3.11.81 \(\int \frac {e^x (1-2 x-4 x^2)+e^x (-4-5 x) \log (x)+(e^x (x^2+x^3)+e^x (x+x^2) \log (x)) \log (\frac {10 x^4+10 x^5}{x+\log (x)})}{(x^2+x^3+(x+x^2) \log (x)) \log ^2(\frac {10 x^4+10 x^5}{x+\log (x)})} \, dx\) [1081]

3.11.81.1 Optimal result

Integrand size = 108, antiderivative size = 23 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {e^x}{\log \left (\frac {10 x^3 \left (x+x^2\right )}{x+\log (x)}\right )} \end {dmath*}

output

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

3.11.81.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {e^x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \end {dmath*}

input

Integrate[(E^x*(1 - 2*x - 4*x^2) + E^x*(-4 - 5*x)*Log[x] + (E^x*(x^2 + x^3 
) + E^x*(x + x^2)*Log[x])*Log[(10*x^4 + 10*x^5)/(x + Log[x])])/((x^2 + x^3 
 + (x + x^2)*Log[x])*Log[(10*x^4 + 10*x^5)/(x + Log[x])]^2),x]
 

output

E^x/Log[(10*x^4*(1 + x))/(x + Log[x])]
 

3.11.81.3 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.

\(\displaystyle \int \frac {e^x \left (-4 x^2-2 x+1\right )+\left (e^x \left (x^2+x\right ) \log (x)+e^x \left (x^3+x^2\right )\right ) \log \left (\frac {10 x^5+10 x^4}{x+\log (x)}\right )+e^x (-5 x-4) \log (x)}{\left (x^3+x^2+\left (x^2+x\right ) \log (x)\right ) \log ^2\left (\frac {10 x^5+10 x^4}{x+\log (x)}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (-4 x^2-2 x+1\right )+\left (e^x \left (x^2+x\right ) \log (x)+e^x \left (x^3+x^2\right )\right ) \log \left (\frac {10 x^5+10 x^4}{x+\log (x)}\right )+e^x (-5 x-4) \log (x)}{x (x+1) (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^x \left (x \log (x) \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )-4 x^2+x^3 \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )+x^2 \log (x) \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )+x^2 \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )-2 x-5 x \log (x)-4 \log (x)+1\right )}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}-\frac {e^x \left (x \log (x) \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )-4 x^2+x^3 \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )+x^2 \log (x) \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )+x^2 \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )-2 x-5 x \log (x)-4 \log (x)+1\right )}{(x+1) (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^x \left (\log (x) \left ((x+1) x \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )-5 x-4\right )-4 x^2+(x+1) x^2 \log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )-2 x+1\right )}{x (x+1) (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^x}{\log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}+\frac {e^x \left (-4 x^2-2 x-5 x \log (x)-4 \log (x)+1\right )}{x (x+1) (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \int \frac {e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx+\int \frac {e^x}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx+\int \frac {e^x}{(x+1) (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx-4 \int \frac {e^x \log (x)}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx-\int \frac {e^x \log (x)}{(x+1) (x+\log (x)) \log ^2\left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx+\int \frac {e^x}{\log \left (\frac {10 x^4 (x+1)}{x+\log (x)}\right )}dx\)

input

Int[(E^x*(1 - 2*x - 4*x^2) + E^x*(-4 - 5*x)*Log[x] + (E^x*(x^2 + x^3) + E^ 
x*(x + x^2)*Log[x])*Log[(10*x^4 + 10*x^5)/(x + Log[x])])/((x^2 + x^3 + (x 
+ x^2)*Log[x])*Log[(10*x^4 + 10*x^5)/(x + Log[x])]^2),x]
 

output

$Aborted
 

3.11.81.3.1 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 7292

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

rule 7293

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

3.11.81.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 51.94 (sec) , antiderivative size = 441, normalized size of antiderivative = 19.17

method	result	size
risch	\(\frac {2 i {\mathrm e}^{x}}{\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right )+2 i \ln \left (1+x \right )+2 i \ln \left (2\right )+2 i \ln \left (5\right )+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\pi \,\operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x +\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{4} \left (1+x \right )}{x +\ln \left (x \right )}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x +\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x +\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left (1+x \right )\right )-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}-\pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )+8 i \ln \left (x \right )+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}+\pi \operatorname {csgn}\left (\frac {i x^{4} \left (1+x \right )}{x +\ln \left (x \right )}\right )^{3}-\pi \,\operatorname {csgn}\left (\frac {i}{x +\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x +\ln \left (x \right )}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x +\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i \left (1+x \right )\right )-\pi \,\operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (\frac {i x^{4} \left (1+x \right )}{x +\ln \left (x \right )}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i \left (1+x \right )}{x +\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{4} \left (1+x \right )}{x +\ln \left (x \right )}\right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x +\ln \left (x \right )}\right )^{3}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-2 i \ln \left (x +\ln \left (x \right )\right )+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}}\)	\(441\)

input

int((((x^2+x)*exp(x)*ln(x)+(x^3+x^2)*exp(x))*ln((10*x^5+10*x^4)/(x+ln(x))) 
+(-5*x-4)*exp(x)*ln(x)+(-4*x^2-2*x+1)*exp(x))/((x^2+x)*ln(x)+x^3+x^2)/ln(( 
10*x^5+10*x^4)/(x+ln(x)))^2,x,method=_RETURNVERBOSE)
 

output

2*I*exp(x)/(Pi*csgn(I*x^3)*csgn(I*x^4)*csgn(I*x)+2*I*ln(1+x)+2*I*ln(2)+2*I 
*ln(5)+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^4)*csgn(I*(1+x)/(x+ln(x)))*csgn(I*x^4/ 
(x+ln(x))*(1+x))+Pi*csgn(I/(x+ln(x)))*csgn(I*(1+x)/(x+ln(x)))*csgn(I*(1+x) 
)-Pi*csgn(I*x^3)*csgn(I*x^4)^2-Pi*csgn(I*x^4)^2*csgn(I*x)+8*I*ln(x)+Pi*csg 
n(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^4)^3+Pi*csgn 
(I*x^4/(x+ln(x))*(1+x))^3-Pi*csgn(I/(x+ln(x)))*csgn(I*(1+x)/(x+ln(x)))^2-P 
i*csgn(I*(1+x)/(x+ln(x)))^2*csgn(I*(1+x))-Pi*csgn(I*x^4)*csgn(I*x^4/(x+ln( 
x))*(1+x))^2-Pi*csgn(I*(1+x)/(x+ln(x)))*csgn(I*x^4/(x+ln(x))*(1+x))^2-Pi*c 
sgn(I*x)*csgn(I*x^3)^2-Pi*csgn(I*x^2)*csgn(I*x^3)^2+Pi*csgn(I*(1+x)/(x+ln( 
x)))^3+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-2*I*ln(x+ln(x))+Pi*csgn(I*x^3) 
^3)
 

3.11.81.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {e^{x}}{\log \left (\frac {10 \, {\left (x^{5} + x^{4}\right )}}{x + \log \left (x\right )}\right )} \end {dmath*}

input

integrate((((x^2+x)*exp(x)*log(x)+(x^3+x^2)*exp(x))*log((10*x^5+10*x^4)/(x 
+log(x)))+(-5*x-4)*exp(x)*log(x)+(-4*x^2-2*x+1)*exp(x))/((x^2+x)*log(x)+x^ 
3+x^2)/log((10*x^5+10*x^4)/(x+log(x)))^2,x, algorithm=\
 

output

e^x/log(10*(x^5 + x^4)/(x + log(x)))
 

3.11.81.6 Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {e^{x}}{\log {\left (\frac {10 x^{5} + 10 x^{4}}{x + \log {\left (x \right )}} \right )}} \end {dmath*}

input

integrate((((x**2+x)*exp(x)*ln(x)+(x**3+x**2)*exp(x))*ln((10*x**5+10*x**4) 
/(x+ln(x)))+(-5*x-4)*exp(x)*ln(x)+(-4*x**2-2*x+1)*exp(x))/((x**2+x)*ln(x)+ 
x**3+x**2)/ln((10*x**5+10*x**4)/(x+ln(x)))**2,x)
 

output

exp(x)/log((10*x**5 + 10*x**4)/(x + log(x)))
 

3.11.81.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {e^{x}}{\log \left (5\right ) + \log \left (2\right ) - \log \left (x + \log \left (x\right )\right ) + \log \left (x + 1\right ) + 4 \, \log \left (x\right )} \end {dmath*}

input

integrate((((x^2+x)*exp(x)*log(x)+(x^3+x^2)*exp(x))*log((10*x^5+10*x^4)/(x 
+log(x)))+(-5*x-4)*exp(x)*log(x)+(-4*x^2-2*x+1)*exp(x))/((x^2+x)*log(x)+x^ 
3+x^2)/log((10*x^5+10*x^4)/(x+log(x)))^2,x, algorithm=\
 

output

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

3.11.81.8 Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {e^{x}}{\log \left (10 \, x + 10\right ) - \log \left (x + \log \left (x\right )\right ) + 4 \, \log \left (x\right )} \end {dmath*}

input

integrate((((x^2+x)*exp(x)*log(x)+(x^3+x^2)*exp(x))*log((10*x^5+10*x^4)/(x 
+log(x)))+(-5*x-4)*exp(x)*log(x)+(-4*x^2-2*x+1)*exp(x))/((x^2+x)*log(x)+x^ 
3+x^2)/log((10*x^5+10*x^4)/(x+log(x)))^2,x, algorithm=\
 

output

e^x/(log(10*x + 10) - log(x + log(x)) + 4*log(x))
 

3.11.81.9 Mupad [B] (verification not implemented)

Time = 17.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx=\frac {{\mathrm {e}}^x}{\ln \left (\frac {10\,x^5+10\,x^4}{x+\ln \left (x\right )}\right )} \end {dmath*}

input

int(-(exp(x)*(2*x + 4*x^2 - 1) - log((10*x^4 + 10*x^5)/(x + log(x)))*(exp( 
x)*(x^2 + x^3) + exp(x)*log(x)*(x + x^2)) + exp(x)*log(x)*(5*x + 4))/(log( 
(10*x^4 + 10*x^5)/(x + log(x)))^2*(x^2 + x^3 + log(x)*(x + x^2))),x)
 

output

exp(x)/log((10*x^4 + 10*x^5)/(x + log(x)))