\(\int \frac {-90 e^x+e^{2 x} (-180-72 x)+(180 e^{2 x}+e^x (90+90 x)) \log (x)+(50 e^x-40 e^{2 x}) \log ^2(x)}{(25+e^x (100+40 x)+e^{2 x} (100+80 x+16 x^2)) \log ^2(x)} \, dx\) [6784]

Optimal result

Integrand size = 89, antiderivative size = 26 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {5+\frac {9 x}{\log (x)}}{5+\frac {5 e^{-x}}{2}+2 x} \]

[Out]

(5+9*x/ln(x))/(5+2*x+5/2/exp(x))

Rubi [F]

\[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx \]

[In]

Int[(-90*E^x + E^(2*x)*(-180 - 72*x) + (180*E^(2*x) + E^x*(90 + 90*x))*Log[x] + (50*E^x - 40*E^(2*x))*Log[x]^2
)/((25 + E^x*(100 + 40*x) + E^(2*x)*(100 + 80*x + 16*x^2))*Log[x]^2),x]

[Out]

50*Defer[Int][E^x/(5 + 10*E^x + 4*E^x*x)^2, x] + 100*Defer[Int][E^x/((5 + 2*x)*(5 + 10*E^x + 4*E^x*x)^2), x] -
20*Defer[Int][E^x/((5 + 2*x)*(5 + 10*E^x + 4*E^x*x)), x] - 18*Defer[Int][E^x/((5 + 10*E^x + 4*E^x*x)*Log[x]^2
), x] + 90*Defer[Int][E^x/((5 + 10*E^x + 4*E^x*x)^2*Log[x]), x] + 90*Defer[Int][(E^x*x)/((5 + 10*E^x + 4*E^x*x
)^2*Log[x]), x] - 450*Defer[Int][E^x/((5 + 2*x)*(5 + 10*E^x + 4*E^x*x)^2*Log[x]), x] + 90*Defer[Int][E^x/((5 +
2*x)*(5 + 10*E^x + 4*E^x*x)*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^x \left (-45-18 e^x (5+2 x)+45 \left (1+2 e^x+x\right ) \log (x)-5 \left (-5+4 e^x\right ) \log ^2(x)\right )}{\left (5+2 e^x (5+2 x)\right )^2 \log ^2(x)} \, dx \\ & = 2 \int \frac {e^x \left (-45-18 e^x (5+2 x)+45 \left (1+2 e^x+x\right ) \log (x)-5 \left (-5+4 e^x\right ) \log ^2(x)\right )}{\left (5+2 e^x (5+2 x)\right )^2 \log ^2(x)} \, dx \\ & = 2 \int \left (\frac {5 e^x (7+2 x) (9 x+5 \log (x))}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)}-\frac {e^x \left (45+18 x-45 \log (x)+10 \log ^2(x)\right )}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)}\right ) \, dx \\ & = -\left (2 \int \frac {e^x \left (45+18 x-45 \log (x)+10 \log ^2(x)\right )}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx\right )+10 \int \frac {e^x (7+2 x) (9 x+5 \log (x))}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx \\ & = -\left (2 \int \left (\frac {10 e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )}+\frac {45 e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)}+\frac {18 e^x x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)}-\frac {45 e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log (x)}\right ) \, dx\right )+10 \int \left (\frac {e^x (9 x+5 \log (x))}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)}+\frac {2 e^x (9 x+5 \log (x))}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)}\right ) \, dx \\ & = 10 \int \frac {e^x (9 x+5 \log (x))}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx-20 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )} \, dx+20 \int \frac {e^x (9 x+5 \log (x))}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx-36 \int \frac {e^x x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx-90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx+90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log (x)} \, dx \\ & = 10 \int \left (\frac {5 e^x}{\left (5+10 e^x+4 e^x x\right )^2}+\frac {9 e^x x}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)}\right ) \, dx-20 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )} \, dx+20 \int \left (\frac {5 e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2}+\frac {9 e^x x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)}\right ) \, dx-36 \int \left (\frac {e^x}{2 \left (5+10 e^x+4 e^x x\right ) \log ^2(x)}-\frac {5 e^x}{2 (5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)}\right ) \, dx-90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx+90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log (x)} \, dx \\ & = -\left (18 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx\right )-20 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )} \, dx+50 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right )^2} \, dx+90 \int \frac {e^x x}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx+90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log (x)} \, dx+100 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2} \, dx+180 \int \frac {e^x x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx \\ & = -\left (18 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx\right )-20 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )} \, dx+50 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right )^2} \, dx+90 \int \frac {e^x x}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx+90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log (x)} \, dx+100 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2} \, dx+180 \int \left (\frac {e^x}{2 \left (5+10 e^x+4 e^x x\right )^2 \log (x)}-\frac {5 e^x}{2 (5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)}\right ) \, dx \\ & = -\left (18 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right ) \log ^2(x)} \, dx\right )-20 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )} \, dx+50 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right )^2} \, dx+90 \int \frac {e^x}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx+90 \int \frac {e^x x}{\left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx+90 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right ) \log (x)} \, dx+100 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2} \, dx-450 \int \frac {e^x}{(5+2 x) \left (5+10 e^x+4 e^x x\right )^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 e^x (9 x+5 \log (x))}{\left (5+2 e^x (5+2 x)\right ) \log (x)} \]

[In]

Integrate[(-90*E^x + E^(2*x)*(-180 - 72*x) + (180*E^(2*x) + E^x*(90 + 90*x))*Log[x] + (50*E^x - 40*E^(2*x))*Lo
g[x]^2)/((25 + E^x*(100 + 40*x) + E^(2*x)*(100 + 80*x + 16*x^2))*Log[x]^2),x]

[Out]

(2*E^x*(9*x + 5*Log[x]))/((5 + 2*E^x*(5 + 2*x))*Log[x])

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23

method	result	size
parallelrisch	\(-\frac {-72 \,{\mathrm e}^{x} x -40 \,{\mathrm e}^{x} \ln \left (x \right )}{4 \ln \left (x \right ) \left (4 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+5\right )}\)	\(32\)
risch	\(\frac {10 \,{\mathrm e}^{x}}{4 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+5}+\frac {18 x \,{\mathrm e}^{x}}{\left (4 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+5\right ) \ln \left (x \right )}\)	\(41\)

[In]

int(((-40*exp(x)^2+50*exp(x))*ln(x)^2+(180*exp(x)^2+(90*x+90)*exp(x))*ln(x)+(-72*x-180)*exp(x)^2-90*exp(x))/((
16*x^2+80*x+100)*exp(x)^2+(40*x+100)*exp(x)+25)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*(-72*exp(x)*x-40*exp(x)*ln(x))/ln(x)/(4*exp(x)*x+10*exp(x)+5)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (9 \, x e^{x} + 5 \, e^{x} \log \left (x\right )\right )}}{{\left (2 \, {\left (2 \, x + 5\right )} e^{x} + 5\right )} \log \left (x\right )} \]

[In]

integrate(((-40*exp(x)^2+50*exp(x))*log(x)^2+(180*exp(x)^2+(90*x+90)*exp(x))*log(x)+(-72*x-180)*exp(x)^2-90*ex
p(x))/((16*x^2+80*x+100)*exp(x)^2+(40*x+100)*exp(x)+25)/log(x)^2,x, algorithm="fricas")

[Out]

2*(9*x*e^x + 5*e^x*log(x))/((2*(2*x + 5)*e^x + 5)*log(x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (19) = 38\).

Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {9 x}{\left (2 x + 5\right ) \log {\left (x \right )}} + \frac {- 45 x - 25 \log {\left (x \right )}}{10 x \log {\left (x \right )} + \left (8 x^{2} \log {\left (x \right )} + 40 x \log {\left (x \right )} + 50 \log {\left (x \right )}\right ) e^{x} + 25 \log {\left (x \right )}} + \frac {10}{4 x + 10} \]

[In]

integrate(((-40*exp(x)**2+50*exp(x))*ln(x)**2+(180*exp(x)**2+(90*x+90)*exp(x))*ln(x)+(-72*x-180)*exp(x)**2-90*
exp(x))/((16*x**2+80*x+100)*exp(x)**2+(40*x+100)*exp(x)+25)/ln(x)**2,x)

[Out]

9*x/((2*x + 5)*log(x)) + (-45*x - 25*log(x))/(10*x*log(x) + (8*x**2*log(x) + 40*x*log(x) + 50*log(x))*exp(x) +
25*log(x)) + 10/(4*x + 10)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (9 \, x + 5 \, \log \left (x\right )\right )} e^{x}}{2 \, {\left (2 \, x + 5\right )} e^{x} \log \left (x\right ) + 5 \, \log \left (x\right )} \]

[In]

integrate(((-40*exp(x)^2+50*exp(x))*log(x)^2+(180*exp(x)^2+(90*x+90)*exp(x))*log(x)+(-72*x-180)*exp(x)^2-90*ex
p(x))/((16*x^2+80*x+100)*exp(x)^2+(40*x+100)*exp(x)+25)/log(x)^2,x, algorithm="maxima")

[Out]

2*(9*x + 5*log(x))*e^x/(2*(2*x + 5)*e^x*log(x) + 5*log(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (9 \, x e^{x} + 5 \, e^{x} \log \left (x\right )\right )}}{4 \, x e^{x} \log \left (x\right ) + 10 \, e^{x} \log \left (x\right ) + 5 \, \log \left (x\right )} \]

[In]

integrate(((-40*exp(x)^2+50*exp(x))*log(x)^2+(180*exp(x)^2+(90*x+90)*exp(x))*log(x)+(-72*x-180)*exp(x)^2-90*ex
p(x))/((16*x^2+80*x+100)*exp(x)^2+(40*x+100)*exp(x)+25)/log(x)^2,x, algorithm="giac")

[Out]

2*(9*x*e^x + 5*e^x*log(x))/(4*x*e^x*log(x) + 10*e^x*log(x) + 5*log(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {-90 e^x+e^{2 x} (-180-72 x)+\left (180 e^{2 x}+e^x (90+90 x)\right ) \log (x)+\left (50 e^x-40 e^{2 x}\right ) \log ^2(x)}{\left (25+e^x (100+40 x)+e^{2 x} \left (100+80 x+16 x^2\right )\right ) \log ^2(x)} \, dx=\int -\frac {\left (40\,{\mathrm {e}}^{2\,x}-50\,{\mathrm {e}}^x\right )\,{\ln \left (x\right )}^2+\left (-180\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (90\,x+90\right )\right )\,\ln \left (x\right )+90\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (72\,x+180\right )}{{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+80\,x+100\right )+{\mathrm {e}}^x\,\left (40\,x+100\right )+25\right )} \,d x \]

[In]

int(-(90*exp(x) - log(x)*(180*exp(2*x) + exp(x)*(90*x + 90)) + exp(2*x)*(72*x + 180) + log(x)^2*(40*exp(2*x) -
50*exp(x)))/(log(x)^2*(exp(2*x)*(80*x + 16*x^2 + 100) + exp(x)*(40*x + 100) + 25)),x)

[Out]

int(-(90*exp(x) - log(x)*(180*exp(2*x) + exp(x)*(90*x + 90)) + exp(2*x)*(72*x + 180) + log(x)^2*(40*exp(2*x) -
50*exp(x)))/(log(x)^2*(exp(2*x)*(80*x + 16*x^2 + 100) + exp(x)*(40*x + 100) + 25)), x)