3.4.43 \(\int \frac {(2 e^5 x+e^x (-e^5 x+x^3)) \log ^3(x)+(-4 x^2+2 e^x x^2) \log (2-e^x) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} (16-8 e^x+(-32+16 e^x) \log (x))}{(-2 x+e^x x) \log ^3(x)} \, dx\)

Optimal. Leaf size=34 \[ x \left (-e^5+\frac {e^{4 \left (-2+\frac {1}{\log (x)}\right )^2}}{x}+x \log \left (2-e^x\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 1.60, antiderivative size = 35, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 10, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6688, 14, 6742, 2184, 2190, 2531, 2282, 6589, 2532, 6706} \begin {gather*} x^2 \log \left (2-e^x\right )-e^5 x+e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2*E^5*x + E^x*(-(E^5*x) + x^3))*Log[x]^3 + (-4*x^2 + 2*E^x*x^2)*Log[2 - E^x]*Log[x]^3 + E^((4 - 16*Log[x
] + 16*Log[x]^2)/Log[x]^2)*(16 - 8*E^x + (-32 + 16*E^x)*Log[x]))/((-2*x + E^x*x)*Log[x]^3),x]

[Out]

E^((4*(1 - 2*Log[x])^2)/Log[x]^2) - E^5*x + x^2*Log[2 - E^x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 2532

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

Rule 6589

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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {2 e^5 x+e^x \left (-e^5 x+x^3\right )}{-2+e^x}+2 x^2 \log \left (2-e^x\right )+\frac {8 e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}} (-1+2 \log (x))}{\log ^3(x)}}{x} \, dx\\ &=\int \left (-\frac {-2 e^5+e^{5+x}-e^x x^2+4 x \log \left (2-e^x\right )-2 e^x x \log \left (2-e^x\right )}{-2+e^x}+\frac {8 e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}} (-1+2 \log (x))}{x \log ^3(x)}\right ) \, dx\\ &=8 \int \frac {e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}} (-1+2 \log (x))}{x \log ^3(x)} \, dx-\int \frac {-2 e^5+e^{5+x}-e^x x^2+4 x \log \left (2-e^x\right )-2 e^x x \log \left (2-e^x\right )}{-2+e^x} \, dx\\ &=8 \operatorname {Subst}\left (\int \frac {e^{\frac {4 (1-2 x)^2}{x^2}} (-1+2 x)}{x^3} \, dx,x,\log (x)\right )-\int \left (e^5-x^2-\frac {2 x^2}{-2+e^x}-2 x \log \left (2-e^x\right )\right ) \, dx\\ &=e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}-e^5 x+\frac {x^3}{3}+2 \int \frac {x^2}{-2+e^x} \, dx+2 \int x \log \left (2-e^x\right ) \, dx\\ &=e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}-e^5 x+x^2 \log \left (2-e^x\right )-x^2 \log \left (1-\frac {e^x}{2}\right )+2 \int x \log \left (1-\frac {e^x}{2}\right ) \, dx+\int \frac {e^x x^2}{-2+e^x} \, dx\\ &=e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}-e^5 x+x^2 \log \left (2-e^x\right )-2 x \text {Li}_2\left (\frac {e^x}{2}\right )-2 \int x \log \left (1-\frac {e^x}{2}\right ) \, dx+2 \int \text {Li}_2\left (\frac {e^x}{2}\right ) \, dx\\ &=e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}-e^5 x+x^2 \log \left (2-e^x\right )-2 \int \text {Li}_2\left (\frac {e^x}{2}\right ) \, dx+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,e^x\right )\\ &=e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}-e^5 x+x^2 \log \left (2-e^x\right )+2 \text {Li}_3\left (\frac {e^x}{2}\right )-2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,e^x\right )\\ &=e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}-e^5 x+x^2 \log \left (2-e^x\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.50, size = 35, normalized size = 1.03 \begin {gather*} e^{16+\frac {4}{\log ^2(x)}-\frac {16}{\log (x)}}-e^5 x+x^2 \log \left (2-e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2*E^5*x + E^x*(-(E^5*x) + x^3))*Log[x]^3 + (-4*x^2 + 2*E^x*x^2)*Log[2 - E^x]*Log[x]^3 + E^((4 - 16
*Log[x] + 16*Log[x]^2)/Log[x]^2)*(16 - 8*E^x + (-32 + 16*E^x)*Log[x]))/((-2*x + E^x*x)*Log[x]^3),x]

[Out]

E^(16 + 4/Log[x]^2 - 16/Log[x]) - E^5*x + x^2*Log[2 - E^x]

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 36, normalized size = 1.06 \begin {gather*} x^{2} \log \left (-e^{x} + 2\right ) - x e^{5} + e^{\left (\frac {4 \, {\left (4 \, \log \relax (x)^{2} - 4 \, \log \relax (x) + 1\right )}}{\log \relax (x)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-4*x^2)*log(x)^3*log(-exp(x)+2)+((16*exp(x)-32)*log(x)-8*exp(x)+16)*exp((16*log(x)^2-1
6*log(x)+4)/log(x)^2)+((-x*exp(5)+x^3)*exp(x)+2*x*exp(5))*log(x)^3)/(exp(x)*x-2*x)/log(x)^3,x, algorithm="fric
as")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-4*x^2)*log(x)^3*log(-exp(x)+2)+((16*exp(x)-32)*log(x)-8*exp(x)+16)*exp((16*log(x)^2-1
6*log(x)+4)/log(x)^2)+((-x*exp(5)+x^3)*exp(x)+2*x*exp(5))*log(x)^3)/(exp(x)*x-2*x)/log(x)^3,x, algorithm="giac
")

[Out]

undef

________________________________________________________________________________________

maple [A]  time = 0.06, size = 33, normalized size = 0.97

method	result	size
risch	\(\ln \left (-{\mathrm e}^{x}+2\right ) x^{2}-x \,{\mathrm e}^{5}+{\mathrm e}^{\frac {4 \left (2 \ln \relax (x )-1\right )^{2}}{\ln \relax (x )^{2}}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x^2-4*x^2)*ln(x)^3*ln(-exp(x)+2)+((16*exp(x)-32)*ln(x)-8*exp(x)+16)*exp((16*ln(x)^2-16*ln(x)+4)
/ln(x)^2)+((-x*exp(5)+x^3)*exp(x)+2*x*exp(5))*ln(x)^3)/(exp(x)*x-2*x)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.65, size = 48, normalized size = 1.41 \begin {gather*} {\left (x^{2} e^{\frac {16}{\log \relax (x)}} \log \left (-e^{x} + 2\right ) - x e^{\left (\frac {16}{\log \relax (x)} + 5\right )} + e^{\left (\frac {4}{\log \relax (x)^{2}} + 16\right )}\right )} e^{\left (-\frac {16}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-4*x^2)*log(x)^3*log(-exp(x)+2)+((16*exp(x)-32)*log(x)-8*exp(x)+16)*exp((16*log(x)^2-1
6*log(x)+4)/log(x)^2)+((-x*exp(5)+x^3)*exp(x)+2*x*exp(5))*log(x)^3)/(exp(x)*x-2*x)/log(x)^3,x, algorithm="maxi
ma")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.62, size = 34, normalized size = 1.00 \begin {gather*} x^2\,\ln \left (2-{\mathrm {e}}^x\right )-x\,{\mathrm {e}}^5+{\mathrm {e}}^{\frac {4}{{\ln \relax (x)}^2}}\,{\mathrm {e}}^{-\frac {16}{\ln \relax (x)}}\,{\mathrm {e}}^{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((16*log(x)^2 - 16*log(x) + 4)/log(x)^2)*(log(x)*(16*exp(x) - 32) - 8*exp(x) + 16) - log(x)^3*(exp(x)
*(x*exp(5) - x^3) - 2*x*exp(5)) + log(2 - exp(x))*log(x)^3*(2*x^2*exp(x) - 4*x^2))/(log(x)^3*(2*x - x*exp(x)))
,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 2.10, size = 34, normalized size = 1.00 \begin {gather*} x^{2} \log {\left (2 - e^{x} \right )} - x e^{5} + e^{\frac {16 \log {\relax (x )}^{2} - 16 \log {\relax (x )} + 4}{\log {\relax (x )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x**2-4*x**2)*ln(x)**3*ln(-exp(x)+2)+((16*exp(x)-32)*ln(x)-8*exp(x)+16)*exp((16*ln(x)**2-1
6*ln(x)+4)/ln(x)**2)+((-x*exp(5)+x**3)*exp(x)+2*x*exp(5))*ln(x)**3)/(exp(x)*x-2*x)/ln(x)**3,x)

[Out]

x**2*log(2 - exp(x)) - x*exp(5) + exp((16*log(x)**2 - 16*log(x) + 4)/log(x)**2)

________________________________________________________________________________________