\(\int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx\) [6538]

Optimal result

Integrand size = 33, antiderivative size = 26 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=-1+\frac {81 x}{\log ^4(4)}-\frac {25-x}{5+e^2 \log (5)} \]

[Out]

81/16*x/ln(2)^4-(-x+25)/(exp(2)*ln(5)+5)-1

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {8} \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {x \left (405+\log ^4(4)+81 e^2 \log (5)\right )}{\log ^4(4) \left (5+e^2 \log (5)\right )} \]

[In]

Int[(405 + Log[4]^4 + 81*E^2*Log[5])/(5*Log[4]^4 + E^2*Log[4]^4*Log[5]),x]

[Out]

(x*(405 + Log[4]^4 + 81*E^2*Log[5]))/(Log[4]^4*(5 + E^2*Log[5]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \frac {x \left (405+\log ^4(4)+81 e^2 \log (5)\right )}{\log ^4(4) \left (5+e^2 \log (5)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(26)=52\).

Time = 0.00 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {405 x}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)}+\frac {x \log ^4(4)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)}+\frac {81 e^2 x \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \]

[In]

Integrate[(405 + Log[4]^4 + 81*E^2*Log[5])/(5*Log[4]^4 + E^2*Log[4]^4*Log[5]),x]

[Out]

(405*x)/(5*Log[4]^4 + E^2*Log[4]^4*Log[5]) + (x*Log[4]^4)/(5*Log[4]^4 + E^2*Log[4]^4*Log[5]) + (81*E^2*x*Log[5
])/(5*Log[4]^4 + E^2*Log[4]^4*Log[5])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19

method	result	size
norman	\(\frac {\left (81 \,{\mathrm e}^{2} \ln \left (5\right )+16 \ln \left (2\right )^{4}+405\right ) x}{16 \ln \left (2\right )^{4} \left ({\mathrm e}^{2} \ln \left (5\right )+5\right )}\)	\(31\)
default	\(\frac {\left (81 \,{\mathrm e}^{2} \ln \left (5\right )+16 \ln \left (2\right )^{4}+405\right ) x}{16 \,{\mathrm e}^{2} \ln \left (2\right )^{4} \ln \left (5\right )+80 \ln \left (2\right )^{4}}\)	\(36\)
parallelrisch	\(\frac {\left (81 \,{\mathrm e}^{2} \ln \left (5\right )+16 \ln \left (2\right )^{4}+405\right ) x}{16 \,{\mathrm e}^{2} \ln \left (2\right )^{4} \ln \left (5\right )+80 \ln \left (2\right )^{4}}\)	\(36\)
risch	\(\frac {16 x \ln \left (2\right )^{4}}{16 \,{\mathrm e}^{2} \ln \left (2\right )^{4} \ln \left (5\right )+80 \ln \left (2\right )^{4}}+\frac {81 x \,{\mathrm e}^{2} \ln \left (5\right )}{16 \,{\mathrm e}^{2} \ln \left (2\right )^{4} \ln \left (5\right )+80 \ln \left (2\right )^{4}}+\frac {405 x}{16 \,{\mathrm e}^{2} \ln \left (2\right )^{4} \ln \left (5\right )+80 \ln \left (2\right )^{4}}\)	\(76\)

[In]

int((81*exp(2)*ln(5)+16*ln(2)^4+405)/(16*exp(2)*ln(2)^4*ln(5)+80*ln(2)^4),x,method=_RETURNVERBOSE)

[Out]

1/16*(81*exp(2)*ln(5)+16*ln(2)^4+405)/ln(2)^4/(exp(2)*ln(5)+5)*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {16 \, x \log \left (2\right )^{4} + 81 \, x e^{2} \log \left (5\right ) + 405 \, x}{16 \, {\left (e^{2} \log \left (5\right ) \log \left (2\right )^{4} + 5 \, \log \left (2\right )^{4}\right )}} \]

[In]

integrate((81*exp(2)*log(5)+16*log(2)^4+405)/(16*exp(2)*log(2)^4*log(5)+80*log(2)^4),x, algorithm="fricas")

[Out]

1/16*(16*x*log(2)^4 + 81*x*e^2*log(5) + 405*x)/(e^2*log(5)*log(2)^4 + 5*log(2)^4)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {x \left (16 \log {\left (2 \right )}^{4} + 405 + 81 e^{2} \log {\left (5 \right )}\right )}{80 \log {\left (2 \right )}^{4} + 16 e^{2} \log {\left (2 \right )}^{4} \log {\left (5 \right )}} \]

[In]

integrate((81*exp(2)*ln(5)+16*ln(2)**4+405)/(16*exp(2)*ln(2)**4*ln(5)+80*ln(2)**4),x)

[Out]

x*(16*log(2)**4 + 405 + 81*exp(2)*log(5))/(80*log(2)**4 + 16*exp(2)*log(2)**4*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {{\left (16 \, \log \left (2\right )^{4} + 81 \, e^{2} \log \left (5\right ) + 405\right )} x}{16 \, {\left (e^{2} \log \left (5\right ) \log \left (2\right )^{4} + 5 \, \log \left (2\right )^{4}\right )}} \]

[In]

integrate((81*exp(2)*log(5)+16*log(2)^4+405)/(16*exp(2)*log(2)^4*log(5)+80*log(2)^4),x, algorithm="maxima")

[Out]

1/16*(16*log(2)^4 + 81*e^2*log(5) + 405)*x/(e^2*log(5)*log(2)^4 + 5*log(2)^4)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {{\left (16 \, \log \left (2\right )^{4} + 81 \, e^{2} \log \left (5\right ) + 405\right )} x}{16 \, {\left (e^{2} \log \left (5\right ) \log \left (2\right )^{4} + 5 \, \log \left (2\right )^{4}\right )}} \]

[In]

integrate((81*exp(2)*log(5)+16*log(2)^4+405)/(16*exp(2)*log(2)^4*log(5)+80*log(2)^4),x, algorithm="giac")

[Out]

1/16*(16*log(2)^4 + 81*e^2*log(5) + 405)*x/(e^2*log(5)*log(2)^4 + 5*log(2)^4)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {405+\log ^4(4)+81 e^2 \log (5)}{5 \log ^4(4)+e^2 \log ^4(4) \log (5)} \, dx=\frac {x\,\left (81\,{\mathrm {e}}^2\,\ln \left (5\right )+16\,{\ln \left (2\right )}^4+405\right )}{80\,{\ln \left (2\right )}^4+16\,{\mathrm {e}}^2\,{\ln \left (2\right )}^4\,\ln \left (5\right )} \]

[In]

int((81*exp(2)*log(5) + 16*log(2)^4 + 405)/(80*log(2)^4 + 16*exp(2)*log(2)^4*log(5)),x)

[Out]

(x*(81*exp(2)*log(5) + 16*log(2)^4 + 405))/(80*log(2)^4 + 16*exp(2)*log(2)^4*log(5))