\(\int \frac {20+8 x+8 x \log (x)+e^{14-x} (25 x+20 x^2+4 x^3) \log ^2(x)}{(25 x+20 x^2+4 x^3) \log ^2(x)} \, dx\) [9365]

Optimal result

Integrand size = 57, antiderivative size = 24 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=14-e^{14-x}-\frac {4}{(5+2 x) \log (x)} \]

[Out]

14-4/(5+2*x)/ln(x)-exp(-x+14)

Rubi [F]

\[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=\int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx \]

[In]

Int[(20 + 8*x + 8*x*Log[x] + E^(14 - x)*(25*x + 20*x^2 + 4*x^3)*Log[x]^2)/((25*x + 20*x^2 + 4*x^3)*Log[x]^2),x
]

[Out]

-E^(14 - x) + 4*Defer[Int][1/(x*(5 + 2*x)*Log[x]^2), x] + 8*Defer[Int][1/((5 + 2*x)^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{x \left (25+20 x+4 x^2\right ) \log ^2(x)} \, dx \\ & = \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{x (5+2 x)^2 \log ^2(x)} \, dx \\ & = \int \left (e^{14-x}+\frac {4 (5+2 x+2 x \log (x))}{x (5+2 x)^2 \log ^2(x)}\right ) \, dx \\ & = 4 \int \frac {5+2 x+2 x \log (x)}{x (5+2 x)^2 \log ^2(x)} \, dx+\int e^{14-x} \, dx \\ & = -e^{14-x}+4 \int \left (\frac {1}{x (5+2 x) \log ^2(x)}+\frac {2}{(5+2 x)^2 \log (x)}\right ) \, dx \\ & = -e^{14-x}+4 \int \frac {1}{x (5+2 x) \log ^2(x)} \, dx+8 \int \frac {1}{(5+2 x)^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=-e^{14-x}-\frac {4}{(5+2 x) \log (x)} \]

[In]

Integrate[(20 + 8*x + 8*x*Log[x] + E^(14 - x)*(25*x + 20*x^2 + 4*x^3)*Log[x]^2)/((25*x + 20*x^2 + 4*x^3)*Log[x
]^2),x]

[Out]

-E^(14 - x) - 4/((5 + 2*x)*Log[x])

Maple [A] (verified)

Time = 5.44 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
risch	\(-{\mathrm e}^{-x +14}-\frac {4}{\left (5+2 x \right ) \ln \left (x \right )}\)	\(23\)
parallelrisch	\(-\frac {20 \,{\mathrm e}^{-x +14} \ln \left (x \right ) x +40+50 \,{\mathrm e}^{-x +14} \ln \left (x \right )}{10 \ln \left (x \right ) \left (5+2 x \right )}\)	\(37\)

[In]

int(((4*x^3+20*x^2+25*x)*exp(-x+14)*ln(x)^2+8*x*ln(x)+8*x+20)/(4*x^3+20*x^2+25*x)/ln(x)^2,x,method=_RETURNVERB
OSE)

[Out]

-exp(-x+14)-4/(5+2*x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=-\frac {{\left (2 \, x + 5\right )} e^{\left (-x + 14\right )} \log \left (x\right ) + 4}{{\left (2 \, x + 5\right )} \log \left (x\right )} \]

[In]

integrate(((4*x^3+20*x^2+25*x)*exp(-x+14)*log(x)^2+8*x*log(x)+8*x+20)/(4*x^3+20*x^2+25*x)/log(x)^2,x, algorith
m="fricas")

[Out]

-((2*x + 5)*e^(-x + 14)*log(x) + 4)/((2*x + 5)*log(x))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=- e^{14 - x} - \frac {4}{\left (2 x + 5\right ) \log {\left (x \right )}} \]

[In]

integrate(((4*x**3+20*x**2+25*x)*exp(-x+14)*ln(x)**2+8*x*ln(x)+8*x+20)/(4*x**3+20*x**2+25*x)/ln(x)**2,x)

[Out]

-exp(14 - x) - 4/((2*x + 5)*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=-\frac {{\left ({\left (2 \, x e^{14} + 5 \, e^{14}\right )} \log \left (x\right ) + 4 \, e^{x}\right )} e^{\left (-x\right )}}{{\left (2 \, x + 5\right )} \log \left (x\right )} \]

[In]

integrate(((4*x^3+20*x^2+25*x)*exp(-x+14)*log(x)^2+8*x*log(x)+8*x+20)/(4*x^3+20*x^2+25*x)/log(x)^2,x, algorith
m="maxima")

[Out]

-((2*x*e^14 + 5*e^14)*log(x) + 4*e^x)*e^(-x)/((2*x + 5)*log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=-\frac {2 \, x e^{\left (-x + 14\right )} \log \left (x\right ) + 5 \, e^{\left (-x + 14\right )} \log \left (x\right ) + 4}{2 \, x \log \left (x\right ) + 5 \, \log \left (x\right )} \]

[In]

integrate(((4*x^3+20*x^2+25*x)*exp(-x+14)*log(x)^2+8*x*log(x)+8*x+20)/(4*x^3+20*x^2+25*x)/log(x)^2,x, algorith
m="giac")

[Out]

-(2*x*e^(-x + 14)*log(x) + 5*e^(-x + 14)*log(x) + 4)/(2*x*log(x) + 5*log(x))

Mupad [B] (verification not implemented)

Time = 14.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx=-\frac {4}{\ln \left (x\right )\,\left (2\,x+5\right )}-\frac {5\,{\mathrm {e}}^{14-x}}{2\,x+5}-\frac {2\,x\,{\mathrm {e}}^{14-x}}{2\,x+5} \]

[In]

int((8*x + 8*x*log(x) + exp(14 - x)*log(x)^2*(25*x + 20*x^2 + 4*x^3) + 20)/(log(x)^2*(25*x + 20*x^2 + 4*x^3)),
x)

[Out]

- 4/(log(x)*(2*x + 5)) - (5*exp(14 - x))/(2*x + 5) - (2*x*exp(14 - x))/(2*x + 5)