\(\int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx\) [496]

Optimal result

Integrand size = 39, antiderivative size = 14 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{2 x}}{\log ^2(2-x)} \]

[Out]

exp(x)^2/ln(2-x)^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(39\) vs. \(2(14)=28\).

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.79, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6873, 12, 2326} \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{2 x} (2 \log (2-x)-x \log (2-x))}{(2-x) \log ^3(2-x)} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

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 6873

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{2 x}}{\log ^2(2-x)} \]

[In]

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

[Out]

E^(2*x)/Log[2 - x]^2

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{2 x}}{\ln \left (2-x \right )^{2}}\)	\(14\)
parallelrisch	\(\frac {{\mathrm e}^{2 x}}{\ln \left (2-x \right )^{2}}\)	\(14\)

[In]

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

[Out]

exp(2*x)/ln(2-x)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{\left (2 \, x\right )}}{\log \left (-x + 2\right )^{2}} \]

[In]

integrate(((2*x-4)*exp(x)^2*log(2-x)-2*exp(x)^2)/(-2+x)/log(2-x)^3,x, algorithm="fricas")

[Out]

e^(2*x)/log(-x + 2)^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{2 x}}{\log {\left (2 - x \right )}^{2}} \]

[In]

integrate(((2*x-4)*exp(x)**2*ln(2-x)-2*exp(x)**2)/(-2+x)/ln(2-x)**3,x)

[Out]

exp(2*x)/log(2 - x)**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{\left (2 \, x\right )}}{\log \left (-x + 2\right )^{2}} \]

[In]

integrate(((2*x-4)*exp(x)^2*log(2-x)-2*exp(x)^2)/(-2+x)/log(2-x)^3,x, algorithm="maxima")

[Out]

e^(2*x)/log(-x + 2)^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {e^{\left (2 \, x\right )}}{\log \left (-x + 2\right )^{2}} \]

[In]

integrate(((2*x-4)*exp(x)^2*log(2-x)-2*exp(x)^2)/(-2+x)/log(2-x)^3,x, algorithm="giac")

[Out]

e^(2*x)/log(-x + 2)^2

Mupad [B] (verification not implemented)

Time = 7.69 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-2 e^{2 x}+e^{2 x} (-4+2 x) \log (2-x)}{(-2+x) \log ^3(2-x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{{\ln \left (2-x\right )}^2} \]

[In]

int(-(2*exp(2*x) - exp(2*x)*log(2 - x)*(2*x - 4))/(log(2 - x)^3*(x - 2)),x)

[Out]

exp(2*x)/log(2 - x)^2