3.4.10 \(\int \frac {(8+2 x-2 x^2+e (16 x+4 x^2-4 x^3)) \log (x)+(-4+8 x+e (-4 x+8 x^2)) \log ^2(x) \log (\frac {x+e x^2}{e})+(-8-2 x+2 x^2+e (-8 x-2 x^2+2 x^3)+(8+4 x-6 x^2+e (8 x+4 x^2-6 x^3)) \log (x)) \log (\frac {x+e x^2}{e}) \log (\log (\frac {x+e x^2}{e}))}{(1+e x) \log ^2(x) \log (\frac {x+e x^2}{e})} \, dx\) [310]

3.4.10.1 Optimal result

Integrand size = 170, antiderivative size = 29 \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\left (4+x-x^2\right ) \left (-4+\frac {2 x \log \left (\log \left (\frac {x}{e}+x^2\right )\right )}{\log (x)}\right ) \end {dmath*}

output

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

3.4.10.2 Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=-4 x+4 x^2-\frac {2 x \left (-4-x+x^2\right ) \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \end {dmath*}

input

Integrate[((8 + 2*x - 2*x^2 + E*(16*x + 4*x^2 - 4*x^3))*Log[x] + (-4 + 8*x 
 + E*(-4*x + 8*x^2))*Log[x]^2*Log[(x + E*x^2)/E] + (-8 - 2*x + 2*x^2 + E*( 
-8*x - 2*x^2 + 2*x^3) + (8 + 4*x - 6*x^2 + E*(8*x + 4*x^2 - 6*x^3))*Log[x] 
)*Log[(x + E*x^2)/E]*Log[Log[(x + E*x^2)/E]])/((1 + E*x)*Log[x]^2*Log[(x + 
 E*x^2)/E]),x]
 

output

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

3.4.10.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (e \left (8 x^2-4 x\right )+8 x-4\right ) \log \left (\frac {e x^2+x}{e}\right ) \log ^2(x)+\left (-2 x^2+e \left (-4 x^3+4 x^2+16 x\right )+2 x+8\right ) \log (x)+\left (2 x^2+e \left (2 x^3-2 x^2-8 x\right )+\left (-6 x^2+e \left (-6 x^3+4 x^2+8 x\right )+4 x+8\right ) \log (x)-2 x-8\right ) \log \left (\frac {e x^2+x}{e}\right ) \log \left (\log \left (\frac {e x^2+x}{e}\right )\right )}{(e x+1) \log ^2(x) \log \left (\frac {e x^2+x}{e}\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {-\frac {2 (e x+1) \left (-x^2+\left (3 x^2-2 x-4\right ) \log (x)+x+4\right ) \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}-\frac {2 (2 e x+1) \left (x^2-x-4\right )}{\log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}+4 e (2 x-1) x+8 x-4}{e x+1}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 \left (-2 e x^3-(1-2 e) x^2+4 e x^2 \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )+(1+8 e) x+4 \left (1-\frac {e}{2}\right ) x \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )-2 \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )+4\right )}{(e x+1) \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}-\frac {2 \left (-x^2+3 x^2 \log (x)+x-2 x \log (x)-4 \log (x)+4\right ) \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int \frac {x^2 \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}dx-4 \int \frac {x^2}{\log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}dx-6 \int \frac {x^2 \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log (x)}dx-8 \int \frac {\log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}dx-2 \int \frac {x \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}dx-\frac {2 \left (1+e-8 e^2\right ) \int \frac {1}{\log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}dx}{e^2}+\frac {2 \left (1+e-4 e^2\right ) \int \frac {1}{(e x+1) \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}dx}{e^2}-\frac {2 (1+2 e) \int \frac {x}{\log (x) (1-\log (x (e x+1)))}dx}{e}+8 \int \frac {\log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log (x)}dx+4 \int \frac {x \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log (x)}dx+4 x^2-4 x\)

input

Int[((8 + 2*x - 2*x^2 + E*(16*x + 4*x^2 - 4*x^3))*Log[x] + (-4 + 8*x + E*( 
-4*x + 8*x^2))*Log[x]^2*Log[(x + E*x^2)/E] + (-8 - 2*x + 2*x^2 + E*(-8*x - 
 2*x^2 + 2*x^3) + (8 + 4*x - 6*x^2 + E*(8*x + 4*x^2 - 6*x^3))*Log[x])*Log[ 
(x + E*x^2)/E]*Log[Log[(x + E*x^2)/E]])/((1 + E*x)*Log[x]^2*Log[(x + E*x^2 
)/E]),x]
 

output

$Aborted
 

3.4.10.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.4.10.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(76\) vs. \(2(30)=60\).

Time = 54.72 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.66

method	result	size
parallelrisch	\(\frac {-2 x^{3} \ln \left (\ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{-1}\right )\right )+4 x^{2} \ln \left (x \right )+2 \ln \left (\ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{-1}\right )\right ) x^{2}-4 x \ln \left (x \right )+8 x \ln \left (\ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{-1}\right )\right )}{\ln \left (x \right )}\)	\(77\)
risch	\(-\frac {2 x \left (x^{2}-x -4\right ) \ln \left (-1+\ln \left (x \right )+\ln \left (x \,{\mathrm e}+1\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right ) \left (-\operatorname {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right )+\operatorname {csgn}\left (i \left (x \,{\mathrm e}+1\right )\right )\right )}{2}\right )}{\ln \left (x \right )}+4 x^{2}-4 x\)	\(95\)

input

int(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*ln(x)+(2*x^3-2*x^2-8*x)*exp( 
1)+2*x^2-2*x-8)*ln((x^2*exp(1)+x)/exp(1))*ln(ln((x^2*exp(1)+x)/exp(1)))+(( 
8*x^2-4*x)*exp(1)+8*x-4)*ln(x)^2*ln((x^2*exp(1)+x)/exp(1))+((-4*x^3+4*x^2+ 
16*x)*exp(1)-2*x^2+2*x+8)*ln(x))/(x*exp(1)+1)/ln(x)^2/ln((x^2*exp(1)+x)/ex 
p(1)),x,method=_RETURNVERBOSE)
 

output

(-2*x^3*ln(ln((x^2*exp(1)+x)/exp(1)))+4*x^2*ln(x)+2*ln(ln((x^2*exp(1)+x)/e 
xp(1)))*x^2-4*x*ln(x)+8*x*ln(ln((x^2*exp(1)+x)/exp(1))))/ln(x)
 

3.4.10.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\frac {2 \, {\left (2 \, {\left (x^{2} - x\right )} \log \left (x\right ) - {\left (x^{3} - x^{2} - 4 \, x\right )} \log \left (\log \left ({\left (x^{2} e + x\right )} e^{\left (-1\right )}\right )\right )\right )}}{\log \left (x\right )} \end {dmath*}

input

integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8* 
x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/e 
xp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((- 
4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(x*exp(1)+1)/log(x)^2/log((x 
^2*exp(1)+x)/exp(1)),x, algorithm=\
 

output

2*(2*(x^2 - x)*log(x) - (x^3 - x^2 - 4*x)*log(log((x^2*e + x)*e^(-1))))/lo 
g(x)
 

3.4.10.6 Sympy [F(-2)]

Exception generated. \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\text {Exception raised: TypeError} \end {dmath*}

input

integrate(((((-6*x**3+4*x**2+8*x)*exp(1)-6*x**2+4*x+8)*ln(x)+(2*x**3-2*x** 
2-8*x)*exp(1)+2*x**2-2*x-8)*ln((x**2*exp(1)+x)/exp(1))*ln(ln((x**2*exp(1)+ 
x)/exp(1)))+((8*x**2-4*x)*exp(1)+8*x-4)*ln(x)**2*ln((x**2*exp(1)+x)/exp(1) 
)+((-4*x**3+4*x**2+16*x)*exp(1)-2*x**2+2*x+8)*ln(x))/(x*exp(1)+1)/ln(x)**2 
/ln((x**2*exp(1)+x)/exp(1)),x)
 

output

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

3.4.10.7 Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\frac {2 \, {\left (2 \, {\left (x^{2} - x\right )} \log \left (x\right ) - {\left (x^{3} - x^{2} - 4 \, x\right )} \log \left (\log \left (x e + 1\right ) + \log \left (x\right ) - 1\right )\right )}}{\log \left (x\right )} \end {dmath*}

input

integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8* 
x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/e 
xp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((- 
4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(x*exp(1)+1)/log(x)^2/log((x 
^2*exp(1)+x)/exp(1)),x, algorithm=\
 

output

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

3.4.10.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).

Time = 0.51 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31 \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=-\frac {2 \, {\left (x^{3} \log \left (\log \left (x^{2} e + x\right ) - 1\right ) - 2 \, x^{2} \log \left (x\right ) - x^{2} \log \left (\log \left (x^{2} e + x\right ) - 1\right ) + 2 \, x \log \left (x\right ) - 4 \, x \log \left (\log \left (x^{2} e + x\right ) - 1\right )\right )}}{\log \left (x\right )} \end {dmath*}

input

integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8* 
x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/e 
xp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((- 
4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(x*exp(1)+1)/log(x)^2/log((x 
^2*exp(1)+x)/exp(1)),x, algorithm=\
 

output

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

3.4.10.9 Mupad [B] (verification not implemented)

Time = 14.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \begin {dmath*} \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=4\,x^2-4\,x+\frac {2\,x^2\,\ln \left (\ln \left ({\mathrm {e}}^{-1}\,\left (\mathrm {e}\,x^2+x\right )\right )\right )\,\left (x\,\mathrm {e}+1\right )\,\left (-x^2+x+4\right )}{\ln \left (x\right )\,\left (\mathrm {e}\,x^2+x\right )} \end {dmath*}

input

int(-(log(log(exp(-1)*(x + x^2*exp(1))))*log(exp(-1)*(x + x^2*exp(1)))*(2* 
x - log(x)*(4*x + exp(1)*(8*x + 4*x^2 - 6*x^3) - 6*x^2 + 8) + exp(1)*(8*x 
+ 2*x^2 - 2*x^3) - 2*x^2 + 8) - log(x)*(2*x + exp(1)*(16*x + 4*x^2 - 4*x^3 
) - 2*x^2 + 8) + log(exp(-1)*(x + x^2*exp(1)))*log(x)^2*(exp(1)*(4*x - 8*x 
^2) - 8*x + 4))/(log(exp(-1)*(x + x^2*exp(1)))*log(x)^2*(x*exp(1) + 1)),x)
 

output

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