\(\int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx\) [5574]

Optimal result

Integrand size = 52, antiderivative size = 22 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 e^9 \left (e^x+\log \left (\frac {2}{5 x}+\log ^2(x)\right )\right ) \]

[Out]

exp(9)*(16*ln(ln(x)^2+2/5/x)+16*exp(x))

Rubi [F]

\[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=\int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx \]

[In]

Int[(-32*E^9 + 32*E^(9 + x)*x + 160*E^9*x*Log[x] + 80*E^(9 + x)*x^2*Log[x]^2)/(2*x + 5*x^2*Log[x]^2),x]

[Out]

16*E^(9 + x) - 32*E^9*Defer[Int][1/(x*(2 + 5*x*Log[x]^2)), x] + 160*E^9*Defer[Int][Log[x]/(2 + 5*x*Log[x]^2),
x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{x^2 \left (\frac {2}{x}+5 \log ^2(x)\right )} \, dx \\ & = \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = \int \frac {16 e^9 \left (-2+2 e^x x+10 x \log (x)+5 e^x x^2 \log ^2(x)\right )}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = \left (16 e^9\right ) \int \frac {-2+2 e^x x+10 x \log (x)+5 e^x x^2 \log ^2(x)}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = \left (16 e^9\right ) \int \left (e^x+\frac {2 (-1+5 x \log (x))}{x \left (2+5 x \log ^2(x)\right )}\right ) \, dx \\ & = \left (16 e^9\right ) \int e^x \, dx+\left (32 e^9\right ) \int \frac {-1+5 x \log (x)}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = 16 e^{9+x}+\left (32 e^9\right ) \int \left (-\frac {1}{x \left (2+5 x \log ^2(x)\right )}+\frac {5 \log (x)}{2+5 x \log ^2(x)}\right ) \, dx \\ & = 16 e^{9+x}-\left (32 e^9\right ) \int \frac {1}{x \left (2+5 x \log ^2(x)\right )} \, dx+\left (160 e^9\right ) \int \frac {\log (x)}{2+5 x \log ^2(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 e^9 \left (e^x-\log (x)+\log \left (2+5 x \log ^2(x)\right )\right ) \]

[In]

Integrate[(-32*E^9 + 32*E^(9 + x)*x + 160*E^9*x*Log[x] + 80*E^(9 + x)*x^2*Log[x]^2)/(2*x + 5*x^2*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
risch	\(16 \,{\mathrm e}^{x +9}+16 \,{\mathrm e}^{9} \ln \left (\ln \left (x \right )^{2}+\frac {2}{5 x}\right )\)	\(23\)
parallelrisch	\(16 \,{\mathrm e}^{9} \ln \left (x \ln \left (x \right )^{2}+\frac {2}{5}\right )-16 \,{\mathrm e}^{9} \ln \left (x \right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\)	\(27\)
norman	\(-16 \,{\mathrm e}^{9} \ln \left (x \right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}+16 \,{\mathrm e}^{9} \ln \left (5 x \ln \left (x \right )^{2}+2\right )\)	\(28\)
default	\(32 \,{\mathrm e}^{9} \left (-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (5 x \ln \left (x \right )^{2}+2\right )}{2}\right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\)	\(29\)
parts	\(32 \,{\mathrm e}^{9} \left (-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (5 x \ln \left (x \right )^{2}+2\right )}{2}\right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\)	\(29\)

[In]

int((80*x^2*exp(9)*exp(x)*ln(x)^2+160*x*exp(9)*ln(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*ln(x)^2+2*x),x,metho
d=_RETURNVERBOSE)

[Out]

16*exp(x+9)+16*exp(9)*ln(ln(x)^2+2/5/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 \, e^{9} \log \left (\frac {5 \, x \log \left (x\right )^{2} + 2}{x}\right ) + 16 \, e^{\left (x + 9\right )} \]

[In]

integrate((80*x^2*exp(9)*exp(x)*log(x)^2+160*x*exp(9)*log(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*log(x)^2+2*x
),x, algorithm="fricas")

[Out]

16*e^9*log((5*x*log(x)^2 + 2)/x) + 16*e^(x + 9)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 e^{9} e^{x} + 16 e^{9} \log {\left (\log {\left (x \right )}^{2} + \frac {2}{5 x} \right )} \]

[In]

integrate((80*x**2*exp(9)*exp(x)*ln(x)**2+160*x*exp(9)*ln(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x**2*ln(x)**2+2*
x),x)

[Out]

16*exp(9)*exp(x) + 16*exp(9)*log(log(x)**2 + 2/(5*x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 \, e^{9} \log \left (\frac {5 \, x \log \left (x\right )^{2} + 2}{5 \, x}\right ) + 16 \, e^{\left (x + 9\right )} \]

[In]

integrate((80*x^2*exp(9)*exp(x)*log(x)^2+160*x*exp(9)*log(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*log(x)^2+2*x
),x, algorithm="maxima")

[Out]

16*e^9*log(1/5*(5*x*log(x)^2 + 2)/x) + 16*e^(x + 9)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 \, e^{9} \log \left (5 \, x \log \left (x\right )^{2} + 2\right ) - 16 \, e^{9} \log \left (x\right ) + 16 \, e^{\left (x + 9\right )} \]

[In]

integrate((80*x^2*exp(9)*exp(x)*log(x)^2+160*x*exp(9)*log(x)+32*x*exp(9)*exp(x)-32*exp(9))/(5*x^2*log(x)^2+2*x
),x, algorithm="giac")

[Out]

16*e^9*log(5*x*log(x)^2 + 2) - 16*e^9*log(x) + 16*e^(x + 9)

Mupad [B] (verification not implemented)

Time = 11.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16\,{\mathrm {e}}^9\,{\mathrm {e}}^x+16\,\ln \left (\frac {5\,x\,{\ln \left (x\right )}^2+2}{x}\right )\,{\mathrm {e}}^9 \]

[In]

int((32*x*exp(9)*exp(x) - 32*exp(9) + 160*x*exp(9)*log(x) + 80*x^2*exp(9)*exp(x)*log(x)^2)/(2*x + 5*x^2*log(x)
^2),x)

[Out]

16*exp(9)*exp(x) + 16*log((5*x*log(x)^2 + 2)/x)*exp(9)