3.85.71 \(\int \frac {-10+6 e^4 \log (5)}{3 e^4 \log (5)} \, dx\)

Optimal. Leaf size=17 \[ 4+2 x-\frac {10 x}{3 e^4 \log (5)} \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {8} \begin {gather*} \frac {2}{3} x \left (3-\frac {5}{e^4 \log (5)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + 6*E^4*Log[5])/(3*E^4*Log[5]),x]

[Out]

(2*x*(3 - 5/(E^4*Log[5])))/3

Rule 8

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {2}{3} x \left (3-\frac {5}{e^4 \log (5)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.94 \begin {gather*} 2 x-\frac {10 x}{3 e^4 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 6*E^4*Log[5])/(3*E^4*Log[5]),x]

[Out]

2*x - (10*x)/(3*E^4*Log[5])

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fricas [A]  time = 0.63, size = 19, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (3 \, x e^{4} \log \relax (5) - 5 \, x\right )} e^{\left (-4\right )}}{3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(6*exp(4)*log(5)-10)/exp(4)/log(5),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 17, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (3 \, e^{4} \log \relax (5) - 5\right )} x e^{\left (-4\right )}}{3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(6*exp(4)*log(5)-10)/exp(4)/log(5),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 18, normalized size = 1.06

method	result	size
risch	\(2 \,{\mathrm e}^{-4} x \,{\mathrm e}^{4}-\frac {10 \,{\mathrm e}^{-4} x}{3 \ln \relax (5)}\)	\(18\)
default	\(\frac {\left (6 \,{\mathrm e}^{4} \ln \relax (5)-10\right ) {\mathrm e}^{-4} x}{3 \ln \relax (5)}\)	\(20\)
norman	\(\frac {2 \left (3 \,{\mathrm e}^{4} \ln \relax (5)-5\right ) {\mathrm e}^{-4} x}{3 \ln \relax (5)}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(6*exp(4)*ln(5)-10)/exp(4)/ln(5),x,method=_RETURNVERBOSE)

[Out]

2*exp(-4)*x*exp(4)-10/3*exp(-4)/ln(5)*x

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maxima [A]  time = 0.45, size = 17, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (3 \, e^{4} \log \relax (5) - 5\right )} x e^{\left (-4\right )}}{3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(6*exp(4)*log(5)-10)/exp(4)/log(5),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.00, size = 16, normalized size = 0.94 \begin {gather*} \frac {x\,{\mathrm {e}}^{-4}\,\left (2\,{\mathrm {e}}^4\,\ln \relax (5)-\frac {10}{3}\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(2*exp(4)*log(5) - 10/3))/log(5),x)

[Out]

(x*exp(-4)*(2*exp(4)*log(5) - 10/3))/log(5)

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sympy [A]  time = 0.05, size = 19, normalized size = 1.12 \begin {gather*} \frac {x \left (- \frac {10}{3} + 2 e^{4} \log {\relax (5 )}\right )}{e^{4} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(6*exp(4)*ln(5)-10)/exp(4)/ln(5),x)

[Out]

x*(-10/3 + 2*exp(4)*log(5))*exp(-4)/log(5)

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