\(\int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx\) [9128]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {14, 2228} \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {e^{\frac {14}{5}-2 x} \log ^2(5)}{x}+\frac {\log ^2(4)}{x}-\frac {2 e^{\frac {7}{5}-x} \log (4) \log (5)}{x} \]

[In]

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]

[Out]

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

Rule 14

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]]

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\log ^2(4)}{x^2}+\frac {2 e^{\frac {7}{5}-x} (1+x) \log (4) \log (5)}{x^2}-\frac {e^{\frac {14}{5}-2 x} (1+2 x) \log ^2(5)}{x^2}\right ) \, dx \\ & = \frac {\log ^2(4)}{x}+(2 \log (4) \log (5)) \int \frac {e^{\frac {7}{5}-x} (1+x)}{x^2} \, dx-\log ^2(5) \int \frac {e^{\frac {14}{5}-2 x} (1+2 x)}{x^2} \, dx \\ & = \frac {\log ^2(4)}{x}-\frac {2 e^{\frac {7}{5}-x} \log (4) \log (5)}{x}+\frac {e^{\frac {14}{5}-2 x} \log ^2(5)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {2 \log ^2(4)+e^{\frac {14}{5}-2 x} \log (5) \log (25)-e^{\frac {7}{5}-x} \log (16) \log (25)}{2 x} \]

[In]

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]

[Out]

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

Maple [A] (verified)

Time = 0.60 (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	\(\frac {19 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{5 x}-\frac {28 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{5}+\frac {4 \ln \left (2\right )^{2}}{x}-10 \left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )+50 \ln \left (\ln \left (5\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )-240 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )+100 \ln \left (2\right ) \left (\frac {\left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )}{5}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )-100 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )\)	\(234\)
default	\(\frac {19 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{5 x}-\frac {28 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{5}+\frac {4 \ln \left (2\right )^{2}}{x}-10 \left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )+50 \ln \left (\ln \left (5\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )-240 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )+100 \ln \left (2\right ) \left (\frac {\left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )}{5}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )-100 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )\)	\(234\)

[In]

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=_RETURNVERBO
SE)

[Out]

(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)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-\frac {4 \, e^{\left (-x + \log \left (\log \left (5\right )\right ) + \frac {7}{5}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{\left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) + \frac {14}{5}\right )}}{x} \]

[In]

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, algori
thm="fricas")

[Out]

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

[In]

integrate(((-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)

[Out]

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 = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-2 \, {\rm Ei}\left (-2 \, x\right ) e^{\frac {14}{5}} \log \left (5\right )^{2} + 2 \, e^{\frac {14}{5}} \Gamma \left (-1, 2 \, x\right ) \log \left (5\right )^{2} + 4 \, {\rm Ei}\left (-x\right ) e^{\frac {7}{5}} \log \left (5\right ) \log \left (2\right ) - 4 \, e^{\frac {7}{5}} \Gamma \left (-1, x\right ) \log \left (5\right ) \log \left (2\right ) + \frac {4 \, \log \left (2\right )^{2}}{x} \]

[In]

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, algori
thm="maxima")

[Out]

-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)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-\frac {4 \, e^{\left (-x + \log \left (\log \left (5\right )\right ) + \frac {7}{5}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{\left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) + \frac {14}{5}\right )}}{x} \]

[In]

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, algori
thm="giac")

[Out]

-(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 = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,{\left ({\mathrm {e}}^{7/5}\,\ln \left (5\right )-2\,{\mathrm {e}}^x\,\ln \left (2\right )\right )}^2}{x} \]

[In]

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

[Out]

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