3.57.79 \(\int \frac {x+(-5+19 x+4 x^2) \log (5+x)+(5+x) \log (5+x) \log (\log (5+x))}{(5+x) \log (5+x)} \, dx\)

Optimal. Leaf size=16 \[ 4-x+x (2 x+\log (\log (5+x))) \]

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Rubi [A]  time = 0.17, antiderivative size = 25, normalized size of antiderivative = 1.56, number of steps used = 11, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6688, 2411, 2353, 2298, 2302, 29, 2520} \begin {gather*} 2 x^2-x+(x+5) \log (\log (x+5))-5 \log (\log (x+5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + (-5 + 19*x + 4*x^2)*Log[5 + x] + (5 + x)*Log[5 + x]*Log[Log[5 + x]])/((5 + x)*Log[5 + x]),x]

[Out]

-x + 2*x^2 - 5*Log[Log[5 + x]] + (5 + x)*Log[Log[5 + x]]

Rule 29

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

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 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 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

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

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Mathematica [C]  time = 0.05, size = 27, normalized size = 1.69 \begin {gather*} -x+2 x^2-\text {Ei}(\log (5+x))+x \log (\log (5+x))+\text {li}(5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + (-5 + 19*x + 4*x^2)*Log[5 + x] + (5 + x)*Log[5 + x]*Log[Log[5 + x]])/((5 + x)*Log[5 + x]),x]

[Out]

-x + 2*x^2 - ExpIntegralEi[Log[5 + x]] + x*Log[Log[5 + x]] + LogIntegral[5 + x]

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fricas [A]  time = 0.76, size = 16, normalized size = 1.00 \begin {gather*} 2 \, x^{2} + x \log \left (\log \left (x + 5\right )\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*log(5+x)*log(log(5+x))+(4*x^2+19*x-5)*log(5+x)+x)/(5+x)/log(5+x),x, algorithm="fricas")

[Out]

2*x^2 + x*log(log(x + 5)) - x

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giac [A]  time = 0.16, size = 16, normalized size = 1.00 \begin {gather*} 2 \, x^{2} + x \log \left (\log \left (x + 5\right )\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*log(5+x)*log(log(5+x))+(4*x^2+19*x-5)*log(5+x)+x)/(5+x)/log(5+x),x, algorithm="giac")

[Out]

2*x^2 + x*log(log(x + 5)) - x

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maple [A]  time = 0.25, size = 17, normalized size = 1.06

method	result	size
default	\(-x +2 x^{2}+\ln \left (\ln \left (5+x \right )\right ) x\)	\(17\)
norman	\(-x +2 x^{2}+\ln \left (\ln \left (5+x \right )\right ) x\)	\(17\)
risch	\(-x +2 x^{2}+\ln \left (\ln \left (5+x \right )\right ) x\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5+x)*ln(5+x)*ln(ln(5+x))+(4*x^2+19*x-5)*ln(5+x)+x)/(5+x)/ln(5+x),x,method=_RETURNVERBOSE)

[Out]

-x+2*x^2+ln(ln(5+x))*x

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maxima [A]  time = 0.39, size = 16, normalized size = 1.00 \begin {gather*} 2 \, x^{2} + x \log \left (\log \left (x + 5\right )\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*log(5+x)*log(log(5+x))+(4*x^2+19*x-5)*log(5+x)+x)/(5+x)/log(5+x),x, algorithm="maxima")

[Out]

2*x^2 + x*log(log(x + 5)) - x

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mupad [B]  time = 3.68, size = 12, normalized size = 0.75 \begin {gather*} x\,\left (2\,x+\ln \left (\ln \left (x+5\right )\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(x + 5)*(19*x + 4*x^2 - 5) + log(x + 5)*log(log(x + 5))*(x + 5))/(log(x + 5)*(x + 5)),x)

[Out]

x*(2*x + log(log(x + 5)) - 1)

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sympy [A]  time = 0.46, size = 27, normalized size = 1.69 \begin {gather*} 2 x^{2} - x + \left (x + \frac {5}{2}\right ) \log {\left (\log {\left (x + 5 \right )} \right )} - \frac {5 \log {\left (\log {\left (x + 5 \right )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*ln(5+x)*ln(ln(5+x))+(4*x**2+19*x-5)*ln(5+x)+x)/(5+x)/ln(5+x),x)

[Out]

2*x**2 - x + (x + 5/2)*log(log(x + 5)) - 5*log(log(x + 5))/2

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