\(\int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx\) [6669]

Optimal result

Integrand size = 35, antiderivative size = 20 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=e^3-e^{2 e^x}+e^x+\frac {1}{\log ^4(x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6820, 2225, 2320, 2339, 30} \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=-e^{2 e^x}+e^x+\frac {1}{\log ^4(x)} \]

[In]

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

[Out]

-E^(2*E^x) + E^x + Log[x]^(-4)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

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 2339

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 6820

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=-e^{2 e^x}+e^x+\frac {1}{\log ^4(x)} \]

[In]

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

[Out]

-E^(2*E^x) + E^x + Log[x]^(-4)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
default	\({\mathrm e}^{x}-{\mathrm e}^{2 \,{\mathrm e}^{x}}+\frac {1}{\ln \left (x \right )^{4}}\)	\(15\)
risch	\({\mathrm e}^{x}-{\mathrm e}^{2 \,{\mathrm e}^{x}}+\frac {1}{\ln \left (x \right )^{4}}\)	\(15\)
parts	\({\mathrm e}^{x}-{\mathrm e}^{2 \,{\mathrm e}^{x}}+\frac {1}{\ln \left (x \right )^{4}}\)	\(15\)
parallelrisch	\(-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}} \ln \left (x \right )^{4}-{\mathrm e}^{x} \ln \left (x \right )^{4}-1}{\ln \left (x \right )^{4}}\)	\(27\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (16) = 32\).

Time = 0.41 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=\frac {{\left (e^{\left (2 \, x\right )} \log \left (x\right )^{4} - e^{\left (x + 2 \, e^{x}\right )} \log \left (x\right )^{4} + e^{x}\right )} e^{\left (-x\right )}}{\log \left (x\right )^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=e^{x} - e^{2 e^{x}} + \frac {1}{\log {\left (x \right )}^{4}} \]

[In]

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

[Out]

exp(x) - exp(2*exp(x)) + log(x)**(-4)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=\frac {1}{\log \left (x\right )^{4}} + e^{x} - e^{\left (2 \, e^{x}\right )} \]

[In]

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

[Out]

1/log(x)^4 + e^x - e^(2*e^x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (16) = 32\).

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx=\frac {{\left (e^{\left (2 \, x\right )} \log \left (x\right )^{4} - e^{\left (x + 2 \, e^{x}\right )} \log \left (x\right )^{4} + e^{x}\right )} e^{\left (-x\right )}}{\log \left (x\right )^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.48 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-4+e^x x \log ^5(x)-2 e^{2 e^x+x} x \log ^5(x)}{x \log ^5(x)} \, dx={\mathrm {e}}^x-{\mathrm {e}}^{2\,{\mathrm {e}}^x}+\frac {1}{{\ln \left (x\right )}^4} \]

[In]

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

[Out]

exp(x) - exp(2*exp(x)) + 1/log(x)^4