\(\int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+(1088+64 x^2) \log ^2(5)}{1+(544-8 x-32 x^2) \log (5)+16 e^{2 x} \log ^2(5)+(73984-2176 x-8688 x^2+128 x^3+256 x^4) \log ^2(5)+e^x (8 \log (5)+(2176-32 x-128 x^2) \log ^2(5))} \, dx\) [9187]

Optimal result

Integrand size = 108, antiderivative size = 29 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {x}{4+e^x-x+4 \left (16-x^2\right )+\frac {1}{4 \log (5)}} \]

[Out]

x/(-4*x^2+68+1/4/ln(5)-x+exp(x))

Rubi [F]

\[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx \]

[In]

Int[(4*Log[5] + E^x*(16 - 16*x)*Log[5]^2 + (1088 + 64*x^2)*Log[5]^2)/(1 + (544 - 8*x - 32*x^2)*Log[5] + 16*E^(
2*x)*Log[5]^2 + (73984 - 2176*x - 8688*x^2 + 128*x^3 + 256*x^4)*Log[5]^2 + E^x*(8*Log[5] + (2176 - 32*x - 128*
x^2)*Log[5]^2)),x]

[Out]

4*Log[5]*Defer[Int][(1 + 272*Log[5] + 4*E^x*Log[5] - 4*x*Log[5] - 16*x^2*Log[5])^(-1), x] + 4*Log[5]*(1 + 276*
Log[5])*Defer[Int][x/(-1 - 272*Log[5] - 4*E^x*Log[5] + 4*x*Log[5] + 16*x^2*Log[5])^2, x] + 112*Log[5]^2*Defer[
Int][x^2/(-1 - 272*Log[5] - 4*E^x*Log[5] + 4*x*Log[5] + 16*x^2*Log[5])^2, x] - 64*Log[5]^2*Defer[Int][x^3/(-1
- 272*Log[5] - 4*E^x*Log[5] + 4*x*Log[5] + 16*x^2*Log[5])^2, x] + 4*Log[5]*Defer[Int][x/(-1 - 272*Log[5] - 4*E
^x*Log[5] + 4*x*Log[5] + 16*x^2*Log[5]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \log (5) \left (1+272 \log (5)-4 e^x (-1+x) \log (5)+16 x^2 \log (5)\right )}{\left (1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)\right )^2} \, dx \\ & = (4 \log (5)) \int \frac {1+272 \log (5)-4 e^x (-1+x) \log (5)+16 x^2 \log (5)}{\left (1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)\right )^2} \, dx \\ & = (4 \log (5)) \int \left (-\frac {x \left (-1-276 \log (5)-28 x \log (5)+16 x^2 \log (5)\right )}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}+\frac {-1+x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)}\right ) \, dx \\ & = -\left ((4 \log (5)) \int \frac {x \left (-1-276 \log (5)-28 x \log (5)+16 x^2 \log (5)\right )}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx\right )+(4 \log (5)) \int \frac {-1+x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)} \, dx \\ & = -\left ((4 \log (5)) \int \left (-\frac {28 x^2 \log (5)}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}+\frac {16 x^3 \log (5)}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}-\frac {x (1+276 \log (5))}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2}\right ) \, dx\right )+(4 \log (5)) \int \left (\frac {1}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)}+\frac {x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)}\right ) \, dx \\ & = (4 \log (5)) \int \frac {1}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)} \, dx+(4 \log (5)) \int \frac {x}{-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)} \, dx-\left (64 \log ^2(5)\right ) \int \frac {x^3}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx+\left (112 \log ^2(5)\right ) \int \frac {x^2}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx+(4 \log (5) (1+276 \log (5))) \int \frac {x}{\left (-1-272 \log (5)-4 e^x \log (5)+4 x \log (5)+16 x^2 \log (5)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4 x \log (5)}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)} \]

[In]

Integrate[(4*Log[5] + E^x*(16 - 16*x)*Log[5]^2 + (1088 + 64*x^2)*Log[5]^2)/(1 + (544 - 8*x - 32*x^2)*Log[5] +
16*E^(2*x)*Log[5]^2 + (73984 - 2176*x - 8688*x^2 + 128*x^3 + 256*x^4)*Log[5]^2 + E^x*(8*Log[5] + (2176 - 32*x
- 128*x^2)*Log[5]^2)),x]

[Out]

(4*x*Log[5])/(1 + 272*Log[5] + 4*E^x*Log[5] - 4*x*Log[5] - 16*x^2*Log[5])

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10

method	result	size
risch	\(-\frac {4 x \ln \left (5\right )}{16 x^{2} \ln \left (5\right )+4 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )-272 \ln \left (5\right )-1}\)	\(32\)
parallelrisch	\(-\frac {4 x \ln \left (5\right )}{16 x^{2} \ln \left (5\right )+4 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )-272 \ln \left (5\right )-1}\)	\(32\)
norman	\(\frac {-4 \,{\mathrm e}^{x} \ln \left (5\right )+16 x^{2} \ln \left (5\right )-1-272 \ln \left (5\right )}{16 x^{2} \ln \left (5\right )+4 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )-272 \ln \left (5\right )-1}\)	\(47\)

[In]

int(((-16*x+16)*ln(5)^2*exp(x)+(64*x^2+1088)*ln(5)^2+4*ln(5))/(16*ln(5)^2*exp(x)^2+((-128*x^2-32*x+2176)*ln(5)
^2+8*ln(5))*exp(x)+(256*x^4+128*x^3-8688*x^2-2176*x+73984)*ln(5)^2+(-32*x^2-8*x+544)*ln(5)+1),x,method=_RETURN
VERBOSE)

[Out]

-4*x*ln(5)/(16*x^2*ln(5)+4*x*ln(5)-4*exp(x)*ln(5)-272*ln(5)-1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=-\frac {4 \, x \log \left (5\right )}{4 \, {\left (4 \, x^{2} + x - 68\right )} \log \left (5\right ) - 4 \, e^{x} \log \left (5\right ) - 1} \]

[In]

integrate(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16*log(5)^2*exp(x)^2+((-128*x^2-32*x+2
176)*log(5)^2+8*log(5))*exp(x)+(256*x^4+128*x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x,
algorithm="fricas")

[Out]

-4*x*log(5)/(4*(4*x^2 + x - 68)*log(5) - 4*e^x*log(5) - 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4 x \log {\left (5 \right )}}{- 16 x^{2} \log {\left (5 \right )} - 4 x \log {\left (5 \right )} + 4 e^{x} \log {\left (5 \right )} + 1 + 272 \log {\left (5 \right )}} \]

[In]

integrate(((-16*x+16)*ln(5)**2*exp(x)+(64*x**2+1088)*ln(5)**2+4*ln(5))/(16*ln(5)**2*exp(x)**2+((-128*x**2-32*x
+2176)*ln(5)**2+8*ln(5))*exp(x)+(256*x**4+128*x**3-8688*x**2-2176*x+73984)*ln(5)**2+(-32*x**2-8*x+544)*ln(5)+1
),x)

[Out]

4*x*log(5)/(-16*x**2*log(5) - 4*x*log(5) + 4*exp(x)*log(5) + 1 + 272*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=-\frac {4 \, x \log \left (5\right )}{16 \, x^{2} \log \left (5\right ) + 4 \, x \log \left (5\right ) - 4 \, e^{x} \log \left (5\right ) - 272 \, \log \left (5\right ) - 1} \]

[In]

integrate(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16*log(5)^2*exp(x)^2+((-128*x^2-32*x+2
176)*log(5)^2+8*log(5))*exp(x)+(256*x^4+128*x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x,
algorithm="maxima")

[Out]

-4*x*log(5)/(16*x^2*log(5) + 4*x*log(5) - 4*e^x*log(5) - 272*log(5) - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=-\frac {4 \, x \log \left (5\right )}{16 \, x^{2} \log \left (5\right ) + 4 \, x \log \left (5\right ) - 4 \, e^{x} \log \left (5\right ) - 272 \, \log \left (5\right ) - 1} \]

[In]

integrate(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16*log(5)^2*exp(x)^2+((-128*x^2-32*x+2
176)*log(5)^2+8*log(5))*exp(x)+(256*x^4+128*x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x,
algorithm="giac")

[Out]

-4*x*log(5)/(16*x^2*log(5) + 4*x*log(5) - 4*e^x*log(5) - 272*log(5) - 1)

Mupad [B] (verification not implemented)

Time = 14.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4\,x\,\ln \left (5\right )}{272\,\ln \left (5\right )-4\,x\,\ln \left (5\right )-16\,x^2\,\ln \left (5\right )+4\,{\mathrm {e}}^x\,\ln \left (5\right )+1} \]

[In]

int((4*log(5) + log(5)^2*(64*x^2 + 1088) - exp(x)*log(5)^2*(16*x - 16))/(exp(x)*(8*log(5) - log(5)^2*(32*x + 1
28*x^2 - 2176)) + log(5)^2*(128*x^3 - 8688*x^2 - 2176*x + 256*x^4 + 73984) - log(5)*(8*x + 32*x^2 - 544) + 16*
exp(2*x)*log(5)^2 + 1),x)

[Out]

(4*x*log(5))/(272*log(5) - 4*x*log(5) - 16*x^2*log(5) + 4*exp(x)*log(5) + 1)