3.8.87 \(\int \frac {-x+\frac {e^{3-x-30 \log (5) \log (x^2)-3 \log (5) \log ^2(x^2)} (-x-60 \log (5)-12 \log (5) \log (x^2))}{26469779601696885595885078146238811314105987548828125}}{x} \, dx\) [787]

3.8.87.1 Optimal result

Integrand size = 53, antiderivative size = 23 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=e^{3-x-3 \log (5) \left (5+\log \left (x^2\right )\right )^2}-x \]

output

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

3.8.87.2 Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x-30 \log (5) \log \left (x^2\right )}-x \]

input

Integrate[(-x + (E^(3 - x - 30*Log[5]*Log[x^2] - 3*Log[5]*Log[x^2]^2)*(-x 
- 60*Log[5] - 12*Log[5]*Log[x^2]))/264697796016968855958850781462388113141 
05987548828125)/x,x]
 

output

E^(3 - x - 30*Log[5]*Log[x^2])/5^(3*(25 + Log[x^2]^2)) - x
 

3.8.87.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2010, 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.

\(\displaystyle \int \frac {\frac {\left (-12 \log (5) \log \left (x^2\right )-x-60 \log (5)\right ) \exp \left (-3 \log (5) \log ^2\left (x^2\right )-30 \log (5) \log \left (x^2\right )-x+3\right )}{26469779601696885595885078146238811314105987548828125}-x}{x} \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (\frac {5^{-3 \left (\log ^2\left (x^2\right )+25\right )} e^{-30 \log (5) \log \left (x^2\right )-x+3} \left (-12 \log (5) \log \left (x^2\right )-x-60 \log (5)\right )}{x}-1\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle e^{3-x} 5^{-3 \left (\log ^2\left (x^2\right )+25\right )} \left (x^2\right )^{-30 \log (5)}-x\)

input

Int[(-x + (E^(3 - x - 30*Log[5]*Log[x^2] - 3*Log[5]*Log[x^2]^2)*(-x - 60*L 
og[5] - 12*Log[5]*Log[x^2]))/264697796016968855958850781462388113141059875 
48828125)/x,x]
 

output

-x + E^(3 - x)/(5^(3*(25 + Log[x^2]^2))*(x^2)^(30*Log[5]))
 

3.8.87.3.1 Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

3.8.87.4 Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26

method	result	size
risch	\(-x +\frac {\left (\frac {1}{125}\right )^{\ln \left (x^{2}\right )^{2}} \left (x^{2}\right )^{-30 \ln \left (5\right )} {\mathrm e}^{-x +3}}{26469779601696885595885078146238811314105987548828125}\)	\(29\)
default	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)
norman	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)
parallelrisch	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)
parts	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)

input

int(((-12*ln(5)*ln(x^2)-60*ln(5)-x)*exp(-3*ln(5)*ln(x^2)^2-30*ln(5)*ln(x^2 
)-75*ln(5)+3-x)-x)/x,x,method=_RETURNVERBOSE)
 

output

-x+1/26469779601696885595885078146238811314105987548828125*(1/125)^(ln(x^2 
)^2)*(x^2)^(-30*ln(5))*exp(-x+3)
 

3.8.87.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=-x + e^{\left (-3 \, \log \left (5\right ) \log \left (x^{2}\right )^{2} - 30 \, \log \left (5\right ) \log \left (x^{2}\right ) - x - 75 \, \log \left (5\right ) + 3\right )} \]

input

integrate(((-12*log(5)*log(x^2)-60*log(5)-x)*exp(-3*log(5)*log(x^2)^2-30*l 
og(5)*log(x^2)-75*log(5)+3-x)-x)/x,x, algorithm=\
 

output

-x + e^(-3*log(5)*log(x^2)^2 - 30*log(5)*log(x^2) - x - 75*log(5) + 3)
 

3.8.87.6 Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=- x + \frac {e^{- x - 3 \log {\left (5 \right )} \log {\left (x^{2} \right )}^{2} - 30 \log {\left (5 \right )} \log {\left (x^{2} \right )} + 3}}{26469779601696885595885078146238811314105987548828125} \]

input

integrate(((-12*ln(5)*ln(x**2)-60*ln(5)-x)*exp(-3*ln(5)*ln(x**2)**2-30*ln( 
5)*ln(x**2)-75*ln(5)+3-x)-x)/x,x)
 

output

-x + exp(-x - 3*log(5)*log(x**2)**2 - 30*log(5)*log(x**2) + 3)/26469779601 
696885595885078146238811314105987548828125
 

3.8.87.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=-x + \frac {1}{26469779601696885595885078146238811314105987548828125} \, e^{\left (-12 \, \log \left (5\right ) \log \left (x\right )^{2} - 60 \, \log \left (5\right ) \log \left (x\right ) - x + 3\right )} \]

input

integrate(((-12*log(5)*log(x^2)-60*log(5)-x)*exp(-3*log(5)*log(x^2)^2-30*l 
og(5)*log(x^2)-75*log(5)+3-x)-x)/x,x, algorithm=\
 

output

-x + 1/26469779601696885595885078146238811314105987548828125*e^(-12*log(5) 
*log(x)^2 - 60*log(5)*log(x) - x + 3)
 

3.8.87.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=-x + \frac {1}{26469779601696885595885078146238811314105987548828125} \, e^{\left (-3 \, \log \left (5\right ) \log \left (x^{2}\right )^{2} - 30 \, \log \left (5\right ) \log \left (x^{2}\right ) - x + 3\right )} \]

input

integrate(((-12*log(5)*log(x^2)-60*log(5)-x)*exp(-3*log(5)*log(x^2)^2-30*l 
og(5)*log(x^2)-75*log(5)+3-x)-x)/x,x, algorithm=\
 

output

-x + 1/26469779601696885595885078146238811314105987548828125*e^(-3*log(5)* 
log(x^2)^2 - 30*log(5)*log(x^2) - x + 3)
 

3.8.87.9 Mupad [B] (verification not implemented)

Time = 11.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=\frac {{\mathrm {e}}^{3-x}}{26469779601696885595885078146238811314105987548828125\,5^{3\,{\ln \left (x^2\right )}^2}\,{\left (x^2\right )}^{30\,\ln \left (5\right )}}-x \]

input

int(-(x + exp(3 - 75*log(5) - 30*log(x^2)*log(5) - 3*log(x^2)^2*log(5) - x 
)*(x + 60*log(5) + 12*log(x^2)*log(5)))/x,x)
 

output

exp(3 - x)/(26469779601696885595885078146238811314105987548828125*5^(3*log 
(x^2)^2)*(x^2)^(30*log(5))) - x