\(\int \frac {e^{-2 x} (-8 x+(8 x-8 x^2) \log (-\frac {2 x}{5}))}{25 \log ^3(-\frac {2 x}{5})} \, dx\) [3938]

Optimal result

Integrand size = 37, antiderivative size = 22 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=3+\frac {4 e^{-2 x} x^2}{25 \log ^2\left (-\frac {2 x}{5}\right )} \]

[Out]

4/25/ln(-2/5*x)^2/exp(x)^2*x^2+3

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 6873, 2326} \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=\frac {4 e^{-2 x} x^2}{25 \log ^2\left (-\frac {2 x}{5}\right )} \]

[In]

Int[(-8*x + (8*x - 8*x^2)*Log[(-2*x)/5])/(25*E^(2*x)*Log[(-2*x)/5]^3),x]

[Out]

(4*x^2)/(25*E^(2*x)*Log[(-2*x)/5]^2)

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}& = \frac {1}{25} \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{\log ^3\left (-\frac {2 x}{5}\right )} \, dx \\ & = \frac {1}{25} \int \frac {8 e^{-2 x} x \left (-1+\log \left (-\frac {2 x}{5}\right )-x \log \left (-\frac {2 x}{5}\right )\right )}{\log ^3\left (-\frac {2 x}{5}\right )} \, dx \\ & = \frac {8}{25} \int \frac {e^{-2 x} x \left (-1+\log \left (-\frac {2 x}{5}\right )-x \log \left (-\frac {2 x}{5}\right )\right )}{\log ^3\left (-\frac {2 x}{5}\right )} \, dx \\ & = \frac {4 e^{-2 x} x^2}{25 \log ^2\left (-\frac {2 x}{5}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=\frac {4 e^{-2 x} x^2}{25 \log ^2\left (-\frac {2 x}{5}\right )} \]

[In]

Integrate[(-8*x + (8*x - 8*x^2)*Log[(-2*x)/5])/(25*E^(2*x)*Log[(-2*x)/5]^3),x]

[Out]

(4*x^2)/(25*E^(2*x)*Log[(-2*x)/5]^2)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-2 x} x^{2}}{25 \ln \left (-\frac {2 x}{5}\right )^{2}}\)	\(16\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{-2 x} x^{2}}{25 \ln \left (-\frac {2 x}{5}\right )^{2}}\)	\(16\)

[In]

int(1/25*((-8*x^2+8*x)*ln(-2/5*x)-8*x)/exp(x)^2/ln(-2/5*x)^3,x,method=_RETURNVERBOSE)

[Out]

4/25/ln(-2/5*x)^2*exp(-2*x)*x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=\frac {4 \, x^{2} e^{\left (-2 \, x\right )}}{25 \, \log \left (-\frac {2}{5} \, x\right )^{2}} \]

[In]

integrate(1/25*((-8*x^2+8*x)*log(-2/5*x)-8*x)/exp(x)^2/log(-2/5*x)^3,x, algorithm="fricas")

[Out]

4/25*x^2*e^(-2*x)/log(-2/5*x)^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=\frac {4 x^{2} e^{- 2 x}}{25 \log {\left (- \frac {2 x}{5} \right )}^{2}} \]

[In]

integrate(1/25*((-8*x**2+8*x)*ln(-2/5*x)-8*x)/exp(x)**2/ln(-2/5*x)**3,x)

[Out]

4*x**2*exp(-2*x)/(25*log(-2*x/5)**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.64 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=-\frac {4 \, x^{2}}{25 \, {\left (2 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} e^{\left (2 \, x\right )} \log \left (-x\right ) - e^{\left (2 \, x\right )} \log \left (-x\right )^{2} - {\left (\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2}\right )} e^{\left (2 \, x\right )}\right )}} \]

[In]

integrate(1/25*((-8*x^2+8*x)*log(-2/5*x)-8*x)/exp(x)^2/log(-2/5*x)^3,x, algorithm="maxima")

[Out]

-4/25*x^2/(2*(log(5) - log(2))*e^(2*x)*log(-x) - e^(2*x)*log(-x)^2 - (log(5)^2 - 2*log(5)*log(2) + log(2)^2)*e
^(2*x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=\frac {4 \, x^{2} e^{\left (-2 \, x\right )}}{25 \, \log \left (-\frac {2}{5} \, x\right )^{2}} \]

[In]

integrate(1/25*((-8*x^2+8*x)*log(-2/5*x)-8*x)/exp(x)^2/log(-2/5*x)^3,x, algorithm="giac")

[Out]

4/25*x^2*e^(-2*x)/log(-2/5*x)^2

Mupad [B] (verification not implemented)

Time = 9.80 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-2 x} \left (-8 x+\left (8 x-8 x^2\right ) \log \left (-\frac {2 x}{5}\right )\right )}{25 \log ^3\left (-\frac {2 x}{5}\right )} \, dx=\frac {4\,x^2\,{\mathrm {e}}^{-2\,x}}{25\,{\ln \left (-\frac {2\,x}{5}\right )}^2} \]

[In]

int(-(exp(-2*x)*((8*x)/25 - (log(-(2*x)/5)*(8*x - 8*x^2))/25))/log(-(2*x)/5)^3,x)

[Out]

(4*x^2*exp(-2*x))/(25*log(-(2*x)/5)^2)