\(\int \frac {(-15-5 x+e^4 (10+5 x)+e^{2 x} (-3-5 x-2 x^2+e^4 (2+5 x+2 x^2))+e^{2 x} (-4-2 x) \log (2+x)) \log (\log (i \pi +\log (2)))}{2+x} \, dx\) [3798]

Optimal result

Integrand size = 75, antiderivative size = 33 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\left (5+e^{2 x}\right ) \left (-x+e^4 x-\log (2+x)\right ) \log (\log (i \pi +\log (2))) \]

[Out]

(5+exp(2*x))*ln(ln(ln(2)+I*Pi))*(x*exp(4)-x-ln(2+x))

Rubi [C] (verified)

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

Time = 0.54 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.76, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 6874, 45, 6820, 2230, 2225, 2209, 2207, 2634} \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=-\frac {\log (\log (\log (2)+i \pi )) \operatorname {ExpIntegralEi}(2 (x+2))}{e^4}+\frac {\log (\log (\log (2)+i \pi )) \operatorname {ExpIntegralEi}(2 x+4)}{e^4}+\left (1-e^4\right ) \left (-e^{2 x}\right ) x \log (\log (\log (2)+i \pi ))-5 \left (1-e^4\right ) x \log (\log (\log (2)+i \pi ))-e^{2 x} \log (\log (\log (2)+i \pi )) \log (x+2)-5 \log (\log (\log (2)+i \pi )) \log (x+2) \]

[In]

Int[((-15 - 5*x + E^4*(10 + 5*x) + E^(2*x)*(-3 - 5*x - 2*x^2 + E^4*(2 + 5*x + 2*x^2)) + E^(2*x)*(-4 - 2*x)*Log
[2 + x])*Log[Log[I*Pi + Log[2]]])/(2 + x),x]

[Out]

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

Rule 12

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2207

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

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}& = \log (\log (i \pi +\log (2))) \int \frac {-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)}{2+x} \, dx \\ & = \log (\log (i \pi +\log (2))) \int \left (\frac {5 \left (-3+2 e^4-\left (1-e^4\right ) x\right )}{2+x}+\frac {e^{2 x} \left (-3 \left (1-\frac {2 e^4}{3}\right )-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2-4 \log (2+x)-2 x \log (2+x)\right )}{2+x}\right ) \, dx \\ & = \log (\log (i \pi +\log (2))) \int \frac {e^{2 x} \left (-3 \left (1-\frac {2 e^4}{3}\right )-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2-4 \log (2+x)-2 x \log (2+x)\right )}{2+x} \, dx+(5 \log (\log (i \pi +\log (2)))) \int \frac {-3+2 e^4-\left (1-e^4\right ) x}{2+x} \, dx \\ & = \log (\log (i \pi +\log (2))) \int \frac {e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )-2 (2+x) \log (2+x)\right )}{2+x} \, dx+(5 \log (\log (i \pi +\log (2)))) \int \left (-1+e^4+\frac {1}{-2-x}\right ) \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-5 \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \left (\frac {e^{2 x} \left (-3+2 e^4-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2\right )}{2+x}-2 e^{2 x} \log (2+x)\right ) \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-5 \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \frac {e^{2 x} \left (-3+2 e^4-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2\right )}{2+x} \, dx-(2 \log (\log (i \pi +\log (2)))) \int e^{2 x} \log (2+x) \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \left (e^{2 x} \left (-1+e^4\right )+\frac {e^{2 x}}{-2-x}+2 e^{2 x} \left (-1+e^4\right ) x\right ) \, dx+(2 \log (\log (i \pi +\log (2)))) \int \frac {e^{2 x}}{4+2 x} \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))+\frac {\text {Ei}(4+2 x) \log (\log (i \pi +\log (2)))}{e^4}-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \frac {e^{2 x}}{-2-x} \, dx-\left (\left (1-e^4\right ) \log (\log (i \pi +\log (2)))\right ) \int e^{2 x} \, dx-\left (2 \left (1-e^4\right ) \log (\log (i \pi +\log (2)))\right ) \int e^{2 x} x \, dx \\ & = -\frac {1}{2} e^{2 x} \left (1-e^4\right ) \log (\log (i \pi +\log (2)))-5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-e^{2 x} \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-\frac {\text {Ei}(2 (2+x)) \log (\log (i \pi +\log (2)))}{e^4}+\frac {\text {Ei}(4+2 x) \log (\log (i \pi +\log (2)))}{e^4}-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2)))+\left (\left (1-e^4\right ) \log (\log (i \pi +\log (2)))\right ) \int e^{2 x} \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-e^{2 x} \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-\frac {\text {Ei}(2 (2+x)) \log (\log (i \pi +\log (2)))}{e^4}+\frac {\text {Ei}(4+2 x) \log (\log (i \pi +\log (2)))}{e^4}-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2))) \\ \end{align*}

Mathematica [A] (verified)

Time = 2.57 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\left (\left (-1+e^4\right ) \left (10+\left (5+e^{2 x}\right ) x\right )-\left (5+e^{2 x}\right ) \log (2+x)\right ) \log (\log (i \pi +\log (2))) \]

[In]

Integrate[((-15 - 5*x + E^4*(10 + 5*x) + E^(2*x)*(-3 - 5*x - 2*x^2 + E^4*(2 + 5*x + 2*x^2)) + E^(2*x)*(-4 - 2*
x)*Log[2 + x])*Log[Log[I*Pi + Log[2]]])/(2 + x),x]

[Out]

((-1 + E^4)*(10 + (5 + E^(2*x))*x) - (5 + E^(2*x))*Log[2 + x])*Log[Log[I*Pi + Log[2]]]

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39

method	result	size
default	\(\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left (\left ({\mathrm e}^{4}-1\right ) x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )+5 x \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )\right )\)	\(46\)
parallelrisch	\(\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left ({\mathrm e}^{4} {\mathrm e}^{2 x} x +5 x \,{\mathrm e}^{4}-x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )-20 \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )+20\right )\)	\(56\)
parts	\(\left (\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left (x \,{\mathrm e}^{4}-x -\ln \left (2+x \right )\right )\)	\(76\)
norman	\(\left (5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x -5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \ln \left (2+x \right )+\left (\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )\)	\(93\)
risch	\(-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{4+2 x}+5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{2 x}-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \ln \left (2+x \right )-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x\)	\(95\)

[In]

int(((-2*x-4)*exp(2*x)*ln(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*exp(2*x)+(5*x+10)*exp(4)-5*x-15)*ln(ln(ln(2)
+I*Pi))/(2+x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(ln(2)+I*Pi))*((exp(4)-1)*x*exp(2*x)-exp(2*x)*ln(2+x)+5*x*exp(4)-5*x-5*ln(2+x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx={\left (5 \, x e^{4} + {\left (x e^{4} - x\right )} e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} + 5\right )} \log \left (x + 2\right ) - 5 \, x\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right ) \]

[In]

integrate(((-2*x-4)*exp(2*x)*log(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*exp(2*x)+(5*x+10)*exp(4)-5*x-15)*log(
log(log(2)+I*pi))/(2+x),x, algorithm="fricas")

[Out]

(5*x*e^4 + (x*e^4 - x)*e^(2*x) - (e^(2*x) + 5)*log(x + 2) - 5*x)*log(log(I*pi + log(2)))

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (27) = 54\).

Time = 0.51 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.79 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=x \left (- 5 \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} + 5 e^{4} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )}\right ) + \left (- x \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} + x e^{4} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} - \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )}\right ) e^{2 x} - 5 \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} \]

[In]

integrate(((-2*x-4)*exp(2*x)*ln(2+x)+((2*x**2+5*x+2)*exp(4)-2*x**2-5*x-3)*exp(2*x)+(5*x+10)*exp(4)-5*x-15)*ln(
ln(ln(2)+I*pi))/(2+x),x)

[Out]

x*(-5*log(log(log(2) + I*pi)) + 5*exp(4)*log(log(log(2) + I*pi))) + (-x*log(log(log(2) + I*pi)) + x*exp(4)*log
(log(log(2) + I*pi)) - log(x + 2)*log(log(log(2) + I*pi)))*exp(2*x) - 5*log(x + 2)*log(log(log(2) + I*pi))

Maxima [F]

\[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\int { -\frac {{\left (2 \, {\left (x + 2\right )} e^{\left (2 \, x\right )} \log \left (x + 2\right ) - 5 \, {\left (x + 2\right )} e^{4} + {\left (2 \, x^{2} - {\left (2 \, x^{2} + 5 \, x + 2\right )} e^{4} + 5 \, x + 3\right )} e^{\left (2 \, x\right )} + 5 \, x + 15\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right )}{x + 2} \,d x } \]

[In]

integrate(((-2*x-4)*exp(2*x)*log(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*exp(2*x)+(5*x+10)*exp(4)-5*x-15)*log(
log(log(2)+I*pi))/(2+x),x, algorithm="maxima")

[Out]

(x*(e^4 - 1)*e^(2*x) + 5*(x - 2*log(x + 2))*e^4 + 3*e^(-4)*exp_integral_e(1, -2*x - 4) + 10*e^4*log(x + 2) - e
^(2*x)*log(x + 2) - 5*x + 3*integrate(e^(2*x)/(x + 2), x) - 5*log(x + 2))*log(log(I*pi + log(2)))

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (29) = 58\).

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.52 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx={\left (5 \, {\left (x + 2\right )} e^{8} - 5 \, {\left (x + 2\right )} e^{4} + {\left (x + 2\right )} e^{\left (2 \, x + 8\right )} - {\left (x + 2\right )} e^{\left (2 \, x + 4\right )} - 5 \, e^{4} \log \left (x + 2\right ) - e^{\left (2 \, x + 4\right )} \log \left (x + 2\right ) - 2 \, e^{\left (2 \, x + 8\right )} + 2 \, e^{\left (2 \, x + 4\right )}\right )} e^{\left (-4\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right ) \]

[In]

integrate(((-2*x-4)*exp(2*x)*log(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*exp(2*x)+(5*x+10)*exp(4)-5*x-15)*log(
log(log(2)+I*pi))/(2+x),x, algorithm="giac")

[Out]

(5*(x + 2)*e^8 - 5*(x + 2)*e^4 + (x + 2)*e^(2*x + 8) - (x + 2)*e^(2*x + 4) - 5*e^4*log(x + 2) - e^(2*x + 4)*lo
g(x + 2) - 2*e^(2*x + 8) + 2*e^(2*x + 4))*e^(-4)*log(log(I*pi + log(2)))

Mupad [B] (verification not implemented)

Time = 9.78 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=-\ln \left (\ln \left (\ln \left (2\right )+\Pi \,1{}\mathrm {i}\right )\right )\,\left (5\,x+5\,\ln \left (x+2\right )+x\,{\mathrm {e}}^{2\,x}-5\,x\,{\mathrm {e}}^4-x\,{\mathrm {e}}^{2\,x+4}+\ln \left (x+2\right )\,{\mathrm {e}}^{2\,x}\right ) \]

[In]

int(-(log(log(Pi*1i + log(2)))*(5*x + exp(2*x)*(5*x - exp(4)*(5*x + 2*x^2 + 2) + 2*x^2 + 3) - exp(4)*(5*x + 10
) + log(x + 2)*exp(2*x)*(2*x + 4) + 15))/(x + 2),x)

[Out]

-log(log(Pi*1i + log(2)))*(5*x + 5*log(x + 2) + x*exp(2*x) - 5*x*exp(4) - x*exp(2*x + 4) + log(x + 2)*exp(2*x)
)