\(\int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx\) [5648]

Optimal result

Integrand size = 67, antiderivative size = 19 \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\frac {1}{-16 e^{-x (4-\log (2))}+3 x} \]

[Out]

1/(3*x-16/exp(x)/exp((4-ln(2))*x-x))

Rubi [F]

\[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx \]

[In]

Int[(-3*E^(8*x - 2*x*Log[2]) + E^(4*x - x*Log[2])*(-64 + 16*Log[2]))/(256 - 96*E^(4*x - x*Log[2])*x + 9*E^(8*x
- 2*x*Log[2])*x^2),x]

[Out]

1/(3*x) - Defer[Int][4^(4 + x)/(x^2*(-2^(4 + x) + 3*E^(4*x)*x)^2), x]/3 - ((4 - Log[2])*Defer[Int][4^(4 + x)/(
x*(-2^(4 + x) + 3*E^(4*x)*x)^2), x])/3 - Defer[Int][2^(5 + x)/(x^2*(-2^(4 + x) + 3*E^(4*x)*x)), x]/3 - ((4 - L
og[2])*Defer[Int][2^(4 + x)/(x*(-2^(4 + x) + 3*E^(4*x)*x)), x])/3

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

Mathematica [A] (verified)

Time = 1.67 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=-\frac {e^{4 x}}{2^{4+x}-3 e^{4 x} x} \]

[In]

Integrate[(-3*E^(8*x - 2*x*Log[2]) + E^(4*x - x*Log[2])*(-64 + 16*Log[2]))/(256 - 96*E^(4*x - x*Log[2])*x + 9*
E^(8*x - 2*x*Log[2])*x^2),x]

[Out]

-(E^(4*x)/(2^(4 + x) - 3*E^(4*x)*x))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {1}{3 x}+\frac {16}{3 x \left (3 \left (\frac {1}{2}\right )^{x} x \,{\mathrm e}^{4 x}-16\right )}\)	\(26\)
norman	\(\frac {{\mathrm e}^{-x \ln \left (2\right )+3 x} {\mathrm e}^{x}}{3 x \,{\mathrm e}^{-x \ln \left (2\right )+3 x} {\mathrm e}^{x}-16}\)	\(33\)
parallelrisch	\(\frac {{\mathrm e}^{-x \ln \left (2\right )+3 x} {\mathrm e}^{x}}{3 x \,{\mathrm e}^{-x \ln \left (2\right )+3 x} {\mathrm e}^{x}-16}\)	\(33\)

[In]

int((-3*exp(-x*ln(2)+3*x)^2*exp(x)^2+(16*ln(2)-64)*exp(-x*ln(2)+3*x)*exp(x))/(9*x^2*exp(-x*ln(2)+3*x)^2*exp(x)
^2-96*x*exp(-x*ln(2)+3*x)*exp(x)+256),x,method=_RETURNVERBOSE)

[Out]

1/3/x+16/3/x/(3*(1/2)^x*x*exp(4*x)-16)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\frac {e^{\left (-x \log \left (2\right ) + 4 \, x\right )}}{3 \, x e^{\left (-x \log \left (2\right ) + 4 \, x\right )} - 16} \]

[In]

integrate((-3*exp(-x*log(2)+3*x)^2*exp(x)^2+(16*log(2)-64)*exp(-x*log(2)+3*x)*exp(x))/(9*x^2*exp(-x*log(2)+3*x
)^2*exp(x)^2-96*x*exp(-x*log(2)+3*x)*exp(x)+256),x, algorithm="fricas")

[Out]

e^(-x*log(2) + 4*x)/(3*x*e^(-x*log(2) + 4*x) - 16)

Sympy [F(-1)]

Timed out. \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\text {Timed out} \]

[In]

integrate((-3*exp(-x*ln(2)+3*x)**2*exp(x)**2+(16*ln(2)-64)*exp(-x*ln(2)+3*x)*exp(x))/(9*x**2*exp(-x*ln(2)+3*x)
**2*exp(x)**2-96*x*exp(-x*ln(2)+3*x)*exp(x)+256),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\frac {e^{\left (4 \, x\right )}}{3 \, x e^{\left (4 \, x\right )} - 16 \cdot 2^{x}} \]

[In]

integrate((-3*exp(-x*log(2)+3*x)^2*exp(x)^2+(16*log(2)-64)*exp(-x*log(2)+3*x)*exp(x))/(9*x^2*exp(-x*log(2)+3*x
)^2*exp(x)^2-96*x*exp(-x*log(2)+3*x)*exp(x)+256),x, algorithm="maxima")

[Out]

e^(4*x)/(3*x*e^(4*x) - 16*2^x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\frac {e^{\left (-x \log \left (2\right ) + 4 \, x\right )}}{3 \, x e^{\left (-x \log \left (2\right ) + 4 \, x\right )} - 16} \]

[In]

integrate((-3*exp(-x*log(2)+3*x)^2*exp(x)^2+(16*log(2)-64)*exp(-x*log(2)+3*x)*exp(x))/(9*x^2*exp(-x*log(2)+3*x
)^2*exp(x)^2-96*x*exp(-x*log(2)+3*x)*exp(x)+256),x, algorithm="giac")

[Out]

e^(-x*log(2) + 4*x)/(3*x*e^(-x*log(2) + 4*x) - 16)

Mupad [F(-1)]

Timed out. \[ \int \frac {-3 e^{8 x-2 x \log (2)}+e^{4 x-x \log (2)} (-64+16 \log (2))}{256-96 e^{4 x-x \log (2)} x+9 e^{8 x-2 x \log (2)} x^2} \, dx=\int -\frac {3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{6\,x-2\,x\,\ln \left (2\right )}-{\mathrm {e}}^x\,{\mathrm {e}}^{3\,x-x\,\ln \left (2\right )}\,\left (16\,\ln \left (2\right )-64\right )}{9\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{6\,x-2\,x\,\ln \left (2\right )}-96\,x\,{\mathrm {e}}^x\,{\mathrm {e}}^{3\,x-x\,\ln \left (2\right )}+256} \,d x \]

[In]

int(-(3*exp(2*x)*exp(6*x - 2*x*log(2)) - exp(x)*exp(3*x - x*log(2))*(16*log(2) - 64))/(9*x^2*exp(2*x)*exp(6*x
- 2*x*log(2)) - 96*x*exp(x)*exp(3*x - x*log(2)) + 256),x)

[Out]

int(-(3*exp(2*x)*exp(6*x - 2*x*log(2)) - exp(x)*exp(3*x - x*log(2))*(16*log(2) - 64))/(9*x^2*exp(2*x)*exp(6*x
- 2*x*log(2)) - 96*x*exp(x)*exp(3*x - x*log(2)) + 256), x)