3.11.56 \(\int \frac {-1+6 e^x+e^x \log (\frac {5 e^x}{-e^5+\log (5)})}{5 e^x-x+e^x \log (\frac {5 e^x}{-e^5+\log (5)})} \, dx\)

Optimal. Leaf size=27 \[ \log \left (-x+e^x \left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )\right ) \]

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Rubi [F]  time = 2.46, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+6 e^x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + 6*E^x + E^x*Log[(5*E^x)/(-E^5 + Log[5])])/(5*E^x - x + E^x*Log[(5*E^x)/(-E^5 + Log[5])]),x]

[Out]

x + Log[5 + Log[(-5*E^x)/(E^5 - Log[5])]] - 6*Defer[Int][x/((5 + Log[(5*E^x)/(-E^5 + Log[5])])*(-5*E^x + x - E
^x*Log[(5*E^x)/(-E^5 + Log[5])])), x] - Defer[Int][(x*Log[(5*E^x)/(-E^5 + Log[5])])/((5 + Log[(5*E^x)/(-E^5 +
Log[5])])*(-5*E^x + x - E^x*Log[(5*E^x)/(-E^5 + Log[5])])), x] - 5*Defer[Int][1/((5 + Log[(5*E^x)/(-E^5 + Log[
5])])*(5*E^x - x + E^x*Log[(5*E^x)/(-E^5 + Log[5])])), x] - Defer[Int][Log[(5*E^x)/(-E^5 + Log[5])]/((5 + Log[
(5*E^x)/(-E^5 + Log[5])])*(5*E^x - x + E^x*Log[(5*E^x)/(-E^5 + Log[5])])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {6+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}+\frac {-5+6 x-\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )+x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}\right ) \, dx\\ &=\int \frac {6+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )} \, dx+\int \frac {-5+6 x-\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )+x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx\\ &=\int \frac {-5+6 x+(-1+x) \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx+\operatorname {Subst}\left (\int \frac {6+\log \left (\frac {5 x}{-e^5+\log (5)}\right )}{x \left (5+\log \left (\frac {5 x}{-e^5+\log (5)}\right )\right )} \, dx,x,e^x\right )\\ &=\int \left (-\frac {6 x}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}-\frac {x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}-\frac {5}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}-\frac {\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}\right ) \, dx+\operatorname {Subst}\left (\int \frac {6+x}{5+x} \, dx,x,\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )\\ &=-\left (5 \int \frac {1}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx\right )-6 \int \frac {x}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx+\operatorname {Subst}\left (\int \left (1+\frac {1}{5+x}\right ) \, dx,x,\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )\\ &=x+\log \left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )-5 \int \frac {1}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-6 \int \frac {x}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 30, normalized size = 1.11 \begin {gather*} \log \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 6*E^x + E^x*Log[(5*E^x)/(-E^5 + Log[5])])/(5*E^x - x + E^x*Log[(5*E^x)/(-E^5 + Log[5])]),x]

[Out]

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

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fricas [B]  time = 0.81, size = 172, normalized size = 6.37 \begin {gather*} \frac {1}{2} \, \log \left (\pi ^{2} + x^{2} + 2 \, {\left (x + 5\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right ) + \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )^{2} + 10 \, x + 25\right ) + \frac {1}{2} \, \log \left (\frac {e^{\left (2 \, x\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )^{2} + x^{2} + {\left (\pi ^{2} + x^{2} + 10 \, x + 25\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{x} + 2 \, {\left ({\left (x + 5\right )} e^{\left (2 \, x\right )} - x e^{x}\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )}{\pi ^{2} + x^{2} + 2 \, {\left (x + 5\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right ) + \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )^{2} + 10 \, x + 25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(pi^2 + x^2 + 2*(x + 5)*log(5/(e^5 - log(5))) + log(5/(e^5 - log(5)))^2 + 10*x + 25) + 1/2*log((e^(2*x)
*log(5/(e^5 - log(5)))^2 + x^2 + (pi^2 + x^2 + 10*x + 25)*e^(2*x) - 2*(x^2 + 5*x)*e^x + 2*((x + 5)*e^(2*x) - x
*e^x)*log(5/(e^5 - log(5))))/(pi^2 + x^2 + 2*(x + 5)*log(5/(e^5 - log(5))) + log(5/(e^5 - log(5)))^2 + 10*x +
25))

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giac [B]  time = 0.58, size = 154, normalized size = 5.70 \begin {gather*} \frac {1}{2} \, \log \left (\pi ^{2} e^{\left (2 \, x\right )} + x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + 2 \, x e^{\left (2 \, x\right )} \log \relax (5) - 2 \, x e^{x} \log \relax (5) + e^{\left (2 \, x\right )} \log \relax (5)^{2} - 2 \, x e^{\left (2 \, x\right )} \log \left (e^{5} - \log \relax (5)\right ) + 2 \, x e^{x} \log \left (e^{5} - \log \relax (5)\right ) - 2 \, e^{\left (2 \, x\right )} \log \relax (5) \log \left (e^{5} - \log \relax (5)\right ) + e^{\left (2 \, x\right )} \log \left (e^{5} - \log \relax (5)\right )^{2} + x^{2} + 10 \, x e^{\left (2 \, x\right )} - 10 \, x e^{x} + 10 \, e^{\left (2 \, x\right )} \log \relax (5) - 10 \, e^{\left (2 \, x\right )} \log \left (e^{5} - \log \relax (5)\right ) + 25 \, e^{\left (2 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(pi^2*e^(2*x) + x^2*e^(2*x) - 2*x^2*e^x + 2*x*e^(2*x)*log(5) - 2*x*e^x*log(5) + e^(2*x)*log(5)^2 - 2*x*
e^(2*x)*log(e^5 - log(5)) + 2*x*e^x*log(e^5 - log(5)) - 2*e^(2*x)*log(5)*log(e^5 - log(5)) + e^(2*x)*log(e^5 -
log(5))^2 + x^2 + 10*x*e^(2*x) - 10*x*e^x + 10*e^(2*x)*log(5) - 10*e^(2*x)*log(e^5 - log(5)) + 25*e^(2*x))

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maple [A]  time = 0.06, size = 26, normalized size = 0.96

method	result	size
norman	\(\ln \left (-{\mathrm e}^{x} \ln \left (\frac {5 \,{\mathrm e}^{x}}{\ln \relax (5)-{\mathrm e}^{5}}\right )+x -5 \,{\mathrm e}^{x}\right )\)	\(26\)
derivativedivides	\(\ln \left ({\mathrm e}^{x} \ln \left (\frac {5 \,{\mathrm e}^{x}}{\ln \relax (5)-{\mathrm e}^{5}}\right )+5 \,{\mathrm e}^{x}-x \right )\)	\(27\)
default	\(\ln \left ({\mathrm e}^{x} \ln \left (\frac {5 \,{\mathrm e}^{x}}{\ln \relax (5)-{\mathrm e}^{5}}\right )+5 \,{\mathrm e}^{x}-x \right )\)	\(27\)
risch	\(x +\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (2 \,{\mathrm e}^{x} \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}-2 \,{\mathrm e}^{x} \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3}+2 i {\mathrm e}^{x} \ln \relax (5)-2 i {\mathrm e}^{x} \ln \left (-\ln \relax (5)+{\mathrm e}^{5}\right )-2 \,{\mathrm e}^{x} \pi -2 i x +10 i {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{2}\right )\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*ln(5*exp(x)/(ln(5)-exp(5)))+6*exp(x)-1)/(exp(x)*ln(5*exp(x)/(ln(5)-exp(5)))+5*exp(x)-x),x,method=_
RETURNVERBOSE)

[Out]

ln(-exp(x)*ln(5*exp(x)/(ln(5)-exp(5)))+x-5*exp(x))

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maxima [A]  time = 0.40, size = 26, normalized size = 0.96 \begin {gather*} \log \left (e^{x} \log \left (-\frac {5 \, e^{x}}{e^{5} - \log \relax (5)}\right ) - x + 5 \, e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^x*log(-5*e^x/(e^5 - log(5))) - x + 5*e^x)

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mupad [B]  time = 0.90, size = 33, normalized size = 1.22 \begin {gather*} \ln \left (5\,{\mathrm {e}}^x-x+{\mathrm {e}}^x\,\left (\ln \relax (5)-\ln \left ({\mathrm {e}}^5-\ln \relax (5)\right )+\pi \,1{}\mathrm {i}\right )+x\,{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*exp(x) + exp(x)*log(-(5*exp(x))/(exp(5) - log(5))) - 1)/(5*exp(x) - x + exp(x)*log(-(5*exp(x))/(exp(5)
- log(5)))),x)

[Out]

log(5*exp(x) - x + exp(x)*(pi*1i + log(5) - log(exp(5) - log(5))) + x*exp(x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*ln(5*exp(x)/(ln(5)-exp(5)))+6*exp(x)-1)/(exp(x)*ln(5*exp(x)/(ln(5)-exp(5)))+5*exp(x)-x),x)

[Out]

Timed out

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