3.20.5 \(\int -\frac {e^{25+\frac {1}{5} (-1+\log ^{-\frac {e^{25} x}{5+2 x}}(5))} \log ^{-\frac {e^{25} x}{5+2 x}}(5) \log (\log (5))}{25+20 x+4 x^2} \, dx\)

Optimal. Leaf size=23 \[ e^{\frac {1}{5} \left (-1+\log ^{\frac {e^{25} x}{-5-2 x}}(5)\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.44, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 27, 6706} \begin {gather*} e^{\frac {1}{5} \left (\log ^{-\frac {e^{25} x}{2 x+5}}(5)-1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-((E^(25 + (-1 + Log[5]^(-((E^25*x)/(5 + 2*x))))/5)*Log[Log[5]])/((25 + 20*x + 4*x^2)*Log[5]^((E^25*x)/(5
+ 2*x)))),x]

[Out]

E^((-1 + Log[5]^(-((E^25*x)/(5 + 2*x))))/5)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\log (\log (5)) \int \frac {e^{25+\frac {1}{5} \left (-1+\log ^{-\frac {e^{25} x}{5+2 x}}(5)\right )} \log ^{-\frac {e^{25} x}{5+2 x}}(5)}{25+20 x+4 x^2} \, dx\right )\\ &=-\left (\log (\log (5)) \int \frac {e^{25+\frac {1}{5} \left (-1+\log ^{-\frac {e^{25} x}{5+2 x}}(5)\right )} \log ^{-\frac {e^{25} x}{5+2 x}}(5)}{(5+2 x)^2} \, dx\right )\\ &=e^{\frac {1}{5} \left (-1+\log ^{-\frac {e^{25} x}{5+2 x}}(5)\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 24, normalized size = 1.04 \begin {gather*} e^{\frac {1}{5} \left (-1+\log ^{-\frac {e^{25} x}{5+2 x}}(5)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-((E^(25 + (-1 + Log[5]^(-((E^25*x)/(5 + 2*x))))/5)*Log[Log[5]])/((25 + 20*x + 4*x^2)*Log[5]^((E^25*
x)/(5 + 2*x)))),x]

[Out]

E^((-1 + Log[5]^(-((E^25*x)/(5 + 2*x))))/5)

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 20, normalized size = 0.87 \begin {gather*} e^{\left (\frac {1}{5} \, \log \relax (5)^{-\frac {x e^{25}}{2 \, x + 5}} - \frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(25)*log(log(5))*exp(-x*exp(25)*log(log(5))/(5+2*x))*exp(1/5*exp(-x*exp(25)*log(log(5))/(5+2*x))
-1/5)/(4*x^2+20*x+25),x, algorithm="fricas")

[Out]

e^(1/5*log(5)^(-x*e^25/(2*x + 5)) - 1/5)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 20, normalized size = 0.87 \begin {gather*} e^{\left (\frac {1}{5} \, \log \relax (5)^{-\frac {x e^{25}}{2 \, x + 5}} - \frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(25)*log(log(5))*exp(-x*exp(25)*log(log(5))/(5+2*x))*exp(1/5*exp(-x*exp(25)*log(log(5))/(5+2*x))
-1/5)/(4*x^2+20*x+25),x, algorithm="giac")

[Out]

e^(1/5*log(5)^(-x*e^25/(2*x + 5)) - 1/5)

________________________________________________________________________________________

maple [A]  time = 0.36, size = 21, normalized size = 0.91

method	result	size
risch	\({\mathrm e}^{\frac {\ln \relax (5)^{-\frac {x \,{\mathrm e}^{25}}{5+2 x}}}{5}-\frac {1}{5}}\)	\(21\)
norman	\(\frac {2 x \,{\mathrm e}^{\frac {{\mathrm e}^{-\frac {x \,{\mathrm e}^{25} \ln \left (\ln \relax (5)\right )}{5+2 x}}}{5}-\frac {1}{5}}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{-\frac {x \,{\mathrm e}^{25} \ln \left (\ln \relax (5)\right )}{5+2 x}}}{5}-\frac {1}{5}}}{5+2 x}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(25)*ln(ln(5))*exp(-x*exp(25)*ln(ln(5))/(5+2*x))*exp(1/5*exp(-x*exp(25)*ln(ln(5))/(5+2*x))-1/5)/(4*x^2
+20*x+25),x,method=_RETURNVERBOSE)

[Out]

exp(1/5*ln(5)^(-x*exp(25)/(5+2*x))-1/5)

________________________________________________________________________________________

maxima [A]  time = 0.77, size = 26, normalized size = 1.13 \begin {gather*} e^{\left (\frac {1}{5} \, \log \relax (5)^{-\frac {1}{2} \, e^{25}} \log \relax (5)^{\frac {5 \, e^{25}}{2 \, {\left (2 \, x + 5\right )}}} - \frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(25)*log(log(5))*exp(-x*exp(25)*log(log(5))/(5+2*x))*exp(1/5*exp(-x*exp(25)*log(log(5))/(5+2*x))
-1/5)/(4*x^2+20*x+25),x, algorithm="maxima")

[Out]

e^(1/5*log(5)^(-1/2*e^25)*log(5)^(5/2*e^25/(2*x + 5)) - 1/5)

________________________________________________________________________________________

mupad [B]  time = 1.48, size = 21, normalized size = 0.91 \begin {gather*} {\mathrm {e}}^{\frac {1}{5\,{\ln \relax (5)}^{\frac {x\,{\mathrm {e}}^{25}}{2\,x+5}}}-\frac {1}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x*exp(25)*log(log(5)))/(2*x + 5))*exp(exp(-(x*exp(25)*log(log(5)))/(2*x + 5))/5 - 1/5)*exp(25)*log
(log(5)))/(20*x + 4*x^2 + 25),x)

[Out]

exp(1/(5*log(5)^((x*exp(25))/(2*x + 5))) - 1/5)

________________________________________________________________________________________

sympy [A]  time = 0.40, size = 24, normalized size = 1.04 \begin {gather*} e^{- \frac {1}{5} + \frac {e^{- \frac {x e^{25} \log {\left (\log {\relax (5 )} \right )}}{2 x + 5}}}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(25)*ln(ln(5))*exp(-x*exp(25)*ln(ln(5))/(5+2*x))*exp(1/5*exp(-x*exp(25)*ln(ln(5))/(5+2*x))-1/5)/
(4*x**2+20*x+25),x)

[Out]

exp(-1/5 + exp(-x*exp(25)*log(log(5))/(2*x + 5))/5)

________________________________________________________________________________________