3.33.20 $\int \frac{-5e^x{x}^2\log (5)+(-24-5x^2)\log (5)+20\log (5)\log (x)}{x^2+5e^x{x}^2\log (5)+(-4x+15x^2+5x^3)\log (5)+20x\log (5)\log (x)}dx$

Optimal. Leaf size=36 $$\log \left(\frac{3}{4(3+e^x+x-\frac{\frac{1}{5}(4-\frac{x}{\log (5)})-4\log (x)}{x})}\right)$$

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Rubi [F]  time = 3.06, 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 = {} $$\begin{array}{c}\int \frac{-5e^x{x}^2\log (5)+(-24-5x^2)\log (5)+20\log (5)\log (x)}{x^2+5e^x{x}^2\log (5)+(-4x+15x^2+5x^3)\log (5)+20x\log (5)\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-5*E^x*x^2*Log[5] + (-24 - 5*x^2)*Log[5] + 20*Log[5]*Log[x])/(x^2 + 5*E^x*x^2*Log[5] + (-4*x + 15*x^2 + 5
*x^3)*Log[5] + 20*x*Log[5]*Log[x]),x]

[Out]

-x + 4*Log[5]*Defer[Int][(4*Log[5] - 5*E^x*x*Log[5] - 5*x^2*Log[5] - x*(1 + 15*Log[5]) - 20*Log[5]*Log[x])^(-1
), x] + 24*Log[5]*Defer[Int][1/(x*(4*Log[5] - 5*E^x*x*Log[5] - 5*x^2*Log[5] - x*(1 + 15*Log[5]) - 20*Log[5]*Lo
g[x])), x] - (1 + 10*Log[5])*Defer[Int][x/(4*Log[5] - 5*E^x*x*Log[5] - 5*x^2*Log[5] - x*(1 + 15*Log[5]) - 20*L
og[5]*Log[x]), x] + 5*Log[5]*Defer[Int][x^2/(-4*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + x*(1 + 15*Log[5]) + 2
0*Log[5]*Log[x]), x] + 20*Log[5]*Defer[Int][Log[x]/(-4*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + x*(1 + 15*Log[
5]) + 20*Log[5]*Log[x]), x] + 20*Log[5]*Defer[Int][Log[x]/(x*(-4*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + x*(1
+ 15*Log[5]) + 20*Log[5]*Log[x])), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{\log (5)(-24-5x^2-5e^x{x}^2+20\log (x))}{x^2+5e^x{x}^2\log (5)+(-4x+15x^2+5x^3)\log (5)+20x\log (5)\log (x)}dx\\ & =\log (5)\int \frac{-24-5x^2-5e^x{x}^2+20\log (x)}{x^2+5e^x{x}^2\log (5)+(-4x+15x^2+5x^3)\log (5)+20x\log (5)\log (x)}dx\\ & =\log (5)\int (-\frac{1}{\log (5)}+\frac{24\log (5)+4x\log (5)-5x^3\log (5)-x^2(1+10\log (5))-20\log (5)\log (x)-20x\log (5)\log (x)}{x\log (5)(4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x))})dx\\ & =-x+\int \frac{24\log (5)+4x\log (5)-5x^3\log (5)-x^2(1+10\log (5))-20\log (5)\log (x)-20x\log (5)\log (x)}{x(4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x))}dx\\ & =-x+\int (\frac{x(-1-10\log (5))}{4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x)}+\frac{4\log (5)}{4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x)}+\frac{24\log (5)}{x(4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x))}+\frac{5x^2\log (5)}{-4\log (5)+5e^{x}x\log (5)+5x^2\log (5)+x(1+15\log (5))+20\log (5)\log (x)}+\frac{20\log (5)\log (x)}{-4\log (5)+5e^{x}x\log (5)+5x^2\log (5)+x(1+15\log (5))+20\log (5)\log (x)}+\frac{20\log (5)\log (x)}{x(-4\log (5)+5e^{x}x\log (5)+5x^2\log (5)+x(1+15\log (5))+20\log (5)\log (x))})dx\\ & =-x+(-1-10\log (5))\int \frac{x}{4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x)}dx+(4\log (5))\int \frac{1}{4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x)}dx+(5\log (5))\int \frac{x^2}{-4\log (5)+5e^{x}x\log (5)+5x^2\log (5)+x(1+15\log (5))+20\log (5)\log (x)}dx+(20\log (5))\int \frac{\log (x)}{-4\log (5)+5e^{x}x\log (5)+5x^2\log (5)+x(1+15\log (5))+20\log (5)\log (x)}dx+(20\log (5))\int \frac{\log (x)}{x(-4\log (5)+5e^{x}x\log (5)+5x^2\log (5)+x(1+15\log (5))+20\log (5)\log (x))}dx+(24\log (5))\int \frac{1}{x(4\log (5)-5e^{x}x\log (5)-5x^2\log (5)-x(1+15\log (5))-20\log (5)\log (x))}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.67, size = 51, normalized size = 1.42 $$\begin{array}{c}-\log (5)(-\frac{\log (x)}{\log (5)}+\frac{\log (x-4\log (5)+15x\log (5)+5e^{x}x\log (5)+5x^2\log (5)+20\log (5)\log (x))}{\log (5)})\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^x*x^2*Log[5] + (-24 - 5*x^2)*Log[5] + 20*Log[5]*Log[x])/(x^2 + 5*E^x*x^2*Log[5] + (-4*x + 15*x
^2 + 5*x^3)*Log[5] + 20*x*Log[5]*Log[x]),x]

[Out]

-(Log[5]*(-(Log[x]/Log[5]) + Log[x - 4*Log[5] + 15*x*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + 20*Log[5]*Log[x]
]/Log[5]))

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fricas [A]  time = 0.55, size = 34, normalized size = 0.94 $$\begin{array}{c}-\log (5x{e}^x\log (5)+(5x^2+15x-4)\log (5)+20\log (5)\log (x)+x)+\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5)*log(x)+5*x^2*log(5)*exp(x)+(5
*x^3+15*x^2-4*x)*log(5)+x^2),x, algorithm="fricas")

[Out]

-log(5*x*e^x*log(5) + (5*x^2 + 15*x - 4)*log(5) + 20*log(5)*log(x) + x) + log(x)

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giac [A]  time = 0.14, size = 37, normalized size = 1.03 $$\begin{array}{c}-\log (5x^2\log (5)+5x{e}^x\log (5)+15x\log (5)+20\log (5)\log (x)+x-4\log (5))+\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5)*log(x)+5*x^2*log(5)*exp(x)+(5
*x^3+15*x^2-4*x)*log(5)+x^2),x, algorithm="giac")

[Out]

-log(5*x^2*log(5) + 5*x*e^x*log(5) + 15*x*log(5) + 20*log(5)*log(x) + x - 4*log(5)) + log(x)

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maple [A]  time = 0.10, size = 38, normalized size = 1.06

method	result	size
norman	$\ln (x)-\ln (5x{\mathrm{e}}^x\ln (5)+5x^2\ln (5)+20\ln (5)\ln (x)+15x\ln (5)-4\ln (5)+x)$	$38$
risch	$\ln (x)-\ln (\ln (x)+\frac{5x^2\ln (5)+5x{\mathrm{e}}^x\ln (5)+15x\ln (5)-4\ln (5)+x}{20\ln (5)})$	$41$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*ln(5)*ln(x)-5*x^2*ln(5)*exp(x)+(-5*x^2-24)*ln(5))/(20*x*ln(5)*ln(x)+5*x^2*ln(5)*exp(x)+(5*x^3+15*x^2-4
*x)*ln(5)+x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x)-ln(5*x*exp(x)*ln(5)+5*x^2*ln(5)+20*ln(5)*ln(x)+15*x*ln(5)-4*ln(5)+x)

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maxima [A]  time = 0.68, size = 45, normalized size = 1.25 $$\begin{array}{c}-\log \left(\frac{5x^2\log (5)+5x{e}^x\log (5)+x(15\log (5)+1)+20\log (5)\log (x)-4\log (5)}{5x\log (5)}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5)*log(x)+5*x^2*log(5)*exp(x)+(5
*x^3+15*x^2-4*x)*log(5)+x^2),x, algorithm="maxima")

[Out]

-log(1/5*(5*x^2*log(5) + 5*x*e^x*log(5) + x*(15*log(5) + 1) + 20*log(5)*log(x) - 4*log(5))/(x*log(5)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}\int -\frac{\ln (5)(5x^2+24)-20\ln (5)\ln (x)+5x^2{\mathrm{e}}^x\ln (5)}{\ln (5)(5x^3+15x^2-4x)+x^2+20x\ln (5)\ln (x)+5x^2{\mathrm{e}}^x\ln (5)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(5*x^2 + 24) - 20*log(5)*log(x) + 5*x^2*exp(x)*log(5))/(log(5)*(15*x^2 - 4*x + 5*x^3) + x^2 + 20*
x*log(5)*log(x) + 5*x^2*exp(x)*log(5)),x)

[Out]

int(-(log(5)*(5*x^2 + 24) - 20*log(5)*log(x) + 5*x^2*exp(x)*log(5))/(log(5)*(15*x^2 - 4*x + 5*x^3) + x^2 + 20*
x*log(5)*log(x) + 5*x^2*exp(x)*log(5)), x)

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sympy [A]  time = 0.50, size = 42, normalized size = 1.17 $$\begin{array}{c}-\log (e^x+\frac{5x^2\log (5)+x+15x\log (5)+20\log (5)\log (x)-4\log (5)}{5x\log (5)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*ln(5)*ln(x)-5*x**2*ln(5)*exp(x)+(-5*x**2-24)*ln(5))/(20*x*ln(5)*ln(x)+5*x**2*ln(5)*exp(x)+(5*x**
3+15*x**2-4*x)*ln(5)+x**2),x)

[Out]

-log(exp(x) + (5*x**2*log(5) + x + 15*x*log(5) + 20*log(5)*log(x) - 4*log(5))/(5*x*log(5)))

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