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

Optimal. Leaf size=20 \[ e^3-e^{2 e^x}+e^x+\frac {1}{\log ^4(x)} \]

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Rubi [A]  time = 0.23, antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6688, 2194, 2282, 2302, 30} \begin {gather*} -e^{2 e^x}+e^x+\frac {1}{\log ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[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 2194

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 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 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.85 \begin {gather*} -e^{2 e^x}+e^x+\frac {1}{\log ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.51, size = 34, normalized size = 1.70 \begin {gather*} \frac {{\left (e^{\left (2 \, x\right )} \log \relax (x)^{4} - e^{\left (x + 2 \, e^{x}\right )} \log \relax (x)^{4} + e^{x}\right )} e^{\left (-x\right )}}{\log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.25, size = 34, normalized size = 1.70 \begin {gather*} \frac {{\left (e^{\left (2 \, x\right )} \log \relax (x)^{4} - e^{\left (x + 2 \, e^{x}\right )} \log \relax (x)^{4} + e^{x}\right )} e^{\left (-x\right )}}{\log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 15, normalized size = 0.75

method	result	size
default	\({\mathrm e}^{x}-{\mathrm e}^{2 \,{\mathrm e}^{x}}+\frac {1}{\ln \relax (x )^{4}}\)	\(15\)
risch	\({\mathrm e}^{x}-{\mathrm e}^{2 \,{\mathrm e}^{x}}+\frac {1}{\ln \relax (x )^{4}}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.39, size = 14, normalized size = 0.70 \begin {gather*} \frac {1}{\log \relax (x)^{4}} + e^{x} - e^{\left (2 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.19, size = 14, normalized size = 0.70 \begin {gather*} {\mathrm {e}}^x-{\mathrm {e}}^{2\,{\mathrm {e}}^x}+\frac {1}{{\ln \relax (x)}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.35, size = 15, normalized size = 0.75 \begin {gather*} e^{x} - e^{2 e^{x}} + \frac {1}{\log {\relax (x )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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