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

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

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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} $$\begin{array}{c}\frac{x^2\log (x\log (4))}{8\log (x)}+\frac{e^{x}x\log (x\log (4))}{8\log (x)}+\frac{x\log (x\log (4))}{8\log (x)}\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{8}\int \frac{(1+e^x+x)\log (x)+(-1-e^x-x+(1+2x+e^x(1+x))\log (x))\log (x\log (4))}{{\log }^2(x)}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)}+\frac{\log (x)+x\log (x)-\log (x\log (4))-x\log (x\log (4))+\log (x)\log (x\log (4))+2x\log (x)\log (x\log (4))}{{\log }^2(x)})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))+2x\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+2x)\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)}+\frac{(-1-x+\log (x)+2x\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)}dx+\frac{1}{8}\int \frac{(-1-x+\log (x)+2x\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}{\log (x)}+\frac{x}{\log (x)})dx+\frac{1}{8}\int (-\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{2x\log (x\log (4))}{\log (x)})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 (\frac{2\text{Ei}(2\log (x))}{x}-\frac{x}{\log (x)})dx-\frac{1}{8}\int \frac{\text{li}(x)}{x}dx+\frac{1}{8}\int (-\frac{1}{\log (x)}+\frac{\text{li}(x)}{x})dx+\frac{1}{8}\mathrm{Subst}(\int \frac{e^{2x}}{x}dx,x,\log (x))-\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}\mathrm{Subst}(\int \text{Ei}(2x)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}\mathrm{Subst}(\int \frac{e^{2x}}{x}dx,x,\log (x))+\frac{1}{4}\mathrm{Subst}(\int \text{Ei}(2x)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{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{(x^2+x{e}^x+x)\log (x)+(x^2+x{e}^x+x)\log (2\log (2))}{8\log (x)}\end{array}$$

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 $$\begin{array}{c}\frac{x^2\log (2)+x{e}^x\log (2)+x^2\log (x)+x{e}^x\log (x)+x^2\log (\log (2))+x{e}^x\log (\log (2))+x\log (2)+x\log (x)+x\log (\log (2))}{8\log (x)}\end{array}$$

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	$\frac{x^2}{8}+\frac{{\mathrm{e}}^{x}x}{8}+\frac{x}{8}+\frac{x(2x\ln (2)+2{\mathrm{e}}^x\ln (2)+2x\ln (\ln (2))+2{\mathrm{e}}^x\ln (\ln (2))+2\ln (2)+2\ln (\ln (2)))}{16\ln (x)}$	$56$
default	$\frac{x}{8}+\frac{(\ln (2)+\ln (\ln (2)))x{\mathrm{e}}^x+x{\mathrm{e}}^x\ln (x)}{8\ln (x)}+\frac{x^2}{8}+\frac{\ln (2)x^2}{8\ln (x)}+\frac{\ln (2)x}{8\ln (x)}+\frac{\ln (\ln (2))x^2}{8\ln (x)}+\frac{\ln (\ln (2))x}{8\ln (x)}$	$75$

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 $$\begin{array}{c}\frac{x^2(\log (2)+\log (\log (2)))+x(\log (2)+\log (\log (2)))+(x(\log (2)+\log (\log (2)))+x\log (x))e^x+(x^2+x)\log (x)}{8\log (x)}+\frac{1}{8}\mathrm{E}\mathrm{i}(2\log (x))+\frac{1}{8}\mathrm{E}\mathrm{i}(\log (x))-\frac{1}{8}\int \frac{x+1}{\log (x)}dx\end{array}$$

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 $$\begin{array}{c}\frac{x(\ln (2\ln (2))+\ln (x))(x+{\mathrm{e}}^x+1)}{8\ln (x)}\end{array}$$

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 $$\begin{array}{c}\frac{x^2}{8}+\frac{x}{8}+\frac{(x\log (x)+x\log (\log (2))+x\log (2))e^x}{8\log (x)}+\frac{x^2\log (\log (2))+x^2\log (2)+x\log (\log (2))+x\log (2)}{8\log (x)}\end{array}$$

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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