\(\int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx\) [8835]

Optimal result

Integrand size = 41, antiderivative size = 17 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=-1+\frac {2 x}{e^{25 e^2}+\log (x)} \]

[Out]

2/(ln(x)+exp(25/4*exp(1)^2)^4)*x-1

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6820, 12, 2407, 2334, 2336, 2209} \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2 x}{\log (x)+e^{25 e^2}} \]

[In]

Int[(-2 + 2*E^(25*E^2) + 2*Log[x])/(E^(50*E^2) + 2*E^(25*E^2)*Log[x] + Log[x]^2),x]

[Out]

(2*x)/(E^(25*E^2) + Log[x])

Rule 12

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2336

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2407

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (-1+e^{25 e^2}+\log (x)\right )}{\left (e^{25 e^2}+\log (x)\right )^2} \, dx \\ & = 2 \int \frac {-1+e^{25 e^2}+\log (x)}{\left (e^{25 e^2}+\log (x)\right )^2} \, dx \\ & = 2 \int \left (-\frac {1}{\left (e^{25 e^2}+\log (x)\right )^2}+\frac {1}{e^{25 e^2}+\log (x)}\right ) \, dx \\ & = -\left (2 \int \frac {1}{\left (e^{25 e^2}+\log (x)\right )^2} \, dx\right )+2 \int \frac {1}{e^{25 e^2}+\log (x)} \, dx \\ & = \frac {2 x}{e^{25 e^2}+\log (x)}-2 \int \frac {1}{e^{25 e^2}+\log (x)} \, dx+2 \text {Subst}\left (\int \frac {e^x}{e^{25 e^2}+x} \, dx,x,\log (x)\right ) \\ & = 2 e^{-e^{25 e^2}} \operatorname {ExpIntegralEi}\left (e^{25 e^2}+\log (x)\right )+\frac {2 x}{e^{25 e^2}+\log (x)}-2 \text {Subst}\left (\int \frac {e^x}{e^{25 e^2}+x} \, dx,x,\log (x)\right ) \\ & = \frac {2 x}{e^{25 e^2}+\log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2 x}{e^{25 e^2}+\log (x)} \]

[In]

Integrate[(-2 + 2*E^(25*E^2) + 2*Log[x])/(E^(50*E^2) + 2*E^(25*E^2)*Log[x] + Log[x]^2),x]

[Out]

(2*x)/(E^(25*E^2) + Log[x])

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \left (x \right )}\)	\(14\)
default	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \left (x \right )}\)	\(16\)
norman	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \left (x \right )}\)	\(18\)
parallelrisch	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \left (x \right )}\)	\(18\)

[In]

int((2*ln(x)+2*exp(25/4*exp(1)^2)^4-2)/(ln(x)^2+2*exp(25/4*exp(1)^2)^4*ln(x)+exp(25/4*exp(1)^2)^8),x,method=_R
ETURNVERBOSE)

[Out]

2*x/(exp(25*exp(2))+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2 \, x}{e^{\left (25 \, e^{2}\right )} + \log \left (x\right )} \]

[In]

integrate((2*log(x)+2*exp(25/4*exp(1)^2)^4-2)/(log(x)^2+2*exp(25/4*exp(1)^2)^4*log(x)+exp(25/4*exp(1)^2)^8),x,
algorithm="fricas")

[Out]

2*x/(e^(25*e^2) + log(x))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2 x}{\log {\left (x \right )} + e^{25 e^{2}}} \]

[In]

integrate((2*ln(x)+2*exp(25/4*exp(1)**2)**4-2)/(ln(x)**2+2*exp(25/4*exp(1)**2)**4*ln(x)+exp(25/4*exp(1)**2)**8
),x)

[Out]

2*x/(log(x) + exp(25*exp(2)))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2 \, x}{e^{\left (25 \, e^{2}\right )} + \log \left (x\right )} \]

[In]

integrate((2*log(x)+2*exp(25/4*exp(1)^2)^4-2)/(log(x)^2+2*exp(25/4*exp(1)^2)^4*log(x)+exp(25/4*exp(1)^2)^8),x,
algorithm="maxima")

[Out]

2*x/(e^(25*e^2) + log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2 \, x}{e^{\left (25 \, e^{2}\right )} + \log \left (x\right )} \]

[In]

integrate((2*log(x)+2*exp(25/4*exp(1)^2)^4-2)/(log(x)^2+2*exp(25/4*exp(1)^2)^4*log(x)+exp(25/4*exp(1)^2)^8),x,
algorithm="giac")

[Out]

2*x/(e^(25*e^2) + log(x))

Mupad [B] (verification not implemented)

Time = 11.82 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx=\frac {2\,x}{{\mathrm {e}}^{25\,{\mathrm {e}}^2}+\ln \left (x\right )} \]

[In]

int((2*exp(25*exp(2)) + 2*log(x) - 2)/(exp(50*exp(2)) + log(x)^2 + 2*exp(25*exp(2))*log(x)),x)

[Out]

(2*x)/(exp(25*exp(2)) + log(x))