3.63.65 \(\int \frac {x+e^x (x+x^2)+(x+e^x (x+x^2)) \log (x)+\log (e^5 x)+(1+\log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx\)

Optimal. Leaf size=19 \[ x+e^x x+\log \left (e^5 x\right ) \log (1+\log (x)) \]

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Rubi [A]  time = 0.68, antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 16, number of rules used = 9, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {2561, 6742, 2176, 2194, 43, 2302, 29, 2389, 2295} \begin {gather*} x-e^x+e^x (x+1)+(\log (x)+1) \log (\log (x)+1)+4 \log (\log (x)+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + E^x*(x + x^2) + (x + E^x*(x + x^2))*Log[x] + Log[E^5*x] + (1 + Log[x])*Log[1 + Log[x]])/(x + x*Log[x]
),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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 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 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x+e^x \left (x+x^2\right )+\left (x+e^x \left (x+x^2\right )\right ) \log (x)+\log \left (e^5 x\right )+(1+\log (x)) \log (1+\log (x))}{x (1+\log (x))} \, dx\\ &=\int \left (e^x (1+x)+\frac {5+x+\log (x)+x \log (x)+\log (1+\log (x))+\log (x) \log (1+\log (x))}{x (1+\log (x))}\right ) \, dx\\ &=\int e^x (1+x) \, dx+\int \frac {5+x+\log (x)+x \log (x)+\log (1+\log (x))+\log (x) \log (1+\log (x))}{x (1+\log (x))} \, dx\\ &=e^x (1+x)-\int e^x \, dx+\int \left (\frac {5+x+\log (x)+x \log (x)}{x (1+\log (x))}+\frac {\log (1+\log (x))}{x}\right ) \, dx\\ &=-e^x+e^x (1+x)+\int \frac {5+x+\log (x)+x \log (x)}{x (1+\log (x))} \, dx+\int \frac {\log (1+\log (x))}{x} \, dx\\ &=-e^x+e^x (1+x)+\int \left (\frac {1+x}{x}+\frac {4}{x (1+\log (x))}\right ) \, dx+\operatorname {Subst}(\int \log (1+x) \, dx,x,\log (x))\\ &=-e^x+e^x (1+x)+4 \int \frac {1}{x (1+\log (x))} \, dx+\int \frac {1+x}{x} \, dx+\operatorname {Subst}(\int \log (x) \, dx,x,1+\log (x))\\ &=-e^x+e^x (1+x)-\log (x)+(1+\log (x)) \log (1+\log (x))+4 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,1+\log (x)\right )+\int \left (1+\frac {1}{x}\right ) \, dx\\ &=-e^x+x+e^x (1+x)+4 \log (1+\log (x))+(1+\log (x)) \log (1+\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 18, normalized size = 0.95 \begin {gather*} \left (1+e^x\right ) x+(5+\log (x)) \log (1+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + E^x*(x + x^2) + (x + E^x*(x + x^2))*Log[x] + Log[E^5*x] + (1 + Log[x])*Log[1 + Log[x]])/(x + x*
Log[x]),x]

[Out]

(1 + E^x)*x + (5 + Log[x])*Log[1 + Log[x]]

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fricas [A]  time = 1.04, size = 16, normalized size = 0.84 \begin {gather*} x e^{x} + {\left (\log \relax (x) + 5\right )} \log \left (\log \relax (x) + 1\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)+1)*log(log(x)+1)+log(x*exp(5))+((x^2+x)*exp(x)+x)*log(x)+(x^2+x)*exp(x)+x)/(x*log(x)+x),x,
algorithm="fricas")

[Out]

x*e^x + (log(x) + 5)*log(log(x) + 1) + x

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giac [A]  time = 0.15, size = 21, normalized size = 1.11 \begin {gather*} x e^{x} + \log \relax (x) \log \left (\log \relax (x) + 1\right ) + x + 5 \, \log \left (\log \relax (x) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)+1)*log(log(x)+1)+log(x*exp(5))+((x^2+x)*exp(x)+x)*log(x)+(x^2+x)*exp(x)+x)/(x*log(x)+x),x,
algorithm="giac")

[Out]

x*e^x + log(x)*log(log(x) + 1) + x + 5*log(log(x) + 1)

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maple [A]  time = 0.13, size = 22, normalized size = 1.16

method	result	size
risch	\(\ln \left (\ln \relax (x )+1\right ) \ln \relax (x )+{\mathrm e}^{x} x +x +5 \ln \left (\ln \relax (x )+1\right )\)	\(22\)
default	\(\left (\ln \relax (x )+1\right ) \ln \left (\ln \relax (x )+1\right )-1+x +4 \ln \left (\ln \relax (x )+1\right )+{\mathrm e}^{x} x\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x)+1)*ln(ln(x)+1)+ln(x*exp(5))+((x^2+x)*exp(x)+x)*ln(x)+(x^2+x)*exp(x)+x)/(x*ln(x)+x),x,method=_RETUR
NVERBOSE)

[Out]

ln(ln(x)+1)*ln(x)+exp(x)*x+x+5*ln(ln(x)+1)

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maxima [A]  time = 0.42, size = 16, normalized size = 0.84 \begin {gather*} x e^{x} + {\left (\log \relax (x) + 5\right )} \log \left (\log \relax (x) + 1\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)+1)*log(log(x)+1)+log(x*exp(5))+((x^2+x)*exp(x)+x)*log(x)+(x^2+x)*exp(x)+x)/(x*log(x)+x),x,
algorithm="maxima")

[Out]

x*e^x + (log(x) + 5)*log(log(x) + 1) + x

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mupad [B]  time = 4.25, size = 21, normalized size = 1.11 \begin {gather*} x+5\,\ln \left (\ln \relax (x)+1\right )+x\,{\mathrm {e}}^x+\ln \left (\ln \relax (x)+1\right )\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(x*exp(5)) + log(x)*(x + exp(x)*(x + x^2)) + log(log(x) + 1)*(log(x) + 1) + exp(x)*(x + x^2))/(x +
x*log(x)),x)

[Out]

x + 5*log(log(x) + 1) + x*exp(x) + log(log(x) + 1)*log(x)

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sympy [A]  time = 0.47, size = 24, normalized size = 1.26 \begin {gather*} x e^{x} + x + \log {\relax (x )} \log {\left (\log {\relax (x )} + 1 \right )} + 5 \log {\left (\log {\relax (x )} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x)+1)*ln(ln(x)+1)+ln(x*exp(5))+((x**2+x)*exp(x)+x)*ln(x)+(x**2+x)*exp(x)+x)/(x*ln(x)+x),x)

[Out]

x*exp(x) + x + log(x)*log(log(x) + 1) + 5*log(log(x) + 1)

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