3.98.84 \(\int \frac {e^{5 x} (-45 x+15 x^2+e^x (15 x^2-5 x^3))+(36-24 x+e^x (-12 x+8 x^2)) \log (e^{-x} (-3+e^x x))+(-36 x+12 x^2+e^x (-12 x+4 x^2)) \log (\frac {1}{3} (-3 x+x^2))}{9 x-3 x^2+e^x (-3 x^2+x^3)} \, dx\)

Optimal. Leaf size=32 \[ -e^{5 x}+4 \log \left (-3 e^{-x}+x\right ) \log \left (-x+\frac {x^2}{3}\right ) \]

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Rubi [F]  time = 1.75, 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^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(5*x)*(-45*x + 15*x^2 + E^x*(15*x^2 - 5*x^3)) + (36 - 24*x + E^x*(-12*x + 8*x^2))*Log[(-3 + E^x*x)/E^x]
+ (-36*x + 12*x^2 + E^x*(-12*x + 4*x^2))*Log[(-3*x + x^2)/3])/(9*x - 3*x^2 + E^x*(-3*x^2 + x^3)),x]

[Out]

-E^(5*x) - 4*Log[3]*Log[-3 + x] - 2*Log[x]^2 + 4*Log[x]*Log[-1/3*((3 - x)*x)] + 4*PolyLog[2, 1 - x/3] + 12*Log
[-1/3*((3 - x)*x)]*Defer[Int][(-3 + E^x*x)^(-1), x] + 12*Log[-1/3*((3 - x)*x)]*Defer[Int][1/(x*(-3 + E^x*x)),
x] + 4*Defer[Int][Log[-3/E^x + x]/(-3 + x), x] + 4*Defer[Int][Log[-3/E^x + x]/x, x] - 12*Defer[Int][Defer[Int]
[(-3 + E^x*x)^(-1), x]/(-3 + x), x] - 12*Defer[Int][Defer[Int][(-3 + E^x*x)^(-1), x]/x, x] - 12*Defer[Int][Def
er[Int][1/(x*(-3 + E^x*x)), x]/(-3 + x), x] - 12*Defer[Int][Defer[Int][1/(x*(-3 + E^x*x)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-5 e^{5 x}+\frac {4 \left (3+e^x\right ) \log \left (\frac {1}{3} (-3+x) x\right )}{-3+e^x x}+\frac {4 (-3+2 x) \log \left (-3 e^{-x}+x\right )}{(-3+x) x}\right ) \, dx\\ &=4 \int \frac {\left (3+e^x\right ) \log \left (\frac {1}{3} (-3+x) x\right )}{-3+e^x x} \, dx+4 \int \frac {(-3+2 x) \log \left (-3 e^{-x}+x\right )}{(-3+x) x} \, dx-5 \int e^{5 x} \, dx\\ &=-e^{5 x}+4 \int \left (\frac {\log \left (\frac {1}{3} (-3+x) x\right )}{x}+\frac {3 (1+x) \log \left (\frac {1}{3} (-3+x) x\right )}{x \left (-3+e^x x\right )}\right ) \, dx+4 \int \left (\frac {\log \left (-3 e^{-x}+x\right )}{-3+x}+\frac {\log \left (-3 e^{-x}+x\right )}{x}\right ) \, dx\\ &=-e^{5 x}+4 \int \frac {\log \left (\frac {1}{3} (-3+x) x\right )}{x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx+12 \int \frac {(1+x) \log \left (\frac {1}{3} (-3+x) x\right )}{x \left (-3+e^x x\right )} \, dx\\ &=-e^{5 x}+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )-4 \int \frac {\log (x)}{-3+x} \, dx-4 \int \frac {\log (x)}{x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \frac {(3-2 x) \left (\int \frac {1}{-3+e^x x} \, dx+\int \frac {1}{x \left (-3+e^x x\right )} \, dx\right )}{(3-x) x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx\\ &=-e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )-4 \int \frac {\log \left (\frac {x}{3}\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \left (\frac {(-3+2 x) \int \frac {1}{-3+e^x x} \, dx}{(-3+x) x}+\frac {(-3+2 x) \int \frac {1}{x \left (-3+e^x x\right )} \, dx}{(-3+x) x}\right ) \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx\\ &=-e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )+4 \text {Li}_2\left (1-\frac {x}{3}\right )+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \frac {(-3+2 x) \int \frac {1}{-3+e^x x} \, dx}{(-3+x) x} \, dx-12 \int \frac {(-3+2 x) \int \frac {1}{x \left (-3+e^x x\right )} \, dx}{(-3+x) x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx\\ &=-e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )+4 \text {Li}_2\left (1-\frac {x}{3}\right )+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \left (\frac {\int \frac {1}{-3+e^x x} \, dx}{-3+x}+\frac {\int \frac {1}{-3+e^x x} \, dx}{x}\right ) \, dx-12 \int \left (\frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{-3+x}+\frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{x}\right ) \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx\\ &=-e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )+4 \text {Li}_2\left (1-\frac {x}{3}\right )+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \frac {\int \frac {1}{-3+e^x x} \, dx}{-3+x} \, dx-12 \int \frac {\int \frac {1}{-3+e^x x} \, dx}{x} \, dx-12 \int \frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{-3+x} \, dx-12 \int \frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.34, size = 95, normalized size = 2.97 \begin {gather*} -e^{5 x}-4 x \log \left (\frac {1}{3} (-3+x) x\right )+4 \log (3-x) \left (x+\log \left (-3 e^{-x}+x\right )-\log \left (-3+e^x x\right )\right )+4 \log (x) \left (x+\log \left (-3 e^{-x}+x\right )-\log \left (-3+e^x x\right )\right )+4 \log \left (\frac {1}{3} (-3+x) x\right ) \log \left (-3+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5*x)*(-45*x + 15*x^2 + E^x*(15*x^2 - 5*x^3)) + (36 - 24*x + E^x*(-12*x + 8*x^2))*Log[(-3 + E^x*x
)/E^x] + (-36*x + 12*x^2 + E^x*(-12*x + 4*x^2))*Log[(-3*x + x^2)/3])/(9*x - 3*x^2 + E^x*(-3*x^2 + x^3)),x]

[Out]

-E^(5*x) - 4*x*Log[((-3 + x)*x)/3] + 4*Log[3 - x]*(x + Log[-3/E^x + x] - Log[-3 + E^x*x]) + 4*Log[x]*(x + Log[
-3/E^x + x] - Log[-3 + E^x*x]) + 4*Log[((-3 + x)*x)/3]*Log[-3 + E^x*x]

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fricas [A]  time = 0.79, size = 31, normalized size = 0.97 \begin {gather*} 4 \, \log \left (\frac {1}{3} \, x^{2} - x\right ) \log \left ({\left (x e^{x} - 3\right )} e^{\left (-x\right )}\right ) - e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2-12*x)*exp(x)+12*x^2-36*x)*log(1/3*x^
2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 59, normalized size = 1.84 \begin {gather*} 4 \, x \log \relax (3) - 4 \, \log \relax (3) \log \left (x e^{x} - 3\right ) - 4 \, x \log \left (x - 3\right ) + 4 \, \log \left (x e^{x} - 3\right ) \log \left (x - 3\right ) - 4 \, x \log \relax (x) + 4 \, \log \left (x e^{x} - 3\right ) \log \relax (x) - e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2-12*x)*exp(x)+12*x^2-36*x)*log(1/3*x^
2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x, algorithm="giac")

[Out]

4*x*log(3) - 4*log(3)*log(x*e^x - 3) - 4*x*log(x - 3) + 4*log(x*e^x - 3)*log(x - 3) - 4*x*log(x) + 4*log(x*e^x
- 3)*log(x) - e^(5*x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (8 x^{2}-12 x \right ) {\mathrm e}^{x}-24 x +36\right ) \ln \left (\left ({\mathrm e}^{x} x -3\right ) {\mathrm e}^{-x}\right )+\left (\left (4 x^{2}-12 x \right ) {\mathrm e}^{x}+12 x^{2}-36 x \right ) \ln \left (\frac {1}{3} x^{2}-x \right )+\left (\left (-5 x^{3}+15 x^{2}\right ) {\mathrm e}^{x}+15 x^{2}-45 x \right ) {\mathrm e}^{5 x}}{\left (x^{3}-3 x^{2}\right ) {\mathrm e}^{x}-3 x^{2}+9 x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2-12*x)*exp(x)-24*x+36)*ln((exp(x)*x-3)/exp(x))+((4*x^2-12*x)*exp(x)+12*x^2-36*x)*ln(1/3*x^2-x)+((-
5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x)

[Out]

int((((8*x^2-12*x)*exp(x)-24*x+36)*ln((exp(x)*x-3)/exp(x))+((4*x^2-12*x)*exp(x)+12*x^2-36*x)*ln(1/3*x^2-x)+((-
5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x)

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maxima [B]  time = 0.51, size = 58, normalized size = 1.81 \begin {gather*} 4 \, x \log \relax (3) + 4 \, {\left (\log \left (x - 3\right ) + \log \relax (x)\right )} \log \left (x e^{x} - 3\right ) - 4 \, x \log \left (x - 3\right ) - 4 \, {\left (x + \log \relax (3)\right )} \log \relax (x) - 4 \, \log \relax (3) \log \left (\frac {x e^{x} - 3}{x}\right ) - e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2-12*x)*exp(x)+12*x^2-36*x)*log(1/3*x^
2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x, algorithm="maxima")

[Out]

4*x*log(3) + 4*(log(x - 3) + log(x))*log(x*e^x - 3) - 4*x*log(x - 3) - 4*(x + log(3))*log(x) - 4*log(3)*log((x
*e^x - 3)/x) - e^(5*x)

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mupad [B]  time = 6.04, size = 28, normalized size = 0.88 \begin {gather*} 4\,\ln \left (\frac {x^2}{3}-x\right )\,\ln \left (x-3\,{\mathrm {e}}^{-x}\right )-{\mathrm {e}}^{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2/3 - x)*(36*x + exp(x)*(12*x - 4*x^2) - 12*x^2) - exp(5*x)*(exp(x)*(15*x^2 - 5*x^3) - 45*x + 15*x^
2) + log(exp(-x)*(x*exp(x) - 3))*(24*x + exp(x)*(12*x - 8*x^2) - 36))/(exp(x)*(3*x^2 - x^3) - 9*x + 3*x^2),x)

[Out]

4*log(x^2/3 - x)*log(x - 3*exp(-x)) - exp(5*x)

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sympy [A]  time = 0.85, size = 26, normalized size = 0.81 \begin {gather*} - e^{5 x} + 4 \log {\left (\left (x e^{x} - 3\right ) e^{- x} \right )} \log {\left (\frac {x^{2}}{3} - x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2-12*x)*exp(x)-24*x+36)*ln((exp(x)*x-3)/exp(x))+((4*x**2-12*x)*exp(x)+12*x**2-36*x)*ln(1/3*x
**2-x)+((-5*x**3+15*x**2)*exp(x)+15*x**2-45*x)*exp(5*x))/((x**3-3*x**2)*exp(x)-3*x**2+9*x),x)

[Out]

-exp(5*x) + 4*log((x*exp(x) - 3)*exp(-x))*log(x**2/3 - x)

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