\(\int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} (-64+120 x-48 x^2-8 x^3+e^4 (-8+16 x-8 x^2))}{4-8 x+4 x^2} \, dx\) [2141]

Optimal result

Integrand size = 67, antiderivative size = 31 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=1-e^{\left (8+e^4+x\right )^2}+\frac {1}{4} \left (4+\frac {x}{-x+x^2}\right ) \]

[Out]

2+1/4*x/(x^2-x)-exp((8+exp(4)+x)^2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {27, 12, 6820, 2259, 2240} \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=-e^{\left (x+e^4+8\right )^2}-\frac {1}{4 (1-x)} \]

[In]

Int[(-1 + E^(64 + E^8 + 16*x + x^2 + E^4*(16 + 2*x))*(-64 + 120*x - 48*x^2 - 8*x^3 + E^4*(-8 + 16*x - 8*x^2)))
/(4 - 8*x + 4*x^2),x]

[Out]

-E^(8 + E^4 + x)^2 - 1/(4*(1 - x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2259

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4 (-1+x)^2} \, dx \\ & = \frac {1}{4} \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{(-1+x)^2} \, dx \\ & = \frac {1}{4} \int \left (-\frac {1}{(-1+x)^2}-8 e^{\left (8+e^4\right )^2+2 \left (8+e^4\right ) x+x^2} \left (8+e^4+x\right )\right ) \, dx \\ & = -\frac {1}{4 (1-x)}-2 \int e^{\left (8+e^4\right )^2+2 \left (8+e^4\right ) x+x^2} \left (8+e^4+x\right ) \, dx \\ & = -\frac {1}{4 (1-x)}-2 \int e^{\left (8+e^4+x\right )^2} \left (8+e^4+x\right ) \, dx \\ & = -e^{\left (8+e^4+x\right )^2}-\frac {1}{4 (1-x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=-e^{\left (8+e^4+x\right )^2}+\frac {1}{4 (-1+x)} \]

[In]

Integrate[(-1 + E^(64 + E^8 + 16*x + x^2 + E^4*(16 + 2*x))*(-64 + 120*x - 48*x^2 - 8*x^3 + E^4*(-8 + 16*x - 8*
x^2)))/(4 - 8*x + 4*x^2),x]

[Out]

-E^(8 + E^4 + x)^2 + 1/(4*(-1 + x))

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {1}{-4+4 x}-{\mathrm e}^{2 x \,{\mathrm e}^{4}+x^{2}+16 \,{\mathrm e}^{4}+{\mathrm e}^{8}+16 x +64}\)	\(31\)
norman	\(\frac {-{\mathrm e}^{{\mathrm e}^{8}+\left (2 x +16\right ) {\mathrm e}^{4}+x^{2}+16 x +64} x +\frac {1}{4}+{\mathrm e}^{{\mathrm e}^{8}+\left (2 x +16\right ) {\mathrm e}^{4}+x^{2}+16 x +64}}{-1+x}\)	\(54\)
parallelrisch	\(-\frac {4 \,{\mathrm e}^{{\mathrm e}^{8}+\left (2 x +16\right ) {\mathrm e}^{4}+x^{2}+16 x +64} x -1-4 \,{\mathrm e}^{{\mathrm e}^{8}+\left (2 x +16\right ) {\mathrm e}^{4}+x^{2}+16 x +64}}{4 \left (-1+x \right )}\)	\(57\)
default	\(-{\mathrm e}^{x^{2}+\left (2 \,{\mathrm e}^{4}+16\right ) x +{\mathrm e}^{8}+16 \,{\mathrm e}^{4}+64}-\frac {i \left (2 \,{\mathrm e}^{4}+16\right ) \sqrt {\pi }\, {\mathrm e}^{{\mathrm e}^{8}+16 \,{\mathrm e}^{4}+64-\frac {\left (2 \,{\mathrm e}^{4}+16\right )^{2}}{4}} \operatorname {erf}\left (i x +\frac {i \left (2 \,{\mathrm e}^{4}+16\right )}{2}\right )}{2}+i {\mathrm e}^{4} \sqrt {\pi }\, {\mathrm e}^{{\mathrm e}^{8}+16 \,{\mathrm e}^{4}+64-\frac {\left (2 \,{\mathrm e}^{4}+16\right )^{2}}{4}} \operatorname {erf}\left (i x +\frac {i \left (2 \,{\mathrm e}^{4}+16\right )}{2}\right )+8 i \sqrt {\pi }\, {\mathrm e}^{{\mathrm e}^{8}+16 \,{\mathrm e}^{4}+64-\frac {\left (2 \,{\mathrm e}^{4}+16\right )^{2}}{4}} \operatorname {erf}\left (i x +\frac {i \left (2 \,{\mathrm e}^{4}+16\right )}{2}\right )+\frac {1}{-4+4 x}\)	\(167\)
parts	\(8 i {\mathrm e}^{{\mathrm e}^{8}} {\mathrm e}^{16 \,{\mathrm e}^{4}} {\mathrm e}^{64} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (2 \,{\mathrm e}^{4}+16\right )^{2}}{4}} \operatorname {erf}\left (i x +\frac {i \left (2 \,{\mathrm e}^{4}+16\right )}{2}\right )-2 \,{\mathrm e}^{{\mathrm e}^{8}} {\mathrm e}^{16 \,{\mathrm e}^{4}} {\mathrm e}^{64} \left (\frac {{\mathrm e}^{x^{2}+\left (2 \,{\mathrm e}^{4}+16\right ) x}}{2}+\frac {i \left (2 \,{\mathrm e}^{4}+16\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (2 \,{\mathrm e}^{4}+16\right )^{2}}{4}} \operatorname {erf}\left (i x +\frac {i \left (2 \,{\mathrm e}^{4}+16\right )}{2}\right )}{4}\right )+i {\mathrm e}^{{\mathrm e}^{8}} {\mathrm e}^{16 \,{\mathrm e}^{4}} {\mathrm e}^{64} {\mathrm e}^{4} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (2 \,{\mathrm e}^{4}+16\right )^{2}}{4}} \operatorname {erf}\left (i x +\frac {i \left (2 \,{\mathrm e}^{4}+16\right )}{2}\right )+\frac {1}{-4+4 x}\)	\(167\)

[In]

int((((-8*x^2+16*x-8)*exp(4)-8*x^3-48*x^2+120*x-64)*exp(exp(4)^2+(2*x+16)*exp(4)+x^2+16*x+64)-1)/(4*x^2-8*x+4)
,x,method=_RETURNVERBOSE)

[Out]

1/4/(-1+x)-exp(2*x*exp(4)+x^2+16*exp(4)+exp(8)+16*x+64)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=-\frac {4 \, {\left (x - 1\right )} e^{\left (x^{2} + 2 \, {\left (x + 8\right )} e^{4} + 16 \, x + e^{8} + 64\right )} - 1}{4 \, {\left (x - 1\right )}} \]

[In]

integrate((((-8*x^2+16*x-8)*exp(4)-8*x^3-48*x^2+120*x-64)*exp(exp(4)^2+(2*x+16)*exp(4)+x^2+16*x+64)-1)/(4*x^2-
8*x+4),x, algorithm="fricas")

[Out]

-1/4*(4*(x - 1)*e^(x^2 + 2*(x + 8)*e^4 + 16*x + e^8 + 64) - 1)/(x - 1)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=- e^{x^{2} + 16 x + \left (2 x + 16\right ) e^{4} + 64 + e^{8}} + \frac {1}{4 x - 4} \]

[In]

integrate((((-8*x**2+16*x-8)*exp(4)-8*x**3-48*x**2+120*x-64)*exp(exp(4)**2+(2*x+16)*exp(4)+x**2+16*x+64)-1)/(4
*x**2-8*x+4),x)

[Out]

-exp(x**2 + 16*x + (2*x + 16)*exp(4) + 64 + exp(8)) + 1/(4*x - 4)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=\frac {1}{4 \, {\left (x - 1\right )}} - e^{\left (x^{2} + 2 \, x e^{4} + 16 \, x + e^{8} + 16 \, e^{4} + 64\right )} \]

[In]

integrate((((-8*x^2+16*x-8)*exp(4)-8*x^3-48*x^2+120*x-64)*exp(exp(4)^2+(2*x+16)*exp(4)+x^2+16*x+64)-1)/(4*x^2-
8*x+4),x, algorithm="maxima")

[Out]

1/4/(x - 1) - e^(x^2 + 2*x*e^4 + 16*x + e^8 + 16*e^4 + 64)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=-\frac {4 \, x e^{\left (x^{2} + 2 \, x e^{4} + 16 \, x + e^{8} + 16 \, e^{4} + 64\right )} - 4 \, e^{\left (x^{2} + 2 \, x e^{4} + 16 \, x + e^{8} + 16 \, e^{4} + 64\right )} - 1}{4 \, {\left (x - 1\right )}} \]

[In]

integrate((((-8*x^2+16*x-8)*exp(4)-8*x^3-48*x^2+120*x-64)*exp(exp(4)^2+(2*x+16)*exp(4)+x^2+16*x+64)-1)/(4*x^2-
8*x+4),x, algorithm="giac")

[Out]

-1/4*(4*x*e^(x^2 + 2*x*e^4 + 16*x + e^8 + 16*e^4 + 64) - 4*e^(x^2 + 2*x*e^4 + 16*x + e^8 + 16*e^4 + 64) - 1)/(
x - 1)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-1+e^{64+e^8+16 x+x^2+e^4 (16+2 x)} \left (-64+120 x-48 x^2-8 x^3+e^4 \left (-8+16 x-8 x^2\right )\right )}{4-8 x+4 x^2} \, dx=\frac {1}{4\,\left (x-1\right )}-{\mathrm {e}}^{16\,{\mathrm {e}}^4}\,{\mathrm {e}}^{16\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{64}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^4}\,{\mathrm {e}}^{{\mathrm {e}}^8} \]

[In]

int(-(exp(16*x + exp(8) + x^2 + exp(4)*(2*x + 16) + 64)*(exp(4)*(8*x^2 - 16*x + 8) - 120*x + 48*x^2 + 8*x^3 +
64) + 1)/(4*x^2 - 8*x + 4),x)

[Out]

1/(4*(x - 1)) - exp(16*exp(4))*exp(16*x)*exp(x^2)*exp(64)*exp(2*x*exp(4))*exp(exp(8))