3.60.81 \(\int \frac {4 e^x x+(10-2 e^4-2 e^x) \log (\frac {4 (5-e^4-e^x)^2}{e^{10}})}{(-5 x+e^4 x+e^x x) \log (\frac {4 (5-e^4-e^x)^2}{e^{10}})} \, dx\)

Optimal. Leaf size=28 \[ 2 \log \left (\frac {3 \log \left (\frac {4 \left (5-e^4-e^x\right )^2}{e^{10}}\right )}{x}\right ) \]

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Rubi [A]  time = 1.01, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.101, Rules used = {6, 6741, 6688, 2282, 2390, 12, 2302, 29} \begin {gather*} 2 \log \left (-\log \left (\left (e^x-5+e^4\right )^2\right )+10-\log (4)\right )-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^x*x + (10 - 2*E^4 - 2*E^x)*Log[(4*(5 - E^4 - E^x)^2)/E^10])/((-5*x + E^4*x + E^x*x)*Log[(4*(5 - E^4 -
E^x)^2)/E^10]),x]

[Out]

-2*Log[x] + 2*Log[10 - Log[4] - Log[(-5 + E^4 + E^x)^2]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6688

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

Rule 6741

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

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 27, normalized size = 0.96 \begin {gather*} -2 \log (x)+2 \log \left (10-\log (4)-\log \left (\left (-5+e^4+e^x\right )^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^x*x + (10 - 2*E^4 - 2*E^x)*Log[(4*(5 - E^4 - E^x)^2)/E^10])/((-5*x + E^4*x + E^x*x)*Log[(4*(5 -
E^4 - E^x)^2)/E^10]),x]

[Out]

-2*Log[x] + 2*Log[10 - Log[4] - Log[(-5 + E^4 + E^x)^2]]

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fricas [A]  time = 0.67, size = 33, normalized size = 1.18 \begin {gather*} -2 \, \log \relax (x) + 2 \, \log \left (\log \left (4 \, {\left (2 \, {\left (e^{4} - 5\right )} e^{x} + e^{8} - 10 \, e^{4} + e^{\left (2 \, x\right )} + 25\right )} e^{\left (-10\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)-2*exp(4)+10)*log(4*exp(log(-exp(x)+5-exp(4))-5)^2)+4*exp(x)*x)/(exp(x)*x+x*exp(4)-5*x)/l
og(4*exp(log(-exp(x)+5-exp(4))-5)^2),x, algorithm="fricas")

[Out]

-2*log(x) + 2*log(log(4*(2*(e^4 - 5)*e^x + e^8 - 10*e^4 + e^(2*x) + 25)*e^(-10)))

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giac [A]  time = 0.40, size = 47, normalized size = 1.68 \begin {gather*} -2 \, \log \relax (x) + 2 \, \log \left (-2 \, \log \relax (2) - 2 \, \log \left (e^{4} \mathrm {sgn}\left (e^{4} + e^{x} - 5\right ) + e^{x} \mathrm {sgn}\left (e^{4} + e^{x} - 5\right ) - 5 \, \mathrm {sgn}\left (e^{4} + e^{x} - 5\right )\right ) + 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)-2*exp(4)+10)*log(4*exp(log(-exp(x)+5-exp(4))-5)^2)+4*exp(x)*x)/(exp(x)*x+x*exp(4)-5*x)/l
og(4*exp(log(-exp(x)+5-exp(4))-5)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.37, size = 28, normalized size = 1.00

method	result	size
default	\(-2 \ln \relax (x )+2 \ln \left (\ln \left (4 \left (-{\mathrm e}^{x}+5-{\mathrm e}^{4}\right )^{2} {\mathrm e}^{-10}\right )\right )\)	\(28\)
norman	\(-2 \ln \relax (x )+2 \ln \left (\ln \left (4 \left (-{\mathrm e}^{x}+5-{\mathrm e}^{4}\right )^{2} {\mathrm e}^{-10}\right )\right )\)	\(28\)
risch	\(-2 \ln \relax (x )+2 \ln \left (\ln \left ({\mathrm e}^{x}+{\mathrm e}^{4}-5\right )-\frac {i \left (\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{4}-5\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{4}-5\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{4}-5\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{4}-5\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{4}-5\right )^{2}\right )^{3}+4 i \ln \relax (2)-20 i\right )}{4}\right )\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(x)-2*exp(4)+10)*ln(4*exp(ln(-exp(x)+5-exp(4))-5)^2)+4*exp(x)*x)/(exp(x)*x+x*exp(4)-5*x)/ln(4*exp(
ln(-exp(x)+5-exp(4))-5)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.55, size = 19, normalized size = 0.68 \begin {gather*} -2 \, \log \relax (x) + 2 \, \log \left (\log \relax (2) + \log \left (e^{4} + e^{x} - 5\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)-2*exp(4)+10)*log(4*exp(log(-exp(x)+5-exp(4))-5)^2)+4*exp(x)*x)/(exp(x)*x+x*exp(4)-5*x)/l
og(4*exp(log(-exp(x)+5-exp(4))-5)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.33, size = 21, normalized size = 0.75 \begin {gather*} 2\,\ln \left (\ln \left (4\,{\mathrm {e}}^{-10}\,{\left ({\mathrm {e}}^4+{\mathrm {e}}^x-5\right )}^2\right )\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(4*exp(2*log(5 - exp(x) - exp(4)) - 10))*(2*exp(4) + 2*exp(x) - 10) - 4*x*exp(x))/(log(4*exp(2*log(5
- exp(x) - exp(4)) - 10))*(x*exp(4) - 5*x + x*exp(x))),x)

[Out]

2*log(log(4*exp(-10)*(exp(4) + exp(x) - 5)^2)) - 2*log(x)

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sympy [A]  time = 0.24, size = 24, normalized size = 0.86 \begin {gather*} - 2 \log {\relax (x )} + 2 \log {\left (\log {\left (\frac {4 \left (- e^{x} - e^{4} + 5\right )^{2}}{e^{10}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)-2*exp(4)+10)*ln(4*exp(ln(-exp(x)+5-exp(4))-5)**2)+4*exp(x)*x)/(exp(x)*x+x*exp(4)-5*x)/ln
(4*exp(ln(-exp(x)+5-exp(4))-5)**2),x)

[Out]

-2*log(x) + 2*log(log(4*(-exp(x) - exp(4) + 5)**2*exp(-10)))

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