3.33.33 $\int \frac{e^x(1+(1-x)\log (5))+e^x(1-x)\log (x)}{-e^{x}x\log (5)+x^2{\log }^2(5)+(-e^{x}x+2x^2\log (5))\log (x)+x^2{\log }^2(x)}dx$

Optimal. Leaf size=20 $$\log (-95(-2+\frac{2e^x}{x(\log (5)+\log (x))}))$$

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Rubi [F]  time = 2.95, 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 = {} $$\begin{array}{c}\int \frac{e^x(1+(1-x)\log (5))+e^x(1-x)\log (x)}{-e^{x}x\log (5)+x^2{\log }^2(5)+(-e^{x}x+2x^2\log (5))\log (x)+x^2{\log }^2(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*Log[5]
)*Log[x] + x^2*Log[x]^2),x]

[Out]

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^x(-1+x\log (5)+x\log (x)-\log (5x))}{x(e^x-x\log (5)-x\log (x))\log (5x)}dx\\ & =\int (\frac{e^x}{x(-e^x+x\log (5)+x\log (x))}+\frac{e^x}{x(-e^x+x\log (5)+x\log (x))\log (5x)}-\frac{e^x\log (5)}{(-e^x+x\log (5)+x\log (x))\log (5x)}-\frac{e^x\log (x)}{(-e^x+x\log (5)+x\log (x))\log (5x)})dx\\ & =-(\log (5)\int \frac{e^x}{(-e^x+x\log (5)+x\log (x))\log (5x)}dx)+\int \frac{e^x}{x(-e^x+x\log (5)+x\log (x))}dx+\int \frac{e^x}{x(-e^x+x\log (5)+x\log (x))\log (5x)}dx-\int \frac{e^x\log (x)}{(-e^x+x\log (5)+x\log (x))\log (5x)}dx\end{array}\end{array}$$

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Mathematica [F]  time = 0.81, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{e^x(1+(1-x)\log (5))+e^x(1-x)\log (x)}{-e^{x}x\log (5)+x^2{\log }^2(5)+(-e^{x}x+2x^2\log (5))\log (x)+x^2{\log }^2(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Integrate[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*
Log[5])*Log[x] + x^2*Log[x]^2),x]

[Out]

Integrate[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*
Log[5])*Log[x] + x^2*Log[x]^2), x]

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fricas [A]  time = 0.54, size = 27, normalized size = 1.35 $$\begin{array}{c}\log \left(\frac{x\log (5)+x\log (x)-e^x}{x}\right)-\log (\log (5)+\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*log(x)+((-x+1)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x+2*x^2*log(5))*log(x)-x*exp(
x)*log(5)+x^2*log(5)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 27, normalized size = 1.35 $$\begin{array}{c}\log (-x\log (5)-x\log (x)+e^x)-\log (x)-\log (\log (5)+\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*log(x)+((-x+1)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x+2*x^2*log(5))*log(x)-x*exp(
x)*log(5)+x^2*log(5)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.21, size = 27, normalized size = 1.35

method	result	size
risch	$\ln (\ln (x)+\frac{x\ln (5)-{\mathrm{e}}^x}{x})-\ln (\ln (5)+\ln (x))$	$27$
norman	$-\ln (x)-\ln (\ln (5)+\ln (x))+\ln (x\ln (5)+x\ln (x)-{\mathrm{e}}^x)$	$28$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-x)*exp(x)*ln(x)+((1-x)*ln(5)+1)*exp(x))/(x^2*ln(x)^2+(-exp(x)*x+2*x^2*ln(5))*ln(x)-x*exp(x)*ln(5)+x^2*
ln(5)^2),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x)+1/x*(x*ln(5)-exp(x)))-ln(ln(5)+ln(x))

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maxima [A]  time = 0.91, size = 27, normalized size = 1.35 $$\begin{array}{c}\log (-x\log (5)-x\log (x)+e^x)-\log (x)-\log (\log (5)+\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*log(x)+((-x+1)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x+2*x^2*log(5))*log(x)-x*exp(
x)*log(5)+x^2*log(5)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 $$\begin{array}{c}-\int \frac{{\mathrm{e}}^x(\ln (5)(x-1)-1)+{\mathrm{e}}^x\ln (x)(x-1)}{x^2{\ln (5)}^2+x^2{\ln (x)}^2+\ln (x)(2x^2\ln (5)-x{\mathrm{e}}^x)-x{\mathrm{e}}^x\ln (5)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(log(5)*(x - 1) - 1) + exp(x)*log(x)*(x - 1))/(x^2*log(5)^2 + x^2*log(x)^2 + log(x)*(2*x^2*log(5)
- x*exp(x)) - x*exp(x)*log(5)),x)

[Out]

-int((exp(x)*(log(5)*(x - 1) - 1) + exp(x)*log(x)*(x - 1))/(x^2*log(5)^2 + x^2*log(x)^2 + log(x)*(2*x^2*log(5)
- x*exp(x)) - x*exp(x)*log(5)), x)

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sympy [A]  time = 0.39, size = 26, normalized size = 1.30 $$\begin{array}{c}-\log (x)-\log (\log (x)+\log (5))+\log (-x\log (x)-x\log (5)+e^x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*ln(x)+((-x+1)*ln(5)+1)*exp(x))/(x**2*ln(x)**2+(-exp(x)*x+2*x**2*ln(5))*ln(x)-x*exp(x)
*ln(5)+x**2*ln(5)**2),x)

[Out]

-log(x) - log(log(x) + log(5)) + log(-x*log(x) - x*log(5) + exp(x))

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