\(\int \frac {e^x (1-2 x-2560 x^2)+1280 e^x x \log (2 x)+(-2 e^x x^2+e^x x \log (2 x)) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx\) [3628]

Optimal result

Integrand size = 71, antiderivative size = 21 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=e^x \left (5+\frac {1}{256} \log (2 x-\log (2 x))\right ) \]

[Out]

exp(x)*(1/256*ln(2*x-ln(2*x))+5)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(21)=42\).

Time = 0.81 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.29, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2641, 6873, 12, 2326} \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {e^x \left (2560 x^2+2 x^2 \log (2 x-\log (2 x))-1280 x \log (2 x)-x \log (2 x) \log (2 x-\log (2 x))\right )}{256 x (2 x-\log (2 x))} \]

[In]

Int[(E^x*(1 - 2*x - 2560*x^2) + 1280*E^x*x*Log[2*x] + (-2*E^x*x^2 + E^x*x*Log[2*x])*Log[2*x - Log[2*x]])/(-512
*x^2 + 256*x*Log[2*x]),x]

[Out]

(E^x*(2560*x^2 - 1280*x*Log[2*x] + 2*x^2*Log[2*x - Log[2*x]] - x*Log[2*x]*Log[2*x - Log[2*x]]))/(256*x*(2*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 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 2641

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{x (-512 x+256 \log (2 x))} \, dx \\ & = \int \frac {e^x \left (-1+2 x+2560 x^2-1280 x \log (2 x)+2 x^2 \log (2 x-\log (2 x))-x \log (2 x) \log (2 x-\log (2 x))\right )}{256 x (2 x-\log (2 x))} \, dx \\ & = \frac {1}{256} \int \frac {e^x \left (-1+2 x+2560 x^2-1280 x \log (2 x)+2 x^2 \log (2 x-\log (2 x))-x \log (2 x) \log (2 x-\log (2 x))\right )}{x (2 x-\log (2 x))} \, dx \\ & = \frac {e^x \left (2560 x^2-1280 x \log (2 x)+2 x^2 \log (2 x-\log (2 x))-x \log (2 x) \log (2 x-\log (2 x))\right )}{256 x (2 x-\log (2 x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {1}{256} e^x (1280+\log (2 x-\log (2 x))) \]

[In]

Integrate[(E^x*(1 - 2*x - 2560*x^2) + 1280*E^x*x*Log[2*x] + (-2*E^x*x^2 + E^x*x*Log[2*x])*Log[2*x - Log[2*x]])
/(-512*x^2 + 256*x*Log[2*x]),x]

[Out]

(E^x*(1280 + Log[2*x - Log[2*x]]))/256

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{x} \ln \left (2 x -\ln \left (2 x \right )\right )}{256}+5 \,{\mathrm e}^{x}\)	\(21\)
parallelrisch	\(\frac {{\mathrm e}^{x} \ln \left (2 x -\ln \left (2 x \right )\right )}{256}+5 \,{\mathrm e}^{x}\)	\(21\)

[In]

int(((x*exp(x)*ln(2*x)-2*exp(x)*x^2)*ln(2*x-ln(2*x))+1280*x*exp(x)*ln(2*x)+(-2560*x^2-2*x+1)*exp(x))/(256*x*ln
(2*x)-512*x^2),x,method=_RETURNVERBOSE)

[Out]

1/256*exp(x)*ln(2*x-ln(2*x))+5*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {1}{256} \, e^{x} \log \left (2 \, x - \log \left (2 \, x\right )\right ) + 5 \, e^{x} \]

[In]

integrate(((x*exp(x)*log(2*x)-2*exp(x)*x^2)*log(2*x-log(2*x))+1280*x*exp(x)*log(2*x)+(-2560*x^2-2*x+1)*exp(x))
/(256*x*log(2*x)-512*x^2),x, algorithm="fricas")

[Out]

1/256*e^x*log(2*x - log(2*x)) + 5*e^x

Sympy [A] (verification not implemented)

Time = 10.74 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {\left (\log {\left (2 x - \log {\left (2 x \right )} \right )} + 1280\right ) e^{x}}{256} \]

[In]

integrate(((x*exp(x)*ln(2*x)-2*exp(x)*x**2)*ln(2*x-ln(2*x))+1280*x*exp(x)*ln(2*x)+(-2560*x**2-2*x+1)*exp(x))/(
256*x*ln(2*x)-512*x**2),x)

[Out]

(log(2*x - log(2*x)) + 1280)*exp(x)/256

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {1}{256} \, e^{x} \log \left (2 \, x - \log \left (2\right ) - \log \left (x\right )\right ) + 5 \, e^{x} \]

[In]

integrate(((x*exp(x)*log(2*x)-2*exp(x)*x^2)*log(2*x-log(2*x))+1280*x*exp(x)*log(2*x)+(-2560*x^2-2*x+1)*exp(x))
/(256*x*log(2*x)-512*x^2),x, algorithm="maxima")

[Out]

1/256*e^x*log(2*x - log(2) - log(x)) + 5*e^x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {1}{256} \, e^{x} \log \left (2 \, x - \log \left (2 \, x\right )\right ) + 5 \, e^{x} \]

[In]

integrate(((x*exp(x)*log(2*x)-2*exp(x)*x^2)*log(2*x-log(2*x))+1280*x*exp(x)*log(2*x)+(-2560*x^2-2*x+1)*exp(x))
/(256*x*log(2*x)-512*x^2),x, algorithm="giac")

[Out]

1/256*e^x*log(2*x - log(2*x)) + 5*e^x

Mupad [B] (verification not implemented)

Time = 9.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (1-2 x-2560 x^2\right )+1280 e^x x \log (2 x)+\left (-2 e^x x^2+e^x x \log (2 x)\right ) \log (2 x-\log (2 x))}{-512 x^2+256 x \log (2 x)} \, dx=\frac {{\mathrm {e}}^x\,\left (\ln \left (2\,x-\ln \left (2\,x\right )\right )+1280\right )}{256} \]

[In]

int(-(log(2*x - log(2*x))*(2*x^2*exp(x) - x*log(2*x)*exp(x)) + exp(x)*(2*x + 2560*x^2 - 1) - 1280*x*log(2*x)*e
xp(x))/(256*x*log(2*x) - 512*x^2),x)

[Out]

(exp(x)*(log(2*x - log(2*x)) + 1280))/256