3.58.62 \(\int \frac {5-15 e^{4+e^5+2 x}+(-1+e^{4+e^5+2 x}) \log (e^{1-e^5} (1-e^{4+e^5+2 x}))}{25-25 e^{4+e^5+2 x}+(-5+5 e^{4+e^5+2 x}) \log (e^{1-e^5} (1-e^{4+e^5+2 x}))} \, dx\)

Optimal. Leaf size=33 \[ \log \left (\frac {e^{x/5}}{-5+\log \left (e^{1-e^5}-e^{5+2 x}\right )}\right ) \]

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Rubi [A]  time = 0.53, antiderivative size = 31, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {2282, 12, 6742, 2390, 2302, 29} \begin {gather*} \frac {x}{5}-\log \left (-\log \left (1-e^{2 x+e^5+4}\right )+e^5+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 15*E^(4 + E^5 + 2*x) + (-1 + E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))])/(25 - 25*E^
(4 + E^5 + 2*x) + (-5 + 5*E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))]),x]

[Out]

x/5 - Log[4 + E^5 - Log[1 - E^(4 + E^5 + 2*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2302

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

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {4-e^5 (-1+x)-14 x+(-1+x) \log (1-x)}{5 (1-x) x \left (4 \left (1+\frac {e^5}{4}\right )-\log (1-x)\right )} \, dx,x,e^{4+e^5+2 x}\right )\\ &=\frac {1}{10} \operatorname {Subst}\left (\int \frac {4-e^5 (-1+x)-14 x+(-1+x) \log (1-x)}{(1-x) x \left (4 \left (1+\frac {e^5}{4}\right )-\log (1-x)\right )} \, dx,x,e^{4+e^5+2 x}\right )\\ &=\frac {1}{10} \operatorname {Subst}\left (\int \left (\frac {1}{x}+\frac {10}{(-1+x) \left (4 \left (1+\frac {e^5}{4}\right )-\log (1-x)\right )}\right ) \, dx,x,e^{4+e^5+2 x}\right )\\ &=\frac {x}{5}+\operatorname {Subst}\left (\int \frac {1}{(-1+x) \left (4 \left (1+\frac {e^5}{4}\right )-\log (1-x)\right )} \, dx,x,e^{4+e^5+2 x}\right )\\ &=\frac {x}{5}+\operatorname {Subst}\left (\int \frac {1}{x \left (4 \left (1+\frac {e^5}{4}\right )-\log (x)\right )} \, dx,x,1-e^{4+e^5+2 x}\right )\\ &=\frac {x}{5}-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,4 \left (1+\frac {e^5}{4}\right )-\log \left (1-e^{4+e^5+2 x}\right )\right )\\ &=\frac {x}{5}-\log \left (4 \left (1+\frac {e^5}{4}\right )-\log \left (1-e^{4+e^5+2 x}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 31, normalized size = 0.94 \begin {gather*} \frac {1}{5} \left (x-5 \log \left (4+e^5-\log \left (1-e^{4+e^5+2 x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 15*E^(4 + E^5 + 2*x) + (-1 + E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))])/(25 -
25*E^(4 + E^5 + 2*x) + (-5 + 5*E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))]),x]

[Out]

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

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fricas [A]  time = 0.49, size = 29, normalized size = 0.88 \begin {gather*} \frac {1}{5} \, x - \log \left (\log \left (-{\left (e^{\left (2 \, x + e^{5} + 4\right )} - 1\right )} e^{\left (-e^{5} + 1\right )}\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-15*exp(5+2*x)*exp(exp
(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1)-5)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(e
xp(5)-1)+25),x, algorithm="fricas")

[Out]

1/5*x - log(log(-(e^(2*x + e^5 + 4) - 1)*e^(-e^5 + 1)) - 5)

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giac [B]  time = 0.24, size = 59, normalized size = 1.79 \begin {gather*} \frac {1}{5} \, x + \frac {1}{10} \, \log \left (e^{\left (2 \, x + e^{5} + 4\right )} - 1\right ) - \frac {1}{10} \, \log \left (-e^{\left (2 \, x + e^{5} + 4\right )} + 1\right ) - \log \left (\log \left ({\left (e^{5} - e^{\left (2 \, x + e^{5} + 9\right )}\right )} e^{\left (-e^{5} - 4\right )}\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-15*exp(5+2*x)*exp(exp
(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1)-5)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(e
xp(5)-1)+25),x, algorithm="giac")

[Out]

1/5*x + 1/10*log(e^(2*x + e^5 + 4) - 1) - 1/10*log(-e^(2*x + e^5 + 4) + 1) - log(log((e^5 - e^(2*x + e^5 + 9))
*e^(-e^5 - 4)) - 5)

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maple [A]  time = 0.33, size = 31, normalized size = 0.94

method	result	size
risch	\(\frac {x}{5}-\ln \left (\ln \left (\left (-{\mathrm e}^{{\mathrm e}^{5}+2 x +4}+1\right ) {\mathrm e}^{1-{\mathrm e}^{5}}\right )-5\right )\)	\(31\)
norman	\(\frac {x}{5}-\ln \left (\ln \left (\left (-{\mathrm e}^{5+2 x} {\mathrm e}^{{\mathrm e}^{5}-1}+1\right ) {\mathrm e}^{1-{\mathrm e}^{5}}\right )-5\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(5+2*x)*exp(exp(5)-1)-1)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-15*exp(5+2*x)*exp(exp(5)-1)+
5)/((5*exp(5+2*x)*exp(exp(5)-1)-5)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp(5)-1)
+25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 31, normalized size = 0.94 \begin {gather*} \frac {1}{5} \, x + \frac {1}{10} \, e^{5} - \log \left (-e^{5} + \log \left (-e^{\left (2 \, x + e^{5} + 4\right )} + 1\right ) - 4\right ) + \frac {2}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-15*exp(5+2*x)*exp(exp
(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1)-5)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(e
xp(5)-1)+25),x, algorithm="maxima")

[Out]

1/5*x + 1/10*e^5 - log(-e^5 + log(-e^(2*x + e^5 + 4) + 1) - 4) + 2/5

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mupad [B]  time = 0.97, size = 27, normalized size = 0.82 \begin {gather*} \frac {x}{5}-\ln \left (\ln \left ({\mathrm {e}}^{-{\mathrm {e}}^5}\,\mathrm {e}-{\mathrm {e}}^{2\,x+5}\right )-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-exp(1 - exp(5))*(exp(exp(5) - 1)*exp(2*x + 5) - 1))*(exp(exp(5) - 1)*exp(2*x + 5) - 1) - 15*exp(exp(
5) - 1)*exp(2*x + 5) + 5)/(log(-exp(1 - exp(5))*(exp(exp(5) - 1)*exp(2*x + 5) - 1))*(5*exp(exp(5) - 1)*exp(2*x
+ 5) - 5) - 25*exp(exp(5) - 1)*exp(2*x + 5) + 25),x)

[Out]

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

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sympy [A]  time = 0.26, size = 29, normalized size = 0.88 \begin {gather*} \frac {x}{5} - \log {\left (\log {\left (\frac {- e^{-1 + e^{5}} e^{2 x + 5} + 1}{e^{-1 + e^{5}}} \right )} - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-15*exp(5+2*x)*exp(exp(
5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1)-5)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp
(5)-1)+25),x)

[Out]

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

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