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

Optimal. Leaf size=31 \[ 6-\frac {3-e^{-1+2 x} x \left (e^{-2-x} x+\log (5)\right )}{x} \]

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Rubi [A]  time = 0.24, antiderivative size = 35, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 3, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6688, 2176, 2194} \begin {gather*} e^{x-3} (x+1)-e^{x-3}-\frac {3}{x}+\frac {1}{2} e^{2 x-1} \log (25) \end {gather*}

Antiderivative was successfully verified.

[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 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 0.84 \begin {gather*} -\frac {3}{x}+e^{-3+x} x+\frac {1}{2} e^{-1+2 x} \log (25) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.94, size = 29, normalized size = 0.94 \begin {gather*} \frac {{\left (x^{2} e^{\left (x + 2\right )} + x e^{\left (2 \, x + 4\right )} \log \relax (5) - 3 \, e^{5}\right )} e^{\left (-5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.29, size = 29, normalized size = 0.94 \begin {gather*} \frac {{\left (x^{2} e^{\left (x + 1\right )} + x e^{\left (2 \, x + 3\right )} \log \relax (5) - 3 \, e^{4}\right )} e^{\left (-4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*e^(x + 1) + x*e^(2*x + 3)*log(5) - 3*e^4)*e^(-4)/x

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maple [A]  time = 0.15, size = 22, normalized size = 0.71

method	result	size
risch	\(-\frac {3}{x}+\ln \relax (5) {\mathrm e}^{2 x -1}+x \,{\mathrm e}^{x -3}\)	\(22\)
default	\({\mathrm e}^{-1} \left ({\mathrm e}^{x} {\mathrm e}^{-2}+{\mathrm e}^{-2} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-\frac {3 \,{\mathrm e}}{x}+\ln \relax (5) {\mathrm e}^{2 x}\right )\)	\(42\)
norman	\(\frac {\left ({\mathrm e}^{-1} {\mathrm e}^{-2} x^{2} {\mathrm e}^{2 x}+{\mathrm e}^{-1} \ln \relax (5) x \,{\mathrm e}^{3 x}-3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\)	\(42\)
meijerg	\(\frac {2 \ln \relax (5) {\mathrm e}^{-x +1+x \,{\mathrm e}^{-2}} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{-2} \left (1-{\mathrm e}^{2}\right ) \left (1-\frac {2 \,{\mathrm e}^{2}}{1-{\mathrm e}^{2}}\right )}\right )}{\left (1-{\mathrm e}^{2}\right ) \left (1-\frac {2 \,{\mathrm e}^{2}}{1-{\mathrm e}^{2}}\right )}+3 \,{\mathrm e}^{-x -2+x \,{\mathrm e}^{-2}} \left (1-{\mathrm e}^{2}\right ) \left (\frac {{\mathrm e}^{2} \left (2-2 x \,{\mathrm e}^{-2} \left (1-{\mathrm e}^{2}\right )\right )}{2 x \left (1-{\mathrm e}^{2}\right )}-\frac {{\mathrm e}^{2-x \,{\mathrm e}^{-2} \left (1-{\mathrm e}^{2}\right )}}{x \left (1-{\mathrm e}^{2}\right )}+\ln \left (x \,{\mathrm e}^{-2} \left (1-{\mathrm e}^{2}\right )\right )+\expIntegralEi \left (1, x \,{\mathrm e}^{-2} \left (1-{\mathrm e}^{2}\right )\right )+3-\ln \relax (x )-\ln \left (1-{\mathrm e}^{2}\right )-\frac {{\mathrm e}^{2}}{x \left (1-{\mathrm e}^{2}\right )}\right )+\frac {{\mathrm e}^{-3-x +x \,{\mathrm e}^{-2}} \left (1-\frac {\left (2-2 x \left (2-{\mathrm e}^{-2}\right )\right ) {\mathrm e}^{x \left (2-{\mathrm e}^{-2}\right )}}{2}\right )}{\left (2-{\mathrm e}^{-2}\right )^{2}}-\frac {{\mathrm e}^{-3-x +x \,{\mathrm e}^{-2}} \left (1-{\mathrm e}^{x \left (2-{\mathrm e}^{-2}\right )}\right )}{2-{\mathrm e}^{-2}}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[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(2*x-1)+x*exp(x-3)

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maxima [A]  time = 0.35, size = 27, normalized size = 0.87 \begin {gather*} {\left (x - 1\right )} e^{\left (x - 3\right )} + e^{\left (2 \, x - 1\right )} \log \relax (5) - \frac {3}{x} + e^{\left (x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.08, size = 21, normalized size = 0.68 \begin {gather*} x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x-\frac {3}{x}+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-1}\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.18, size = 31, normalized size = 1.00 \begin {gather*} \frac {e x \sqrt {e^{2 x}} + e^{3} e^{2 x} \log {\relax (5 )}}{e^{4}} - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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