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

Optimal. Leaf size=14 \[ \frac {e^{2 x}}{\log ^2(2-x)} \]

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Rubi [B]  time = 0.31, antiderivative size = 39, normalized size of antiderivative = 2.79, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6741, 12, 2288} \begin {gather*} \frac {e^{2 x} (2 \log (2-x)-x \log (2-x))}{(2-x) \log ^3(2-x)} \end {gather*}

Antiderivative was successfully verified.

[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 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 6741

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} \frac {e^{2 x}}{\log ^2(2-x)} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.58, size = 13, normalized size = 0.93 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{\log \left (-x + 2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x)/log(-x + 2)^2

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giac [A]  time = 0.68, size = 13, normalized size = 0.93 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{\log \left (-x + 2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x)/log(-x + 2)^2

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maple [A]  time = 0.28, size = 14, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.72, size = 13, normalized size = 0.93 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{\log \left (-x + 2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x)/log(-x + 2)^2

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mupad [B]  time = 0.79, size = 13, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}}{{\ln \left (2-x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.28, size = 10, normalized size = 0.71 \begin {gather*} \frac {e^{2 x}}{\log {\left (2 - x \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*x)/log(2 - x)**2

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