3.19.52 \(\int \frac {e^{2+x} (-4 e^2-2 e^{2+x})+(-25 e^x+e^{2+2 x}) \log (-25+e^{2+x})}{(-100 e^2-50 e^{2+x}+e^{2+x} (4 e^2+2 e^{2+x})) \log (-25+e^{2+x})} \, dx\)

Optimal. Leaf size=26 \[ 3+\frac {\log \left (2+e^x\right )}{2 e^2}+\log \left (\frac {1}{\log \left (-25+e^{2+x}\right )}\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2282, 12, 2390, 2302, 29} \begin {gather*} \frac {\log \left (e^x+2\right )}{2 e^2}-\log \left (\log \left (e^{x+2}-25\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2 + x)*(-4*E^2 - 2*E^(2 + x)) + (-25*E^x + E^(2 + 2*x))*Log[-25 + E^(2 + x)])/((-100*E^2 - 50*E^(2 + x
) + E^(2 + x)*(4*E^2 + 2*E^(2 + x)))*Log[-25 + E^(2 + x)]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (\frac {\log \left (2+e^x\right )}{e^2}-2 \log \left (\log \left (-25+e^{2+x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + x)*(-4*E^2 - 2*E^(2 + x)) + (-25*E^x + E^(2 + 2*x))*Log[-25 + E^(2 + x)])/((-100*E^2 - 50*E^
(2 + x) + E^(2 + x)*(4*E^2 + 2*E^(2 + x)))*Log[-25 + E^(2 + x)]),x]

[Out]

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

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fricas [A]  time = 0.73, size = 29, normalized size = 1.12 \begin {gather*} -\frac {1}{2} \, {\left (2 \, e^{2} \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) - \log \left (2 \, e^{2} + e^{\left (x + 2\right )}\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp(2))*exp(2+x))/((2*exp(2)*exp(x
)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-100*exp(2))/log(exp(2+x)-25),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.43, size = 56, normalized size = 2.15 \begin {gather*} \frac {2 \, e^{2} \log \left (-e^{x} - 2\right ) - 4 \, e^{4} \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) - 50 \, e^{2} \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) + 25 \, \log \left (e^{x} + 2\right )}{2 \, {\left (2 \, e^{4} + 25 \, e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp(2))*exp(2+x))/((2*exp(2)*exp(x
)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-100*exp(2))/log(exp(2+x)-25),x, algorithm="giac")

[Out]

1/2*(2*e^2*log(-e^x - 2) - 4*e^4*log(log(e^(x + 2) - 25)) - 50*e^2*log(log(e^(x + 2) - 25)) + 25*log(e^x + 2))
/(2*e^4 + 25*e^2)

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maple [A]  time = 0.26, size = 21, normalized size = 0.81

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{x}+2\right ) {\mathrm e}^{-2}}{2}-\ln \left (\ln \left ({\mathrm e}^{2+x}-25\right )\right )\)	\(21\)
norman	\(\frac {\ln \left ({\mathrm e}^{x}+2\right ) {\mathrm e}^{-2}}{2}-\ln \left (\ln \left ({\mathrm e}^{2} {\mathrm e}^{x}-25\right )\right )\)	\(24\)
default	\(\frac {25 \,{\mathrm e}^{-2} \ln \left ({\mathrm e}^{2+x}-25\right )}{2 \left (2 \,{\mathrm e}^{2}+25\right )}+\frac {\ln \left ({\mathrm e}^{2+x}+2 \,{\mathrm e}^{2}\right )}{2 \,{\mathrm e}^{2}+25}-\ln \left (\ln \left ({\mathrm e}^{2+x}-25\right )\right )-\frac {25 \,{\mathrm e}^{-2} \ln \left ({\mathrm e}^{2} {\mathrm e}^{x}-25\right )}{2 \left (2 \,{\mathrm e}^{2}+25\right )}+\frac {25 \,{\mathrm e}^{-2} \ln \left ({\mathrm e}^{x}+2\right )}{2 \left (2 \,{\mathrm e}^{2}+25\right )}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*exp(2+x)-25*exp(x))*ln(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp(2))*exp(2+x))/((2*exp(2)*exp(x)+4*exp
(2))*exp(2+x)-50*exp(2)*exp(x)-100*exp(2))/ln(exp(2+x)-25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.58, size = 25, normalized size = 0.96 \begin {gather*} \frac {1}{2} \, e^{\left (-2\right )} \log \left (2 \, e^{2} + e^{\left (x + 2\right )}\right ) - \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp(2))*exp(2+x))/((2*exp(2)*exp(x
)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-100*exp(2))/log(exp(2+x)-25),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.65, size = 21, normalized size = 0.81 \begin {gather*} \frac {{\mathrm {e}}^{-2}\,\ln \left ({\mathrm {e}}^x+2\right )}{2}-\ln \left (\ln \left ({\mathrm {e}}^2\,{\mathrm {e}}^x-25\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + 2)*(4*exp(2) + 2*exp(2)*exp(x)) + log(exp(x + 2) - 25)*(25*exp(x) - exp(x + 2)*exp(x)))/(log(exp(
x + 2) - 25)*(100*exp(2) - exp(x + 2)*(4*exp(2) + 2*exp(2)*exp(x)) + 50*exp(2)*exp(x))),x)

[Out]

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

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sympy [A]  time = 0.22, size = 22, normalized size = 0.85 \begin {gather*} \frac {\log {\left (e^{x} + 2 \right )}}{2 e^{2}} - \log {\left (\log {\left (e^{2} e^{x} - 25 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(2+x)-25*exp(x))*ln(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp(2))*exp(2+x))/((2*exp(2)*exp(x)
+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-100*exp(2))/ln(exp(2+x)-25),x)

[Out]

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

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