3.11.99 \(\int \frac {(-16+32 x-16 x^2+e^x (-26-24 x+25 x^2)) \log (x)+e^x (-1-x+x^2) \log ^2(x)+(-8+8 x) \log (\log (x))-4 \log (x) \log ^2(\log (x))}{(8-16 x+8 x^2) \log (x)} \, dx\)

Optimal. Leaf size=29 \[ x \left (-2+\frac {\frac {1}{4} e^x (25+\log (x))+\log ^2(\log (x))}{-2+2 x}\right ) \]

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Rubi [F]  time = 1.33, 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 (-16+32 x-16 x^2+e^x \left (-26-24 x+25 x^2\right )\right ) \log (x)+e^x \left (-1-x+x^2\right ) \log ^2(x)+(-8+8 x) \log (\log (x))-4 \log (x) \log ^2(\log (x))}{\left (8-16 x+8 x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-16 + 32*x - 16*x^2 + E^x*(-26 - 24*x + 25*x^2))*Log[x] + E^x*(-1 - x + x^2)*Log[x]^2 + (-8 + 8*x)*Log[L
og[x]] - 4*Log[x]*Log[Log[x]]^2)/((8 - 16*x + 8*x^2)*Log[x]),x]

[Out]

(25*E^x)/8 - (25*E^x)/(8*(1 - x)) - 2*x + (E^x*Log[x])/8 - (E^x*Log[x])/(8*(1 - x)) + Defer[Int][Log[Log[x]]/(
(-1 + x)*Log[x]), x] - Defer[Int][Log[Log[x]]^2/(-1 + x)^2, x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-16+32 x-16 x^2+e^x \left (-26-24 x+25 x^2\right )\right ) \log (x)+e^x \left (-1-x+x^2\right ) \log ^2(x)+(-8+8 x) \log (\log (x))-4 \log (x) \log ^2(\log (x))}{8 (-1+x)^2 \log (x)} \, dx\\ &=\frac {1}{8} \int \frac {\left (-16+32 x-16 x^2+e^x \left (-26-24 x+25 x^2\right )\right ) \log (x)+e^x \left (-1-x+x^2\right ) \log ^2(x)+(-8+8 x) \log (\log (x))-4 \log (x) \log ^2(\log (x))}{(-1+x)^2 \log (x)} \, dx\\ &=\frac {1}{8} \int \left (-\frac {16}{(-1+x)^2}+\frac {32 x}{(-1+x)^2}-\frac {16 x^2}{(-1+x)^2}+\frac {e^x \left (-26-24 x+25 x^2-\log (x)-x \log (x)+x^2 \log (x)\right )}{(-1+x)^2}+\frac {8 \log (\log (x))}{(-1+x) \log (x)}-\frac {4 \log ^2(\log (x))}{(-1+x)^2}\right ) \, dx\\ &=-\frac {2}{1-x}+\frac {1}{8} \int \frac {e^x \left (-26-24 x+25 x^2-\log (x)-x \log (x)+x^2 \log (x)\right )}{(-1+x)^2} \, dx-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx-2 \int \frac {x^2}{(-1+x)^2} \, dx+4 \int \frac {x}{(-1+x)^2} \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ &=-\frac {2}{1-x}+\frac {1}{8} \int \left (\frac {e^x \left (-26-24 x+25 x^2\right )}{(-1+x)^2}+\frac {e^x \left (-1-x+x^2\right ) \log (x)}{(-1+x)^2}\right ) \, dx-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx-2 \int \left (1+\frac {1}{(-1+x)^2}+\frac {2}{-1+x}\right ) \, dx+4 \int \left (\frac {1}{(-1+x)^2}+\frac {1}{-1+x}\right ) \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ &=-2 x+\frac {1}{8} \int \frac {e^x \left (-26-24 x+25 x^2\right )}{(-1+x)^2} \, dx+\frac {1}{8} \int \frac {e^x \left (-1-x+x^2\right ) \log (x)}{(-1+x)^2} \, dx-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ &=-2 x+\frac {1}{8} e^x \log (x)-\frac {e^x \log (x)}{8 (1-x)}+\frac {1}{8} \int \left (25 e^x-\frac {25 e^x}{(-1+x)^2}+\frac {26 e^x}{-1+x}\right ) \, dx-\frac {1}{8} \int \frac {e^x}{-1+x} \, dx-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ &=-2 x-\frac {1}{8} e \text {Ei}(-1+x)+\frac {1}{8} e^x \log (x)-\frac {e^x \log (x)}{8 (1-x)}-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx+\frac {25 \int e^x \, dx}{8}-\frac {25}{8} \int \frac {e^x}{(-1+x)^2} \, dx+\frac {13}{4} \int \frac {e^x}{-1+x} \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ &=\frac {25 e^x}{8}-\frac {25 e^x}{8 (1-x)}-2 x+\frac {25}{8} e \text {Ei}(-1+x)+\frac {1}{8} e^x \log (x)-\frac {e^x \log (x)}{8 (1-x)}-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx-\frac {25}{8} \int \frac {e^x}{-1+x} \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ &=\frac {25 e^x}{8}-\frac {25 e^x}{8 (1-x)}-2 x+\frac {1}{8} e^x \log (x)-\frac {e^x \log (x)}{8 (1-x)}-\frac {1}{2} \int \frac {\log ^2(\log (x))}{(-1+x)^2} \, dx+\int \frac {\log (\log (x))}{(-1+x) \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 33, normalized size = 1.14 \begin {gather*} \frac {x \left (16+25 e^x-16 x+e^x \log (x)+4 \log ^2(\log (x))\right )}{8 (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-16 + 32*x - 16*x^2 + E^x*(-26 - 24*x + 25*x^2))*Log[x] + E^x*(-1 - x + x^2)*Log[x]^2 + (-8 + 8*x)
*Log[Log[x]] - 4*Log[x]*Log[Log[x]]^2)/((8 - 16*x + 8*x^2)*Log[x]),x]

[Out]

(x*(16 + 25*E^x - 16*x + E^x*Log[x] + 4*Log[Log[x]]^2))/(8*(-1 + x))

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fricas [A]  time = 0.58, size = 35, normalized size = 1.21 \begin {gather*} \frac {x e^{x} \log \relax (x) + 4 \, x \log \left (\log \relax (x)\right )^{2} - 16 \, x^{2} + 25 \, x e^{x} + 16 \, x}{8 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)*log(log(x))^2+(8*x-8)*log(log(x))+(x^2-x-1)*exp(x)*log(x)^2+((25*x^2-24*x-26)*exp(x)-16*x
^2+32*x-16)*log(x))/(8*x^2-16*x+8)/log(x),x, algorithm="fricas")

[Out]

1/8*(x*e^x*log(x) + 4*x*log(log(x))^2 - 16*x^2 + 25*x*e^x + 16*x)/(x - 1)

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giac [A]  time = 0.27, size = 35, normalized size = 1.21 \begin {gather*} \frac {x e^{x} \log \relax (x) + 4 \, x \log \left (\log \relax (x)\right )^{2} - 16 \, x^{2} + 25 \, x e^{x} + 16 \, x}{8 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)*log(log(x))^2+(8*x-8)*log(log(x))+(x^2-x-1)*exp(x)*log(x)^2+((25*x^2-24*x-26)*exp(x)-16*x
^2+32*x-16)*log(x))/(8*x^2-16*x+8)/log(x),x, algorithm="giac")

[Out]

1/8*(x*e^x*log(x) + 4*x*log(log(x))^2 - 16*x^2 + 25*x*e^x + 16*x)/(x - 1)

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maple [A]  time = 0.23, size = 38, normalized size = 1.31

method	result	size
risch	\(\frac {x \ln \left (\ln \relax (x )\right )^{2}}{2 x -2}-\frac {x \left (-{\mathrm e}^{x} \ln \relax (x )+16 x -25 \,{\mathrm e}^{x}-16\right )}{8 \left (x -1\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(x)*ln(ln(x))^2+(8*x-8)*ln(ln(x))+(x^2-x-1)*exp(x)*ln(x)^2+((25*x^2-24*x-26)*exp(x)-16*x^2+32*x-16)*
ln(x))/(8*x^2-16*x+8)/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/2*x/(x-1)*ln(ln(x))^2-1/8*x*(-exp(x)*ln(x)+16*x-25*exp(x)-16)/(x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {13 \, e E_{2}\left (-x + 1\right )}{4 \, {\left (x - 1\right )}} + \frac {x e^{x} \log \relax (x) + 4 \, x \log \left (\log \relax (x)\right )^{2}}{8 \, {\left (x - 1\right )}} - \frac {2 \, {\left (x^{2} - x - 1\right )}}{x - 1} - \frac {2}{x - 1} + \frac {1}{8} \, \int \frac {{\left (25 \, x^{2} - 25 \, x + 1\right )} e^{x}}{x^{2} - 2 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)*log(log(x))^2+(8*x-8)*log(log(x))+(x^2-x-1)*exp(x)*log(x)^2+((25*x^2-24*x-26)*exp(x)-16*x
^2+32*x-16)*log(x))/(8*x^2-16*x+8)/log(x),x, algorithm="maxima")

[Out]

13/4*e*exp_integral_e(2, -x + 1)/(x - 1) + 1/8*(x*e^x*log(x) + 4*x*log(log(x))^2)/(x - 1) - 2*(x^2 - x - 1)/(x
- 1) - 2/(x - 1) + 1/8*integrate((25*x^2 - 25*x + 1)*e^x/(x^2 - 2*x + 1), x)

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mupad [B]  time = 1.07, size = 66, normalized size = 2.28 \begin {gather*} {\mathrm {e}}^x\,\ln \relax (x)\,\left (\frac {1}{8\,\left (x-1\right )}+\frac {\frac {x}{8}-\frac {1}{8}}{x-1}\right )-{\ln \left (\ln \relax (x)\right )}^2\,\left (\frac {x-x^2}{2\,x\,{\left (x-1\right )}^2}-\frac {1}{2}\right )-2\,x+\frac {25\,x\,{\mathrm {e}}^x}{8\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(exp(x)*(24*x - 25*x^2 + 26) - 32*x + 16*x^2 + 16) - log(log(x))*(8*x - 8) + 4*log(log(x))^2*log(
x) + exp(x)*log(x)^2*(x - x^2 + 1))/(log(x)*(8*x^2 - 16*x + 8)),x)

[Out]

exp(x)*log(x)*(1/(8*(x - 1)) + (x/8 - 1/8)/(x - 1)) - log(log(x))^2*((x - x^2)/(2*x*(x - 1)^2) - 1/2) - 2*x +
(25*x*exp(x))/(8*(x - 1))

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sympy [A]  time = 0.57, size = 32, normalized size = 1.10 \begin {gather*} - 2 x + \frac {x \log {\left (\log {\relax (x )} \right )}^{2}}{2 x - 2} + \frac {\left (x \log {\relax (x )} + 25 x\right ) e^{x}}{8 x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(x)*ln(ln(x))**2+(8*x-8)*ln(ln(x))+(x**2-x-1)*exp(x)*ln(x)**2+((25*x**2-24*x-26)*exp(x)-16*x**
2+32*x-16)*ln(x))/(8*x**2-16*x+8)/ln(x),x)

[Out]

-2*x + x*log(log(x))**2/(2*x - 2) + (x*log(x) + 25*x)*exp(x)/(8*x - 8)

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