\(\int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx\) [4206]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3 e^2}{16 x^2 \log ^2(5)} \]

[Out]

-3/16*exp(2)/ln(5)^2/x^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 30} \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3 e^2}{16 x^2 \log ^2(5)} \]

[In]

Int[(3*E^2)/(8*x^3*Log[5]^2),x]

[Out]

(-3*E^2)/(16*x^2*Log[5]^2)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (3 e^2\right ) \int \frac {1}{x^3} \, dx}{8 \log ^2(5)} \\ & = -\frac {3 e^2}{16 x^2 \log ^2(5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3 e^2}{16 x^2 \log ^2(5)} \]

[In]

Integrate[(3*E^2)/(8*x^3*Log[5]^2),x]

[Out]

(-3*E^2)/(16*x^2*Log[5]^2)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
gosper	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \left (5\right )^{2} x^{2}}\)	\(12\)
default	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \left (5\right )^{2} x^{2}}\)	\(12\)
norman	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \left (5\right )^{2} x^{2}}\)	\(12\)
risch	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \left (5\right )^{2} x^{2}}\)	\(12\)
parallelrisch	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \left (5\right )^{2} x^{2}}\)	\(12\)

[In]

int(3/8*exp(2)/x^3/ln(5)^2,x,method=_RETURNVERBOSE)

[Out]

-3/16*exp(2)/ln(5)^2/x^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3 \, e^{2}}{16 \, x^{2} \log \left (5\right )^{2}} \]

[In]

integrate(3/8*exp(2)/x^3/log(5)^2,x, algorithm="fricas")

[Out]

-3/16*e^2/(x^2*log(5)^2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=- \frac {3 e^{2}}{16 x^{2} \log {\left (5 \right )}^{2}} \]

[In]

integrate(3/8*exp(2)/x**3/ln(5)**2,x)

[Out]

-3*exp(2)/(16*x**2*log(5)**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3 \, e^{2}}{16 \, x^{2} \log \left (5\right )^{2}} \]

[In]

integrate(3/8*exp(2)/x^3/log(5)^2,x, algorithm="maxima")

[Out]

-3/16*e^2/(x^2*log(5)^2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3 \, e^{2}}{16 \, x^{2} \log \left (5\right )^{2}} \]

[In]

integrate(3/8*exp(2)/x^3/log(5)^2,x, algorithm="giac")

[Out]

-3/16*e^2/(x^2*log(5)^2)

Mupad [B] (verification not implemented)

Time = 10.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx=-\frac {3\,{\mathrm {e}}^2}{16\,x^2\,{\ln \left (5\right )}^2} \]

[In]

int((3*exp(2))/(8*x^3*log(5)^2),x)

[Out]

-(3*exp(2))/(16*x^2*log(5)^2)