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

Optimal. Leaf size=33 \[ \left (5+e^{2 x}\right ) \left (-x+e^4 x-\log (2+x)\right ) \log (\log (i \pi +\log (2))) \]

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Rubi [C]  time = 0.77, antiderivative size = 124, normalized size of antiderivative = 3.76, number of steps used = 16, number of rules used = 9, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 6742, 43, 6688, 2199, 2194, 2178, 2176, 2554} \begin {gather*} -\frac {\log (\log (\log (2)+i \pi )) \text {Ei}(2 (x+2))}{e^4}+\frac {\log (\log (\log (2)+i \pi )) \text {Ei}(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) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2176

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] &&  !$UseGamma === True

Rule 2178

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

Rule 2194

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 2199

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] &&  !$UseGamma === True

Rule 2554

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 42, normalized size = 1.27 \begin {gather*} \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))) \end {gather*}

Antiderivative was successfully verified.

[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]]]

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fricas [A]  time = 1.17, size = 43, normalized size = 1.30 \begin {gather*} {\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 \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [A]  time = 0.26, size = 49, normalized size = 1.48 \begin {gather*} {\left (5 \, x e^{4} - x e^{\left (2 \, x\right )} + x e^{\left (2 \, x + 4\right )} - e^{\left (2 \, x\right )} \log \left (x + 2\right ) - 5 \, x - 5 \, \log \left (x + 2\right )\right )} \log \left (\log \left (i \, \pi + \log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [A]  time = 0.13, size = 46, normalized size = 1.39

method	result	size
default	\(\ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right ) \left (\left ({\mathrm e}^{4}-1\right ) x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )+5 x \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )\right )\)	\(46\)
norman	\(\left (5 \ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right ) {\mathrm e}^{4}-5 \ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right )\right ) x -5 \ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right ) \ln \left (2+x \right )+\left (\ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \relax (2)+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )\)	\(93\)
risch	\(-\ln \left (\ln \left (-i \left (i \ln \relax (2)-\pi \right )\right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+\ln \left (\ln \left (-i \left (i \ln \relax (2)-\pi \right )\right )\right ) x \,{\mathrm e}^{2 x +4}+5 \ln \left (\ln \left (-i \left (i \ln \relax (2)-\pi \right )\right )\right ) x \,{\mathrm e}^{4}-\ln \left (\ln \left (-i \left (i \ln \relax (2)-\pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-5 \ln \left (\ln \left (-i \left (i \ln \relax (2)-\pi \right )\right )\right ) \ln \left (2+x \right )-5 \ln \left (\ln \left (-i \left (i \ln \relax (2)-\pi \right )\right )\right ) x\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} {\left (x {\left (e^{4} - 1\right )} e^{\left (2 \, x\right )} + 5 \, {\left (x - 2 \, \log \left (x + 2\right )\right )} e^{4} + 3 \, e^{\left (-4\right )} E_{1}\left (-2 \, x - 4\right ) + 10 \, e^{4} \log \left (x + 2\right ) - e^{\left (2 \, x\right )} \log \left (x + 2\right ) - 5 \, x + 3 \, \int \frac {e^{\left (2 \, x\right )}}{x + 2}\,{d x} - 5 \, \log \left (x + 2\right )\right )} \log \left (\log \left (i \, \pi + \log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 2.58, size = 50, normalized size = 1.52 \begin {gather*} -\ln \left (\ln \left (\ln \relax (2)+\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)
)

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sympy [B]  time = 0.70, size = 92, normalized size = 2.79 \begin {gather*} x \left (- 5 \log {\left (\log {\left (\log {\relax (2 )} + i \pi \right )} \right )} + 5 e^{4} \log {\left (\log {\left (\log {\relax (2 )} + i \pi \right )} \right )}\right ) + \left (- x \log {\left (\log {\left (\log {\relax (2 )} + i \pi \right )} \right )} + x e^{4} \log {\left (\log {\left (\log {\relax (2 )} + i \pi \right )} \right )} - \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\relax (2 )} + i \pi \right )} \right )}\right ) e^{2 x} - 5 \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\relax (2 )} + i \pi \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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