3.3.2 \(\int \frac {16+8 x+e^{5 x^2} (4+4 x-40 x^2-40 x^3)+(4+4 x) \log (\frac {1+2 x+x^2}{9 x^4})}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log (\frac {1+2 x+x^2}{9 x^4})+(1+x) \log ^2(\frac {1+2 x+x^2}{9 x^4})} \, dx\) [202]

3.3.2.1 Optimal result

Integrand size = 116, antiderivative size = 26 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 x}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \]

output

4/(ln(1/9*(1+x)^2/x^4)+exp(5*x^2))*x
 

3.3.2.2 Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 x}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \]

input

Integrate[(16 + 8*x + E^(5*x^2)*(4 + 4*x - 40*x^2 - 40*x^3) + (4 + 4*x)*Lo 
g[(1 + 2*x + x^2)/(9*x^4)])/(E^(10*x^2)*(1 + x) + E^(5*x^2)*(2 + 2*x)*Log[ 
(1 + 2*x + x^2)/(9*x^4)] + (1 + x)*Log[(1 + 2*x + x^2)/(9*x^4)]^2),x]
 

output

(4*x)/(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])
 

3.3.2.3 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.

\(\displaystyle \int \frac {(4 x+4) \log \left (\frac {x^2+2 x+1}{9 x^4}\right )+e^{5 x^2} \left (-40 x^3-40 x^2+4 x+4\right )+8 x+16}{e^{10 x^2} (x+1)+(x+1) \log ^2\left (\frac {x^2+2 x+1}{9 x^4}\right )+e^{5 x^2} (2 x+2) \log \left (\frac {x^2+2 x+1}{9 x^4}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {(4 x+4) \log \left (\frac {x^2+2 x+1}{9 x^4}\right )+e^{5 x^2} \left (-40 x^3-40 x^2+4 x+4\right )+8 x+16}{(x+1) \left (\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {8 \left (5 x^3 \log \left (\frac {(x+1)^2}{9 x^4}\right )+5 x^2 \log \left (\frac {(x+1)^2}{9 x^4}\right )+x+2\right )}{(x+1) \left (\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}\right )^2}-\frac {4 \left (10 x^2-1\right )}{\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 8 \int \frac {1}{\left (\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}\right )^2}dx+8 \int \frac {1}{(x+1) \left (\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}\right )^2}dx+40 \int \frac {x^2 \log \left (\frac {(x+1)^2}{9 x^4}\right )}{\left (\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}\right )^2}dx+4 \int \frac {1}{\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}}dx-40 \int \frac {x^2}{\log \left (\frac {(x+1)^2}{9 x^4}\right )+e^{5 x^2}}dx\)

input

Int[(16 + 8*x + E^(5*x^2)*(4 + 4*x - 40*x^2 - 40*x^3) + (4 + 4*x)*Log[(1 + 
 2*x + x^2)/(9*x^4)])/(E^(10*x^2)*(1 + x) + E^(5*x^2)*(2 + 2*x)*Log[(1 + 2 
*x + x^2)/(9*x^4)] + (1 + x)*Log[(1 + 2*x + x^2)/(9*x^4)]^2),x]
 

output

$Aborted
 

3.3.2.3.1 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] 
]
 

3.3.2.4 Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {4 x}{{\mathrm e}^{5 x^{2}}+\ln \left (\frac {x^{2}+2 x +1}{9 x^{4}}\right )}\)	\(27\)
risch	\(-\frac {8 i x}{8 i \ln \left (x \right )+4 i \ln \left (3\right )-4 i \ln \left (1+x \right )-2 i {\mathrm e}^{5 x^{2}}+\pi \,\operatorname {csgn}\left (\frac {i}{x^{4}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}-\pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i \left (1+x \right )\right )^{2} \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )-\pi \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )^{3}+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-\pi \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x^{4}}\right ) \operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )+\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right )}\)	\(366\)

input

int(((4+4*x)*ln(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x 
+16)/((1+x)*ln(1/9*(x^2+2*x+1)/x^4)^2+(2+2*x)*exp(5*x^2)*ln(1/9*(x^2+2*x+1 
)/x^4)+(1+x)*exp(5*x^2)^2),x,method=_RETURNVERBOSE)
 

output

4*x/(exp(5*x^2)+ln(1/9*(x^2+2*x+1)/x^4))
 

3.3.2.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 \, x}{e^{\left (5 \, x^{2}\right )} + \log \left (\frac {x^{2} + 2 \, x + 1}{9 \, x^{4}}\right )} \]

input

integrate(((4+4*x)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x 
^2)+8*x+16)/((1+x)*log(1/9*(x^2+2*x+1)/x^4)^2+(2+2*x)*exp(5*x^2)*log(1/9*( 
x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x, algorithm=\
 

output

4*x/(e^(5*x^2) + log(1/9*(x^2 + 2*x + 1)/x^4))
 

3.3.2.6 Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 x}{e^{5 x^{2}} + \log {\left (\frac {\frac {x^{2}}{9} + \frac {2 x}{9} + \frac {1}{9}}{x^{4}} \right )}} \]

input

integrate(((4+4*x)*ln(1/9*(x**2+2*x+1)/x**4)+(-40*x**3-40*x**2+4*x+4)*exp( 
5*x**2)+8*x+16)/((1+x)*ln(1/9*(x**2+2*x+1)/x**4)**2+(2+2*x)*exp(5*x**2)*ln 
(1/9*(x**2+2*x+1)/x**4)+(1+x)*exp(5*x**2)**2),x)
 

output

4*x/(exp(5*x**2) + log((x**2/9 + 2*x/9 + 1/9)/x**4))
 

3.3.2.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 \, x}{e^{\left (5 \, x^{2}\right )} - 2 \, \log \left (3\right ) + 2 \, \log \left (x + 1\right ) - 4 \, \log \left (x\right )} \]

input

integrate(((4+4*x)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x 
^2)+8*x+16)/((1+x)*log(1/9*(x^2+2*x+1)/x^4)^2+(2+2*x)*exp(5*x^2)*log(1/9*( 
x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x, algorithm=\
 

output

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

3.3.2.8 Giac [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 \, x}{e^{\left (5 \, x^{2}\right )} + \log \left (\frac {x^{2} + 2 \, x + 1}{9 \, x^{4}}\right )} \]

input

integrate(((4+4*x)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x 
^2)+8*x+16)/((1+x)*log(1/9*(x^2+2*x+1)/x^4)^2+(2+2*x)*exp(5*x^2)*log(1/9*( 
x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x, algorithm=\
 

output

4*x/(e^(5*x^2) + log(1/9*(x^2 + 2*x + 1)/x^4))
 

3.3.2.9 Mupad [F(-1)]

Timed out. \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\int \frac {8\,x+{\mathrm {e}}^{5\,x^2}\,\left (-40\,x^3-40\,x^2+4\,x+4\right )+\ln \left (\frac {\frac {x^2}{9}+\frac {2\,x}{9}+\frac {1}{9}}{x^4}\right )\,\left (4\,x+4\right )+16}{\left (x+1\right )\,{\ln \left (\frac {\frac {x^2}{9}+\frac {2\,x}{9}+\frac {1}{9}}{x^4}\right )}^2+{\mathrm {e}}^{5\,x^2}\,\left (2\,x+2\right )\,\ln \left (\frac {\frac {x^2}{9}+\frac {2\,x}{9}+\frac {1}{9}}{x^4}\right )+{\mathrm {e}}^{10\,x^2}\,\left (x+1\right )} \,d x \]

input

int((8*x + exp(5*x^2)*(4*x - 40*x^2 - 40*x^3 + 4) + log(((2*x)/9 + x^2/9 + 
 1/9)/x^4)*(4*x + 4) + 16)/(log(((2*x)/9 + x^2/9 + 1/9)/x^4)^2*(x + 1) + e 
xp(10*x^2)*(x + 1) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*exp(5*x^2)*(2*x + 2) 
),x)
 

output

int((8*x + exp(5*x^2)*(4*x - 40*x^2 - 40*x^3 + 4) + log(((2*x)/9 + x^2/9 + 
 1/9)/x^4)*(4*x + 4) + 16)/(log(((2*x)/9 + x^2/9 + 1/9)/x^4)^2*(x + 1) + e 
xp(10*x^2)*(x + 1) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*exp(5*x^2)*(2*x + 2) 
), x)