}\)	\(32\)

[In]

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

[Out]

x+exp(exp(1))/x+1/ln(5)^2/x*exp(4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=\frac {x^{2} \log \left (5\right )^{2} + e^{e} \log \left (5\right )^{2} + e^{4}}{x \log \left (5\right )^{2}} \]

[In]

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

[Out]

(x^2*log(5)^2 + e^e*log(5)^2 + e^4)/(x*log(5)^2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=\frac {x \log {\left (5 \right )}^{2} + \frac {e^{e} \log {\left (5 \right )}^{2} + e^{4}}{x}}{\log {\left (5 \right )}^{2}} \]

[In]

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

[Out]

(x*log(5)**2 + (exp(E)*log(5)**2 + exp(4))/x)/log(5)**2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=\frac {x \log \left (5\right )^{2} + \frac {e^{e} \log \left (5\right )^{2} + e^{4}}{x}}{\log \left (5\right )^{2}} \]

[In]

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

[Out]

(x*log(5)^2 + (e^e*log(5)^2 + e^4)/x)/log(5)^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=\frac {x \log \left (5\right )^{2} + \frac {e^{e} \log \left (5\right )^{2} + e^{4}}{x}}{\log \left (5\right )^{2}} \]

[In]

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

[Out]

(x*log(5)^2 + (e^e*log(5)^2 + e^4)/x)/log(5)^2

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-e^4-e^e \log ^2(5)+x^2 \log ^2(5)}{x^2 \log ^2(5)} \, dx=x+\frac {{\mathrm {e}}^4+{\mathrm {e}}^{\mathrm {e}}\,{\ln \left (5\right )}^2}{x\,{\ln \left (5\right )}^2} \]

[In]

int(-(exp(4) - x^2*log(5)^2 + exp(exp(1))*log(5)^2)/(x^2*log(5)^2),x)

[Out]

x + (exp(4) + exp(exp(1))*log(5)^2)/(x*log(5)^2)