3.6.16 \(\int \frac {16 x^5-8 x^6+e^{\frac {-16 x+\log ^2(-2+x)}{x}} (16 x-8 x^2+8 x \log (-2+x)+(8-4 x) \log ^2(-2+x))}{-10 x^4+5 x^5} \, dx\)

Optimal. Leaf size=28 \[ \frac {4}{5} \left (\frac {e^{-16+\frac {\log ^2(-2+x)}{x}}}{x^2}-x^2\right ) \]

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Rubi [B]  time = 1.08, antiderivative size = 90, normalized size of antiderivative = 3.21, number of steps used = 4, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1593, 6742, 2288} \begin {gather*} \frac {4 e^{\frac {\log ^2(x-2)}{x}-16} \left (-x \log ^2(x-2)+2 \log ^2(x-2)+2 x \log (x-2)\right )}{5 (2-x) x^4 \left (\frac {\log ^2(x-2)}{x^2}+\frac {2 \log (x-2)}{(2-x) x}\right )}-\frac {4 x^2}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x^5 - 8*x^6 + E^((-16*x + Log[-2 + x]^2)/x)*(16*x - 8*x^2 + 8*x*Log[-2 + x] + (8 - 4*x)*Log[-2 + x]^2)
)/(-10*x^4 + 5*x^5),x]

[Out]

(-4*x^2)/5 + (4*E^(-16 + Log[-2 + x]^2/x)*(2*x*Log[-2 + x] + 2*Log[-2 + x]^2 - x*Log[-2 + x]^2))/(5*(2 - x)*x^
4*((2*Log[-2 + x])/((2 - x)*x) + Log[-2 + x]^2/x^2))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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 {16 x^5-8 x^6+e^{\frac {-16 x+\log ^2(-2+x)}{x}} \left (16 x-8 x^2+8 x \log (-2+x)+(8-4 x) \log ^2(-2+x)\right )}{x^4 (-10+5 x)} \, dx\\ &=\int \left (-\frac {8 x}{5}-\frac {4 e^{-16+\frac {\log ^2(-2+x)}{x}} \left (-4 x+2 x^2-2 x \log (-2+x)-2 \log ^2(-2+x)+x \log ^2(-2+x)\right )}{5 (-2+x) x^4}\right ) \, dx\\ &=-\frac {4 x^2}{5}-\frac {4}{5} \int \frac {e^{-16+\frac {\log ^2(-2+x)}{x}} \left (-4 x+2 x^2-2 x \log (-2+x)-2 \log ^2(-2+x)+x \log ^2(-2+x)\right )}{(-2+x) x^4} \, dx\\ &=-\frac {4 x^2}{5}+\frac {4 e^{-16+\frac {\log ^2(-2+x)}{x}} \left (2 x \log (-2+x)+2 \log ^2(-2+x)-x \log ^2(-2+x)\right )}{5 (2-x) x^4 \left (\frac {2 \log (-2+x)}{(2-x) x}+\frac {\log ^2(-2+x)}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 29, normalized size = 1.04 \begin {gather*} \frac {1}{5} \left (\frac {4 e^{-16+\frac {\log ^2(-2+x)}{x}}}{x^2}-4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x^5 - 8*x^6 + E^((-16*x + Log[-2 + x]^2)/x)*(16*x - 8*x^2 + 8*x*Log[-2 + x] + (8 - 4*x)*Log[-2 +
x]^2))/(-10*x^4 + 5*x^5),x]

[Out]

((4*E^(-16 + Log[-2 + x]^2/x))/x^2 - 4*x^2)/5

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fricas [A]  time = 0.72, size = 26, normalized size = 0.93 \begin {gather*} -\frac {4 \, {\left (x^{4} - e^{\left (\frac {\log \left (x - 2\right )^{2} - 16 \, x}{x}\right )}\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*log(x-2)^2+8*x*log(x-2)-8*x^2+16*x)*exp((log(x-2)^2-16*x)/x)-8*x^6+16*x^5)/(5*x^5-10*x^4)
,x, algorithm="fricas")

[Out]

-4/5*(x^4 - e^((log(x - 2)^2 - 16*x)/x))/x^2

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giac [A]  time = 0.44, size = 26, normalized size = 0.93 \begin {gather*} -\frac {4 \, {\left (x^{4} - e^{\left (\frac {\log \left (x - 2\right )^{2} - 16 \, x}{x}\right )}\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*log(x-2)^2+8*x*log(x-2)-8*x^2+16*x)*exp((log(x-2)^2-16*x)/x)-8*x^6+16*x^5)/(5*x^5-10*x^4)
,x, algorithm="giac")

[Out]

-4/5*(x^4 - e^((log(x - 2)^2 - 16*x)/x))/x^2

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maple [A]  time = 0.27, size = 27, normalized size = 0.96

method	result	size
risch	\(-\frac {4 x^{2}}{5}+\frac {4 \,{\mathrm e}^{\frac {\ln \left (x -2\right )^{2}-16 x}{x}}}{5 x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x+8)*ln(x-2)^2+8*x*ln(x-2)-8*x^2+16*x)*exp((ln(x-2)^2-16*x)/x)-8*x^6+16*x^5)/(5*x^5-10*x^4),x,method
=_RETURNVERBOSE)

[Out]

-4/5*x^2+4/5/x^2*exp((ln(x-2)^2-16*x)/x)

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maxima [A]  time = 1.23, size = 24, normalized size = 0.86 \begin {gather*} -\frac {4}{5} \, x^{2} + \frac {4 \, e^{\left (\frac {\log \left (x - 2\right )^{2}}{x} - 16\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*log(x-2)^2+8*x*log(x-2)-8*x^2+16*x)*exp((log(x-2)^2-16*x)/x)-8*x^6+16*x^5)/(5*x^5-10*x^4)
,x, algorithm="maxima")

[Out]

-4/5*x^2 + 4/5*e^(log(x - 2)^2/x - 16)/x^2

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mupad [B]  time = 0.59, size = 24, normalized size = 0.86 \begin {gather*} \frac {4\,{\mathrm {e}}^{\frac {{\ln \left (x-2\right )}^2}{x}}\,{\mathrm {e}}^{-16}}{5\,x^2}-\frac {4\,x^2}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x^5 - 8*x^6 + exp(-(16*x - log(x - 2)^2)/x)*(16*x + 8*x*log(x - 2) - log(x - 2)^2*(4*x - 8) - 8*x^2))
/(10*x^4 - 5*x^5),x)

[Out]

(4*exp(log(x - 2)^2/x)*exp(-16))/(5*x^2) - (4*x^2)/5

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sympy [A]  time = 0.44, size = 26, normalized size = 0.93 \begin {gather*} - \frac {4 x^{2}}{5} + \frac {4 e^{\frac {- 16 x + \log {\left (x - 2 \right )}^{2}}{x}}}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*ln(x-2)**2+8*x*ln(x-2)-8*x**2+16*x)*exp((ln(x-2)**2-16*x)/x)-8*x**6+16*x**5)/(5*x**5-10*x
**4),x)

[Out]

-4*x**2/5 + 4*exp((-16*x + log(x - 2)**2)/x)/(5*x**2)

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