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

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

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Rubi [A]  time = 0.69, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.159, Rules used = {6742, 2282, 2194, 2176, 6688, 2302, 29} \begin {gather*} -\frac {1}{x^4}+x+e^{x+e^x-3}-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3 + E^x + x) - x^(-4) + x - Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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]]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-3+e^x+x}+e^{-3+e^x+2 x}+\frac {-x^4+4 \log (x)+x^5 \log (x)}{x^5 \log (x)}\right ) \, dx\\ &=\int e^{-3+e^x+x} \, dx+\int e^{-3+e^x+2 x} \, dx+\int \frac {-x^4+4 \log (x)+x^5 \log (x)}{x^5 \log (x)} \, dx\\ &=\int \left (1+\frac {4}{x^5}-\frac {1}{x \log (x)}\right ) \, dx+\operatorname {Subst}\left (\int e^{-3+x} \, dx,x,e^x\right )+\operatorname {Subst}\left (\int e^{-3+x} x \, dx,x,e^x\right )\\ &=e^{-3+e^x}+e^{-3+e^x+x}-\frac {1}{x^4}+x-\int \frac {1}{x \log (x)} \, dx-\operatorname {Subst}\left (\int e^{-3+x} \, dx,x,e^x\right )\\ &=e^{-3+e^x+x}-\frac {1}{x^4}+x-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=e^{-3+e^x+x}-\frac {1}{x^4}+x-\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 1.00 \begin {gather*} e^{-3+e^x+x}-\frac {1}{x^4}+x-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3 + E^x + x) - x^(-4) + x - Log[Log[x]]

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fricas [A]  time = 0.64, size = 27, normalized size = 1.35 \begin {gather*} \frac {x^{5} + x^{4} e^{\left (x + e^{x} - 3\right )} - x^{4} \log \left (\log \relax (x)\right ) - 1}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 54, normalized size = 2.70 \begin {gather*} \frac {{\left (x^{5} e^{\left (2 \, x + 3\right )} - x^{4} e^{\left (2 \, x + 3\right )} \log \left (\log \relax (x)\right ) + x^{4} e^{\left (3 \, x + e^{x}\right )} - e^{\left (2 \, x + 3\right )}\right )} e^{\left (-2 \, x - 3\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 22, normalized size = 1.10

method	result	size
risch	\(\frac {x^{5}-1}{x^{4}}-\ln \left (\ln \relax (x )\right )+{\mathrm e}^{{\mathrm e}^{x}-3+x}\)	\(22\)
default	\(x -\frac {1}{x^{4}}+{\mathrm e}^{-3} \left ({\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{{\mathrm e}^{x}}\right )-\ln \left (\ln \relax (x )\right )+{\mathrm e}^{-3} {\mathrm e}^{{\mathrm e}^{x}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.73, size = 27, normalized size = 1.35 \begin {gather*} {\left (e^{x} - 1\right )} e^{\left (e^{x} - 3\right )} + x - \frac {1}{x^{4}} + e^{\left (e^{x} - 3\right )} - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.27, size = 18, normalized size = 0.90 \begin {gather*} x+{\mathrm {e}}^{x+{\mathrm {e}}^x-3}-\ln \left (\ln \relax (x)\right )-\frac {1}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(x^5 + 4) - x^4 + exp(x + exp(x) - 3)*log(x)*(x^5*exp(x) + x^5))/(x^5*log(x)),x)

[Out]

x + exp(x + exp(x) - 3) - log(log(x)) - 1/x^4

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sympy [A]  time = 0.36, size = 19, normalized size = 0.95 \begin {gather*} x + e^{x + e^{x} - 3} - \log {\left (\log {\relax (x )} \right )} - \frac {1}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(x + exp(x) - 3) - log(log(x)) - 1/x**4

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