3.80.83 \(\int \frac {7+x \log (x)}{x+2 x \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=10 \[ \frac {-7+x}{1+\log (x)} \]

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Rubi [A]  time = 0.21, antiderivative size = 17, normalized size of antiderivative = 1.70, number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6688, 6742, 2353, 2297, 2299, 2178, 2302, 30} \begin {gather*} \frac {x}{\log (x)+1}-\frac {7}{\log (x)+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(7 + x*Log[x])/(x + 2*x*Log[x] + x*Log[x]^2),x]

[Out]

-7/(1 + Log[x]) + x/(1 + Log[x])

Rule 30

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

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 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 2299

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

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 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} &=\int \frac {7+x \log (x)}{x (1+\log (x))^2} \, dx\\ &=\int \left (\frac {7-x}{x (1+\log (x))^2}+\frac {1}{1+\log (x)}\right ) \, dx\\ &=\int \frac {7-x}{x (1+\log (x))^2} \, dx+\int \frac {1}{1+\log (x)} \, dx\\ &=\int \left (-\frac {1}{(1+\log (x))^2}+\frac {7}{x (1+\log (x))^2}\right ) \, dx+\operatorname {Subst}\left (\int \frac {e^x}{1+x} \, dx,x,\log (x)\right )\\ &=\frac {\text {Ei}(1+\log (x))}{e}+7 \int \frac {1}{x (1+\log (x))^2} \, dx-\int \frac {1}{(1+\log (x))^2} \, dx\\ &=\frac {\text {Ei}(1+\log (x))}{e}+\frac {x}{1+\log (x)}+7 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,1+\log (x)\right )-\int \frac {1}{1+\log (x)} \, dx\\ &=\frac {\text {Ei}(1+\log (x))}{e}-\frac {7}{1+\log (x)}+\frac {x}{1+\log (x)}-\operatorname {Subst}\left (\int \frac {e^x}{1+x} \, dx,x,\log (x)\right )\\ &=-\frac {7}{1+\log (x)}+\frac {x}{1+\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 10, normalized size = 1.00 \begin {gather*} \frac {-7+x}{1+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(7 + x*Log[x])/(x + 2*x*Log[x] + x*Log[x]^2),x]

[Out]

(-7 + x)/(1 + Log[x])

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fricas [A]  time = 0.48, size = 10, normalized size = 1.00 \begin {gather*} \frac {x - 7}{\log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+7)/(x*log(x)^2+2*x*log(x)+x),x, algorithm="fricas")

[Out]

(x - 7)/(log(x) + 1)

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giac [A]  time = 0.15, size = 10, normalized size = 1.00 \begin {gather*} \frac {x - 7}{\log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+7)/(x*log(x)^2+2*x*log(x)+x),x, algorithm="giac")

[Out]

(x - 7)/(log(x) + 1)

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maple [A]  time = 0.02, size = 11, normalized size = 1.10

method	result	size
norman	\(\frac {x -7}{\ln \relax (x )+1}\)	\(11\)
risch	\(\frac {x -7}{\ln \relax (x )+1}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x)+7)/(x*ln(x)^2+2*x*ln(x)+x),x,method=_RETURNVERBOSE)

[Out]

(x-7)/(ln(x)+1)

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maxima [A]  time = 0.37, size = 17, normalized size = 1.70 \begin {gather*} \frac {x}{\log \relax (x) + 1} - \frac {7}{\log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+7)/(x*log(x)^2+2*x*log(x)+x),x, algorithm="maxima")

[Out]

x/(log(x) + 1) - 7/(log(x) + 1)

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mupad [B]  time = 5.84, size = 10, normalized size = 1.00 \begin {gather*} \frac {x-7}{\ln \relax (x)+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x) + 7)/(x + x*log(x)^2 + 2*x*log(x)),x)

[Out]

(x - 7)/(log(x) + 1)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.70 \begin {gather*} \frac {x - 7}{\log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x)+7)/(x*ln(x)**2+2*x*ln(x)+x),x)

[Out]

(x - 7)/(log(x) + 1)

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