3.90.65 \(\int \frac {e^x (4 x-8 x^2)+e (-2 x^2-x^3)+(-4 e^x x+e (-4 x-2 x^2)) \log (\frac {2}{x})+e (-2-x) \log ^2(\frac {2}{x})}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log (\frac {2}{x})+5 e^{1+x} x^3 \log ^2(\frac {2}{x})} \, dx\)

Optimal. Leaf size=38 \[ \frac {2+\frac {\frac {e^{1-x}}{x}+x+\frac {4}{x+\log \left (\frac {2}{x}\right )}}{5 x}}{e} \]

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Rubi [A]  time = 1.26, antiderivative size = 33, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 12, 6742, 2197, 6687} \begin {gather*} \frac {e^{-x}}{5 x^2}+\frac {4}{5 e x \left (x+\log \left (\frac {2}{x}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(4*x - 8*x^2) + E*(-2*x^2 - x^3) + (-4*E^x*x + E*(-4*x - 2*x^2))*Log[2/x] + E*(-2 - x)*Log[2/x]^2)/(5
*E^(1 + x)*x^5 + 10*E^(1 + x)*x^4*Log[2/x] + 5*E^(1 + x)*x^3*Log[2/x]^2),x]

[Out]

1/(5*E^x*x^2) + 4/(5*E*x*(x + Log[2/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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-1-x} \left (-x \left (e x (2+x)+e^x (-4+8 x)\right )-2 x \left (2 e^x+e (2+x)\right ) \log \left (\frac {2}{x}\right )-e (2+x) \log ^2\left (\frac {2}{x}\right )\right )}{5 x^3 \left (x+\log \left (\frac {2}{x}\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-1-x} \left (-x \left (e x (2+x)+e^x (-4+8 x)\right )-2 x \left (2 e^x+e (2+x)\right ) \log \left (\frac {2}{x}\right )-e (2+x) \log ^2\left (\frac {2}{x}\right )\right )}{x^3 \left (x+\log \left (\frac {2}{x}\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {e^{-x} (2+x)}{x^3}-\frac {4 \left (-1+2 x+\log \left (\frac {2}{x}\right )\right )}{e x^2 \left (x+\log \left (\frac {2}{x}\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{-x} (2+x)}{x^3} \, dx\right )-\frac {4 \int \frac {-1+2 x+\log \left (\frac {2}{x}\right )}{x^2 \left (x+\log \left (\frac {2}{x}\right )\right )^2} \, dx}{5 e}\\ &=\frac {e^{-x}}{5 x^2}+\frac {4}{5 e x \left (x+\log \left (\frac {2}{x}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 32, normalized size = 0.84 \begin {gather*} \frac {1}{5} \left (\frac {e^{-x}}{x^2}+\frac {4}{e x \left (x+\log \left (\frac {2}{x}\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(4*x - 8*x^2) + E*(-2*x^2 - x^3) + (-4*E^x*x + E*(-4*x - 2*x^2))*Log[2/x] + E*(-2 - x)*Log[2/x]
^2)/(5*E^(1 + x)*x^5 + 10*E^(1 + x)*x^4*Log[2/x] + 5*E^(1 + x)*x^3*Log[2/x]^2),x]

[Out]

(1/(E^x*x^2) + 4/(E*x*(x + Log[2/x])))/5

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fricas [A]  time = 0.59, size = 48, normalized size = 1.26 \begin {gather*} \frac {x e^{2} + 4 \, x e^{\left (x + 1\right )} + e^{2} \log \left (\frac {2}{x}\right )}{5 \, {\left (x^{3} e^{\left (x + 2\right )} + x^{2} e^{\left (x + 2\right )} \log \left (\frac {2}{x}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log(2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2
)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x)^2+10*x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x, algorithm="fri
cas")

[Out]

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

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giac [A]  time = 0.25, size = 59, normalized size = 1.55 \begin {gather*} \frac {x e^{\left (-x + 1\right )} + e^{\left (-x + 1\right )} \log \relax (2) - e^{\left (-x + 1\right )} \log \relax (x) + 4 \, x}{5 \, {\left (x^{3} e + x^{2} e \log \relax (2) - x^{2} e \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log(2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2
)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x)^2+10*x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x, algorithm="gia
c")

[Out]

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

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maple [C]  time = 1.11, size = 36, normalized size = 0.95

method	result	size
risch	\(\frac {{\mathrm e}^{-x}}{5 x^{2}}+\frac {8 i {\mathrm e}^{-1}}{5 x \left (2 i \ln \relax (2)+2 i x -2 i \ln \relax (x )\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x-2)*exp(1)*ln(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*ln(2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2)*exp(1)
)/(5*x^3*exp(1)*exp(x)*ln(2/x)^2+10*x^4*exp(1)*exp(x)*ln(2/x)+5*x^5*exp(1)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

1/5*exp(-x)/x^2+8/5*I/x*exp(-1)/(2*I*ln(2)+2*I*x-2*I*ln(x))

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maxima [A]  time = 0.55, size = 53, normalized size = 1.39 \begin {gather*} \frac {{\left (x e + e \log \relax (2) - e \log \relax (x)\right )} e^{\left (-x\right )} + 4 \, x}{5 \, {\left (x^{3} e + x^{2} e \log \relax (2) - x^{2} e \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log(2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2
)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x)^2+10*x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 8.01, size = 50, normalized size = 1.32 \begin {gather*} \frac {x^2\,\left (\mathrm {e}+4\,{\mathrm {e}}^x\right )+x\,\mathrm {e}\,\ln \left (\frac {2}{x}\right )}{5\,x^4\,{\mathrm {e}}^{x+1}+5\,x^3\,{\mathrm {e}}^{x+1}\,\ln \left (\frac {2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1)*(2*x^2 + x^3) + log(2/x)*(exp(1)*(4*x + 2*x^2) + 4*x*exp(x)) - exp(x)*(4*x - 8*x^2) + exp(1)*log(
2/x)^2*(x + 2))/(5*x^5*exp(1)*exp(x) + 10*x^4*exp(1)*exp(x)*log(2/x) + 5*x^3*exp(1)*exp(x)*log(2/x)^2),x)

[Out]

(x^2*(exp(1) + 4*exp(x)) + x*exp(1)*log(2/x))/(5*x^4*exp(x + 1) + 5*x^3*exp(x + 1)*log(2/x))

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sympy [A]  time = 0.37, size = 31, normalized size = 0.82 \begin {gather*} \frac {4}{5 e x^{2} + 5 e x \log {\left (\frac {2}{x} \right )}} + \frac {e^{- x}}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(1)*ln(2/x)**2+(-4*exp(x)*x+(-2*x**2-4*x)*exp(1))*ln(2/x)+(-8*x**2+4*x)*exp(x)+(-x**3-2*x
**2)*exp(1))/(5*x**3*exp(1)*exp(x)*ln(2/x)**2+10*x**4*exp(1)*exp(x)*ln(2/x)+5*x**5*exp(1)*exp(x)),x)

[Out]

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

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