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

Optimal. Leaf size=26 \[ -1+\frac {81 x}{\log ^4(4)}-\frac {25-x}{5+e^2 \log (5)} \]

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {8} \begin {gather*} \frac {x \left (405+\log ^4(4)+81 e^2 \log (5)\right )}{\log ^4(4) \left (5+e^2 \log (5)\right )} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [B]  time = 0.00, size = 75, normalized size = 2.88 \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.76, size = 38, normalized size = 1.46 \begin {gather*} \frac {16 \, x \log \relax (2)^{4} + 81 \, x e^{2} \log \relax (5) + 405 \, x}{16 \, {\left (e^{2} \log \relax (5) \log \relax (2)^{4} + 5 \, \log \relax (2)^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.17, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\left (16 \, \log \relax (2)^{4} + 81 \, e^{2} \log \relax (5) + 405\right )} x}{16 \, {\left (e^{2} \log \relax (5) \log \relax (2)^{4} + 5 \, \log \relax (2)^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 31, normalized size = 1.19

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.50, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\left (16 \, \log \relax (2)^{4} + 81 \, e^{2} \log \relax (5) + 405\right )} x}{16 \, {\left (e^{2} \log \relax (5) \log \relax (2)^{4} + 5 \, \log \relax (2)^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.00, size = 35, normalized size = 1.35 \begin {gather*} \frac {x\,\left (81\,{\mathrm {e}}^2\,\ln \relax (5)+16\,{\ln \relax (2)}^4+405\right )}{80\,{\ln \relax (2)}^4+16\,{\mathrm {e}}^2\,{\ln \relax (2)}^4\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.05, size = 37, normalized size = 1.42 \begin {gather*} \frac {x \left (16 \log {\relax (2 )}^{4} + 405 + 81 e^{2} \log {\relax (5 )}\right )}{80 \log {\relax (2 )}^{4} + 16 e^{2} \log {\relax (2 )}^{4} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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