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

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

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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} $$\begin{array}{c}{e}^{x}x+5x-\frac{e^9}{x}-(e^x-\log \left(\frac{e^{e^x}}{5}\right))\log \left(e^x\right)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (e^{x}x+\frac{e^9+5x^2+x^2\log \left(\frac{e^{e^x}}{5}\right)}{x^2})dx\\ & =\int e^{x}xdx+\int \frac{e^9+5x^2+x^2\log \left(\frac{e^{e^x}}{5}\right)}{x^2}dx\\ & =e^{x}x-\int e^{x}dx+\int (\frac{e^9+5x^2}{x^2}+\log \left(\frac{e^{e^x}}{5}\right))dx\\ & =-e^x+e^{x}x+\int \frac{e^9+5x^2}{x^2}dx+\int \log \left(\frac{e^{e^x}}{5}\right)dx\\ & =-e^x+e^{x}x+\int (5+\frac{e^9}{x^2})dx+\mathrm{Subst}(\int \frac{\log \left(\frac{e^x}{5}\right)}{x}dx,x,e^x)\\ & =-\frac{e^9}{x}+5x+e^{x}x-(e^x-\log \left(\frac{e^{e^x}}{5}\right))\mathrm{Subst}(\int \frac{1}{x}dx,x,e^x)\\ & =-\frac{e^9}{x}+5x+e^{x}x-x(e^x-\log \left(\frac{e^{e^x}}{5}\right))\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{x^2{e}^x-x^2\log (5)+5x^2-e^9}{x}\end{array}$$

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 $$\begin{array}{c}\frac{x^2{e}^x-x^2\log (5)+5x^2-e^9}{x}\end{array}$$

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	$x\ln \left({\mathrm{e}}^{{\mathrm{e}}^x}\right)-\frac{2x^2\ln (5)-10x^2+2{\mathrm{e}}^9}{2x}$	$30$
default	$5x-\frac{{\mathrm{e}}^9}{x}+{\mathrm{e}}^{x}x+\ln \left({\mathrm{e}}^x\right)\ln \left(\frac{{\mathrm{e}}^{{\mathrm{e}}^x}}{5}\right)-{\mathrm{e}}^x\ln \left({\mathrm{e}}^x\right)$	$33$

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 $$\begin{array}{c}(x-1)e^x-x{e}^x+x\log \left(\frac{1}{5}{e}^{\left(e^x\right)}\right)+5x-\frac{e^9}{x}+e^x\end{array}$$

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 $$\begin{array}{c}x({\mathrm{e}}^x-\ln (5)+5)-\frac{{\mathrm{e}}^9}{x}\end{array}$$

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 $$\begin{array}{c}x{e}^x+x(5-\log (5))-\frac{e^9}{x}\end{array}$$

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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