3.26.77 \(\int \frac {e^{x^2} (x+2 x^3) \log (x)+e^{x^2} (-6-12 x^2+(1+2 x^2) \log (5)) \log ^2(x)+e^{\log ^2(\frac {x+(-6+\log (5)) \log (x)}{\log (x)})} (-10+10 \log (x)) \log (\frac {x+(-6+\log (5)) \log (x)}{\log (x)})}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx\) [2577]

3.26.77.1 Optimal result

Integrand size = 104, antiderivative size = 26 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+\frac {e^{x^2} x}{5} \]

output

1/5*exp(x^2)*x+exp(ln(x/ln(x)-6+ln(5))^2)
 

3.26.77.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {1}{5} \left (5 e^{\log ^2\left (-6+\log (5)+\frac {x}{\log (x)}\right )}+e^{x^2} x\right ) \]

input

Integrate[(E^x^2*(x + 2*x^3)*Log[x] + E^x^2*(-6 - 12*x^2 + (1 + 2*x^2)*Log 
[5])*Log[x]^2 + E^Log[(x + (-6 + Log[5])*Log[x])/Log[x]]^2*(-10 + 10*Log[x 
])*Log[(x + (-6 + Log[5])*Log[x])/Log[x]])/(5*x*Log[x] + (-30 + 5*Log[5])* 
Log[x]^2),x]
 

output

(5*E^Log[-6 + Log[5] + x/Log[x]]^2 + E^x^2*x)/5
 

3.26.77.3 Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7292, 27, 7293, 2009}

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.

\(\displaystyle \int \frac {e^{x^2} \left (-12 x^2+\left (2 x^2+1\right ) \log (5)-6\right ) \log ^2(x)+e^{x^2} \left (2 x^3+x\right ) \log (x)+e^{\log ^2\left (\frac {x+(\log (5)-6) \log (x)}{\log (x)}\right )} (10 \log (x)-10) \log \left (\frac {x+(\log (5)-6) \log (x)}{\log (x)}\right )}{(5 \log (5)-30) \log ^2(x)+5 x \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{x^2} \left (-12 x^2+\left (2 x^2+1\right ) \log (5)-6\right ) \log ^2(x)+e^{x^2} \left (2 x^3+x\right ) \log (x)+e^{\log ^2\left (\frac {x+(\log (5)-6) \log (x)}{\log (x)}\right )} (10 \log (x)-10) \log \left (\frac {x+(\log (5)-6) \log (x)}{\log (x)}\right )}{5 \log (x) \left (x-6 \left (1-\frac {\log (5)}{6}\right ) \log (x)\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {-e^{x^2} \left (12 x^2-\left (2 x^2+1\right ) \log (5)+6\right ) \log ^2(x)+e^{x^2} \left (2 x^3+x\right ) \log (x)-10 e^{\log ^2\left (\frac {x-(6-\log (5)) \log (x)}{\log (x)}\right )} (1-\log (x)) \log \left (\frac {x-(6-\log (5)) \log (x)}{\log (x)}\right )}{\log (x) (x-(6-\log (5)) \log (x))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{5} \int \left (e^{x^2} \left (2 x^2+1\right )+\frac {10 e^{\log ^2\left (\frac {x}{\log (x)}-6 \left (1-\frac {\log (5)}{6}\right )\right )} (\log (x)-1) \log \left (\frac {x}{\log (x)}-6 \left (1-\frac {\log (5)}{6}\right )\right )}{\log (x) \left (x-6 \left (1-\frac {\log (5)}{6}\right ) \log (x)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (e^{x^2} x+5 e^{\log ^2\left (\frac {x}{\log (x)}-6+\log (5)\right )}\right )\)

input

Int[(E^x^2*(x + 2*x^3)*Log[x] + E^x^2*(-6 - 12*x^2 + (1 + 2*x^2)*Log[5])*L 
og[x]^2 + E^Log[(x + (-6 + Log[5])*Log[x])/Log[x]]^2*(-10 + 10*Log[x])*Log 
[(x + (-6 + Log[5])*Log[x])/Log[x]])/(5*x*Log[x] + (-30 + 5*Log[5])*Log[x] 
^2),x]
 

output

(5*E^Log[-6 + Log[5] + x/Log[x]]^2 + E^x^2*x)/5
 

3.26.77.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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] 
]
 

3.26.77.4 Maple [A] (verified)

Time = 101.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{x^{2}} x}{5}+{\mathrm e}^{\ln \left (\frac {\left (\ln \left (5\right )-6\right ) \ln \left (x \right )+x}{\ln \left (x \right )}\right )^{2}}\)	\(27\)
risch	\(\frac {{\mathrm e}^{x^{2}} x}{5}+{\mathrm e}^{\frac {\left (i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right )^{3} \pi -i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \pi -i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )\right ) \pi +i \operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )\right ) \pi -2 \ln \left (\ln \left (5\right ) \ln \left (x \right )-6 \ln \left (x \right )+x \right )+2 \ln \left (\ln \left (x \right )\right )\right )^{2}}{4}}\)	\(178\)

input

int(((10*ln(x)-10)*ln(((ln(5)-6)*ln(x)+x)/ln(x))*exp(ln(((ln(5)-6)*ln(x)+x 
)/ln(x))^2)+((2*x^2+1)*ln(5)-12*x^2-6)*exp(x^2)*ln(x)^2+(2*x^3+x)*exp(x^2) 
*ln(x))/((5*ln(5)-30)*ln(x)^2+5*x*ln(x)),x,method=_RETURNVERBOSE)
 

output

1/5*exp(x^2)*x+exp(ln(((ln(5)-6)*ln(x)+x)/ln(x))^2)
 

3.26.77.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {1}{5} \, x e^{\left (x^{2}\right )} + e^{\left (\log \left (\frac {{\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )^{2}\right )} \]

input

integrate(((10*log(x)-10)*log(((log(5)-6)*log(x)+x)/log(x))*exp(log(((log( 
5)-6)*log(x)+x)/log(x))^2)+((2*x^2+1)*log(5)-12*x^2-6)*exp(x^2)*log(x)^2+( 
2*x^3+x)*exp(x^2)*log(x))/((5*log(5)-30)*log(x)^2+5*x*log(x)),x, algorithm 
=\
 

output

1/5*x*e^(x^2) + e^(log(((log(5) - 6)*log(x) + x)/log(x))^2)
 

3.26.77.6 Sympy [A] (verification not implemented)

Time = 29.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {x e^{x^{2}}}{5} + e^{\log {\left (\frac {x + \left (-6 + \log {\left (5 \right )}\right ) \log {\left (x \right )}}{\log {\left (x \right )}} \right )}^{2}} \]

input

integrate(((10*ln(x)-10)*ln(((ln(5)-6)*ln(x)+x)/ln(x))*exp(ln(((ln(5)-6)*l 
n(x)+x)/ln(x))**2)+((2*x**2+1)*ln(5)-12*x**2-6)*exp(x**2)*ln(x)**2+(2*x**3 
+x)*exp(x**2)*ln(x))/((5*ln(5)-30)*ln(x)**2+5*x*ln(x)),x)
 

output

x*exp(x**2)/5 + exp(log((x + (-6 + log(5))*log(x))/log(x))**2)
 

3.26.77.7 Maxima [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\frac {1}{5} \, x e^{\left (x^{2}\right )} + e^{\left (\log \left ({\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x\right )^{2} - 2 \, \log \left ({\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right )} \]

input

integrate(((10*log(x)-10)*log(((log(5)-6)*log(x)+x)/log(x))*exp(log(((log( 
5)-6)*log(x)+x)/log(x))^2)+((2*x^2+1)*log(5)-12*x^2-6)*exp(x^2)*log(x)^2+( 
2*x^3+x)*exp(x^2)*log(x))/((5*log(5)-30)*log(x)^2+5*x*log(x)),x, algorithm 
=\
 

output

1/5*x*e^(x^2) + e^(log((log(5) - 6)*log(x) + x)^2 - 2*log((log(5) - 6)*log 
(x) + x)*log(log(x)) + log(log(x))^2)
 

3.26.77.8 Giac [F]

\[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx=\int { -\frac {{\left (12 \, x^{2} - {\left (2 \, x^{2} + 1\right )} \log \left (5\right ) + 6\right )} e^{\left (x^{2}\right )} \log \left (x\right )^{2} - {\left (2 \, x^{3} + x\right )} e^{\left (x^{2}\right )} \log \left (x\right ) - 10 \, {\left (\log \left (x\right ) - 1\right )} e^{\left (\log \left (\frac {{\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )^{2}\right )} \log \left (\frac {{\left (\log \left (5\right ) - 6\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )}{5 \, {\left ({\left (\log \left (5\right ) - 6\right )} \log \left (x\right )^{2} + x \log \left (x\right )\right )}} \,d x } \]

input

integrate(((10*log(x)-10)*log(((log(5)-6)*log(x)+x)/log(x))*exp(log(((log( 
5)-6)*log(x)+x)/log(x))^2)+((2*x^2+1)*log(5)-12*x^2-6)*exp(x^2)*log(x)^2+( 
2*x^3+x)*exp(x^2)*log(x))/((5*log(5)-30)*log(x)^2+5*x*log(x)),x, algorithm 
=\
 

output

undef
 

3.26.77.9 Mupad [B] (verification not implemented)

Time = 11.97 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{x^2} \left (x+2 x^3\right ) \log (x)+e^{x^2} \left (-6-12 x^2+\left (1+2 x^2\right ) \log (5)\right ) \log ^2(x)+e^{\log ^2\left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )} (-10+10 \log (x)) \log \left (\frac {x+(-6+\log (5)) \log (x)}{\log (x)}\right )}{5 x \log (x)+(-30+5 \log (5)) \log ^2(x)} \, dx={\mathrm {e}}^{{\ln \left (\frac {x-6\,\ln \left (x\right )+\ln \left (5\right )\,\ln \left (x\right )}{\ln \left (x\right )}\right )}^2}+\frac {x\,{\mathrm {e}}^{x^2}}{5} \]

input

int((exp(x^2)*log(x)*(x + 2*x^3) + exp(log((x + log(x)*(log(5) - 6))/log(x 
))^2)*log((x + log(x)*(log(5) - 6))/log(x))*(10*log(x) - 10) - exp(x^2)*lo 
g(x)^2*(12*x^2 - log(5)*(2*x^2 + 1) + 6))/(5*x*log(x) + log(x)^2*(5*log(5) 
 - 30)),x)
 

output

exp(log((x - 6*log(x) + log(5)*log(x))/log(x))^2) + (x*exp(x^2))/5