3.89.74 \(\int \frac {(1+e^x+x) \log (x)+(-1-e^x-x+(1+2 x+e^x (1+x)) \log (x)) \log (x \log (4))}{8 \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {x \left (1+e^x+x\right ) \log (x \log (4))}{8 \log (x)} \]

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Rubi [B]  time = 0.77, antiderivative size = 48, normalized size of antiderivative = 2.40, number of steps used = 29, number of rules used = 15, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {12, 6742, 2288, 6688, 2330, 2298, 2309, 2178, 2320, 2297, 2361, 6496, 2306, 2366, 6482} \begin {gather*} \frac {x^2 \log (x \log (4))}{8 \log (x)}+\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {x \log (x \log (4))}{8 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + E^x + x)*Log[x] + (-1 - E^x - x + (1 + 2*x + E^x*(1 + x))*Log[x])*Log[x*Log[4]])/(8*Log[x]^2),x]

[Out]

(x*Log[x*Log[4]])/(8*Log[x]) + (E^x*x*Log[x*Log[4]])/(8*Log[x]) + (x^2*Log[x*Log[4]])/(8*Log[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 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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2320

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

Rule 2330

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

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

Rule 6496

Int[LogIntegral[(b_.)*(x_)]/(x_), x_Symbol] :> -Simp[b*x, x] + Simp[Log[b*x]*LogIntegral[b*x], x] /; FreeQ[b,
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} &=\frac {1}{8} \int \frac {\left (1+e^x+x\right ) \log (x)+\left (-1-e^x-x+\left (1+2 x+e^x (1+x)\right ) \log (x)\right ) \log (x \log (4))}{\log ^2(x)} \, dx\\ &=\frac {1}{8} \int \left (\frac {e^x (\log (x)-\log (x \log (4))+\log (x) \log (x \log (4))+x \log (x) \log (x \log (4)))}{\log ^2(x)}+\frac {\log (x)+x \log (x)-\log (x \log (4))-x \log (x \log (4))+\log (x) \log (x \log (4))+2 x \log (x) \log (x \log (4))}{\log ^2(x)}\right ) \, dx\\ &=\frac {1}{8} \int \frac {e^x (\log (x)-\log (x \log (4))+\log (x) \log (x \log (4))+x \log (x) \log (x \log (4)))}{\log ^2(x)} \, dx+\frac {1}{8} \int \frac {\log (x)+x \log (x)-\log (x \log (4))-x \log (x \log (4))+\log (x) \log (x \log (4))+2 x \log (x) \log (x \log (4))}{\log ^2(x)} \, dx\\ &=\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {1}{8} \int \frac {-((1+x) \log (x \log (4)))+\log (x) (1+x+(1+2 x) \log (x \log (4)))}{\log ^2(x)} \, dx\\ &=\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {1}{8} \int \left (\frac {1+x}{\log (x)}+\frac {(-1-x+\log (x)+2 x \log (x)) \log (x \log (4))}{\log ^2(x)}\right ) \, dx\\ &=\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {1}{8} \int \frac {1+x}{\log (x)} \, dx+\frac {1}{8} \int \frac {(-1-x+\log (x)+2 x \log (x)) \log (x \log (4))}{\log ^2(x)} \, dx\\ &=\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {1}{8} \int \left (\frac {1}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\frac {1}{8} \int \left (-\frac {\log (x \log (4))}{\log ^2(x)}-\frac {x \log (x \log (4))}{\log ^2(x)}+\frac {\log (x \log (4))}{\log (x)}+\frac {2 x \log (x \log (4))}{\log (x)}\right ) \, dx\\ &=\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {1}{8} \int \frac {1}{\log (x)} \, dx+\frac {1}{8} \int \frac {x}{\log (x)} \, dx-\frac {1}{8} \int \frac {\log (x \log (4))}{\log ^2(x)} \, dx-\frac {1}{8} \int \frac {x \log (x \log (4))}{\log ^2(x)} \, dx+\frac {1}{8} \int \frac {\log (x \log (4))}{\log (x)} \, dx+\frac {1}{4} \int \frac {x \log (x \log (4))}{\log (x)} \, dx\\ &=\frac {x \log (x \log (4))}{8 \log (x)}+\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {x^2 \log (x \log (4))}{8 \log (x)}+\frac {\text {li}(x)}{8}+\frac {1}{8} \int \left (\frac {2 \text {Ei}(2 \log (x))}{x}-\frac {x}{\log (x)}\right ) \, dx-\frac {1}{8} \int \frac {\text {li}(x)}{x} \, dx+\frac {1}{8} \int \left (-\frac {1}{\log (x)}+\frac {\text {li}(x)}{x}\right ) \, dx+\frac {1}{8} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {1}{4} \int \frac {\text {Ei}(2 \log (x))}{x} \, dx\\ &=\frac {x}{8}+\frac {1}{8} \text {Ei}(2 \log (x))+\frac {x \log (x \log (4))}{8 \log (x)}+\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {x^2 \log (x \log (4))}{8 \log (x)}+\frac {\text {li}(x)}{8}-\frac {1}{8} \log (x) \text {li}(x)-\frac {1}{8} \int \frac {1}{\log (x)} \, dx-\frac {1}{8} \int \frac {x}{\log (x)} \, dx+\frac {1}{8} \int \frac {\text {li}(x)}{x} \, dx+\frac {1}{4} \int \frac {\text {Ei}(2 \log (x))}{x} \, dx-\frac {1}{4} \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))\\ &=\frac {x^2}{8}+\frac {1}{8} \text {Ei}(2 \log (x))-\frac {1}{4} \text {Ei}(2 \log (x)) \log (x)+\frac {x \log (x \log (4))}{8 \log (x)}+\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {x^2 \log (x \log (4))}{8 \log (x)}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\frac {1}{4} \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))\\ &=\frac {x \log (x \log (4))}{8 \log (x)}+\frac {e^x x \log (x \log (4))}{8 \log (x)}+\frac {x^2 \log (x \log (4))}{8 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 20, normalized size = 1.00 \begin {gather*} \frac {x \left (1+e^x+x\right ) \log (x \log (4))}{8 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + E^x + x)*Log[x] + (-1 - E^x - x + (1 + 2*x + E^x*(1 + x))*Log[x])*Log[x*Log[4]])/(8*Log[x]^2),
x]

[Out]

(x*(1 + E^x + x)*Log[x*Log[4]])/(8*Log[x])

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fricas [A]  time = 0.54, size = 34, normalized size = 1.70 \begin {gather*} \frac {{\left (x^{2} + x e^{x} + x\right )} \log \relax (x) + {\left (x^{2} + x e^{x} + x\right )} \log \left (2 \, \log \relax (2)\right )}{8 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((((x+1)*exp(x)+2*x+1)*log(x)-exp(x)-x-1)*log(2*x*log(2))+(exp(x)+1+x)*log(x))/log(x)^2,x, algor
ithm="fricas")

[Out]

1/8*((x^2 + x*e^x + x)*log(x) + (x^2 + x*e^x + x)*log(2*log(2)))/log(x)

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giac [B]  time = 0.22, size = 58, normalized size = 2.90 \begin {gather*} \frac {x^{2} \log \relax (2) + x e^{x} \log \relax (2) + x^{2} \log \relax (x) + x e^{x} \log \relax (x) + x^{2} \log \left (\log \relax (2)\right ) + x e^{x} \log \left (\log \relax (2)\right ) + x \log \relax (2) + x \log \relax (x) + x \log \left (\log \relax (2)\right )}{8 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((((x+1)*exp(x)+2*x+1)*log(x)-exp(x)-x-1)*log(2*x*log(2))+(exp(x)+1+x)*log(x))/log(x)^2,x, algor
ithm="giac")

[Out]

1/8*(x^2*log(2) + x*e^x*log(2) + x^2*log(x) + x*e^x*log(x) + x^2*log(log(2)) + x*e^x*log(log(2)) + x*log(2) +
x*log(x) + x*log(log(2)))/log(x)

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maple [B]  time = 0.14, size = 56, normalized size = 2.80

method	result	size
risch	\(\frac {x^{2}}{8}+\frac {{\mathrm e}^{x} x}{8}+\frac {x}{8}+\frac {x \left (2 x \ln \relax (2)+2 \,{\mathrm e}^{x} \ln \relax (2)+2 x \ln \left (\ln \relax (2)\right )+2 \,{\mathrm e}^{x} \ln \left (\ln \relax (2)\right )+2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )\right )}{16 \ln \relax (x )}\)	\(56\)
default	\(\frac {x}{8}+\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) x \,{\mathrm e}^{x}+x \,{\mathrm e}^{x} \ln \relax (x )}{8 \ln \relax (x )}+\frac {x^{2}}{8}+\frac {\ln \relax (2) x^{2}}{8 \ln \relax (x )}+\frac {\ln \relax (2) x}{8 \ln \relax (x )}+\frac {\ln \left (\ln \relax (2)\right ) x^{2}}{8 \ln \relax (x )}+\frac {\ln \left (\ln \relax (2)\right ) x}{8 \ln \relax (x )}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((((x+1)*exp(x)+2*x+1)*ln(x)-exp(x)-x-1)*ln(2*x*ln(2))+(exp(x)+1+x)*ln(x))/ln(x)^2,x,method=_RETURNVER
BOSE)

[Out]

1/8*x^2+1/8*exp(x)*x+1/8*x+1/16*x*(2*x*ln(2)+2*exp(x)*ln(2)+2*x*ln(ln(2))+2*exp(x)*ln(ln(2))+2*ln(2)+2*ln(ln(2
)))/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {x^{2} {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + {\left (x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + x \log \relax (x)\right )} e^{x} + {\left (x^{2} + x\right )} \log \relax (x)}{8 \, \log \relax (x)} + \frac {1}{8} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + \frac {1}{8} \, {\rm Ei}\left (\log \relax (x)\right ) - \frac {1}{8} \, \int \frac {x + 1}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((((x+1)*exp(x)+2*x+1)*log(x)-exp(x)-x-1)*log(2*x*log(2))+(exp(x)+1+x)*log(x))/log(x)^2,x, algor
ithm="maxima")

[Out]

1/8*(x^2*(log(2) + log(log(2))) + x*(log(2) + log(log(2))) + (x*(log(2) + log(log(2))) + x*log(x))*e^x + (x^2
+ x)*log(x))/log(x) + 1/8*Ei(2*log(x)) + 1/8*Ei(log(x)) - 1/8*integrate((x + 1)/log(x), x)

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mupad [B]  time = 5.38, size = 20, normalized size = 1.00 \begin {gather*} \frac {x\,\left (\ln \left (2\,\ln \relax (2)\right )+\ln \relax (x)\right )\,\left (x+{\mathrm {e}}^x+1\right )}{8\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(2*x*log(2))*(x + exp(x) - log(x)*(2*x + exp(x)*(x + 1) + 1) + 1))/8 - (log(x)*(x + exp(x) + 1))/8)/
log(x)^2,x)

[Out]

(x*(log(2*log(2)) + log(x))*(x + exp(x) + 1))/(8*log(x))

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sympy [B]  time = 0.36, size = 65, normalized size = 3.25 \begin {gather*} \frac {x^{2}}{8} + \frac {x}{8} + \frac {\left (x \log {\relax (x )} + x \log {\left (\log {\relax (2 )} \right )} + x \log {\relax (2 )}\right ) e^{x}}{8 \log {\relax (x )}} + \frac {x^{2} \log {\left (\log {\relax (2 )} \right )} + x^{2} \log {\relax (2 )} + x \log {\left (\log {\relax (2 )} \right )} + x \log {\relax (2 )}}{8 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/8 + x/8 + (x*log(x) + x*log(log(2)) + x*log(2))*exp(x)/(8*log(x)) + (x**2*log(log(2)) + x**2*log(2) + x*l
og(log(2)) + x*log(2))/(8*log(x))

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