\(\int \frac {-40 x-10 e^{2 x} x+(-160 x-160 x^3) \log (4)+e^x (40 x+(-20+80 x-40 x^2+80 x^3-20 x^4) \log (4))}{4-4 e^x+e^{2 x}} \, dx\) [342]

Optimal result

Integrand size = 70, antiderivative size = 27 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=5 \left (9-x^2+\frac {4 \left (1+x^2\right )^2 \log (4)}{-2+e^x}\right ) \]

[Out]

45+40/(exp(x)-2)*(x^2+1)^2*ln(2)-5*x^2

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89, number of steps used = 62, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {6873, 6874, 2320, 46, 2216, 2215, 2221, 2611, 6724, 2222, 2317, 2438, 6744, 36, 31, 29} \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=-\frac {20 x^4 \log (4)}{2-e^x}-5 x^2-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 \log (4)}{2-e^x} \]

[In]

Int[(-40*x - 10*E^(2*x)*x + (-160*x - 160*x^3)*Log[4] + E^x*(40*x + (-20 + 80*x - 40*x^2 + 80*x^3 - 20*x^4)*Lo
g[4]))/(4 - 4*E^x + E^(2*x)),x]

[Out]

-5*x^2 - (20*Log[4])/(2 - E^x) - (40*x^2*Log[4])/(2 - E^x) - (20*x^4*Log[4])/(2 - E^x)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2215

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

Rule 2216

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2221

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

Rule 2222

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

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

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

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{\left (2-e^x\right )^2} \, dx \\ & = \int \left (-10 x-\frac {40 \left (1+x^2\right )^2 \log (4)}{\left (-2+e^x\right )^2}-\frac {20 \left (1-4 x+2 x^2-4 x^3+x^4\right ) \log (4)}{-2+e^x}\right ) \, dx \\ & = -5 x^2-(20 \log (4)) \int \frac {1-4 x+2 x^2-4 x^3+x^4}{-2+e^x} \, dx-(40 \log (4)) \int \frac {\left (1+x^2\right )^2}{\left (-2+e^x\right )^2} \, dx \\ & = -5 x^2-(20 \log (4)) \int \left (\frac {1}{-2+e^x}-\frac {4 x}{-2+e^x}+\frac {2 x^2}{-2+e^x}-\frac {4 x^3}{-2+e^x}+\frac {x^4}{-2+e^x}\right ) \, dx-(40 \log (4)) \int \left (\frac {1}{\left (-2+e^x\right )^2}+\frac {2 x^2}{\left (-2+e^x\right )^2}+\frac {x^4}{\left (-2+e^x\right )^2}\right ) \, dx \\ & = -5 x^2-(20 \log (4)) \int \frac {1}{-2+e^x} \, dx-(20 \log (4)) \int \frac {x^4}{-2+e^x} \, dx-(40 \log (4)) \int \frac {1}{\left (-2+e^x\right )^2} \, dx-(40 \log (4)) \int \frac {x^2}{-2+e^x} \, dx-(40 \log (4)) \int \frac {x^4}{\left (-2+e^x\right )^2} \, dx+(80 \log (4)) \int \frac {x}{-2+e^x} \, dx-(80 \log (4)) \int \frac {x^2}{\left (-2+e^x\right )^2} \, dx+(80 \log (4)) \int \frac {x^3}{-2+e^x} \, dx \\ & = -5 x^2-20 x^2 \log (4)+\frac {20}{3} x^3 \log (4)-10 x^4 \log (4)+2 x^5 \log (4)-(10 \log (4)) \int \frac {e^x x^4}{-2+e^x} \, dx-(20 \log (4)) \int \frac {e^x x^2}{-2+e^x} \, dx-(20 \log (4)) \int \frac {e^x x^4}{\left (-2+e^x\right )^2} \, dx+(20 \log (4)) \int \frac {x^4}{-2+e^x} \, dx-(20 \log (4)) \text {Subst}\left (\int \frac {1}{(-2+x) x} \, dx,x,e^x\right )+(40 \log (4)) \int \frac {e^x x}{-2+e^x} \, dx-(40 \log (4)) \int \frac {e^x x^2}{\left (-2+e^x\right )^2} \, dx+(40 \log (4)) \int \frac {x^2}{-2+e^x} \, dx+(40 \log (4)) \int \frac {e^x x^3}{-2+e^x} \, dx-(40 \log (4)) \text {Subst}\left (\int \frac {1}{(-2+x)^2 x} \, dx,x,e^x\right ) \\ & = -5 x^2-20 x^2 \log (4)-\frac {40 x^2 \log (4)}{2-e^x}-10 x^4 \log (4)-\frac {20 x^4 \log (4)}{2-e^x}+40 x \log (4) \log \left (1-\frac {e^x}{2}\right )-20 x^2 \log (4) \log \left (1-\frac {e^x}{2}\right )+40 x^3 \log (4) \log \left (1-\frac {e^x}{2}\right )-10 x^4 \log (4) \log \left (1-\frac {e^x}{2}\right )+(10 \log (4)) \int \frac {e^x x^4}{-2+e^x} \, dx-(10 \log (4)) \text {Subst}\left (\int \frac {1}{-2+x} \, dx,x,e^x\right )+(10 \log (4)) \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+(20 \log (4)) \int \frac {e^x x^2}{-2+e^x} \, dx-(40 \log (4)) \int \log \left (1-\frac {e^x}{2}\right ) \, dx+(40 \log (4)) \int x \log \left (1-\frac {e^x}{2}\right ) \, dx+(40 \log (4)) \int x^3 \log \left (1-\frac {e^x}{2}\right ) \, dx-(40 \log (4)) \text {Subst}\left (\int \left (\frac {1}{2 (-2+x)^2}-\frac {1}{4 (-2+x)}+\frac {1}{4 x}\right ) \, dx,x,e^x\right )-(80 \log (4)) \int \frac {x}{-2+e^x} \, dx-(80 \log (4)) \int \frac {x^3}{-2+e^x} \, dx-(120 \log (4)) \int x^2 \log \left (1-\frac {e^x}{2}\right ) \, dx \\ & = -5 x^2-\frac {20 \log (4)}{2-e^x}-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 x^4 \log (4)}{2-e^x}+40 x \log (4) \log \left (1-\frac {e^x}{2}\right )+40 x^3 \log (4) \log \left (1-\frac {e^x}{2}\right )-40 x \log (4) \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right )+120 x^2 \log (4) \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right )-40 x^3 \log (4) \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right )-(40 \log (4)) \int \frac {e^x x}{-2+e^x} \, dx-(40 \log (4)) \int \frac {e^x x^3}{-2+e^x} \, dx-(40 \log (4)) \int x \log \left (1-\frac {e^x}{2}\right ) \, dx-(40 \log (4)) \int x^3 \log \left (1-\frac {e^x}{2}\right ) \, dx+(40 \log (4)) \int \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right ) \, dx-(40 \log (4)) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )+(120 \log (4)) \int x^2 \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right ) \, dx-(240 \log (4)) \int x \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right ) \, dx \\ & = -5 x^2-\frac {20 \log (4)}{2-e^x}-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 x^4 \log (4)}{2-e^x}+40 \log (4) \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right )+120 x^2 \log (4) \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right )-240 x \log (4) \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right )+120 x^2 \log (4) \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right )+(40 \log (4)) \int \log \left (1-\frac {e^x}{2}\right ) \, dx-(40 \log (4)) \int \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right ) \, dx+(40 \log (4)) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\frac {x}{2}\right )}{x} \, dx,x,e^x\right )+(120 \log (4)) \int x^2 \log \left (1-\frac {e^x}{2}\right ) \, dx-(120 \log (4)) \int x^2 \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right ) \, dx+(240 \log (4)) \int \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right ) \, dx-(240 \log (4)) \int x \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right ) \, dx \\ & = -5 x^2-\frac {20 \log (4)}{2-e^x}-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 x^4 \log (4)}{2-e^x}+40 \log (4) \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right )+40 \log (4) \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right )-240 x \log (4) \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right )-240 x \log (4) \operatorname {PolyLog}\left (4,\frac {e^x}{2}\right )+(40 \log (4)) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-(40 \log (4)) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\frac {x}{2}\right )}{x} \, dx,x,e^x\right )+(240 \log (4)) \int x \operatorname {PolyLog}\left (2,\frac {e^x}{2}\right ) \, dx+(240 \log (4)) \int x \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right ) \, dx+(240 \log (4)) \int \operatorname {PolyLog}\left (4,\frac {e^x}{2}\right ) \, dx+(240 \log (4)) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {x}{2}\right )}{x} \, dx,x,e^x\right ) \\ & = -5 x^2-\frac {20 \log (4)}{2-e^x}-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 x^4 \log (4)}{2-e^x}+240 \log (4) \operatorname {PolyLog}\left (4,\frac {e^x}{2}\right )-(240 \log (4)) \int \operatorname {PolyLog}\left (3,\frac {e^x}{2}\right ) \, dx-(240 \log (4)) \int \operatorname {PolyLog}\left (4,\frac {e^x}{2}\right ) \, dx+(240 \log (4)) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (4,\frac {x}{2}\right )}{x} \, dx,x,e^x\right ) \\ & = -5 x^2-\frac {20 \log (4)}{2-e^x}-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 x^4 \log (4)}{2-e^x}+240 \log (4) \operatorname {PolyLog}\left (4,\frac {e^x}{2}\right )+240 \log (4) \operatorname {PolyLog}\left (5,\frac {e^x}{2}\right )-(240 \log (4)) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-(240 \log (4)) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (4,\frac {x}{2}\right )}{x} \, dx,x,e^x\right ) \\ & = -5 x^2-\frac {20 \log (4)}{2-e^x}-\frac {40 x^2 \log (4)}{2-e^x}-\frac {20 x^4 \log (4)}{2-e^x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=-10 \left (\frac {x^2}{2}+\frac {-4 x^2 \log (4)-\log (16)-x^4 \log (16)}{-2+e^x}\right ) \]

[In]

Integrate[(-40*x - 10*E^(2*x)*x + (-160*x - 160*x^3)*Log[4] + E^x*(40*x + (-20 + 80*x - 40*x^2 + 80*x^3 - 20*x
^4)*Log[4]))/(4 - 4*E^x + E^(2*x)),x]

[Out]

-10*(x^2/2 + (-4*x^2*Log[4] - Log[16] - x^4*Log[16])/(-2 + E^x))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
risch	\(-5 x^{2}+\frac {40 \ln \left (2\right ) \left (x^{4}+2 x^{2}+1\right )}{{\mathrm e}^{x}-2}\)	\(27\)
norman	\(\frac {\left (10+80 \ln \left (2\right )\right ) x^{2}+40 x^{4} \ln \left (2\right )-5 \,{\mathrm e}^{x} x^{2}+40 \ln \left (2\right )}{{\mathrm e}^{x}-2}\)	\(37\)
parallelrisch	\(\frac {40 x^{4} \ln \left (2\right )+80 x^{2} \ln \left (2\right )-5 \,{\mathrm e}^{x} x^{2}+10 x^{2}+40 \ln \left (2\right )}{{\mathrm e}^{x}-2}\)	\(39\)

[In]

int((-10*x*exp(x)^2+(2*(-20*x^4+80*x^3-40*x^2+80*x-20)*ln(2)+40*x)*exp(x)+2*(-160*x^3-160*x)*ln(2)-40*x)/(exp(
x)^2-4*exp(x)+4),x,method=_RETURNVERBOSE)

[Out]

-5*x^2+40*ln(2)*(x^4+2*x^2+1)/(exp(x)-2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=-\frac {5 \, {\left (x^{2} e^{x} - 2 \, x^{2} - 8 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (2\right )\right )}}{e^{x} - 2} \]

[In]

integrate((-10*x*exp(x)^2+(2*(-20*x^4+80*x^3-40*x^2+80*x-20)*log(2)+40*x)*exp(x)+2*(-160*x^3-160*x)*log(2)-40*
x)/(exp(x)^2-4*exp(x)+4),x, algorithm="fricas")

[Out]

-5*(x^2*e^x - 2*x^2 - 8*(x^4 + 2*x^2 + 1)*log(2))/(e^x - 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=- 5 x^{2} + \frac {40 x^{4} \log {\left (2 \right )} + 80 x^{2} \log {\left (2 \right )} + 40 \log {\left (2 \right )}}{e^{x} - 2} \]

[In]

integrate((-10*x*exp(x)**2+(2*(-20*x**4+80*x**3-40*x**2+80*x-20)*ln(2)+40*x)*exp(x)+2*(-160*x**3-160*x)*ln(2)-
40*x)/(exp(x)**2-4*exp(x)+4),x)

[Out]

-5*x**2 + (40*x**4*log(2) + 80*x**2*log(2) + 40*log(2))/(exp(x) - 2)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=\frac {5 \, {\left (8 \, x^{4} \log \left (2\right ) + 2 \, x^{2} {\left (8 \, \log \left (2\right ) + 1\right )} - x^{2} e^{x} + 8 \, \log \left (2\right )\right )}}{e^{x} - 2} \]

[In]

integrate((-10*x*exp(x)^2+(2*(-20*x^4+80*x^3-40*x^2+80*x-20)*log(2)+40*x)*exp(x)+2*(-160*x^3-160*x)*log(2)-40*
x)/(exp(x)^2-4*exp(x)+4),x, algorithm="maxima")

[Out]

5*(8*x^4*log(2) + 2*x^2*(8*log(2) + 1) - x^2*e^x + 8*log(2))/(e^x - 2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=\frac {5 \, {\left (8 \, x^{4} \log \left (2\right ) - x^{2} e^{x} + 16 \, x^{2} \log \left (2\right ) + 2 \, x^{2} + 8 \, \log \left (2\right )\right )}}{e^{x} - 2} \]

[In]

integrate((-10*x*exp(x)^2+(2*(-20*x^4+80*x^3-40*x^2+80*x-20)*log(2)+40*x)*exp(x)+2*(-160*x^3-160*x)*log(2)-40*
x)/(exp(x)^2-4*exp(x)+4),x, algorithm="giac")

[Out]

5*(8*x^4*log(2) - x^2*e^x + 16*x^2*log(2) + 2*x^2 + 8*log(2))/(e^x - 2)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-40 x-10 e^{2 x} x+\left (-160 x-160 x^3\right ) \log (4)+e^x \left (40 x+\left (-20+80 x-40 x^2+80 x^3-20 x^4\right ) \log (4)\right )}{4-4 e^x+e^{2 x}} \, dx=\frac {40\,\ln \left (2\right )\,x^4+80\,\ln \left (2\right )\,x^2+40\,\ln \left (2\right )}{{\mathrm {e}}^x-2}-5\,x^2 \]

[In]

int(-(40*x + 10*x*exp(2*x) - exp(x)*(40*x - 2*log(2)*(40*x^2 - 80*x - 80*x^3 + 20*x^4 + 20)) + 2*log(2)*(160*x
+ 160*x^3))/(exp(2*x) - 4*exp(x) + 4),x)

[Out]

(40*log(2) + 80*x^2*log(2) + 40*x^4*log(2))/(exp(x) - 2) - 5*x^2