\(\int \frac {e^{-3-x} (e^{2 x} (x^2+x^3)+e^{2+x} (3 e+2 e^{2 x} x^2 \log (5)))}{x^2} \, dx\) [1467]

Optimal result

Integrand size = 47, antiderivative size = 31 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=6-\frac {3-e^{-1+2 x} x \left (e^{-2-x} x+\log (5)\right )}{x} \]

[Out]

6-(3-exp(x)^2*(ln(5)+x/exp(2+x))/exp(1)*x)/x

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6820, 2207, 2225} \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=e^{x-3} (x+1)-e^{x-3}-\frac {3}{x}+\frac {1}{2} e^{2 x-1} \log (25) \]

[In]

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

[Out]

-E^(-3 + x) - 3/x + E^(-3 + x)*(1 + x) + (E^(-1 + 2*x)*Log[25])/2

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{x^2}+e^{-3+x} (1+x)+e^{-1+2 x} \log (25)\right ) \, dx \\ & = -\frac {3}{x}+\log (25) \int e^{-1+2 x} \, dx+\int e^{-3+x} (1+x) \, dx \\ & = -\frac {3}{x}+e^{-3+x} (1+x)+\frac {1}{2} e^{-1+2 x} \log (25)-\int e^{-3+x} \, dx \\ & = -e^{-3+x}-\frac {3}{x}+e^{-3+x} (1+x)+\frac {1}{2} e^{-1+2 x} \log (25) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=-\frac {3}{x}+e^{-3+x} x+\frac {1}{2} e^{-1+2 x} \log (25) \]

[In]

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

[Out]

-3/x + E^(-3 + x)*x + (E^(-1 + 2*x)*Log[25])/2

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {3}{x}+\ln \left (5\right ) {\mathrm e}^{-1+2 x}+x \,{\mathrm e}^{-3+x}\)	\(22\)
parts	\({\mathrm e}^{x} {\mathrm e}^{-1} {\mathrm e}^{-2} x +{\mathrm e}^{-1} \ln \left (5\right ) {\mathrm e}^{2 x}-\frac {3}{x}\)	\(30\)
default	\({\mathrm e}^{-1} \left ({\mathrm e}^{-2} {\mathrm e}^{x}+{\mathrm e}^{-2} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-\frac {3 \,{\mathrm e}}{x}+\ln \left (5\right ) {\mathrm e}^{2 x}\right )\)	\(42\)
norman	\(\frac {\left ({\mathrm e}^{-1} {\mathrm e}^{-2} x^{2} {\mathrm e}^{2 x}+{\mathrm e}^{-1} \ln \left (5\right ) x \,{\mathrm e}^{3 x}-3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\)	\(42\)
parallelrisch	\(\frac {{\mathrm e}^{-1} \left (\ln \left (5\right ) {\mathrm e}^{2+x} {\mathrm e}^{2 x} x +{\mathrm e}^{2 x} x^{2}-3 \,{\mathrm e} \,{\mathrm e}^{2+x}\right ) {\mathrm e}^{-2-x}}{x}\)	\(44\)

[In]

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

[Out]

-3/x+ln(5)*exp(-1+2*x)+x*exp(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=\frac {{\left (x^{2} e^{\left (x + 2\right )} + x e^{\left (2 \, x + 4\right )} \log \left (5\right ) - 3 \, e^{5}\right )} e^{\left (-5\right )}}{x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=\frac {e x \sqrt {e^{2 x}} + e^{3} e^{2 x} \log {\left (5 \right )}}{e^{4}} - \frac {3}{x} \]

[In]

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

[Out]

(E*x*sqrt(exp(2*x)) + exp(3)*exp(2*x)*log(5))*exp(-4) - 3/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx={\left (x - 1\right )} e^{\left (x - 3\right )} + e^{\left (2 \, x - 1\right )} \log \left (5\right ) - \frac {3}{x} + e^{\left (x - 3\right )} \]

[In]

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

[Out]

(x - 1)*e^(x - 3) + e^(2*x - 1)*log(5) - 3/x + e^(x - 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.13 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=\frac {{\left (x + 2\right )}^{2} e^{\left (x + 2\right )} + {\left (x + 2\right )} e^{\left (2 \, x + 4\right )} \log \left (5\right ) - 4 \, {\left (x + 2\right )} e^{\left (x + 2\right )} - 2 \, e^{\left (2 \, x + 4\right )} \log \left (5\right ) - 3 \, e^{5} + 4 \, e^{\left (x + 2\right )}}{{\left (x + 2\right )} e^{5} - 2 \, e^{5}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x-\frac {3}{x}+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-1}\,\ln \left (5\right ) \]

[In]

int((exp(-1)*exp(- x - 2)*(exp(x + 2)*(3*exp(1) + 2*x^2*exp(2*x)*log(5)) + exp(2*x)*(x^2 + x^3)))/x^2,x)

[Out]

x*exp(-3)*exp(x) - 3/x + exp(2*x)*exp(-1)*log(5)