3.36.7 \(\int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log (\frac {4}{3})}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx\)

Optimal. Leaf size=32 \[ 4-e^x-\frac {\log \left (\frac {4}{3}\right )}{5 \left (-e^{e^4}+2 e^x\right )}+\log (2) \]

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Rubi [A]  time = 0.12, antiderivative size = 27, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {2282, 12, 683} \begin {gather*} \frac {\log \left (\frac {16}{9}\right )}{10 \left (e^{e^4}-2 e^x\right )}-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*E^(3*x) - 5*E^(2*E^4 + x) + 20*E^(E^4 + 2*x) + 2*E^x*Log[4/3])/(5*E^(2*E^4) + 20*E^(2*x) - 20*E^(E^4
+ x)),x]

[Out]

-E^x + Log[16/9]/(10*(E^E^4 - 2*E^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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {-5 e^{2 e^4}+20 e^{e^4} x-20 x^2+\log \left (\frac {16}{9}\right )}{5 \left (e^{e^4}-2 x\right )^2} \, dx,x,e^x\right )\\ &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {-5 e^{2 e^4}+20 e^{e^4} x-20 x^2+\log \left (\frac {16}{9}\right )}{\left (e^{e^4}-2 x\right )^2} \, dx,x,e^x\right )\\ &=\frac {1}{5} \operatorname {Subst}\left (\int \left (-5+\frac {\log \left (\frac {16}{9}\right )}{\left (e^{e^4}-2 x\right )^2}\right ) \, dx,x,e^x\right )\\ &=-e^x+\frac {\log \left (\frac {16}{9}\right )}{10 \left (e^{e^4}-2 e^x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 31, normalized size = 0.97 \begin {gather*} \frac {1}{5} \left (-5 e^x+\frac {\log \left (\frac {16}{9}\right )}{2 \left (e^{e^4}-2 e^x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*E^(3*x) - 5*E^(2*E^4 + x) + 20*E^(E^4 + 2*x) + 2*E^x*Log[4/3])/(5*E^(2*E^4) + 20*E^(2*x) - 20*E
^(E^4 + x)),x]

[Out]

(-5*E^x + Log[16/9]/(2*(E^E^4 - 2*E^x)))/5

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fricas [B]  time = 0.55, size = 50, normalized size = 1.56 \begin {gather*} -\frac {e^{\left (4 \, e^{4}\right )} \log \left (\frac {4}{3}\right ) + 10 \, e^{\left (2 \, x + 4 \, e^{4}\right )} - 5 \, e^{\left (x + 5 \, e^{4}\right )}}{5 \, {\left (2 \, e^{\left (x + 4 \, e^{4}\right )} - e^{\left (5 \, e^{4}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)*exp(exp(4))^2+20*exp(x)^2*exp(exp(4))-20*exp(x)^3+2*log(4/3)*exp(x))/(5*exp(exp(4))^2-20*
exp(x)*exp(exp(4))+20*exp(x)^2),x, algorithm="fricas")

[Out]

-1/5*(e^(4*e^4)*log(4/3) + 10*e^(2*x + 4*e^4) - 5*e^(x + 5*e^4))/(2*e^(x + 4*e^4) - e^(5*e^4))

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giac [A]  time = 0.16, size = 26, normalized size = 0.81 \begin {gather*} \frac {\log \relax (3) - 2 \, \log \relax (2)}{5 \, {\left (2 \, e^{x} - e^{\left (e^{4}\right )}\right )}} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)*exp(exp(4))^2+20*exp(x)^2*exp(exp(4))-20*exp(x)^3+2*log(4/3)*exp(x))/(5*exp(exp(4))^2-20*
exp(x)*exp(exp(4))+20*exp(x)^2),x, algorithm="giac")

[Out]

1/5*(log(3) - 2*log(2))/(2*e^x - e^(e^4)) - e^x

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maple [A]  time = 0.21, size = 34, normalized size = 1.06

method	result	size
norman	\(\frac {2 \,{\mathrm e}^{2 x}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2}+\frac {2 \ln \relax (2)}{5}-\frac {\ln \relax (3)}{5}}{{\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}}\)	\(34\)
risch	\(-{\mathrm e}^{x}+\frac {2 \ln \relax (2)}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}-\frac {\ln \relax (3)}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}\)	\(34\)
default	\(\frac {10 \,{\mathrm e}^{2 x}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{4}}}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-10 \,{\mathrm e}^{x}}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2 \,{\mathrm e}^{{\mathrm e}^{4}}-4 \,{\mathrm e}^{x}}+\frac {2 \ln \relax (2)}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}-\frac {\ln \relax (3)}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*exp(x)*exp(exp(4))^2+20*exp(x)^2*exp(exp(4))-20*exp(x)^3+2*ln(4/3)*exp(x))/(5*exp(exp(4))^2-20*exp(x)*
exp(exp(4))+20*exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

(2*exp(x)^2-1/2*exp(exp(4))^2+2/5*ln(2)-1/5*ln(3))/(exp(exp(4))-2*exp(x))

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maxima [A]  time = 0.58, size = 21, normalized size = 0.66 \begin {gather*} -\frac {\log \left (\frac {4}{3}\right )}{5 \, {\left (2 \, e^{x} - e^{\left (e^{4}\right )}\right )}} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)*exp(exp(4))^2+20*exp(x)^2*exp(exp(4))-20*exp(x)^3+2*log(4/3)*exp(x))/(5*exp(exp(4))^2-20*
exp(x)*exp(exp(4))+20*exp(x)^2),x, algorithm="maxima")

[Out]

-1/5*log(4/3)/(2*e^x - e^(e^4)) - e^x

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mupad [B]  time = 0.20, size = 41, normalized size = 1.28 \begin {gather*} \frac {10\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{x-{\mathrm {e}}^4}\,\left (5\,{\mathrm {e}}^{2\,{\mathrm {e}}^4}-\ln \left (\frac {16}{9}\right )\right )}{5\,{\mathrm {e}}^{{\mathrm {e}}^4}-10\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*exp(3*x) + 5*exp(2*exp(4))*exp(x) - 20*exp(2*x)*exp(exp(4)) - 2*exp(x)*log(4/3))/(5*exp(2*exp(4)) + 2
0*exp(2*x) - 20*exp(exp(4))*exp(x)),x)

[Out]

(10*exp(2*x) - exp(x - exp(4))*(5*exp(2*exp(4)) - log(16/9)))/(5*exp(exp(4)) - 10*exp(x))

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sympy [A]  time = 0.13, size = 22, normalized size = 0.69 \begin {gather*} - e^{x} + \frac {- 2 \log {\relax (2 )} + \log {\relax (3 )}}{10 e^{x} - 5 e^{e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)*exp(exp(4))**2+20*exp(x)**2*exp(exp(4))-20*exp(x)**3+2*ln(4/3)*exp(x))/(5*exp(exp(4))**2-
20*exp(x)*exp(exp(4))+20*exp(x)**2),x)

[Out]

-exp(x) + (-2*log(2) + log(3))/(10*exp(x) - 5*exp(exp(4)))

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