3.91.84 \(\int \frac {e^x (5-9 x+5 x^2)+(30 x^2-30 x^3) \log (5)+(e^x (-1+2 x-x^2)+(-6 x^2+6 x^3) \log (5)) \log (1-x)+(e^x (-5+10 x-5 x^2)+e^x (1-2 x+x^2) \log (1-x)) \log (5-\log (1-x))}{30 x^2-30 x^3+(-6 x^2+6 x^3) \log (1-x)} \, dx\)

Optimal. Leaf size=28 \[ x \log (5)+\frac {e^x (-1+\log (5-\log (1-x)))}{6 x} \]

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Rubi [F]  time = 17.55, 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 (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(5 - 9*x + 5*x^2) + (30*x^2 - 30*x^3)*Log[5] + (E^x*(-1 + 2*x - x^2) + (-6*x^2 + 6*x^3)*Log[5])*Log[1
- x] + (E^x*(-5 + 10*x - 5*x^2) + E^x*(1 - 2*x + x^2)*Log[1 - x])*Log[5 - Log[1 - x]])/(30*x^2 - 30*x^3 + (-6
*x^2 + 6*x^3)*Log[1 - x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{6 (1-x) x^2 (5-\log (1-x))} \, dx\\ &=\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{(1-x) x^2 (5-\log (1-x))} \, dx\\ &=\frac {1}{6} \int \left (6 \log (5)+\frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)-5 \log (5-\log (1-x))+10 x \log (5-\log (1-x))-5 x^2 \log (5-\log (1-x))+\log (1-x) \log (5-\log (1-x))-2 x \log (1-x) \log (5-\log (1-x))+x^2 \log (1-x) \log (5-\log (1-x))\right )}{(-1+x) x^2 (-5+\log (1-x))}\right ) \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)-5 \log (5-\log (1-x))+10 x \log (5-\log (1-x))-5 x^2 \log (5-\log (1-x))+\log (1-x) \log (5-\log (1-x))-2 x \log (1-x) \log (5-\log (1-x))+x^2 \log (1-x) \log (5-\log (1-x))\right )}{(-1+x) x^2 (-5+\log (1-x))} \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2+(-1+x)^2 \log (1-x) (-1+\log (5-\log (1-x)))-5 (-1+x)^2 \log (5-\log (1-x))\right )}{(1-x) x^2 (5-\log (1-x))} \, dx\\ &=x \log (5)+\frac {1}{6} \int \left (\frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)\right )}{(-1+x) x^2 (-5+\log (1-x))}+\frac {e^x (-1+x) \log (5-\log (1-x))}{x^2}\right ) \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)\right )}{(-1+x) x^2 (-5+\log (1-x))} \, dx+\frac {1}{6} \int \frac {e^x (-1+x) \log (5-\log (1-x))}{x^2} \, dx\\ &=x \log (5)+\frac {1}{6} \int \left (\frac {e^x (1-x)}{x^2}+\frac {e^x}{(-1+x) x (-5+\log (1-x))}\right ) \, dx+\frac {1}{6} \int \left (-\frac {e^x \log (5-\log (1-x))}{x^2}+\frac {e^x \log (5-\log (1-x))}{x}\right ) \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x (1-x)}{x^2} \, dx+\frac {1}{6} \int \frac {e^x}{(-1+x) x (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx\\ &=-\frac {e^x}{6 x}+x \log (5)+\frac {1}{6} \int \left (\frac {e^x}{(-1+x) (-5+\log (1-x))}-\frac {e^x}{x (-5+\log (1-x))}\right ) \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx\\ &=-\frac {e^x}{6 x}+x \log (5)+\frac {1}{6} \int \frac {e^x}{(-1+x) (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x}{x (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 36, normalized size = 1.29 \begin {gather*} \frac {1}{6} \left (-\frac {e^x}{x}+6 x \log (5)+\frac {e^x \log (5-\log (1-x))}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(5 - 9*x + 5*x^2) + (30*x^2 - 30*x^3)*Log[5] + (E^x*(-1 + 2*x - x^2) + (-6*x^2 + 6*x^3)*Log[5])
*Log[1 - x] + (E^x*(-5 + 10*x - 5*x^2) + E^x*(1 - 2*x + x^2)*Log[1 - x])*Log[5 - Log[1 - x]])/(30*x^2 - 30*x^3
+ (-6*x^2 + 6*x^3)*Log[1 - x]),x]

[Out]

(-(E^x/x) + 6*x*Log[5] + (E^x*Log[5 - Log[1 - x]])/x)/6

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fricas [A]  time = 0.78, size = 31, normalized size = 1.11 \begin {gather*} \frac {6 \, x^{2} \log \relax (5) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x+1)*exp(x)*log(-x+1)+(-5*x^2+10*x-5)*exp(x))*log(-log(-x+1)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3
-6*x^2)*log(5))*log(-x+1)+(5*x^2-9*x+5)*exp(x)+(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(-x+1)-30*x^3+30*x^2
),x, algorithm="fricas")

[Out]

1/6*(6*x^2*log(5) + e^x*log(-log(-x + 1) + 5) - e^x)/x

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giac [A]  time = 0.23, size = 31, normalized size = 1.11 \begin {gather*} \frac {6 \, x^{2} \log \relax (5) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x+1)*exp(x)*log(-x+1)+(-5*x^2+10*x-5)*exp(x))*log(-log(-x+1)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3
-6*x^2)*log(5))*log(-x+1)+(5*x^2-9*x+5)*exp(x)+(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(-x+1)-30*x^3+30*x^2
),x, algorithm="giac")

[Out]

1/6*(6*x^2*log(5) + e^x*log(-log(-x + 1) + 5) - e^x)/x

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maple [A]  time = 0.08, size = 37, normalized size = 1.32

method	result	size
risch	\(\frac {{\mathrm e}^{x} \ln \left (-\ln \left (1-x \right )+5\right )}{6 x}+\frac {6 x^{2} \ln \relax (5)-{\mathrm e}^{x}}{6 x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-2*x+1)*exp(x)*ln(1-x)+(-5*x^2+10*x-5)*exp(x))*ln(-ln(1-x)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3-6*x^2)*ln(
5))*ln(1-x)+(5*x^2-9*x+5)*exp(x)+(-30*x^3+30*x^2)*ln(5))/((6*x^3-6*x^2)*ln(1-x)-30*x^3+30*x^2),x,method=_RETUR
NVERBOSE)

[Out]

1/6/x*exp(x)*ln(-ln(1-x)+5)+1/6*(6*x^2*ln(5)-exp(x))/x

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maxima [B]  time = 0.48, size = 106, normalized size = 3.79 \begin {gather*} -\log \relax (5) \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right ) - 5\right ) + {\left ({\left (\log \left (-x + 1\right ) - 5\right )} \log \left (\log \left (-x + 1\right ) - 5\right ) - \log \left (-x + 1\right ) + 5\right )} \log \relax (5) + 5 \, \log \relax (5) \log \left (\log \left (-x + 1\right ) - 5\right ) + \frac {6 \, x^{2} \log \relax (5) + 6 \, x \log \relax (5) \log \left (-x + 1\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x+1)*exp(x)*log(-x+1)+(-5*x^2+10*x-5)*exp(x))*log(-log(-x+1)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3
-6*x^2)*log(5))*log(-x+1)+(5*x^2-9*x+5)*exp(x)+(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(-x+1)-30*x^3+30*x^2
),x, algorithm="maxima")

[Out]

-log(5)*log(-x + 1)*log(log(-x + 1) - 5) + ((log(-x + 1) - 5)*log(log(-x + 1) - 5) - log(-x + 1) + 5)*log(5) +
5*log(5)*log(log(-x + 1) - 5) + 1/6*(6*x^2*log(5) + 6*x*log(5)*log(-x + 1) + e^x*log(-log(-x + 1) + 5) - e^x)
/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (5-\ln \left (1-x\right )\right )\,\left ({\mathrm {e}}^x\,\left (5\,x^2-10\,x+5\right )-{\mathrm {e}}^x\,\ln \left (1-x\right )\,\left (x^2-2\,x+1\right )\right )+\ln \left (1-x\right )\,\left ({\mathrm {e}}^x\,\left (x^2-2\,x+1\right )+\ln \relax (5)\,\left (6\,x^2-6\,x^3\right )\right )-\ln \relax (5)\,\left (30\,x^2-30\,x^3\right )-{\mathrm {e}}^x\,\left (5\,x^2-9\,x+5\right )}{\ln \left (1-x\right )\,\left (6\,x^2-6\,x^3\right )-30\,x^2+30\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5 - log(1 - x))*(exp(x)*(5*x^2 - 10*x + 5) - exp(x)*log(1 - x)*(x^2 - 2*x + 1)) + log(1 - x)*(exp(x)*
(x^2 - 2*x + 1) + log(5)*(6*x^2 - 6*x^3)) - log(5)*(30*x^2 - 30*x^3) - exp(x)*(5*x^2 - 9*x + 5))/(log(1 - x)*(
6*x^2 - 6*x^3) - 30*x^2 + 30*x^3),x)

[Out]

int((log(5 - log(1 - x))*(exp(x)*(5*x^2 - 10*x + 5) - exp(x)*log(1 - x)*(x^2 - 2*x + 1)) + log(1 - x)*(exp(x)*
(x^2 - 2*x + 1) + log(5)*(6*x^2 - 6*x^3)) - log(5)*(30*x^2 - 30*x^3) - exp(x)*(5*x^2 - 9*x + 5))/(log(1 - x)*(
6*x^2 - 6*x^3) - 30*x^2 + 30*x^3), x)

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sympy [A]  time = 0.54, size = 20, normalized size = 0.71 \begin {gather*} x \log {\relax (5 )} + \frac {\left (\log {\left (5 - \log {\left (1 - x \right )} \right )} - 1\right ) e^{x}}{6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-2*x+1)*exp(x)*ln(-x+1)+(-5*x**2+10*x-5)*exp(x))*ln(-ln(-x+1)+5)+((-x**2+2*x-1)*exp(x)+(6*x**
3-6*x**2)*ln(5))*ln(-x+1)+(5*x**2-9*x+5)*exp(x)+(-30*x**3+30*x**2)*ln(5))/((6*x**3-6*x**2)*ln(-x+1)-30*x**3+30
*x**2),x)

[Out]

x*log(5) + (log(5 - log(1 - x)) - 1)*exp(x)/(6*x)

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