3.7.38 \(\int \frac {e^{x+x^2} (9+24 x+13 x^2+2 x^3)+e^{x+x^2} (6+14 x+4 x^2) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x))}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx\) [638]

3.7.38.1 Optimal result

Integrand size = 163, antiderivative size = 31 \begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx=-e^{4+e^{1-\frac {16}{3+x+\log (5 x)}} x}+e^{x+x^2} \end {dmath*}

output

exp(x^2+x)-exp(x*exp(1-16/(ln(5*x)+3+x))+4)
 

3.7.38.2 Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx=-e^{4+e^{1-\frac {16}{3+x+\log (5 x)}} x}+e^{x+x^2} \end {dmath*}

input

Integrate[(E^(x + x^2)*(9 + 24*x + 13*x^2 + 2*x^3) + E^(x + x^2)*(6 + 14*x 
 + 4*x^2)*Log[5*x] + E^(x + x^2)*(1 + 2*x)*Log[5*x]^2 + E^(4 + E^((-13 + x 
 + Log[5*x])/(3 + x + Log[5*x]))*x + (-13 + x + Log[5*x])/(3 + x + Log[5*x 
]))*(-25 - 22*x - x^2 + (-6 - 2*x)*Log[5*x] - Log[5*x]^2))/(9 + 6*x + x^2 
+ (6 + 2*x)*Log[5*x] + Log[5*x]^2),x]
 

output

-E^(4 + E^(1 - 16/(3 + x + Log[5*x]))*x) + E^(x + x^2)
 

3.7.38.3 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.

\(\displaystyle \int \frac {\left (-x^2-22 x-\log ^2(5 x)+(-2 x-6) \log (5 x)-25\right ) \exp \left (x e^{\frac {x+\log (5 x)-13}{x+\log (5 x)+3}}+\frac {x+\log (5 x)-13}{x+\log (5 x)+3}+4\right )+e^{x^2+x} (2 x+1) \log ^2(5 x)+e^{x^2+x} \left (4 x^2+14 x+6\right ) \log (5 x)+e^{x^2+x} \left (2 x^3+13 x^2+24 x+9\right )}{x^2+6 x+\log ^2(5 x)+(2 x+6) \log (5 x)+9} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-x^2-22 x-\log ^2(5 x)+(-2 x-6) \log (5 x)-25\right ) \exp \left (x e^{\frac {x+\log (5 x)-13}{x+\log (5 x)+3}}+\frac {x+\log (5 x)-13}{x+\log (5 x)+3}+4\right )+e^{x^2+x} (2 x+1) \log ^2(5 x)+e^{x^2+x} \left (4 x^2+14 x+6\right ) \log (5 x)+e^{x^2+x} \left (2 x^3+13 x^2+24 x+9\right )}{(x+\log (5 x)+3)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {5^{\frac {5}{x+\log (5 x)+3}} \left (x^2+22 x+\log ^2(5 x)+2 x \log (5 x)+6 \log (5 x)+25\right ) x^{\frac {5}{x+\log (5 x)+3}} \exp \left (5^{\frac {1}{x+\log (5 x)+3}} e^{\frac {x}{x+\log (5 x)+3}-\frac {13}{x+\log (5 x)+3}} x^{\frac {1}{x+\log (5 x)+3}+1}+\frac {5 x}{x+\log (5 x)+3}-\frac {1}{x+\log (5 x)+3}\right )}{(x+\log (5 x)+3)^2}+\frac {e^{x^2+x} (2 x+1) \log ^2(5 x)}{(x+\log (5 x)+3)^2}+\frac {e^{x^2+x} (x+3)^2 (2 x+1)}{(x+\log (5 x)+3)^2}+\frac {2 e^{x^2+x} (x+3) (2 x+1) \log (5 x)}{(x+\log (5 x)+3)^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (-\frac {5^{\frac {5}{x+\log (5 x)+3}} \left (x^2+22 x+\log ^2(5 x)+2 x \log (5 x)+6 \log (5 x)+25\right ) x^{\frac {5}{x+\log (5 x)+3}} \exp \left (5^{\frac {1}{x+\log (5 x)+3}} e^{\frac {x}{x+\log (5 x)+3}-\frac {13}{x+\log (5 x)+3}} x^{\frac {1}{x+\log (5 x)+3}+1}+\frac {5 x}{x+\log (5 x)+3}-\frac {1}{x+\log (5 x)+3}\right )}{(x+\log (5 x)+3)^2}+\frac {e^{x^2+x} (2 x+1) \log ^2(5 x)}{(x+\log (5 x)+3)^2}+\frac {e^{x^2+x} (x+3)^2 (2 x+1)}{(x+\log (5 x)+3)^2}+\frac {2 e^{x^2+x} (x+3) (2 x+1) \log (5 x)}{(x+\log (5 x)+3)^2}\right )dx\)

input

Int[(E^(x + x^2)*(9 + 24*x + 13*x^2 + 2*x^3) + E^(x + x^2)*(6 + 14*x + 4*x 
^2)*Log[5*x] + E^(x + x^2)*(1 + 2*x)*Log[5*x]^2 + E^(4 + E^((-13 + x + Log 
[5*x])/(3 + x + Log[5*x]))*x + (-13 + x + Log[5*x])/(3 + x + Log[5*x]))*(- 
25 - 22*x - x^2 + (-6 - 2*x)*Log[5*x] - Log[5*x]^2))/(9 + 6*x + x^2 + (6 + 
 2*x)*Log[5*x] + Log[5*x]^2),x]
 

output

$Aborted
 

3.7.38.3.1 Defintions of rubi rules used

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] 
]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

3.7.38.4 Maple [A] (verified)

Time = 9.72 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method	result	size
risch	\({\mathrm e}^{\left (1+x \right ) x}-{\mathrm e}^{x \,{\mathrm e}^{\frac {\ln \left (5 x \right )+x -13}{\ln \left (5 x \right )+3+x}}+4}\)	\(33\)
parallelrisch	\({\mathrm e}^{x^{2}+x}-{\mathrm e}^{x \,{\mathrm e}^{\frac {\ln \left (5 x \right )+x -13}{\ln \left (5 x \right )+3+x}}+4}\)	\(33\)

input

int(((-ln(5*x)^2+(-2*x-6)*ln(5*x)-x^2-22*x-25)*exp((ln(5*x)+x-13)/(ln(5*x) 
+3+x))*exp(x*exp((ln(5*x)+x-13)/(ln(5*x)+3+x))+4)+(1+2*x)*exp(x^2+x)*ln(5* 
x)^2+(4*x^2+14*x+6)*exp(x^2+x)*ln(5*x)+(2*x^3+13*x^2+24*x+9)*exp(x^2+x))/( 
ln(5*x)^2+(2*x+6)*ln(5*x)+x^2+6*x+9),x,method=_RETURNVERBOSE)
 

output

exp((1+x)*x)-exp(x*exp((ln(5*x)+x-13)/(ln(5*x)+3+x))+4)
 

3.7.38.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (28) = 56\).

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.23 \begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx={\left (e^{\left (x^{2} + x + \frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} - e^{\left (\frac {{\left (x^{2} + x \log \left (5 \, x\right ) + 3 \, x\right )} e^{\left (\frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} + 5 \, x + 5 \, \log \left (5 \, x\right ) - 1}{x + \log \left (5 \, x\right ) + 3}\right )}\right )} e^{\left (-\frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} \end {dmath*}

input

integrate(((-log(5*x)^2+(-2*x-6)*log(5*x)-x^2-22*x-25)*exp((log(5*x)+x-13) 
/(log(5*x)+3+x))*exp(x*exp((log(5*x)+x-13)/(log(5*x)+3+x))+4)+(1+2*x)*exp( 
x^2+x)*log(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*log(5*x)+(2*x^3+13*x^2+24*x+9) 
*exp(x^2+x))/(log(5*x)^2+(2*x+6)*log(5*x)+x^2+6*x+9),x, algorithm=\
 

output

(e^(x^2 + x + (x + log(5*x) - 13)/(x + log(5*x) + 3)) - e^(((x^2 + x*log(5 
*x) + 3*x)*e^((x + log(5*x) - 13)/(x + log(5*x) + 3)) + 5*x + 5*log(5*x) - 
 1)/(x + log(5*x) + 3)))*e^(-(x + log(5*x) - 13)/(x + log(5*x) + 3))
 

3.7.38.6 Sympy [A] (verification not implemented)

Time = 15.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx=e^{x^{2} + x} - e^{x e^{\frac {x + \log {\left (5 x \right )} - 13}{x + \log {\left (5 x \right )} + 3}} + 4} \end {dmath*}

input

integrate(((-ln(5*x)**2+(-2*x-6)*ln(5*x)-x**2-22*x-25)*exp((ln(5*x)+x-13)/ 
(ln(5*x)+3+x))*exp(x*exp((ln(5*x)+x-13)/(ln(5*x)+3+x))+4)+(1+2*x)*exp(x**2 
+x)*ln(5*x)**2+(4*x**2+14*x+6)*exp(x**2+x)*ln(5*x)+(2*x**3+13*x**2+24*x+9) 
*exp(x**2+x))/(ln(5*x)**2+(2*x+6)*ln(5*x)+x**2+6*x+9),x)
 

output

exp(x**2 + x) - exp(x*exp((x + log(5*x) - 13)/(x + log(5*x) + 3)) + 4)
 

3.7.38.7 Maxima [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx=e^{\left (x^{2} + x\right )} - e^{\left (x e^{\left (-\frac {16}{x + \log \left (5\right ) + \log \left (x\right ) + 3} + 1\right )} + 4\right )} \end {dmath*}

input

integrate(((-log(5*x)^2+(-2*x-6)*log(5*x)-x^2-22*x-25)*exp((log(5*x)+x-13) 
/(log(5*x)+3+x))*exp(x*exp((log(5*x)+x-13)/(log(5*x)+3+x))+4)+(1+2*x)*exp( 
x^2+x)*log(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*log(5*x)+(2*x^3+13*x^2+24*x+9) 
*exp(x^2+x))/(log(5*x)^2+(2*x+6)*log(5*x)+x^2+6*x+9),x, algorithm=\
 

output

e^(x^2 + x) - e^(x*e^(-16/(x + log(5) + log(x) + 3) + 1) + 4)
 

3.7.38.8 Giac [F]

\begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx=\int { \frac {{\left (2 \, x + 1\right )} e^{\left (x^{2} + x\right )} \log \left (5 \, x\right )^{2} + 2 \, {\left (2 \, x^{2} + 7 \, x + 3\right )} e^{\left (x^{2} + x\right )} \log \left (5 \, x\right ) + {\left (2 \, x^{3} + 13 \, x^{2} + 24 \, x + 9\right )} e^{\left (x^{2} + x\right )} - {\left (x^{2} + 2 \, {\left (x + 3\right )} \log \left (5 \, x\right ) + \log \left (5 \, x\right )^{2} + 22 \, x + 25\right )} e^{\left (x e^{\left (\frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3}\right )} + \frac {x + \log \left (5 \, x\right ) - 13}{x + \log \left (5 \, x\right ) + 3} + 4\right )}}{x^{2} + 2 \, {\left (x + 3\right )} \log \left (5 \, x\right ) + \log \left (5 \, x\right )^{2} + 6 \, x + 9} \,d x } \end {dmath*}

input

integrate(((-log(5*x)^2+(-2*x-6)*log(5*x)-x^2-22*x-25)*exp((log(5*x)+x-13) 
/(log(5*x)+3+x))*exp(x*exp((log(5*x)+x-13)/(log(5*x)+3+x))+4)+(1+2*x)*exp( 
x^2+x)*log(5*x)^2+(4*x^2+14*x+6)*exp(x^2+x)*log(5*x)+(2*x^3+13*x^2+24*x+9) 
*exp(x^2+x))/(log(5*x)^2+(2*x+6)*log(5*x)+x^2+6*x+9),x, algorithm=\
 

output

integrate(((2*x + 1)*e^(x^2 + x)*log(5*x)^2 + 2*(2*x^2 + 7*x + 3)*e^(x^2 + 
 x)*log(5*x) + (2*x^3 + 13*x^2 + 24*x + 9)*e^(x^2 + x) - (x^2 + 2*(x + 3)* 
log(5*x) + log(5*x)^2 + 22*x + 25)*e^(x*e^((x + log(5*x) - 13)/(x + log(5* 
x) + 3)) + (x + log(5*x) - 13)/(x + log(5*x) + 3) + 4))/(x^2 + 2*(x + 3)*l 
og(5*x) + log(5*x)^2 + 6*x + 9), x)
 

3.7.38.9 Mupad [B] (verification not implemented)

Time = 14.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \begin {dmath*} \int \frac {e^{x+x^2} \left (9+24 x+13 x^2+2 x^3\right )+e^{x+x^2} \left (6+14 x+4 x^2\right ) \log (5 x)+e^{x+x^2} (1+2 x) \log ^2(5 x)+e^{4+e^{\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} x+\frac {-13+x+\log (5 x)}{3+x+\log (5 x)}} \left (-25-22 x-x^2+(-6-2 x) \log (5 x)-\log ^2(5 x)\right )}{9+6 x+x^2+(6+2 x) \log (5 x)+\log ^2(5 x)} \, dx={\mathrm {e}}^{x^2+x}-{\mathrm {e}}^{5^{\frac {1}{x+\ln \left (5\,x\right )+3}}\,x\,x^{\frac {1}{x+\ln \left (5\,x\right )+3}}\,{\mathrm {e}}^{-\frac {13}{x+\ln \left (5\,x\right )+3}}\,{\mathrm {e}}^{\frac {x}{x+\ln \left (5\,x\right )+3}}+4} \end {dmath*}

input

int((exp(x + x^2)*(24*x + 13*x^2 + 2*x^3 + 9) + log(5*x)*exp(x + x^2)*(14* 
x + 4*x^2 + 6) - exp((x + log(5*x) - 13)/(x + log(5*x) + 3))*exp(x*exp((x 
+ log(5*x) - 13)/(x + log(5*x) + 3)) + 4)*(22*x + log(5*x)^2 + x^2 + log(5 
*x)*(2*x + 6) + 25) + log(5*x)^2*exp(x + x^2)*(2*x + 1))/(6*x + log(5*x)^2 
 + x^2 + log(5*x)*(2*x + 6) + 9),x)
 

output

exp(x + x^2) - exp(5^(1/(x + log(5*x) + 3))*x*x^(1/(x + log(5*x) + 3))*exp 
(-13/(x + log(5*x) + 3))*exp(x/(x + log(5*x) + 3)) + 4)