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

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

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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 {gather*} \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {\log (5) \left (-24-5 x^2-5 e^x x^2+20 \log (x)\right )}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx\\ &=\log (5) \int \frac {-24-5 x^2-5 e^x x^2+20 \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx\\ &=\log (5) \int \left (-\frac {1}{\log (5)}+\frac {24 \log (5)+4 x \log (5)-5 x^3 \log (5)-x^2 (1+10 \log (5))-20 \log (5) \log (x)-20 x \log (5) \log (x)}{x \log (5) \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )}\right ) \, dx\\ &=-x+\int \frac {24 \log (5)+4 x \log (5)-5 x^3 \log (5)-x^2 (1+10 \log (5))-20 \log (5) \log (x)-20 x \log (5) \log (x)}{x \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )} \, dx\\ &=-x+\int \left (\frac {x (-1-10 \log (5))}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)}+\frac {4 \log (5)}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)}+\frac {24 \log (5)}{x \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )}+\frac {5 x^2 \log (5)}{-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)}+\frac {20 \log (5) \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)}+\frac {20 \log (5) \log (x)}{x \left (-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)\right )}\right ) \, dx\\ &=-x+(-1-10 \log (5)) \int \frac {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)} \, dx+(4 \log (5)) \int \frac {1}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)} \, dx+(5 \log (5)) \int \frac {x^2}{-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)} \, dx+(20 \log (5)) \int \frac {\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)} \, dx+(20 \log (5)) \int \frac {\log (x)}{x \left (-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)\right )} \, dx+(24 \log (5)) \int \frac {1}{x \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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

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 {gather*} -\log \left (5 \, x e^{x} \log \relax (5) + {\left (5 \, x^{2} + 15 \, x - 4\right )} \log \relax (5) + 20 \, \log \relax (5) \log \relax (x) + x\right ) + \log \relax (x) \end {gather*}

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 {gather*} -\log \left (5 \, x^{2} \log \relax (5) + 5 \, x e^{x} \log \relax (5) + 15 \, x \log \relax (5) + 20 \, \log \relax (5) \log \relax (x) + x - 4 \, \log \relax (5)\right ) + \log \relax (x) \end {gather*}

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	\(\ln \relax (x )-\ln \left (5 x \,{\mathrm e}^{x} \ln \relax (5)+5 x^{2} \ln \relax (5)+20 \ln \relax (5) \ln \relax (x )+15 x \ln \relax (5)-4 \ln \relax (5)+x \right )\)	\(38\)
risch	\(\ln \relax (x )-\ln \left (\ln \relax (x )+\frac {5 x^{2} \ln \relax (5)+5 x \,{\mathrm e}^{x} \ln \relax (5)+15 x \ln \relax (5)-4 \ln \relax (5)+x}{20 \ln \relax (5)}\right )\)	\(41\)

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 {gather*} -\log \left (\frac {5 \, x^{2} \log \relax (5) + 5 \, x e^{x} \log \relax (5) + x {\left (15 \, \log \relax (5) + 1\right )} + 20 \, \log \relax (5) \log \relax (x) - 4 \, \log \relax (5)}{5 \, x \log \relax (5)}\right ) \end {gather*}

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 {gather*} \int -\frac {\ln \relax (5)\,\left (5\,x^2+24\right )-20\,\ln \relax (5)\,\ln \relax (x)+5\,x^2\,{\mathrm {e}}^x\,\ln \relax (5)}{\ln \relax (5)\,\left (5\,x^3+15\,x^2-4\,x\right )+x^2+20\,x\,\ln \relax (5)\,\ln \relax (x)+5\,x^2\,{\mathrm {e}}^x\,\ln \relax (5)} \,d x \end {gather*}

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 {gather*} - \log {\left (e^{x} + \frac {5 x^{2} \log {\relax (5 )} + x + 15 x \log {\relax (5 )} + 20 \log {\relax (5 )} \log {\relax (x )} - 4 \log {\relax (5 )}}{5 x \log {\relax (5 )}} \right )} \end {gather*}

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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