\(\int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3))}{-64 x+48 x^2-12 x^3+x^4} \, dx\) [6638]

Optimal result

Integrand size = 96, antiderivative size = 30 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=-5-e^{-x+\frac {25 \log ^2(3)}{(4-x)^2}}+x+\log \left (\frac {3 x}{2}\right ) \]

[Out]

ln(3/2*x)+x-5-exp(25*ln(3)^2/(-x+4)^2-x)

Rubi [F]

\[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=\int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx \]

[In]

Int[(-64 - 16*x + 36*x^2 - 11*x^3 + x^4 + E^((-16*x + 8*x^2 - x^3 + 25*Log[3]^2)/(16 - 8*x + x^2))*(-64*x + 48
*x^2 - 12*x^3 + x^4 + 50*x*Log[3]^2))/(-64*x + 48*x^2 - 12*x^3 + x^4),x]

[Out]

72/(4 - x)^2 - 36/(4 - x) + x - (9*x^2)/(2*(4 - x)^2) + Log[x] + Defer[Int][E^((-16*x + 8*x^2 - x^3 + 25*Log[3
]^2)/(-4 + x)^2), x] + 50*Log[3]^2*Defer[Int][E^((-16*x + 8*x^2 - x^3 + 25*Log[3]^2)/(-4 + x)^2)/(-4 + x)^3, x
]

Rubi steps \begin{align*} \text {integral}& = \int \frac {64+16 x-36 x^2+11 x^3-x^4-e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{x \left (64-48 x+12 x^2-x^3\right )} \, dx \\ & = \int \left (-\frac {16}{(-4+x)^3}-\frac {64}{(-4+x)^3 x}+\frac {36 x}{(-4+x)^3}-\frac {11 x^2}{(-4+x)^3}+\frac {x^3}{(-4+x)^3}+\frac {e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{(-4+x)^2}} \left (-48 x+12 x^2-x^3+2 \left (32-25 \log ^2(3)\right )\right )}{(4-x)^3}\right ) \, dx \\ & = \frac {8}{(4-x)^2}-11 \int \frac {x^2}{(-4+x)^3} \, dx+36 \int \frac {x}{(-4+x)^3} \, dx-64 \int \frac {1}{(-4+x)^3 x} \, dx+\int \frac {x^3}{(-4+x)^3} \, dx+\int \frac {e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{(-4+x)^2}} \left (-48 x+12 x^2-x^3+2 \left (32-25 \log ^2(3)\right )\right )}{(4-x)^3} \, dx \\ & = \frac {8}{(4-x)^2}-\frac {9 x^2}{2 (4-x)^2}-11 \int \left (\frac {16}{(-4+x)^3}+\frac {8}{(-4+x)^2}+\frac {1}{-4+x}\right ) \, dx-64 \int \left (\frac {1}{4 (-4+x)^3}-\frac {1}{16 (-4+x)^2}+\frac {1}{64 (-4+x)}-\frac {1}{64 x}\right ) \, dx+\int \left (1+\frac {64}{(-4+x)^3}+\frac {48}{(-4+x)^2}+\frac {12}{-4+x}\right ) \, dx+\int \left (e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{(-4+x)^2}}+\frac {50 e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{(-4+x)^2}} \log ^2(3)}{(-4+x)^3}\right ) \, dx \\ & = \frac {72}{(4-x)^2}-\frac {36}{4-x}+x-\frac {9 x^2}{2 (4-x)^2}+\log (x)+\left (50 \log ^2(3)\right ) \int \frac {e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{(-4+x)^2}}}{(-4+x)^3} \, dx+\int e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{(-4+x)^2}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=-e^{-x+\frac {25 \log ^2(3)}{(-4+x)^2}}+x+\log (x) \]

[In]

Integrate[(-64 - 16*x + 36*x^2 - 11*x^3 + x^4 + E^((-16*x + 8*x^2 - x^3 + 25*Log[3]^2)/(16 - 8*x + x^2))*(-64*
x + 48*x^2 - 12*x^3 + x^4 + 50*x*Log[3]^2))/(-64*x + 48*x^2 - 12*x^3 + x^4),x]

[Out]

-E^(-x + (25*Log[3]^2)/(-4 + x)^2) + x + Log[x]

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13

method	result	size
risch	\(x +\ln \left (x \right )-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{\left (x -4\right )^{2}}}\)	\(34\)
parallelrisch	\(x +\ln \left (x \right )-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}+24\)	\(40\)
parts	\(x +\ln \left (x \right )+\frac {8 x \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}} x^{2}-16 \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}}{\left (x -4\right )^{2}}\)	\(118\)
norman	\(\frac {x^{3}-48 x +8 x \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}} x^{2}-16 \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}+128}{\left (x -4\right )^{2}}+\ln \left (x \right )\)	\(124\)

[In]

int(((50*x*ln(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*ln(3)^2-x^3+8*x^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36*x^2-16*
x-64)/(x^4-12*x^3+48*x^2-64*x),x,method=_RETURNVERBOSE)

[Out]

x+ln(x)-exp((25*ln(3)^2-x^3+8*x^2-16*x)/(x-4)^2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - e^{\left (-\frac {x^{3} - 8 \, x^{2} - 25 \, \log \left (3\right )^{2} + 16 \, x}{x^{2} - 8 \, x + 16}\right )} + \log \left (x\right ) \]

[In]

integrate(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36
*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x),x, algorithm="fricas")

[Out]

x - e^(-(x^3 - 8*x^2 - 25*log(3)^2 + 16*x)/(x^2 - 8*x + 16)) + log(x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - e^{\frac {- x^{3} + 8 x^{2} - 16 x + 25 \log {\left (3 \right )}^{2}}{x^{2} - 8 x + 16}} + \log {\left (x \right )} \]

[In]

integrate(((50*x*ln(3)**2+x**4-12*x**3+48*x**2-64*x)*exp((25*ln(3)**2-x**3+8*x**2-16*x)/(x**2-8*x+16))+x**4-11
*x**3+36*x**2-16*x-64)/(x**4-12*x**3+48*x**2-64*x),x)

[Out]

x - exp((-x**3 + 8*x**2 - 16*x + 25*log(3)**2)/(x**2 - 8*x + 16)) + log(x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (25) = 50\).

Time = 0.40 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.37 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - \frac {16 \, {\left (3 \, x - 10\right )}}{x^{2} - 8 \, x + 16} - \frac {36 \, {\left (x - 2\right )}}{x^{2} - 8 \, x + 16} + \frac {88 \, {\left (x - 3\right )}}{x^{2} - 8 \, x + 16} - \frac {4 \, {\left (x - 6\right )}}{x^{2} - 8 \, x + 16} + \frac {8}{x^{2} - 8 \, x + 16} - e^{\left (-x + \frac {25 \, \log \left (3\right )^{2}}{x^{2} - 8 \, x + 16}\right )} + \log \left (x\right ) \]

[In]

integrate(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36
*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x),x, algorithm="maxima")

[Out]

x - 16*(3*x - 10)/(x^2 - 8*x + 16) - 36*(x - 2)/(x^2 - 8*x + 16) + 88*(x - 3)/(x^2 - 8*x + 16) - 4*(x - 6)/(x^
2 - 8*x + 16) + 8/(x^2 - 8*x + 16) - e^(-x + 25*log(3)^2/(x^2 - 8*x + 16)) + log(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).

Time = 0.40 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - e^{\left (\frac {25}{16} \, \log \left (3\right )^{2} - \frac {25 \, x^{2} \log \left (3\right )^{2} + 16 \, x^{3} - 200 \, x \log \left (3\right )^{2} - 128 \, x^{2} + 256 \, x}{16 \, {\left (x^{2} - 8 \, x + 16\right )}}\right )} + \log \left (x\right ) \]

[In]

integrate(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36
*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x),x, algorithm="giac")

[Out]

x - e^(25/16*log(3)^2 - 1/16*(25*x^2*log(3)^2 + 16*x^3 - 200*x*log(3)^2 - 128*x^2 + 256*x)/(x^2 - 8*x + 16)) +
log(x)

Mupad [B] (verification not implemented)

Time = 11.65 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x+\ln \left (x\right )-{\mathrm {e}}^{-\frac {x^3}{x^2-8\,x+16}}\,{\mathrm {e}}^{\frac {8\,x^2}{x^2-8\,x+16}}\,{\mathrm {e}}^{-\frac {16\,x}{x^2-8\,x+16}}\,{\mathrm {e}}^{\frac {25\,{\ln \left (3\right )}^2}{x^2-8\,x+16}} \]

[In]

int((16*x - exp(-(16*x - 25*log(3)^2 - 8*x^2 + x^3)/(x^2 - 8*x + 16))*(50*x*log(3)^2 - 64*x + 48*x^2 - 12*x^3
+ x^4) - 36*x^2 + 11*x^3 - x^4 + 64)/(64*x - 48*x^2 + 12*x^3 - x^4),x)

[Out]

x + log(x) - exp(-x^3/(x^2 - 8*x + 16))*exp((8*x^2)/(x^2 - 8*x + 16))*exp(-(16*x)/(x^2 - 8*x + 16))*exp((25*lo
g(3)^2)/(x^2 - 8*x + 16))