\(\int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx\) [128]

Optimal result

Integrand size = 57, antiderivative size = 26 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {-2 x+\left (\log (4)-e^{\frac {7}{5}-x} \log (5)\right )^2}{x} \] Output:

((2*ln(2)-exp(ln(ln(5))+7/5-x))^2-2*x)/x
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {2 \log ^2(4)+e^{\frac {14}{5}-2 x} \log (5) \log (25)-e^{\frac {7}{5}-x} \log (16) \log (25)}{2 x} \] Input:

Integrate[(E^((2*(7 - 5*x + 5*Log[Log[5]]))/5)*(-1 - 2*x) + E^((7 - 5*x + 
5*Log[Log[5]])/5)*(2 + 2*x)*Log[4] - Log[4]^2)/x^2,x]
 

Output:

(2*Log[4]^2 + E^(14/5 - 2*x)*Log[5]*Log[25] - E^(7/5 - x)*Log[16]*Log[25]) 
/(2*x)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, 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.

\(\displaystyle \int \frac {(-2 x-1) e^{\frac {2}{5} (-5 x+7+5 \log (\log (5)))}+(2 x+2) \log (4) e^{\frac {1}{5} (-5 x+7+5 \log (\log (5)))}-\log ^2(4)}{x^2} \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (-\frac {e^{\frac {14}{5}-2 x} (2 x+1) \log ^2(5)}{x^2}-\frac {\log ^2(4)}{x^2}+\frac {2 e^{\frac {7}{5}-x} (x+1) \log (4) \log (5)}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^{\frac {14}{5}-2 x} \log ^2(5)}{x}+\frac {\log ^2(4)}{x}-\frac {2 e^{\frac {7}{5}-x} \log (4) \log (5)}{x}\)

Input:

Int[(E^((2*(7 - 5*x + 5*Log[Log[5]]))/5)*(-1 - 2*x) + E^((7 - 5*x + 5*Log[ 
Log[5]])/5)*(2 + 2*x)*Log[4] - Log[4]^2)/x^2,x]
 

Output:

Log[4]^2/x - (2*E^(7/5 - x)*Log[4]*Log[5])/x + (E^(14/5 - 2*x)*Log[5]^2)/x
 

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]]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38

method	result	size
norman	\(\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \ln \left (2\right )^{2}-4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x}\)	\(36\)
parallelrisch	\(\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \ln \left (2\right )^{2}-4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x}\)	\(36\)
risch	\(\frac {4 \ln \left (2\right )^{2}}{x}+\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{x}-\frac {4 \ln \left (2\right ) \ln \left (5\right ) {\mathrm e}^{\frac {7}{5}-x}}{x}\)	\(40\)
parts	\(\frac {4 \ln \left (2\right )^{2}}{x}+\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{x}-\frac {4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x}\)	\(42\)
derivativedivides	\(\frac {19 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{5 x}-\frac {28 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {expIntegral}_{1}\left (2 x \right )}{5}+\frac {4 \ln \left (2\right )^{2}}{x}-10 \left (7+5 \ln \left (\ln \left (5\right )\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {expIntegral}_{1}\left (2 x \right )}{25}\right )+50 \ln \left (\ln \left (5\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {expIntegral}_{1}\left (2 x \right )}{25}\right )-240 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )+100 \ln \left (2\right ) \left (\frac {\left (7+5 \ln \left (\ln \left (5\right )\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )}{5}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )-100 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )\)	\(234\)
default	\(\frac {19 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{5 x}-\frac {28 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {expIntegral}_{1}\left (2 x \right )}{5}+\frac {4 \ln \left (2\right )^{2}}{x}-10 \left (7+5 \ln \left (\ln \left (5\right )\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {expIntegral}_{1}\left (2 x \right )}{25}\right )+50 \ln \left (\ln \left (5\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {expIntegral}_{1}\left (2 x \right )}{25}\right )-240 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )+100 \ln \left (2\right ) \left (\frac {\left (7+5 \ln \left (\ln \left (5\right )\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )}{5}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )-100 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {expIntegral}_{1}\left (x \right )}{25}\right )\)	\(234\)
orering	\(\frac {\left (-3-x \right ) \left (\left (-1-2 x \right ) \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+2 \left (2+2 x \right ) \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}-4 \ln \left (2\right )^{2}\right )}{x^{2}}+\left (-\frac {3}{2}-\frac {3 x}{2}\right ) \left (\frac {-2 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}-2 \left (-1-2 x \right ) \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}-2 \left (2+2 x \right ) \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x^{2}}-\frac {2 \left (\left (-1-2 x \right ) \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+2 \left (2+2 x \right ) \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}-4 \ln \left (2\right )^{2}\right )}{x^{3}}\right )-\frac {x \left (\frac {8 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \left (-1-2 x \right ) \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}-8 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}+2 \left (2+2 x \right ) \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x^{2}}-\frac {4 \left (-2 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}-2 \left (-1-2 x \right ) \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}-2 \left (2+2 x \right ) \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}\right )}{x^{3}}+\frac {6 \left (-1-2 x \right ) \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+12 \left (2+2 x \right ) \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}-24 \ln \left (2\right )^{2}}{x^{4}}\right )}{2}\)	\(360\)

Input:

int(((-1-2*x)*exp(ln(ln(5))+7/5-x)^2+2*(2+2*x)*ln(2)*exp(ln(ln(5))+7/5-x)- 
4*ln(2)^2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

(exp(ln(ln(5))+7/5-x)^2+4*ln(2)^2-4*ln(2)*exp(ln(ln(5))+7/5-x))/x
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-\frac {4 \, e^{\left (-x + \log \left (\log \left (5\right )\right ) + \frac {7}{5}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{\left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) + \frac {14}{5}\right )}}{x} \] Input:

integrate(((-1-2*x)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log( 
5))+7/5-x)-4*log(2)^2)/x^2,x, algorithm="fricas")
 

Output:

-(4*e^(-x + log(log(5)) + 7/5)*log(2) - 4*log(2)^2 - e^(-2*x + 2*log(log(5 
)) + 14/5))/x
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {4 \log {\left (2 \right )}^{2}}{x} + \frac {- 4 x e^{\frac {7}{5} - x} \log {\left (2 \right )} \log {\left (5 \right )} + x e^{\frac {14}{5} - 2 x} \log {\left (5 \right )}^{2}}{x^{2}} \] Input:

integrate(((-2*x-1)*exp(ln(ln(5))+7/5-x)**2+2*(2+2*x)*ln(2)*exp(ln(ln(5))+ 
7/5-x)-4*ln(2)**2)/x**2,x)
 

Output:

4*log(2)**2/x + (-4*x*exp(7/5 - x)*log(2)*log(5) + x*exp(14/5 - 2*x)*log(5 
)**2)/x**2
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-2 \, {\rm Ei}\left (-2 \, x\right ) e^{\frac {14}{5}} \log \left (5\right )^{2} + 2 \, e^{\frac {14}{5}} \Gamma \left (-1, 2 \, x\right ) \log \left (5\right )^{2} + 4 \, {\rm Ei}\left (-x\right ) e^{\frac {7}{5}} \log \left (5\right ) \log \left (2\right ) - 4 \, e^{\frac {7}{5}} \Gamma \left (-1, x\right ) \log \left (5\right ) \log \left (2\right ) + \frac {4 \, \log \left (2\right )^{2}}{x} \] Input:

integrate(((-1-2*x)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log( 
5))+7/5-x)-4*log(2)^2)/x^2,x, algorithm="maxima")
 

Output:

-2*Ei(-2*x)*e^(14/5)*log(5)^2 + 2*e^(14/5)*gamma(-1, 2*x)*log(5)^2 + 4*Ei( 
-x)*e^(7/5)*log(5)*log(2) - 4*e^(7/5)*gamma(-1, x)*log(5)*log(2) + 4*log(2 
)^2/x
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-\frac {4 \, e^{\left (-x + \log \left (\log \left (5\right )\right ) + \frac {7}{5}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{\left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) + \frac {14}{5}\right )}}{x} \] Input:

integrate(((-1-2*x)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log( 
5))+7/5-x)-4*log(2)^2)/x^2,x, algorithm="giac")
 

Output:

-(4*e^(-x + log(log(5)) + 7/5)*log(2) - 4*log(2)^2 - e^(-2*x + 2*log(log(5 
)) + 14/5))/x
 

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,{\left ({\mathrm {e}}^{7/5}\,\ln \left (5\right )-2\,{\mathrm {e}}^x\,\ln \left (2\right )\right )}^2}{x} \] Input:

int(-(exp(2*log(log(5)) - 2*x + 14/5)*(2*x + 1) + 4*log(2)^2 - 2*exp(log(l 
og(5)) - x + 7/5)*log(2)*(2*x + 2))/x^2,x)
 

Output:

(exp(-2*x)*(exp(7/5)*log(5) - 2*exp(x)*log(2))^2)/x
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {e^{\frac {14}{5}} \mathrm {log}\left (5\right )^{2}-4 e^{x +\frac {2}{5}} \mathrm {log}\left (5\right ) \mathrm {log}\left (2\right ) e +4 e^{2 x} \mathrm {log}\left (2\right )^{2}}{e^{2 x} x} \] Input:

int(((-2*x-1)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log(5))+7/ 
5-x)-4*log(2)^2)/x^2,x)
 

Output:

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