3.67.53 \(\int \frac {e^9+5 x^2+e^x x^3+x^2 \log (\frac {e^{e^x}}{5})}{x^2} \, dx\)

Optimal. Leaf size=27 \[ x \left (5-\frac {e^9 (1-x)}{x^2}+\log \left (\frac {e^{e^x}}{5}\right )\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 11, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {14, 2176, 2194, 2282, 2158, 29} \begin {gather*} e^x x+5 x-\frac {e^9}{x}-\left (e^x-\log \left (\frac {e^{e^x}}{5}\right )\right ) \log \left (e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^9 + 5*x^2 + E^x*x^3 + x^2*Log[E^E^x/5])/x^2,x]

[Out]

-(E^9/x) + 5*x + E^x*x - (E^x - Log[E^E^x/5])*Log[E^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 29

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

Rule 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, 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]]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.89 \begin {gather*} -\frac {e^9}{x}+5 x+x \log \left (\frac {e^{e^x}}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^9 + 5*x^2 + E^x*x^3 + x^2*Log[E^E^x/5])/x^2,x]

[Out]

-(E^9/x) + 5*x + x*Log[E^E^x/5]

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fricas [A]  time = 0.52, size = 27, normalized size = 1.00 \begin {gather*} \frac {x^{2} e^{x} - x^{2} \log \relax (5) + 5 \, x^{2} - e^{9}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(1/5*exp(exp(x)))+exp(x)*x^3+exp(9)+5*x^2)/x^2,x, algorithm="fricas")

[Out]

(x^2*e^x - x^2*log(5) + 5*x^2 - e^9)/x

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giac [A]  time = 0.57, size = 27, normalized size = 1.00 \begin {gather*} \frac {x^{2} e^{x} - x^{2} \log \relax (5) + 5 \, x^{2} - e^{9}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(1/5*exp(exp(x)))+exp(x)*x^3+exp(9)+5*x^2)/x^2,x, algorithm="giac")

[Out]

(x^2*e^x - x^2*log(5) + 5*x^2 - e^9)/x

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maple [A]  time = 0.06, size = 30, normalized size = 1.11

method	result	size
risch	\(x \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-\frac {2 x^{2} \ln \relax (5)-10 x^{2}+2 \,{\mathrm e}^{9}}{2 x}\)	\(30\)
default	\(5 x -\frac {{\mathrm e}^{9}}{x}+{\mathrm e}^{x} x +\ln \left ({\mathrm e}^{x}\right ) \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{x}}}{5}\right )-{\mathrm e}^{x} \ln \left ({\mathrm e}^{x}\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(1/5*exp(exp(x)))+exp(x)*x^3+exp(9)+5*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x*ln(exp(exp(x)))-1/2*(2*x^2*ln(5)-10*x^2+2*exp(9))/x

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maxima [A]  time = 0.36, size = 32, normalized size = 1.19 \begin {gather*} {\left (x - 1\right )} e^{x} - x e^{x} + x \log \left (\frac {1}{5} \, e^{\left (e^{x}\right )}\right ) + 5 \, x - \frac {e^{9}}{x} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(1/5*exp(exp(x)))+exp(x)*x^3+exp(9)+5*x^2)/x^2,x, algorithm="maxima")

[Out]

(x - 1)*e^x - x*e^x + x*log(1/5*e^(e^x)) + 5*x - e^9/x + e^x

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mupad [B]  time = 0.11, size = 18, normalized size = 0.67 \begin {gather*} x\,\left ({\mathrm {e}}^x-\ln \relax (5)+5\right )-\frac {{\mathrm {e}}^9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(9) + x^3*exp(x) + x^2*log(exp(exp(x))/5) + 5*x^2)/x^2,x)

[Out]

x*(exp(x) - log(5) + 5) - exp(9)/x

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sympy [A]  time = 0.15, size = 15, normalized size = 0.56 \begin {gather*} x e^{x} + x \left (5 - \log {\relax (5 )}\right ) - \frac {e^{9}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(1/5*exp(exp(x)))+exp(x)*x**3+exp(9)+5*x**2)/x**2,x)

[Out]

x*exp(x) + x*(5 - log(5)) - exp(9)/x

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