3.38.79 \(\int \frac {e^{-2 e^3} (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+(8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))) \log ^2(2+x))}{(2+x) \log (4) \log ^2(2+x)} \, dx\)

Optimal. Leaf size=34 \[ 2-x+\frac {2 \left (-x+e^{-2 e^3} x^2\right )}{\log (4)}+\frac {x}{\log (2+x)} \]

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Rubi [A]  time = 0.46, antiderivative size = 46, normalized size of antiderivative = 1.35, number of steps used = 12, number of rules used = 9, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 6688, 2411, 2353, 2297, 2298, 2302, 30, 2389} \begin {gather*} \frac {2 e^{-2 e^3} x^2}{\log (4)}-\frac {x (2+\log (4))}{\log (4)}+\frac {x+2}{\log (x+2)}-\frac {2}{\log (x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^2)/(E^(2*E^3)*Log[4]) - (x*(2 + Log[4]))/Log[4] - 2/Log[2 + x] + (2 + x)/Log[2 + 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2389

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

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 36, normalized size = 1.06 \begin {gather*} \frac {2 e^{-2 e^3} x^2}{\log (4)}-\frac {x (2+\log (4))}{\log (4)}+\frac {x}{\log (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^2)/(E^(2*E^3)*Log[4]) - (x*(2 + Log[4]))/Log[4] + x/Log[2 + x]

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fricas [A]  time = 0.69, size = 48, normalized size = 1.41 \begin {gather*} \frac {{\left (x e^{\left (2 \, e^{3}\right )} \log \relax (2) + {\left (x^{2} - {\left (x \log \relax (2) + x\right )} e^{\left (2 \, e^{3}\right )}\right )} \log \left (x + 2\right )\right )} e^{\left (-2 \, e^{3}\right )}}{\log \relax (2) \log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*(-x-2)*log(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*log(2+x)^2+2*(2+x)*log(2)*exp(exp(3))^2*log(2
+x)-2*x*log(2)*exp(exp(3))^2)/(2+x)/log(2)/exp(exp(3))^2/log(2+x)^2,x, algorithm="fricas")

[Out]

(x*e^(2*e^3)*log(2) + (x^2 - (x*log(2) + x)*e^(2*e^3))*log(x + 2))*e^(-2*e^3)/(log(2)*log(x + 2))

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giac [A]  time = 0.22, size = 61, normalized size = 1.79 \begin {gather*} -\frac {{\left (x e^{\left (2 \, e^{3}\right )} \log \relax (2) \log \left (x + 2\right ) - x e^{\left (2 \, e^{3}\right )} \log \relax (2) - x^{2} \log \left (x + 2\right ) + x e^{\left (2 \, e^{3}\right )} \log \left (x + 2\right )\right )} e^{\left (-2 \, e^{3}\right )}}{\log \relax (2) \log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*e^(2*e^3)*log(2)*log(x + 2) - x*e^(2*e^3)*log(2) - x^2*log(x + 2) + x*e^(2*e^3)*log(x + 2))*e^(-2*e^3)/(lo
g(2)*log(x + 2))

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maple [A]  time = 0.10, size = 33, normalized size = 0.97

method	result	size
risch	\(-x -\frac {x}{\ln \relax (2)}+\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} x^{2}}{\ln \relax (2)}+\frac {x}{\ln \left (2+x \right )}\)	\(33\)
norman	\(\frac {\left (x \,{\mathrm e}^{{\mathrm e}^{3}}+\frac {{\mathrm e}^{-{\mathrm e}^{3}} x^{2} \ln \left (2+x \right )}{\ln \relax (2)}-\frac {{\mathrm e}^{{\mathrm e}^{3}} \left (1+\ln \relax (2)\right ) x \ln \left (2+x \right )}{\ln \relax (2)}\right ) {\mathrm e}^{-{\mathrm e}^{3}}}{\ln \left (2+x \right )}\)	\(54\)
derivativedivides	\(\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} \left (-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \relax (2) \left (2+x \right )-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \left (2+x \right )-2 \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{3}} \expIntegralEi \left (1, -\ln \left (2+x \right )\right )+2 \left (2+x \right )^{2}-2 \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{3}} \left (-\frac {2+x}{\ln \left (2+x \right )}-\expIntegralEi \left (1, -\ln \left (2+x \right )\right )\right )-16-8 x -\frac {4 \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{3}}}{\ln \left (2+x \right )}\right )}{2 \ln \relax (2)}\)	\(109\)
default	\(\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} \left (-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \relax (2) \left (2+x \right )-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \left (2+x \right )-2 \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{3}} \expIntegralEi \left (1, -\ln \left (2+x \right )\right )+2 \left (2+x \right )^{2}-2 \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{3}} \left (-\frac {2+x}{\ln \left (2+x \right )}-\expIntegralEi \left (1, -\ln \left (2+x \right )\right )\right )-16-8 x -\frac {4 \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{3}}}{\ln \left (2+x \right )}\right )}{2 \ln \relax (2)}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((2*(-x-2)*ln(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*ln(2+x)^2+2*(2+x)*ln(2)*exp(exp(3))^2*ln(2+x)-2*x*ln
(2)*exp(exp(3))^2)/(2+x)/ln(2)/exp(exp(3))^2/ln(2+x)^2,x,method=_RETURNVERBOSE)

[Out]

-x-x/ln(2)+1/ln(2)*exp(-2*exp(3))*x^2+x/ln(2+x)

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maxima [A]  time = 0.48, size = 49, normalized size = 1.44 \begin {gather*} \frac {{\left (x e^{\left (2 \, e^{3}\right )} \log \relax (2) - {\left (x {\left (\log \relax (2) + 1\right )} e^{\left (2 \, e^{3}\right )} - x^{2}\right )} \log \left (x + 2\right )\right )} e^{\left (-2 \, e^{3}\right )}}{\log \relax (2) \log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*(-x-2)*log(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*log(2+x)^2+2*(2+x)*log(2)*exp(exp(3))^2*log(2
+x)-2*x*log(2)*exp(exp(3))^2)/(2+x)/log(2)/exp(exp(3))^2/log(2+x)^2,x, algorithm="maxima")

[Out]

(x*e^(2*e^3)*log(2) - (x*(log(2) + 1)*e^(2*e^3) - x^2)*log(x + 2))*e^(-2*e^3)/(log(2)*log(x + 2))

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mupad [B]  time = 0.39, size = 32, normalized size = 0.94 \begin {gather*} \frac {x}{\ln \left (x+2\right )}-x-\frac {x}{\ln \relax (2)}+\frac {x^2\,{\mathrm {e}}^{-2\,{\mathrm {e}}^3}}{\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/log(x + 2) - x - x/log(2) + (x^2*exp(-2*exp(3)))/log(2)

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sympy [A]  time = 0.15, size = 31, normalized size = 0.91 \begin {gather*} \frac {x^{2}}{e^{2 e^{3}} \log {\relax (2 )}} + \frac {x \left (-1 - \log {\relax (2 )}\right )}{\log {\relax (2 )}} + \frac {x}{\log {\left (x + 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*exp(-2*exp(3))/log(2) + x*(-1 - log(2))/log(2) + x/log(x + 2)

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