\(\int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log (4+2 e^x) (1568+784 e^x+(1120+560 e^x) \log (\log (12))+(312+156 e^x) \log ^2(\log (12))+(40+20 e^x) \log ^3(\log (12))+(2+e^x) \log ^4(\log (12)))}{2+e^x} \, dx\) [9229]

Optimal result

Integrand size = 120, antiderivative size = 21 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x \log \left (2 \left (2+e^x\right )\right ) \left (3+(5+\log (\log (12)))^2\right )^2 \]

[Out]

x*ln(2*exp(x)+4)*((5+ln(ln(12)))^2+3)^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(45\) vs. \(2(21)=42\).

Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14, number of steps used = 15, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 6820, 12, 6874, 2215, 2221, 2317, 2438, 2320, 2439} \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x \left (28+\log ^2(\log (12))+10 \log (\log (12))\right )^2 \log \left (\frac {e^x}{2}+1\right )+x \log (4) \left (28+\log ^2(\log (12))+10 \log (\log (12))\right )^2 \]

[In]

Int[(784*E^x*x + 560*E^x*x*Log[Log[12]] + 156*E^x*x*Log[Log[12]]^2 + 20*E^x*x*Log[Log[12]]^3 + E^x*x*Log[Log[1
2]]^4 + Log[4 + 2*E^x]*(1568 + 784*E^x + (1120 + 560*E^x)*Log[Log[12]] + (312 + 156*E^x)*Log[Log[12]]^2 + (40
+ 20*E^x)*Log[Log[12]]^3 + (2 + E^x)*Log[Log[12]]^4))/(2 + E^x),x]

[Out]

x*Log[4]*(28 + 10*Log[Log[12]] + Log[Log[12]]^2)^2 + x*Log[1 + E^x/2]*(28 + 10*Log[Log[12]] + Log[Log[12]]^2)^
2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 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 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 2439

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

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+e^x x (784+560 \log (\log (12)))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx \\ & = \int \frac {e^x x \log ^4(\log (12))+e^x x (784+560 \log (\log (12)))+e^x x \left (156 \log ^2(\log (12))+20 \log ^3(\log (12))\right )+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx \\ & = \int \frac {e^x x \left (156 \log ^2(\log (12))+20 \log ^3(\log (12))\right )+e^x x \left (784+560 \log (\log (12))+\log ^4(\log (12))\right )+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx \\ & = \int \frac {e^x x \left (784+560 \log (\log (12))+156 \log ^2(\log (12))+20 \log ^3(\log (12))+\log ^4(\log (12))\right )+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx \\ & = \int \frac {\left (e^x x+\left (2+e^x\right ) \log \left (2 \left (2+e^x\right )\right )\right ) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2}{2+e^x} \, dx \\ & = \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \int \frac {e^x x+\left (2+e^x\right ) \log \left (2 \left (2+e^x\right )\right )}{2+e^x} \, dx \\ & = \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \int \left (x-\frac {2 x}{2+e^x}+\log \left (2 \left (2+e^x\right )\right )\right ) \, dx \\ & = \frac {1}{2} x^2 \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2+\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \int \log \left (2 \left (2+e^x\right )\right ) \, dx-\left (2 \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2\right ) \int \frac {x}{2+e^x} \, dx \\ & = \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \int \frac {e^x x}{2+e^x} \, dx+\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \text {Subst}\left (\int \frac {\log (4+2 x)}{x} \, dx,x,e^x\right ) \\ & = x \log (4) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2+x \log \left (1+\frac {e^x}{2}\right ) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2-\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \int \log \left (1+\frac {e^x}{2}\right ) \, dx+\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx,x,e^x\right ) \\ & = x \log (4) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2+x \log \left (1+\frac {e^x}{2}\right ) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2-\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \operatorname {PolyLog}\left (2,-\frac {e^x}{2}\right )-\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx,x,e^x\right ) \\ & = x \log (4) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2+x \log \left (1+\frac {e^x}{2}\right ) \left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.90 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \left (\frac {x^2}{2}+x \log (4)+x \log \left (1+2 e^{-x}\right )-\operatorname {PolyLog}\left (2,-2 e^{-x}\right )-\operatorname {PolyLog}\left (2,-\frac {e^x}{2}\right )\right ) \]

[In]

Integrate[(784*E^x*x + 560*E^x*x*Log[Log[12]] + 156*E^x*x*Log[Log[12]]^2 + 20*E^x*x*Log[Log[12]]^3 + E^x*x*Log
[Log[12]]^4 + Log[4 + 2*E^x]*(1568 + 784*E^x + (1120 + 560*E^x)*Log[Log[12]] + (312 + 156*E^x)*Log[Log[12]]^2
+ (40 + 20*E^x)*Log[Log[12]]^3 + (2 + E^x)*Log[Log[12]]^4))/(2 + E^x),x]

[Out]

(28 + 10*Log[Log[12]] + Log[Log[12]]^2)^2*(x^2/2 + x*Log[4] + x*Log[1 + 2/E^x] - PolyLog[2, -2/E^x] - PolyLog[
2, -1/2*E^x])

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71

method	result	size
norman	\(\left (\ln \left (\ln \left (12\right )\right )^{4}+20 \ln \left (\ln \left (12\right )\right )^{3}+156 \ln \left (\ln \left (12\right )\right )^{2}+560 \ln \left (\ln \left (12\right )\right )+784\right ) x \ln \left (2 \,{\mathrm e}^{x}+4\right )\)	\(36\)
risch	\(x \left (\ln \left (\ln \left (3\right )+2 \ln \left (2\right )\right )^{4}+20 \ln \left (\ln \left (3\right )+2 \ln \left (2\right )\right )^{3}+156 \ln \left (\ln \left (3\right )+2 \ln \left (2\right )\right )^{2}+560 \ln \left (\ln \left (3\right )+2 \ln \left (2\right )\right )+784\right ) \ln \left (2 \,{\mathrm e}^{x}+4\right )\)	\(56\)
parallelrisch	\(\ln \left (\ln \left (12\right )\right )^{4} x \ln \left (2 \,{\mathrm e}^{x}+4\right )+20 \ln \left (\ln \left (12\right )\right )^{3} x \ln \left (2 \,{\mathrm e}^{x}+4\right )+156 \ln \left (\ln \left (12\right )\right )^{2} x \ln \left (2 \,{\mathrm e}^{x}+4\right )+560 \ln \left (\ln \left (12\right )\right ) x \ln \left (2 \,{\mathrm e}^{x}+4\right )+784 x \ln \left (2 \,{\mathrm e}^{x}+4\right )\)	\(69\)
default	\(784 x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )+\ln \left (\ln \left (12\right )\right )^{4} \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+20 \ln \left (\ln \left (12\right )\right )^{3} \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+156 \ln \left (\ln \left (12\right )\right )^{2} \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+560 \ln \left (\ln \left (12\right )\right ) \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+784 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )+\ln \left (\ln \left (12\right )\right )^{4} \left (\left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+\ln \left (\ln \left (12\right )\right )^{3} \left (20 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-20 \operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+\ln \left (\ln \left (12\right )\right )^{2} \left (156 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-156 \operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+\ln \left (\ln \left (12\right )\right ) \left (560 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-560 \operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )\)	\(286\)

[In]

int((((exp(x)+2)*ln(ln(12))^4+(20*exp(x)+40)*ln(ln(12))^3+(156*exp(x)+312)*ln(ln(12))^2+(560*exp(x)+1120)*ln(l
n(12))+784*exp(x)+1568)*ln(2*exp(x)+4)+x*exp(x)*ln(ln(12))^4+20*x*exp(x)*ln(ln(12))^3+156*x*exp(x)*ln(ln(12))^
2+560*x*exp(x)*ln(ln(12))+784*exp(x)*x)/(exp(x)+2),x,method=_RETURNVERBOSE)

[Out]

(ln(ln(12))^4+20*ln(ln(12))^3+156*ln(ln(12))^2+560*ln(ln(12))+784)*x*ln(2*exp(x)+4)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx={\left (x \log \left (\log \left (12\right )\right )^{4} + 20 \, x \log \left (\log \left (12\right )\right )^{3} + 156 \, x \log \left (\log \left (12\right )\right )^{2} + 560 \, x \log \left (\log \left (12\right )\right ) + 784 \, x\right )} \log \left (2 \, e^{x} + 4\right ) \]

[In]

integrate((((exp(x)+2)*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*exp(x)+312)*log(log(12))^2+(560*exp(x
)+1120)*log(log(12))+784*exp(x)+1568)*log(2*exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x
*exp(x)*log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(exp(x)+2),x, algorithm="fricas")

[Out]

(x*log(log(12))^4 + 20*x*log(log(12))^3 + 156*x*log(log(12))^2 + 560*x*log(log(12)) + 784*x)*log(2*e^x + 4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=\left (x \log {\left (\log {\left (12 \right )} \right )}^{4} + 20 x \log {\left (\log {\left (12 \right )} \right )}^{3} + 156 x \log {\left (\log {\left (12 \right )} \right )}^{2} + 560 x \log {\left (\log {\left (12 \right )} \right )} + 784 x\right ) \log {\left (2 e^{x} + 4 \right )} \]

[In]

integrate((((exp(x)+2)*ln(ln(12))**4+(20*exp(x)+40)*ln(ln(12))**3+(156*exp(x)+312)*ln(ln(12))**2+(560*exp(x)+1
120)*ln(ln(12))+784*exp(x)+1568)*ln(2*exp(x)+4)+x*exp(x)*ln(ln(12))**4+20*x*exp(x)*ln(ln(12))**3+156*x*exp(x)*
ln(ln(12))**2+560*x*exp(x)*ln(ln(12))+784*exp(x)*x)/(exp(x)+2),x)

[Out]

(x*log(log(12))**4 + 20*x*log(log(12))**3 + 156*x*log(log(12))**2 + 560*x*log(log(12)) + 784*x)*log(2*exp(x) +
4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (20) = 40\).

Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.95 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx={\left (\log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{4} + 20 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{3} + 156 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{2} + 560 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right ) + 784\right )} x \log \left (2\right ) + {\left (\log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{4} + 20 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{3} + 156 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{2} + 560 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right ) + 784\right )} x \log \left (e^{x} + 2\right ) \]

[In]

integrate((((exp(x)+2)*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*exp(x)+312)*log(log(12))^2+(560*exp(x
)+1120)*log(log(12))+784*exp(x)+1568)*log(2*exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x
*exp(x)*log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(exp(x)+2),x, algorithm="maxima")

[Out]

(log(log(3) + 2*log(2))^4 + 20*log(log(3) + 2*log(2))^3 + 156*log(log(3) + 2*log(2))^2 + 560*log(log(3) + 2*lo
g(2)) + 784)*x*log(2) + (log(log(3) + 2*log(2))^4 + 20*log(log(3) + 2*log(2))^3 + 156*log(log(3) + 2*log(2))^2
+ 560*log(log(3) + 2*log(2)) + 784)*x*log(e^x + 2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (20) = 40\).

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.76 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x \log \left (2\right ) \log \left (\log \left (12\right )\right )^{4} + x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right )^{4} + 20 \, x \log \left (2\right ) \log \left (\log \left (12\right )\right )^{3} + 20 \, x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right )^{3} + 156 \, x \log \left (2\right ) \log \left (\log \left (12\right )\right )^{2} + 156 \, x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right )^{2} + 560 \, x \log \left (2\right ) \log \left (\log \left (12\right )\right ) + 560 \, x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right ) + 784 \, x \log \left (2\right ) + 784 \, x \log \left (e^{x} + 2\right ) \]

[In]

integrate((((exp(x)+2)*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*exp(x)+312)*log(log(12))^2+(560*exp(x
)+1120)*log(log(12))+784*exp(x)+1568)*log(2*exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x
*exp(x)*log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(exp(x)+2),x, algorithm="giac")

[Out]

x*log(2)*log(log(12))^4 + x*log(e^x + 2)*log(log(12))^4 + 20*x*log(2)*log(log(12))^3 + 20*x*log(e^x + 2)*log(l
og(12))^3 + 156*x*log(2)*log(log(12))^2 + 156*x*log(e^x + 2)*log(log(12))^2 + 560*x*log(2)*log(log(12)) + 560*
x*log(e^x + 2)*log(log(12)) + 784*x*log(2) + 784*x*log(e^x + 2)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x\,\left (\ln \left (2\right )+\ln \left ({\mathrm {e}}^x+2\right )\right )\,{\left (10\,\ln \left (\ln \left (12\right )\right )+{\ln \left (\ln \left (12\right )\right )}^2+28\right )}^2 \]

[In]

int((log(2*exp(x) + 4)*(784*exp(x) + log(log(12))^3*(20*exp(x) + 40) + log(log(12))^2*(156*exp(x) + 312) + log
(log(12))*(560*exp(x) + 1120) + log(log(12))^4*(exp(x) + 2) + 1568) + 784*x*exp(x) + 560*x*exp(x)*log(log(12))
+ 156*x*exp(x)*log(log(12))^2 + 20*x*exp(x)*log(log(12))^3 + x*exp(x)*log(log(12))^4)/(exp(x) + 2),x)

[Out]

x*(log(2) + log(exp(x) + 2))*(10*log(log(12)) + log(log(12))^2 + 28)^2