\(\int \frac {128-16 e^x}{(-80+4 e^x-32 x+(-20+e^x-8 x) \log (\frac {1}{4} (100-5 e^x+40 x))) \log ^2(4+\log (\frac {1}{4} (100-5 e^x+40 x)))} \, dx\) [8659]

Optimal result

Integrand size = 64, antiderivative size = 22 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\log \left (4+\log \left (5 \left (5-\frac {e^x}{4}+2 x\right )\right )\right )} \]

[Out]

16/ln(ln(-5/4*exp(x)+10*x+25)+4)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6818} \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\log \left (\log \left (\frac {5}{4} \left (8 x-e^x+20\right )\right )+4\right )} \]

[In]

Int[(128 - 16*E^x)/((-80 + 4*E^x - 32*x + (-20 + E^x - 8*x)*Log[(100 - 5*E^x + 40*x)/4])*Log[4 + Log[(100 - 5*
E^x + 40*x)/4]]^2),x]

[Out]

16/Log[4 + Log[(5*(20 - E^x + 8*x))/4]]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {16}{\log \left (4+\log \left (\frac {5}{4} \left (20-e^x+8 x\right )\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\log \left (4+\log \left (25-\frac {5 e^x}{4}+10 x\right )\right )} \]

[In]

Integrate[(128 - 16*E^x)/((-80 + 4*E^x - 32*x + (-20 + E^x - 8*x)*Log[(100 - 5*E^x + 40*x)/4])*Log[4 + Log[(10
0 - 5*E^x + 40*x)/4]]^2),x]

[Out]

16/Log[4 + Log[25 - (5*E^x)/4 + 10*x]]

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {16}{\ln \left (\ln \left (-\frac {5 \,{\mathrm e}^{x}}{4}+10 x +25\right )+4\right )}\)	\(18\)
parallelrisch	\(\frac {16}{\ln \left (\ln \left (-\frac {5 \,{\mathrm e}^{x}}{4}+10 x +25\right )+4\right )}\)	\(18\)

[In]

int((-16*exp(x)+128)/((exp(x)-8*x-20)*ln(-5/4*exp(x)+10*x+25)+4*exp(x)-32*x-80)/ln(ln(-5/4*exp(x)+10*x+25)+4)^
2,x,method=_RETURNVERBOSE)

[Out]

16/ln(ln(-5/4*exp(x)+10*x+25)+4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\log \left (\log \left (10 \, x - \frac {5}{4} \, e^{x} + 25\right ) + 4\right )} \]

[In]

integrate((-16*exp(x)+128)/((exp(x)-8*x-20)*log(-5/4*exp(x)+10*x+25)+4*exp(x)-32*x-80)/log(log(-5/4*exp(x)+10*
x+25)+4)^2,x, algorithm="fricas")

[Out]

16/log(log(10*x - 5/4*e^x + 25) + 4)

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\log {\left (\log {\left (10 x - \frac {5 e^{x}}{4} + 25 \right )} + 4 \right )}} \]

[In]

integrate((-16*exp(x)+128)/((exp(x)-8*x-20)*ln(-5/4*exp(x)+10*x+25)+4*exp(x)-32*x-80)/ln(ln(-5/4*exp(x)+10*x+2
5)+4)**2,x)

[Out]

16/log(log(10*x - 5*exp(x)/4 + 25) + 4)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\log \left (i \, \pi + \log \left (5\right ) - 2 \, \log \left (2\right ) + \log \left (-8 \, x + e^{x} - 20\right ) + 4\right )} \]

[In]

integrate((-16*exp(x)+128)/((exp(x)-8*x-20)*log(-5/4*exp(x)+10*x+25)+4*exp(x)-32*x-80)/log(log(-5/4*exp(x)+10*
x+25)+4)^2,x, algorithm="maxima")

[Out]

16/log(I*pi + log(5) - 2*log(2) + log(-8*x + e^x - 20) + 4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (17) = 34\).

Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.36 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=-\frac {16 \, {\left (\log \left (10 \, x - \frac {5}{4} \, e^{x} + 25\right ) + 4\right )}}{2 \, \log \left (2\right ) \log \left (\log \left (10 \, x - \frac {5}{4} \, e^{x} + 25\right ) + 4\right ) - \log \left (40 \, x - 5 \, e^{x} + 100\right ) \log \left (\log \left (10 \, x - \frac {5}{4} \, e^{x} + 25\right ) + 4\right ) - 4 \, \log \left (\log \left (10 \, x - \frac {5}{4} \, e^{x} + 25\right ) + 4\right )} \]

[In]

integrate((-16*exp(x)+128)/((exp(x)-8*x-20)*log(-5/4*exp(x)+10*x+25)+4*exp(x)-32*x-80)/log(log(-5/4*exp(x)+10*
x+25)+4)^2,x, algorithm="giac")

[Out]

-16*(log(10*x - 5/4*e^x + 25) + 4)/(2*log(2)*log(log(10*x - 5/4*e^x + 25) + 4) - log(40*x - 5*e^x + 100)*log(l
og(10*x - 5/4*e^x + 25) + 4) - 4*log(log(10*x - 5/4*e^x + 25) + 4))

Mupad [B] (verification not implemented)

Time = 14.56 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {128-16 e^x}{\left (-80+4 e^x-32 x+\left (-20+e^x-8 x\right ) \log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right ) \log ^2\left (4+\log \left (\frac {1}{4} \left (100-5 e^x+40 x\right )\right )\right )} \, dx=\frac {16}{\ln \left (\ln \left (10\,x-\frac {5\,{\mathrm {e}}^x}{4}+25\right )+4\right )} \]

[In]

int((16*exp(x) - 128)/(log(log(10*x - (5*exp(x))/4 + 25) + 4)^2*(32*x - 4*exp(x) + log(10*x - (5*exp(x))/4 + 2
5)*(8*x - exp(x) + 20) + 80)),x)

[Out]

16/log(log(10*x - (5*exp(x))/4 + 25) + 4)