3.24.91 \(\int \frac {e^x (6+24 x-6 x^2-6 x \log (3))+6 e^x x \log (x)}{-5 x+e^x (10 x-2 x^2-2 x \log (3))+2 e^x x \log (x)} \, dx\)

Optimal. Leaf size=23 \[ 1+3 \log \left (-5+2 e^x (5-x-\log (3)+\log (x))\right ) \]

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Rubi [F]  time = 2.39, 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 {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(6 + 24*x - 6*x^2 - 6*x*Log[3]) + 6*E^x*x*Log[x])/(-5*x + E^x*(10*x - 2*x^2 - 2*x*Log[3]) + 2*E^x*x*L
og[x]),x]

[Out]

-6*(4 - Log[3])*Defer[Int][E^x/(5 + E^x*(-10 + 2*x + Log[9]) - 2*E^x*Log[x]), x] + 6*Defer[Int][(E^x*x)/(5 + E
^x*(-10 + 2*x + Log[9]) - 2*E^x*Log[x]), x] + 6*Defer[Int][E^x/(x*(-5 - 2*E^x*x + 10*E^x*(1 - Log[3]/5) + 2*E^
x*Log[x])), x] + 6*Defer[Int][(E^x*Log[x])/(-5 - 2*E^x*(-5 + x + Log[3]) + 2*E^x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 e^x \left (-1+x^2-4 x \left (1-\frac {\log (3)}{4}\right )-x \log (x)\right )}{5 x-e^x \left (10 x-2 x^2-2 x \log (3)\right )-2 e^x x \log (x)} \, dx\\ &=6 \int \frac {e^x \left (-1+x^2-4 x \left (1-\frac {\log (3)}{4}\right )-x \log (x)\right )}{5 x-e^x \left (10 x-2 x^2-2 x \log (3)\right )-2 e^x x \log (x)} \, dx\\ &=6 \int \left (\frac {e^x x}{5+2 e^x x-10 e^x \left (1-\frac {\log (3)}{5}\right )-2 e^x \log (x)}+\frac {e^x (-4+\log (3))}{5+2 e^x x-10 e^x \left (1-\frac {\log (3)}{5}\right )-2 e^x \log (x)}+\frac {e^x}{x \left (-5-2 e^x x+10 e^x \left (1-\frac {\log (3)}{5}\right )+2 e^x \log (x)\right )}+\frac {e^x \log (x)}{-5-2 e^x x+10 e^x \left (1-\frac {\log (3)}{5}\right )+2 e^x \log (x)}\right ) \, dx\\ &=6 \int \frac {e^x x}{5+2 e^x x-10 e^x \left (1-\frac {\log (3)}{5}\right )-2 e^x \log (x)} \, dx+6 \int \frac {e^x}{x \left (-5-2 e^x x+10 e^x \left (1-\frac {\log (3)}{5}\right )+2 e^x \log (x)\right )} \, dx+6 \int \frac {e^x \log (x)}{-5-2 e^x x+10 e^x \left (1-\frac {\log (3)}{5}\right )+2 e^x \log (x)} \, dx-(6 (4-\log (3))) \int \frac {e^x}{5+2 e^x x-10 e^x \left (1-\frac {\log (3)}{5}\right )-2 e^x \log (x)} \, dx\\ &=6 \int \frac {e^x x}{5+e^x (-10+2 x+\log (9))-2 e^x \log (x)} \, dx+6 \int \frac {e^x}{x \left (-5-2 e^x x+10 e^x \left (1-\frac {\log (3)}{5}\right )+2 e^x \log (x)\right )} \, dx+6 \int \frac {e^x \log (x)}{-5-2 e^x (-5+x+\log (3))+2 e^x \log (x)} \, dx-(6 (4-\log (3))) \int \frac {e^x}{5+e^x (-10+2 x+\log (9))-2 e^x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^x*(6 + 24*x - 6*x^2 - 6*x*Log[3]) + 6*E^x*x*Log[x])/(-5*x + E^x*(10*x - 2*x^2 - 2*x*Log[3]) + 2*E
^x*x*Log[x]),x]

[Out]

Integrate[(E^x*(6 + 24*x - 6*x^2 - 6*x*Log[3]) + 6*E^x*x*Log[x])/(-5*x + E^x*(10*x - 2*x^2 - 2*x*Log[3]) + 2*E
^x*x*Log[x]), x]

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fricas [A]  time = 1.07, size = 30, normalized size = 1.30 \begin {gather*} 3 \, x + 3 \, \log \left (-{\left (2 \, {\left (x + \log \relax (3) - 5\right )} e^{x} - 2 \, e^{x} \log \relax (x) + 5\right )} e^{\left (-x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x)*log(x)+(-2*x*log(3)-2*x^2+10*x)*ex
p(x)-5*x),x, algorithm="fricas")

[Out]

3*x + 3*log(-(2*(x + log(3) - 5)*e^x - 2*e^x*log(x) + 5)*e^(-x))

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giac [A]  time = 0.25, size = 26, normalized size = 1.13 \begin {gather*} 3 \, \log \left (2 \, x e^{x} + 2 \, e^{x} \log \relax (3) - 2 \, e^{x} \log \relax (x) - 10 \, e^{x} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x)*log(x)+(-2*x*log(3)-2*x^2+10*x)*ex
p(x)-5*x),x, algorithm="giac")

[Out]

3*log(2*x*e^x + 2*e^x*log(3) - 2*e^x*log(x) - 10*e^x + 5)

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maple [A]  time = 0.10, size = 27, normalized size = 1.17

method	result	size
norman	\(3 \ln \left (2 \ln \relax (3) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x} \ln \relax (x )-10 \,{\mathrm e}^{x}+5\right )\)	\(27\)
risch	\(3 x +3 \ln \left (\ln \relax (x )-\frac {\left (2 \ln \relax (3) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} x -10 \,{\mathrm e}^{x}+5\right ) {\mathrm e}^{-x}}{2}\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x*exp(x)*ln(x)+(-6*x*ln(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x)*ln(x)+(-2*x*ln(3)-2*x^2+10*x)*exp(x)-5*x),
x,method=_RETURNVERBOSE)

[Out]

3*ln(2*ln(3)*exp(x)+2*exp(x)*x-2*exp(x)*ln(x)-10*exp(x)+5)

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maxima [B]  time = 0.68, size = 46, normalized size = 2.00 \begin {gather*} 3 \, \log \left (-x - \log \relax (3) + \log \relax (x) + 5\right ) + 3 \, \log \left (\frac {2 \, {\left (x + \log \relax (3) - \log \relax (x) - 5\right )} e^{x} + 5}{2 \, {\left (x + \log \relax (3) - \log \relax (x) - 5\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x)*log(x)+(-2*x*log(3)-2*x^2+10*x)*ex
p(x)-5*x),x, algorithm="maxima")

[Out]

3*log(-x - log(3) + log(x) + 5) + 3*log(1/2*(2*(x + log(3) - log(x) - 5)*e^x + 5)/(x + log(3) - log(x) - 5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^x\,\left (24\,x-6\,x\,\ln \relax (3)-6\,x^2+6\right )+6\,x\,{\mathrm {e}}^x\,\ln \relax (x)}{5\,x+{\mathrm {e}}^x\,\left (2\,x\,\ln \relax (3)-10\,x+2\,x^2\right )-2\,x\,{\mathrm {e}}^x\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(24*x - 6*x*log(3) - 6*x^2 + 6) + 6*x*exp(x)*log(x))/(5*x + exp(x)*(2*x*log(3) - 10*x + 2*x^2) -
2*x*exp(x)*log(x)),x)

[Out]

int(-(exp(x)*(24*x - 6*x*log(3) - 6*x^2 + 6) + 6*x*exp(x)*log(x))/(5*x + exp(x)*(2*x*log(3) - 10*x + 2*x^2) -
2*x*exp(x)*log(x)), x)

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sympy [A]  time = 1.82, size = 36, normalized size = 1.57 \begin {gather*} 3 \log {\left (e^{x} + \frac {5}{2 x - 2 \log {\relax (x )} - 10 + 2 \log {\relax (3 )}} \right )} + 3 \log {\left (- x + \log {\relax (x )} - \log {\relax (3 )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)*ln(x)+(-6*x*ln(3)-6*x**2+24*x+6)*exp(x))/(2*x*exp(x)*ln(x)+(-2*x*ln(3)-2*x**2+10*x)*exp(
x)-5*x),x)

[Out]

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

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