3.18.4 \(\int \frac {128 x+e^x (-8 x+4 x^2)+e^{\frac {1}{4} (-x+4 x^3+4 \log (4))} (16-192 x^2+e^x (-5+12 x^2))}{1024-128 e^x+4 e^{2 x}} \, dx\)

Optimal. Leaf size=27 \[ \frac {-4 e^{-\frac {x}{4}+x^3}+x^2}{16-e^x} \]

________________________________________________________________________________________

Rubi [F]  time = 2.30, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {128 x+e^x \left (-8 x+4 x^2\right )+e^{\frac {1}{4} \left (-x+4 x^3+4 \log (4)\right )} \left (16-192 x^2+e^x \left (-5+12 x^2\right )\right )}{1024-128 e^x+4 e^{2 x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(128*x + E^x*(-8*x + 4*x^2) + E^((-x + 4*x^3 + 4*Log[4])/4)*(16 - 192*x^2 + E^x*(-5 + 12*x^2)))/(1024 - 12
8*E^x + 4*E^(2*x)),x]

[Out]

x^2/(16 - E^x) + 12*Defer[Int][(E^(-1/4*x + x^3)*x^2)/(-16 + E^x), x] - Defer[Subst][Defer[Int][E^(-x + 64*x^3
)/(-2 + E^x)^2, x], x, x/4]/4 - Defer[Subst][Defer[Int][E^(-x + 64*x^3)/(-2 + E^x), x], x, x/4]/4 - Defer[Subs
t][Defer[Int][E^(-x + 64*x^3)/(2 + E^x)^2, x], x, x/4]/4 + Defer[Subst][Defer[Int][E^(-x + 64*x^3)/(2 + E^x),
x], x, x/4]/4 - 4*Defer[Subst][Defer[Int][E^(-x + 64*x^3)/(4 + E^(2*x))^2, x], x, x/4] + (3*Defer[Subst][Defer
[Int][E^(-x + 64*x^3)/(4 + E^(2*x)), x], x, x/4])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {128 x+e^x \left (-8 x+4 x^2\right )+e^{\frac {1}{4} \left (-x+4 x^3+4 \log (4)\right )} \left (16-192 x^2+e^x \left (-5+12 x^2\right )\right )}{4 \left (16-e^x\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {128 x+e^x \left (-8 x+4 x^2\right )+e^{\frac {1}{4} \left (-x+4 x^3+4 \log (4)\right )} \left (16-192 x^2+e^x \left (-5+12 x^2\right )\right )}{\left (16-e^x\right )^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {128 x}{\left (-16+e^x\right )^2}+\frac {4 e^x (-2+x) x}{\left (-16+e^x\right )^2}+\frac {4 e^{-\frac {x}{4}+x^3} \left (16-5 e^x-192 x^2+12 e^x x^2\right )}{\left (-16+e^x\right )^2}\right ) \, dx\\ &=32 \int \frac {x}{\left (-16+e^x\right )^2} \, dx+\int \frac {e^x (-2+x) x}{\left (-16+e^x\right )^2} \, dx+\int \frac {e^{-\frac {x}{4}+x^3} \left (16-5 e^x-192 x^2+12 e^x x^2\right )}{\left (-16+e^x\right )^2} \, dx\\ &=2 \int \frac {e^x x}{\left (-16+e^x\right )^2} \, dx-2 \int \frac {x}{-16+e^x} \, dx+\int \left (-\frac {2 e^x x}{\left (-16+e^x\right )^2}+\frac {e^x x^2}{\left (-16+e^x\right )^2}\right ) \, dx+\int \left (-\frac {64 e^{-\frac {x}{4}+x^3}}{\left (-16+e^x\right )^2}+\frac {e^{-\frac {x}{4}+x^3} \left (-5+12 x^2\right )}{-16+e^x}\right ) \, dx\\ &=\frac {2 x}{16-e^x}+\frac {x^2}{16}-\frac {1}{8} \int \frac {e^x x}{-16+e^x} \, dx+2 \int \frac {1}{-16+e^x} \, dx-2 \int \frac {e^x x}{\left (-16+e^x\right )^2} \, dx-64 \int \frac {e^{-\frac {x}{4}+x^3}}{\left (-16+e^x\right )^2} \, dx+\int \frac {e^x x^2}{\left (-16+e^x\right )^2} \, dx+\int \frac {e^{-\frac {x}{4}+x^3} \left (-5+12 x^2\right )}{-16+e^x} \, dx\\ &=\frac {x^2}{16}+\frac {x^2}{16-e^x}-\frac {1}{8} x \log \left (1-\frac {e^x}{16}\right )+\frac {1}{8} \int \log \left (1-\frac {e^x}{16}\right ) \, dx-2 \int \frac {1}{-16+e^x} \, dx+2 \int \frac {x}{-16+e^x} \, dx+2 \operatorname {Subst}\left (\int \frac {1}{(-16+x) x} \, dx,x,e^x\right )-256 \operatorname {Subst}\left (\int \frac {e^{x \left (-1+64 x^2\right )}}{\left (-16+e^{4 x}\right )^2} \, dx,x,\frac {x}{4}\right )+\int \left (-\frac {5 e^{-\frac {x}{4}+x^3}}{-16+e^x}+\frac {12 e^{-\frac {x}{4}+x^3} x^2}{-16+e^x}\right ) \, dx\\ &=\frac {x^2}{16-e^x}-\frac {1}{8} x \log \left (1-\frac {e^x}{16}\right )+\frac {1}{8} \int \frac {e^x x}{-16+e^x} \, dx+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{-16+x} \, dx,x,e^x\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{16}\right )}{x} \, dx,x,e^x\right )-2 \operatorname {Subst}\left (\int \frac {1}{(-16+x) x} \, dx,x,e^x\right )-5 \int \frac {e^{-\frac {x}{4}+x^3}}{-16+e^x} \, dx+12 \int \frac {e^{-\frac {x}{4}+x^3} x^2}{-16+e^x} \, dx-256 \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (16-e^{4 x}\right )^2} \, dx,x,\frac {x}{4}\right )\\ &=-\frac {x}{8}+\frac {x^2}{16-e^x}+\frac {1}{8} \log \left (16-e^x\right )-\frac {\text {Li}_2\left (\frac {e^x}{16}\right )}{8}-\frac {1}{8} \int \log \left (1-\frac {e^x}{16}\right ) \, dx-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{-16+x} \, dx,x,e^x\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+12 \int \frac {e^{-\frac {x}{4}+x^3} x^2}{-16+e^x} \, dx-20 \operatorname {Subst}\left (\int \frac {e^{x \left (-1+64 x^2\right )}}{-16+e^{4 x}} \, dx,x,\frac {x}{4}\right )-256 \operatorname {Subst}\left (\int \left (\frac {e^{-x+64 x^3}}{1024 \left (-2+e^x\right )^2}-\frac {3 e^{-x+64 x^3}}{2048 \left (-2+e^x\right )}+\frac {e^{-x+64 x^3}}{1024 \left (2+e^x\right )^2}+\frac {3 e^{-x+64 x^3}}{2048 \left (2+e^x\right )}+\frac {e^{-x+64 x^3}}{64 \left (4+e^{2 x}\right )^2}+\frac {e^{-x+64 x^3}}{256 \left (4+e^{2 x}\right )}\right ) \, dx,x,\frac {x}{4}\right )\\ &=\frac {x^2}{16-e^x}-\frac {\text {Li}_2\left (\frac {e^x}{16}\right )}{8}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{16}\right )}{x} \, dx,x,e^x\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (-2+e^x\right )^2} \, dx,x,\frac {x}{4}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (2+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{-2+e^x} \, dx,x,\frac {x}{4}\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{2+e^x} \, dx,x,\frac {x}{4}\right )-4 \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (4+e^{2 x}\right )^2} \, dx,x,\frac {x}{4}\right )+12 \int \frac {e^{-\frac {x}{4}+x^3} x^2}{-16+e^x} \, dx-20 \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{-16+e^{4 x}} \, dx,x,\frac {x}{4}\right )-\operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{4+e^{2 x}} \, dx,x,\frac {x}{4}\right )\\ &=\frac {x^2}{16-e^x}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (-2+e^x\right )^2} \, dx,x,\frac {x}{4}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (2+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{-2+e^x} \, dx,x,\frac {x}{4}\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{2+e^x} \, dx,x,\frac {x}{4}\right )-4 \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (4+e^{2 x}\right )^2} \, dx,x,\frac {x}{4}\right )+12 \int \frac {e^{-\frac {x}{4}+x^3} x^2}{-16+e^x} \, dx-20 \operatorname {Subst}\left (\int \left (\frac {e^{-x+64 x^3}}{32 \left (-2+e^x\right )}-\frac {e^{-x+64 x^3}}{32 \left (2+e^x\right )}-\frac {e^{-x+64 x^3}}{8 \left (4+e^{2 x}\right )}\right ) \, dx,x,\frac {x}{4}\right )-\operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{4+e^{2 x}} \, dx,x,\frac {x}{4}\right )\\ &=\frac {x^2}{16-e^x}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (-2+e^x\right )^2} \, dx,x,\frac {x}{4}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (2+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{-2+e^x} \, dx,x,\frac {x}{4}\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{2+e^x} \, dx,x,\frac {x}{4}\right )-\frac {5}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{-2+e^x} \, dx,x,\frac {x}{4}\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{2+e^x} \, dx,x,\frac {x}{4}\right )+\frac {5}{2} \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{4+e^{2 x}} \, dx,x,\frac {x}{4}\right )-4 \operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{\left (4+e^{2 x}\right )^2} \, dx,x,\frac {x}{4}\right )+12 \int \frac {e^{-\frac {x}{4}+x^3} x^2}{-16+e^x} \, dx-\operatorname {Subst}\left (\int \frac {e^{-x+64 x^3}}{4+e^{2 x}} \, dx,x,\frac {x}{4}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.69, size = 30, normalized size = 1.11 \begin {gather*} -\frac {4 e^{-\frac {x}{4}+x^3}-x^2}{16-e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(128*x + E^x*(-8*x + 4*x^2) + E^((-x + 4*x^3 + 4*Log[4])/4)*(16 - 192*x^2 + E^x*(-5 + 12*x^2)))/(102
4 - 128*E^x + 4*E^(2*x)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 26, normalized size = 0.96 \begin {gather*} -\frac {x^{2} - e^{\left (x^{3} - \frac {1}{4} \, x + 2 \, \log \relax (2)\right )}}{e^{x} - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2-5)*exp(x)-192*x^2+16)*exp(2*log(2)+x^3-1/4*x)+(4*x^2-8*x)*exp(x)+128*x)/(4*exp(x)^2-128*ex
p(x)+1024),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.51, size = 30, normalized size = 1.11 \begin {gather*} -\frac {x^{2} e^{x} - 4 \, e^{\left (x^{3} + \frac {3}{4} \, x\right )}}{e^{\left (2 \, x\right )} - 16 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2-5)*exp(x)-192*x^2+16)*exp(2*log(2)+x^3-1/4*x)+(4*x^2-8*x)*exp(x)+128*x)/(4*exp(x)^2-128*ex
p(x)+1024),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.11, size = 26, normalized size = 0.96

method	result	size
norman	\(\frac {-x^{2}+{\mathrm e}^{2 \ln \relax (2)+x^{3}-\frac {x}{4}}}{{\mathrm e}^{x}-16}\)	\(26\)
risch	\(-\frac {x^{2}}{{\mathrm e}^{x}-16}+\frac {4 \,{\mathrm e}^{\frac {x \left (2 x -1\right ) \left (2 x +1\right )}{4}}}{{\mathrm e}^{x}-16}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*x^2-5)*exp(x)-192*x^2+16)*exp(2*ln(2)+x^3-1/4*x)+(4*x^2-8*x)*exp(x)+128*x)/(4*exp(x)^2-128*exp(x)+10
24),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 11, normalized size = 0.41 \begin {gather*} -\frac {x^{2}}{e^{x} - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2-5)*exp(x)-192*x^2+16)*exp(2*log(2)+x^3-1/4*x)+(4*x^2-8*x)*exp(x)+128*x)/(4*exp(x)^2-128*ex
p(x)+1024),x, algorithm="maxima")

[Out]

-x^2/(e^x - 16)

________________________________________________________________________________________

mupad [B]  time = 1.22, size = 23, normalized size = 0.85 \begin {gather*} \frac {4\,{\mathrm {e}}^{x^3-\frac {x}{4}}-x^2}{{\mathrm {e}}^x-16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((128*x + exp(2*log(2) - x/4 + x^3)*(exp(x)*(12*x^2 - 5) - 192*x^2 + 16) - exp(x)*(8*x - 4*x^2))/(4*exp(2*x
) - 128*exp(x) + 1024),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 22, normalized size = 0.81 \begin {gather*} - \frac {x^{2}}{e^{x} - 16} + \frac {4 e^{x^{3} - \frac {x}{4}}}{e^{x} - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x**2-5)*exp(x)-192*x**2+16)*exp(2*ln(2)+x**3-1/4*x)+(4*x**2-8*x)*exp(x)+128*x)/(4*exp(x)**2-12
8*exp(x)+1024),x)

[Out]

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

________________________________________________________________________________________