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

Optimal. Leaf size=21 \[ e^{e^{x/5}}+x+\frac {\log (5)}{x}-\log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2282, 2194} \begin {gather*} x+e^{e^{x/5}}-\log (x)+\frac {\log (5)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^E^(x/5) + x + Log[5]/x - Log[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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 21, normalized size = 1.00 \begin {gather*} e^{e^{x/5}}+x+\frac {\log (5)}{x}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^E^(x/5) + x + Log[5]/x - Log[x]

________________________________________________________________________________________

fricas [B]  time = 0.95, size = 42, normalized size = 2.00 \begin {gather*} -\frac {{\left (x e^{\left (\frac {1}{5} \, x\right )} \log \relax (x) - {\left (x^{2} + \log \relax (5)\right )} e^{\left (\frac {1}{5} \, x\right )} - x e^{\left (\frac {1}{5} \, x + e^{\left (\frac {1}{5} \, x\right )}\right )}\right )} e^{\left (-\frac {1}{5} \, x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x^2*exp(1/5*x)*exp(exp(1/5*x))-5*log(5)+5*x^2-5*x)/x^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.23, size = 44, normalized size = 2.10 \begin {gather*} \frac {{\left (x^{2} e^{\left (\frac {1}{5} \, x\right )} - x e^{\left (\frac {1}{5} \, x\right )} \log \relax (x) + x e^{\left (\frac {1}{5} \, x + e^{\left (\frac {1}{5} \, x\right )}\right )} + e^{\left (\frac {1}{5} \, x\right )} \log \relax (5)\right )} e^{\left (-\frac {1}{5} \, x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x^2*exp(1/5*x)*exp(exp(1/5*x))-5*log(5)+5*x^2-5*x)/x^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 18, normalized size = 0.86

method	result	size
default	\({\mathrm e}^{{\mathrm e}^{\frac {x}{5}}}-\ln \relax (x )+x +\frac {\ln \relax (5)}{x}\)	\(18\)
risch	\({\mathrm e}^{{\mathrm e}^{\frac {x}{5}}}-\ln \relax (x )+x +\frac {\ln \relax (5)}{x}\)	\(18\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{{\mathrm e}^{\frac {x}{5}}}+\ln \relax (5)}{x}-\ln \relax (x )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(x^2*exp(1/5*x)*exp(exp(1/5*x))-5*ln(5)+5*x^2-5*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(exp(1/5*x))-ln(x)+x+ln(5)/x

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 17, normalized size = 0.81 \begin {gather*} x + \frac {\log \relax (5)}{x} + e^{\left (e^{\left (\frac {1}{5} \, x\right )}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x^2*exp(1/5*x)*exp(exp(1/5*x))-5*log(5)+5*x^2-5*x)/x^2,x, algorithm="maxima")

[Out]

x + log(5)/x + e^(e^(1/5*x)) - log(x)

________________________________________________________________________________________

mupad [B]  time = 5.18, size = 17, normalized size = 0.81 \begin {gather*} x+{\mathrm {e}}^{{\left ({\mathrm {e}}^x\right )}^{1/5}}-\ln \relax (x)+\frac {\ln \relax (5)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(exp(x)^(1/5)) - log(x) + log(5)/x

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 15, normalized size = 0.71 \begin {gather*} x + e^{e^{\frac {x}{5}}} - \log {\relax (x )} + \frac {\log {\relax (5 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x**2*exp(1/5*x)*exp(exp(1/5*x))-5*ln(5)+5*x**2-5*x)/x**2,x)

[Out]

x + exp(exp(x/5)) - log(x) + log(5)/x

________________________________________________________________________________________