3.70.59 \(\int \frac {(-2+16 e^{\frac {1}{5} (5-2 x)} x-2 x \log (x)) \log (\frac {1}{4} (-1+20 e^{\frac {1}{5} (5-2 x)}-x+(1+x) \log (x)))}{-x+20 e^{\frac {1}{5} (5-2 x)} x-x^2+(x+x^2) \log (x)} \, dx\)

Optimal. Leaf size=30 \[ 5-\log ^2\left (5 e^{1-\frac {2 x}{5}}+\frac {1}{4} (1+x) (-1+\log (x))\right ) \]

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Rubi [F]  time = 141.03, 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 {\left (-2+16 e^{\frac {1}{5} (5-2 x)} x-2 x \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{\frac {1}{5} (5-2 x)}-x+(1+x) \log (x)\right )\right )}{-x+20 e^{\frac {1}{5} (5-2 x)} x-x^2+\left (x+x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-2 + 16*E^((5 - 2*x)/5)*x - 2*x*Log[x])*Log[(-1 + 20*E^((5 - 2*x)/5) - x + (1 + x)*Log[x])/4])/(-x + 20*
E^((5 - 2*x)/5)*x - x^2 + (x + x^2)*Log[x]),x]

[Out]

-2*Defer[Int][Log[(-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])/4]/(x*(-1 + Log[x])), x] + 2*Defer[Int][Log[(
-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])/4]/((1 + x)*(-1 + Log[x])), x] - 2*Defer[Int][(Log[x]*Log[(-1 +
20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])/4])/((1 + x)*(-1 + Log[x])), x] - 40*E*Defer[Int][Log[(-1 + 20*E^(1 -
(2*x)/5) - x + (1 + x)*Log[x])/4]/((1 + x)*(-1 + Log[x])*(20*E - E^((2*x)/5) - E^((2*x)/5)*x + E^((2*x)/5)*Lo
g[x] + E^((2*x)/5)*x*Log[x])), x] + 40*E*Defer[Int][(Log[x]*Log[(-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])
/4])/((1 + x)*(-1 + Log[x])*(20*E - E^((2*x)/5) - E^((2*x)/5)*x + E^((2*x)/5)*Log[x] + E^((2*x)/5)*x*Log[x])),
x] - 16*E*Defer[Int][Log[(-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])/4]/((-1 + Log[x])*(20*E - E^((2*x)/5)
*(1 + x) + E^((2*x)/5)*(1 + x)*Log[x])), x] + 40*E*Defer[Int][Log[(-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x
])/4]/(x*(-1 + Log[x])*(20*E - E^((2*x)/5)*(1 + x) + E^((2*x)/5)*(1 + x)*Log[x])), x] + 16*E*Defer[Int][(Log[x
]*Log[(-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])/4])/((-1 + Log[x])*(20*E - E^((2*x)/5)*(1 + x) + E^((2*x)
/5)*(1 + x)*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 (1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x))}+\frac {8 e \left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x))} \, dx\right )+(8 e) \int \frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )} \, dx\\ &=-\left (2 \int \left (\frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x))}-\frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x))}\right ) \, dx\right )+(8 e) \int \left (\frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )}-\frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x))} \, dx\right )+2 \int \frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x))} \, dx+(8 e) \int \frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )} \, dx-(8 e) \int \frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 31, normalized size = 1.03 \begin {gather*} -\log ^2\left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2 + 16*E^((5 - 2*x)/5)*x - 2*x*Log[x])*Log[(-1 + 20*E^((5 - 2*x)/5) - x + (1 + x)*Log[x])/4])/(-x
+ 20*E^((5 - 2*x)/5)*x - x^2 + (x + x^2)*Log[x]),x]

[Out]

-Log[(-1 + 20*E^(1 - (2*x)/5) - x + (1 + x)*Log[x])/4]^2

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fricas [A]  time = 0.59, size = 25, normalized size = 0.83 \begin {gather*} -\log \left (\frac {1}{4} \, {\left (x + 1\right )} \log \relax (x) - \frac {1}{4} \, x + 5 \, e^{\left (-\frac {2}{5} \, x + 1\right )} - \frac {1}{4}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)+16*x*exp(-2/5*x+1)-2)*log(1/4*log(x)*(x+1)+5*exp(-2/5*x+1)-1/4*x-1/4)/((x^2+x)*log(x)+2
0*x*exp(-2/5*x+1)-x^2-x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (8 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - x \log \relax (x) - 1\right )} \log \left (\frac {1}{4} \, {\left (x + 1\right )} \log \relax (x) - \frac {1}{4} \, x + 5 \, e^{\left (-\frac {2}{5} \, x + 1\right )} - \frac {1}{4}\right )}{x^{2} - 20 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - {\left (x^{2} + x\right )} \log \relax (x) + x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)+16*x*exp(-2/5*x+1)-2)*log(1/4*log(x)*(x+1)+5*exp(-2/5*x+1)-1/4*x-1/4)/((x^2+x)*log(x)+2
0*x*exp(-2/5*x+1)-x^2-x),x, algorithm="giac")

[Out]

integrate(-2*(8*x*e^(-2/5*x + 1) - x*log(x) - 1)*log(1/4*(x + 1)*log(x) - 1/4*x + 5*e^(-2/5*x + 1) - 1/4)/(x^2
- 20*x*e^(-2/5*x + 1) - (x^2 + x)*log(x) + x), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 x \ln \relax (x )+16 x \,{\mathrm e}^{-\frac {2 x}{5}+1}-2\right ) \ln \left (\frac {\ln \relax (x ) \left (x +1\right )}{4}+5 \,{\mathrm e}^{-\frac {2 x}{5}+1}-\frac {x}{4}-\frac {1}{4}\right )}{\left (x^{2}+x \right ) \ln \relax (x )+20 x \,{\mathrm e}^{-\frac {2 x}{5}+1}-x^{2}-x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x)+16*x*exp(-2/5*x+1)-2)*ln(1/4*ln(x)*(x+1)+5*exp(-2/5*x+1)-1/4*x-1/4)/((x^2+x)*ln(x)+20*x*exp(-2
/5*x+1)-x^2-x),x)

[Out]

int((-2*x*ln(x)+16*x*exp(-2/5*x+1)-2)*ln(1/4*ln(x)*(x+1)+5*exp(-2/5*x+1)-1/4*x-1/4)/((x^2+x)*ln(x)+20*x*exp(-2
/5*x+1)-x^2-x),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left (8 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - x \log \relax (x) - 1\right )} \log \left (\frac {1}{4} \, {\left (x + 1\right )} \log \relax (x) - \frac {1}{4} \, x + 5 \, e^{\left (-\frac {2}{5} \, x + 1\right )} - \frac {1}{4}\right )}{x^{2} - 20 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - {\left (x^{2} + x\right )} \log \relax (x) + x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)+16*x*exp(-2/5*x+1)-2)*log(1/4*log(x)*(x+1)+5*exp(-2/5*x+1)-1/4*x-1/4)/((x^2+x)*log(x)+2
0*x*exp(-2/5*x+1)-x^2-x),x, algorithm="maxima")

[Out]

-2*integrate((8*x*e^(-2/5*x + 1) - x*log(x) - 1)*log(1/4*(x + 1)*log(x) - 1/4*x + 5*e^(-2/5*x + 1) - 1/4)/(x^2
- 20*x*e^(-2/5*x + 1) - (x^2 + x)*log(x) + x), x)

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mupad [B]  time = 4.39, size = 26, normalized size = 0.87 \begin {gather*} -{\ln \left (5\,{\mathrm {e}}^{-\frac {2\,x}{5}}\,\mathrm {e}-\frac {x}{4}+\ln \left (x^{1/4}\right )\,\left (x+1\right )-\frac {1}{4}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5*exp(1 - (2*x)/5) - x/4 + (log(x)*(x + 1))/4 - 1/4)*(2*x*log(x) - 16*x*exp(1 - (2*x)/5) + 2))/(x - 2
0*x*exp(1 - (2*x)/5) + x^2 - log(x)*(x + x^2)),x)

[Out]

-log(5*exp(-(2*x)/5)*exp(1) - x/4 + log(x^(1/4))*(x + 1) - 1/4)^2

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sympy [A]  time = 1.26, size = 29, normalized size = 0.97 \begin {gather*} - \log {\left (- \frac {x}{4} + \frac {\left (x + 1\right ) \log {\relax (x )}}{4} + 5 e^{1 - \frac {2 x}{5}} - \frac {1}{4} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x)+16*x*exp(-2/5*x+1)-2)*ln(1/4*ln(x)*(x+1)+5*exp(-2/5*x+1)-1/4*x-1/4)/((x**2+x)*ln(x)+20*x
*exp(-2/5*x+1)-x**2-x),x)

[Out]

-log(-x/4 + (x + 1)*log(x)/4 + 5*exp(1 - 2*x/5) - 1/4)**2

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