3.1.4 \(\int \frac {25^{-\frac {1}{\log (\frac {2+x}{2})}} (25 x+e^{2 x} x)^{\frac {1}{\log (\frac {2+x}{2})}} ((50+25 x+e^{2 x} (2+5 x+2 x^2)) \log (\frac {2+x}{2})+(-25 x-e^{2 x} x) \log (\frac {1}{25} (25 x+e^{2 x} x)))}{(50 x+25 x^2+e^{2 x} (2 x+x^2)) \log ^2(\frac {2+x}{2})} \, dx\) [4]

3.1.4.1 Optimal result

Integrand size = 131, antiderivative size = 23 \[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx=\left (x+\frac {1}{25} e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \]

output

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

3.1.4.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx=\left (x+\frac {1}{25} e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \]

input

Integrate[((25*x + E^(2*x)*x)^Log[(2 + x)/2]^(-1)*((50 + 25*x + E^(2*x)*(2 
 + 5*x + 2*x^2))*Log[(2 + x)/2] + (-25*x - E^(2*x)*x)*Log[(25*x + E^(2*x)* 
x)/25]))/(25^Log[(2 + x)/2]^(-1)*(50*x + 25*x^2 + E^(2*x)*(2*x + x^2))*Log 
[(2 + x)/2]^2),x]
 

output

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

3.1.4.3 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.

\(\displaystyle \int \frac {25^{-\frac {1}{\log \left (\frac {x+2}{2}\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x+2}{2}\right )}} \left (\left (e^{2 x} \left (2 x^2+5 x+2\right )+25 x+50\right ) \log \left (\frac {x+2}{2}\right )+\left (-e^{2 x} x-25 x\right ) \log \left (\frac {1}{25} \left (e^{2 x} x+25 x\right )\right )\right )}{\left (25 x^2+e^{2 x} \left (x^2+2 x\right )+50 x\right ) \log ^2\left (\frac {x+2}{2}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {25^{-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (\left (e^{2 x} \left (2 x^2+5 x+2\right )+25 x+50\right ) \log \left (\frac {x+2}{2}\right )+\left (-e^{2 x} x-25 x\right ) \log \left (\frac {1}{25} \left (e^{2 x} x+25 x\right )\right )\right )}{\left (e^{2 x}+25\right ) x (x+2) \log ^2\left (\frac {x}{2}+1\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {25^{-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (2 x^2 \log \left (\frac {x+2}{2}\right )+5 x \log \left (\frac {x+2}{2}\right )-x \log \left (\frac {1}{25} e^{2 x} x+x\right )+2 \log \left (\frac {x+2}{2}\right )\right ) \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}}}{x (x+2) \log ^2\left (\frac {x}{2}+1\right )}+\frac {2\ 25^{1-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}}}{\left (-e^{2 x}-25\right ) \log \left (\frac {x}{2}+1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {25^{-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}} \log \left (\frac {1}{25} \left (25+e^{2 x}\right ) x\right )}{(-x-2) \log ^2\left (\frac {x}{2}+1\right )}dx+2 \int \frac {25^{-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}}}{\log \left (\frac {x}{2}+1\right )}dx+2 \int \frac {25^{1-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}}}{\left (-25-e^{2 x}\right ) \log \left (\frac {x}{2}+1\right )}dx+\int \frac {25^{-\frac {1}{\log \left (\frac {x}{2}+1\right )}} \left (e^{2 x} x+25 x\right )^{\frac {1}{\log \left (\frac {x}{2}+1\right )}}}{x \log \left (\frac {x}{2}+1\right )}dx\)

input

Int[((25*x + E^(2*x)*x)^Log[(2 + x)/2]^(-1)*((50 + 25*x + E^(2*x)*(2 + 5*x 
 + 2*x^2))*Log[(2 + x)/2] + (-25*x - E^(2*x)*x)*Log[(25*x + E^(2*x)*x)/25] 
))/(25^Log[(2 + x)/2]^(-1)*(50*x + 25*x^2 + E^(2*x)*(2*x + x^2))*Log[(2 + 
x)/2]^2),x]
 

output

$Aborted
 

3.1.4.3.1 Defintions of rubi rules used

rule 2009

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

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] 
]
 

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

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

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 5.48

\[{\mathrm e}^{-\frac {-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}+25\right )\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}+25\right )\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right ) \operatorname {csgn}\left (i x \right )+i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right )}^{3}-i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right )}^{2} \operatorname {csgn}\left (i x \right )+4 \ln \left (5\right )-2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{2 x}+25\right )}{2 \ln \left (1+\frac {x}{2}\right )}}\]

input

int(((-x*exp(x)^2-25*x)*ln(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x)^2+25*x 
+50)*ln(1+1/2*x))*exp(ln(1/25*x*exp(x)^2+x)/ln(1+1/2*x))/((x^2+2*x)*exp(x) 
^2+25*x^2+50*x)/ln(1+1/2*x)^2,x)
 

output

exp(-1/2*(-I*Pi*csgn(I*(exp(2*x)+25))*csgn(I*x*(exp(2*x)+25))^2+I*Pi*csgn( 
I*(exp(2*x)+25))*csgn(I*x*(exp(2*x)+25))*csgn(I*x)+I*Pi*csgn(I*x*(exp(2*x) 
+25))^3-I*Pi*csgn(I*x*(exp(2*x)+25))^2*csgn(I*x)+4*ln(5)-2*ln(x)-2*ln(exp( 
2*x)+25))/ln(1+1/2*x))
 

3.1.4.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx={\left (\frac {1}{25} \, x e^{\left (2 \, x\right )} + x\right )}^{\left (\frac {1}{\log \left (\frac {1}{2} \, x + 1\right )}\right )} \]

input

integrate(((-x*exp(x)^2-25*x)*log(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x) 
^2+25*x+50)*log(1+1/2*x))*exp(log(1/25*x*exp(x)^2+x)/log(1+1/2*x))/((x^2+2 
*x)*exp(x)^2+25*x^2+50*x)/log(1+1/2*x)^2,x, algorithm=\
 

output

(1/25*x*e^(2*x) + x)^(1/log(1/2*x + 1))
 

3.1.4.6 Sympy [F(-1)]

Timed out. \[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx=\text {Timed out} \]

input

integrate(((-x*exp(x)**2-25*x)*ln(1/25*x*exp(x)**2+x)+((2*x**2+5*x+2)*exp( 
x)**2+25*x+50)*ln(1+1/2*x))*exp(ln(1/25*x*exp(x)**2+x)/ln(1+1/2*x))/((x**2 
+2*x)*exp(x)**2+25*x**2+50*x)/ln(1+1/2*x)**2,x)
 

output

Timed out
 

3.1.4.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (18) = 36\).

Time = 0.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.26 \[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx=e^{\left (\frac {2 \, \log \left (5\right )}{\log \left (2\right ) - \log \left (x + 2\right )} - \frac {\log \left (x\right )}{\log \left (2\right ) - \log \left (x + 2\right )} - \frac {\log \left (e^{\left (2 \, x\right )} + 25\right )}{\log \left (2\right ) - \log \left (x + 2\right )}\right )} \]

input

integrate(((-x*exp(x)^2-25*x)*log(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x) 
^2+25*x+50)*log(1+1/2*x))*exp(log(1/25*x*exp(x)^2+x)/log(1+1/2*x))/((x^2+2 
*x)*exp(x)^2+25*x^2+50*x)/log(1+1/2*x)^2,x, algorithm=\
 

output

e^(2*log(5)/(log(2) - log(x + 2)) - log(x)/(log(2) - log(x + 2)) - log(e^( 
2*x) + 25)/(log(2) - log(x + 2)))
 

3.1.4.8 Giac [F]

\[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx=\int { -\frac {{\left ({\left (x e^{\left (2 \, x\right )} + 25 \, x\right )} \log \left (\frac {1}{25} \, x e^{\left (2 \, x\right )} + x\right ) - {\left ({\left (2 \, x^{2} + 5 \, x + 2\right )} e^{\left (2 \, x\right )} + 25 \, x + 50\right )} \log \left (\frac {1}{2} \, x + 1\right )\right )} {\left (\frac {1}{25} \, x e^{\left (2 \, x\right )} + x\right )}^{\left (\frac {1}{\log \left (\frac {1}{2} \, x + 1\right )}\right )}}{{\left (25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} e^{\left (2 \, x\right )} + 50 \, x\right )} \log \left (\frac {1}{2} \, x + 1\right )^{2}} \,d x } \]

input

integrate(((-x*exp(x)^2-25*x)*log(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x) 
^2+25*x+50)*log(1+1/2*x))*exp(log(1/25*x*exp(x)^2+x)/log(1+1/2*x))/((x^2+2 
*x)*exp(x)^2+25*x^2+50*x)/log(1+1/2*x)^2,x, algorithm=\
 

output

integrate(-((x*e^(2*x) + 25*x)*log(1/25*x*e^(2*x) + x) - ((2*x^2 + 5*x + 2 
)*e^(2*x) + 25*x + 50)*log(1/2*x + 1))*(1/25*x*e^(2*x) + x)^(1/log(1/2*x + 
 1))/((25*x^2 + (x^2 + 2*x)*e^(2*x) + 50*x)*log(1/2*x + 1)^2), x)
 

3.1.4.9 Mupad [B] (verification not implemented)

Time = 12.89 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {25^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx={\left (x+\frac {x\,{\mathrm {e}}^{2\,x}}{25}\right )}^{\frac {1}{\ln \left (\frac {x}{2}+1\right )}} \]

input

int(-(exp(log(x + (x*exp(2*x))/25)/log(x/2 + 1))*(log(x + (x*exp(2*x))/25) 
*(25*x + x*exp(2*x)) - log(x/2 + 1)*(25*x + exp(2*x)*(5*x + 2*x^2 + 2) + 5 
0)))/(log(x/2 + 1)^2*(50*x + exp(2*x)*(2*x + x^2) + 25*x^2)),x)
 

output

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