\(\int \frac {96+e^4 (-8-2 x)+24 x+(96-8 e^4) \log (x)-10 \log ^2(x)}{(-96 x-24 x^2+e^4 (8 x+2 x^2)) \log (x)+(10 x+3 x^2) \log ^2(x)} \, dx\) [3760]

Optimal result

Integrand size = 73, antiderivative size = 29 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (\frac {3}{2}+\frac {5}{x}-\frac {\left (12-e^4\right ) (4+x)}{x \log (x)}\right ) \]

[Out]

ln(5/x+3/2-(12-exp(2)^2)/ln(x)*(4+x)/x)

Rubi [F]

\[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx \]

[In]

Int[(96 + E^4*(-8 - 2*x) + 24*x + (96 - 8*E^4)*Log[x] - 10*Log[x]^2)/((-96*x - 24*x^2 + E^4*(8*x + 2*x^2))*Log
[x] + (10*x + 3*x^2)*Log[x]^2),x]

[Out]

-Log[x] + Log[10 + 3*x] - Log[Log[x]] + 3*Defer[Int][(2*(-12 + E^4)*(4 + x) + (10 + 3*x)*Log[x])^(-1), x] - 4*
(12 - E^4)*Defer[Int][1/((-10 - 3*x)*(2*(-12 + E^4)*(4 + x) + (10 + 3*x)*Log[x])), x] + 10*Defer[Int][1/(x*(2*
(-12 + E^4)*(4 + x) + (10 + 3*x)*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-96-e^4 (-8-2 x)-24 x-\left (96-8 e^4\right ) \log (x)+10 \log ^2(x)}{x \log (x) \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )} \, dx \\ & = \int \frac {2 \left (-\left (\left (-12+e^4\right ) (4+x)\right )-4 \left (-12+e^4\right ) \log (x)-5 \log ^2(x)\right )}{x \log (x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx \\ & = 2 \int \frac {-\left (\left (-12+e^4\right ) (4+x)\right )-4 \left (-12+e^4\right ) \log (x)-5 \log ^2(x)}{x \log (x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx \\ & = 2 \int \left (-\frac {5}{x (10+3 x)}-\frac {1}{2 x \log (x)}+\frac {-10-3 x}{2 x \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )}+\frac {2 \left (-12+e^4\right )}{(10+3 x) \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )}\right ) \, dx \\ & = -\left (10 \int \frac {1}{x (10+3 x)} \, dx\right )-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(10+3 x) \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )} \, dx-\int \frac {1}{x \log (x)} \, dx+\int \frac {-10-3 x}{x \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )} \, dx \\ & = 3 \int \frac {1}{10+3 x} \, dx-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx-\int \frac {1}{x} \, dx+\int \frac {10+3 x}{x \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -\log (x)+\log (10+3 x)-\log (\log (x))-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx+\int \left (\frac {3}{-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)}+\frac {10}{x \left (-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)\right )}\right ) \, dx \\ & = -\log (x)+\log (10+3 x)-\log (\log (x))+3 \int \frac {1}{-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)} \, dx+10 \int \frac {1}{x \left (-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)\right )} \, dx-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx \\ & = -\log (x)+\log (10+3 x)-\log (\log (x))+3 \int \frac {1}{2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)} \, dx+10 \int \frac {1}{x \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(29)=58\).

Time = 9.91 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=-2 \left (\frac {\log (x)}{2}-\frac {1}{2} \log (10+3 x)+\frac {1}{2} \log (\log (x))\right )-2 \left (\frac {1}{2} \log (10+3 x)-\frac {1}{2} \log \left (96-8 e^4+24 x-2 e^4 x-10 \log (x)-3 x \log (x)\right )\right ) \]

[In]

Integrate[(96 + E^4*(-8 - 2*x) + 24*x + (96 - 8*E^4)*Log[x] - 10*Log[x]^2)/((-96*x - 24*x^2 + E^4*(8*x + 2*x^2
))*Log[x] + (10*x + 3*x^2)*Log[x]^2),x]

[Out]

-2*(Log[x]/2 - Log[10 + 3*x]/2 + Log[Log[x]]/2) - 2*(Log[10 + 3*x]/2 - Log[96 - 8*E^4 + 24*x - 2*E^4*x - 10*Lo
g[x] - 3*x*Log[x]]/2)

Maple [A] (verified)

Time = 4.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31

method	result	size
parallelrisch	\(-\ln \left (\ln \left (x \right )\right )+\ln \left (\frac {2 x \,{\mathrm e}^{4}}{3}+\frac {8 \,{\mathrm e}^{4}}{3}+x \ln \left (x \right )+\frac {10 \ln \left (x \right )}{3}-8 x -32\right )-\ln \left (x \right )\)	\(38\)
default	\(-\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (3 x \ln \left (x \right )+2 x \,{\mathrm e}^{4}+10 \ln \left (x \right )+8 \,{\mathrm e}^{4}-24 x -96\right )\)	\(39\)
norman	\(-\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (3 x \ln \left (x \right )+2 x \,{\mathrm e}^{4}+10 \ln \left (x \right )+8 \,{\mathrm e}^{4}-24 x -96\right )\)	\(39\)
risch	\(\ln \left (3 x +10\right )-\ln \left (x \right )+\ln \left (\ln \left (x \right )+\frac {2 x \,{\mathrm e}^{4}+8 \,{\mathrm e}^{4}-24 x -96}{3 x +10}\right )-\ln \left (\ln \left (x \right )\right )\)	\(43\)

[In]

int((-10*ln(x)^2+(-8*exp(2)^2+96)*ln(x)+(-2*x-8)*exp(2)^2+24*x+96)/((3*x^2+10*x)*ln(x)^2+((2*x^2+8*x)*exp(2)^2
-24*x^2-96*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x))+ln(2/3*x*exp(2)^2+8/3*exp(2)^2+x*ln(x)+10/3*ln(x)-8*x-32)-ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (3 \, x + 10\right ) - \log \left (x\right ) + \log \left (\frac {2 \, {\left (x + 4\right )} e^{4} + {\left (3 \, x + 10\right )} \log \left (x\right ) - 24 \, x - 96}{3 \, x + 10}\right ) - \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-10*log(x)^2+(-8*exp(2)^2+96)*log(x)+(-2*x-8)*exp(2)^2+24*x+96)/((3*x^2+10*x)*log(x)^2+((2*x^2+8*x)
*exp(2)^2-24*x^2-96*x)*log(x)),x, algorithm="fricas")

[Out]

log(3*x + 10) - log(x) + log((2*(x + 4)*e^4 + (3*x + 10)*log(x) - 24*x - 96)/(3*x + 10)) - log(log(x))

Sympy [F(-2)]

Exception generated. \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\text {Exception raised: PolynomialError} \]

[In]

integrate((-10*ln(x)**2+(-8*exp(2)**2+96)*ln(x)+(-2*x-8)*exp(2)**2+24*x+96)/((3*x**2+10*x)*ln(x)**2+((2*x**2+8
*x)*exp(2)**2-24*x**2-96*x)*ln(x)),x)

[Out]

Exception raised: PolynomialError >> 1/(9*x**3 + 60*x**2 + 100*x) contains an element of the set of generators
.

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (3 \, x + 10\right ) - \log \left (x\right ) + \log \left (\frac {2 \, x {\left (e^{4} - 12\right )} + {\left (3 \, x + 10\right )} \log \left (x\right ) + 8 \, e^{4} - 96}{3 \, x + 10}\right ) - \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-10*log(x)^2+(-8*exp(2)^2+96)*log(x)+(-2*x-8)*exp(2)^2+24*x+96)/((3*x^2+10*x)*log(x)^2+((2*x^2+8*x)
*exp(2)^2-24*x^2-96*x)*log(x)),x, algorithm="maxima")

[Out]

log(3*x + 10) - log(x) + log((2*x*(e^4 - 12) + (3*x + 10)*log(x) + 8*e^4 - 96)/(3*x + 10)) - log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (2 \, x e^{4} + 3 \, x \log \left (x\right ) - 24 \, x + 8 \, e^{4} + 10 \, \log \left (x\right ) - 96\right ) - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-10*log(x)^2+(-8*exp(2)^2+96)*log(x)+(-2*x-8)*exp(2)^2+24*x+96)/((3*x^2+10*x)*log(x)^2+((2*x^2+8*x)
*exp(2)^2-24*x^2-96*x)*log(x)),x, algorithm="giac")

[Out]

log(2*x*e^4 + 3*x*log(x) - 24*x + 8*e^4 + 10*log(x) - 96) - log(x) - log(log(x))

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\ln \left (8\,{\mathrm {e}}^4-24\,x+10\,\ln \left (x\right )+2\,x\,{\mathrm {e}}^4+3\,x\,\ln \left (x\right )-96\right )-\ln \left (\ln \left (x\right )\right )-\ln \left (x\right ) \]

[In]

int(-(10*log(x)^2 - 24*x + log(x)*(8*exp(4) - 96) + exp(4)*(2*x + 8) - 96)/(log(x)^2*(10*x + 3*x^2) - log(x)*(
96*x - exp(4)*(8*x + 2*x^2) + 24*x^2)),x)

[Out]

log(8*exp(4) - 24*x + 10*log(x) + 2*x*exp(4) + 3*x*log(x) - 96) - log(log(x)) - log(x)