3.2.59 \(\int \frac {-6 x+e^{3+x^2} (-1-10 x^2)+(-x-2 e^{3+x^2} x^2) \log (x)+(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x)) \log (5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x))}{(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x)) \log ^2(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x))} \, dx\) [159]

3.2.59.1 Optimal result

Integrand size = 146, antiderivative size = 21 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=-5+\frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \]

output

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

3.2.59.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \]

input

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

output

x/Log[(E^(3 + x^2) + x)*(5 + Log[x])]
 

3.2.59.3 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.2.59.3.1 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] 
]
 

3.2.59.4 Maple [A] (verified)

Time = 1.95 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38

method	result	size
parallelrisch	\(\frac {x}{\ln \left (\left ({\mathrm e}^{x^{2}+3}+x \right ) \ln \left (x \right )+5 \,{\mathrm e}^{x^{2}+3}+5 x \right )}\)	\(29\)
risch	\(\frac {2 i x}{\pi \,\operatorname {csgn}\left (i \left (5+\ln \left (x \right )\right )\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )-\pi \,\operatorname {csgn}\left (i \left (5+\ln \left (x \right )\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )}^{2}+\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )}^{3}+2 i \ln \left (5+\ln \left (x \right )\right )+2 i \ln \left ({\mathrm e}^{x^{2}+3}+x \right )}\)	\(148\)

input

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

output

x/ln((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)
 

3.2.59.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left ({\left (x + e^{\left (x^{2} + 3\right )}\right )} \log \left (x\right ) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \]

input

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

output

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

3.2.59.6 Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log {\left (5 x + \left (x + e^{x^{2} + 3}\right ) \log {\left (x \right )} + 5 e^{x^{2} + 3} \right )}} \]

input

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

output

x/log(5*x + (x + exp(x**2 + 3))*log(x) + 5*exp(x**2 + 3))
 

3.2.59.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left (x + e^{\left (x^{2} + 3\right )}\right ) + \log \left (\log \left (x\right ) + 5\right )} \]

input

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

output

x/(log(x + e^(x^2 + 3)) + log(log(x) + 5))
 

3.2.59.8 Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left (x \log \left (x\right ) + e^{\left (x^{2} + 3\right )} \log \left (x\right ) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \]

input

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

output

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

3.2.59.9 Mupad [F(-1)]

Timed out. \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\int -\frac {6\,x+{\mathrm {e}}^{x^2+3}\,\left (10\,x^2+1\right )-\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )+\ln \left (x\right )\,\left (x+2\,x^2\,{\mathrm {e}}^{x^2+3}\right )}{{\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )}^2\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )} \,d x \]

input

int(-(6*x + exp(x^2 + 3)*(10*x^2 + 1) - log(5*x + 5*exp(x^2 + 3) + log(x)* 
(x + exp(x^2 + 3)))*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3))) + l 
og(x)*(x + 2*x^2*exp(x^2 + 3)))/(log(5*x + 5*exp(x^2 + 3) + log(x)*(x + ex 
p(x^2 + 3)))^2*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))),x)
 

output

int(-(6*x + exp(x^2 + 3)*(10*x^2 + 1) - log(5*x + 5*exp(x^2 + 3) + log(x)* 
(x + exp(x^2 + 3)))*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3))) + l 
og(x)*(x + 2*x^2*exp(x^2 + 3)))/(log(5*x + 5*exp(x^2 + 3) + log(x)*(x + ex 
p(x^2 + 3)))^2*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))), x)