\(\int \frac {e^{-4 e^5} (-2 e^{4 e^5} x^2-\log (3))}{x^2 \log (5)} \, dx\) [1114]

Optimal result

Integrand size = 32, antiderivative size = 22 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=\frac {-2 x+\frac {e^{-4 e^5} \log (3)}{x}}{\log (5)} \]

[Out]

(1/x/exp(4*exp(5))*ln(3)-2*x)/ln(5)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14} \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=\frac {e^{-4 e^5} \log (3)}{x \log (5)}-\frac {2 x}{\log (5)} \]

[In]

Int[(-2*E^(4*E^5)*x^2 - Log[3])/(E^(4*E^5)*x^2*Log[5]),x]

[Out]

(-2*x)/Log[5] + Log[3]/(E^(4*E^5)*x*Log[5])

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 {e^{-4 e^5} \int \frac {-2 e^{4 e^5} x^2-\log (3)}{x^2} \, dx}{\log (5)} \\ & = \frac {e^{-4 e^5} \int \left (-2 e^{4 e^5}-\frac {\log (3)}{x^2}\right ) \, dx}{\log (5)} \\ & = -\frac {2 x}{\log (5)}+\frac {e^{-4 e^5} \log (3)}{x \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=-\frac {2 x-\frac {e^{-4 e^5} \log (3)}{x}}{\log (5)} \]

[In]

Integrate[(-2*E^(4*E^5)*x^2 - Log[3])/(E^(4*E^5)*x^2*Log[5]),x]

[Out]

-((2*x - Log[3]/(E^(4*E^5)*x))/Log[5])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09

method	result	size
risch	\(-\frac {2 x}{\ln \left (5\right )}+\frac {{\mathrm e}^{-4 \,{\mathrm e}^{5}} \ln \left (3\right )}{\ln \left (5\right ) x}\)	\(24\)
norman	\(\frac {\frac {{\mathrm e}^{-4 \,{\mathrm e}^{5}} \ln \left (3\right )}{\ln \left (5\right )}-\frac {2 x^{2}}{\ln \left (5\right )}}{x}\)	\(27\)
default	\(\frac {{\mathrm e}^{-4 \,{\mathrm e}^{5}} \left (-2 x \,{\mathrm e}^{4 \,{\mathrm e}^{5}}+\frac {\ln \left (3\right )}{x}\right )}{\ln \left (5\right )}\)	\(28\)
gosper	\(\frac {\left (-2 x^{2} {\mathrm e}^{4 \,{\mathrm e}^{5}}+\ln \left (3\right )\right ) {\mathrm e}^{-4 \,{\mathrm e}^{5}}}{x \ln \left (5\right )}\)	\(29\)
parallelrisch	\(\frac {\left (-2 x^{2} {\mathrm e}^{4 \,{\mathrm e}^{5}}+\ln \left (3\right )\right ) {\mathrm e}^{-4 \,{\mathrm e}^{5}}}{x \ln \left (5\right )}\)	\(29\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=-\frac {{\left (2 \, x^{2} e^{\left (4 \, e^{5}\right )} - \log \left (3\right )\right )} e^{\left (-4 \, e^{5}\right )}}{x \log \left (5\right )} \]

[In]

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

[Out]

-(2*x^2*e^(4*e^5) - log(3))*e^(-4*e^5)/(x*log(5))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=\frac {- 2 x e^{4 e^{5}} + \frac {\log {\left (3 \right )}}{x}}{e^{4 e^{5}} \log {\left (5 \right )}} \]

[In]

integrate((-2*x**2*exp(4*exp(5))-ln(3))/x**2/ln(5)/exp(4*exp(5)),x)

[Out]

(-2*x*exp(4*exp(5)) + log(3)/x)*exp(-4*exp(5))/log(5)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=-\frac {{\left (2 \, x e^{\left (4 \, e^{5}\right )} - \frac {\log \left (3\right )}{x}\right )} e^{\left (-4 \, e^{5}\right )}}{\log \left (5\right )} \]

[In]

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

[Out]

-(2*x*e^(4*e^5) - log(3)/x)*e^(-4*e^5)/log(5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=-\frac {{\left (2 \, x e^{\left (4 \, e^{5}\right )} - \frac {\log \left (3\right )}{x}\right )} e^{\left (-4 \, e^{5}\right )}}{\log \left (5\right )} \]

[In]

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

[Out]

-(2*x*e^(4*e^5) - log(3)/x)*e^(-4*e^5)/log(5)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-4 e^5} \left (-2 e^{4 e^5} x^2-\log (3)\right )}{x^2 \log (5)} \, dx=\frac {{\mathrm {e}}^{-4\,{\mathrm {e}}^5}\,\left (\ln \left (3\right )-2\,x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^5}\right )}{x\,\ln \left (5\right )} \]

[In]

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

[Out]

(exp(-4*exp(5))*(log(3) - 2*x^2*exp(4*exp(5))))/(x*log(5))