3.89.19 $\int \frac{5\log (x)+(10+(-15+x^2)\log (x))\log (2x)-5\log (x)\log (2x)\log (x{\log }^2(x)\log (2x))}{x^2\log (x)\log (2x)}dx$

Optimal. Leaf size=27 $$x-\log (3)+\frac{5(4-x+\log (x{\log }^2(x)\log (2x)))}{x}$$

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Rubi [C]  time = 0.57, antiderivative size = 50, normalized size of antiderivative = 1.85, number of steps used = 17, number of rules used = 8, integrand size = 53, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.151, Rules used = {6688, 14, 2309, 2178, 30, 2555, 2366, 6482} $$\begin{array}{c}10\text{Ei}(-\log (x))+5\log (x)\text{Ei}(-\log (x))-5(\log (x)+2)\text{Ei}(-\log (x))+x+\frac{20}{x}+\frac{5\log (x{\log }^2(x)\log (2x))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(5*Log[x] + (10 + (-15 + x^2)*Log[x])*Log[2*x] - 5*Log[x]*Log[2*x]*Log[x*Log[x]^2*Log[2*x]])/(x^2*Log[x]*L
og[2*x]),x]

[Out]

20/x + x + 10*ExpIntegralEi[-Log[x]] + 5*ExpIntegralEi[-Log[x]]*Log[x] - 5*ExpIntegralEi[-Log[x]]*(2 + Log[x])
+ (5*Log[x*Log[x]^2*Log[2*x]])/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

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 6688

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-15+x^2+\frac{10}{\log (x)}+\frac{5}{\log (2x)}-5\log (x{\log }^2(x)\log (2x))}{x^2}dx\\ & =\int (\frac{5\log (x)+10\log (2x)-15\log (x)\log (2x)+x^2\log (x)\log (2x)}{x^2\log (x)\log (2x)}-\frac{5\log (x{\log }^2(x)\log (2x))}{x^2})dx\\ & =-(5\int \frac{\log (x{\log }^2(x)\log (2x))}{x^2}dx)+\int \frac{5\log (x)+10\log (2x)-15\log (x)\log (2x)+x^2\log (x)\log (2x)}{x^2\log (x)\log (2x)}dx\\ & =\frac{5\log (x{\log }^2(x)\log (2x))}{x}-5\int \frac{1+\frac{2}{\log (x)}+\frac{1}{\log (2x)}}{x^2}dx+\int (1-\frac{15}{x^2}+\frac{10}{x^2\log (x)}+\frac{5}{x^2\log (2x)})dx\\ & =\frac{15}{x}+x+\frac{5\log (x{\log }^2(x)\log (2x))}{x}-5\int (\frac{2+\log (x)}{x^2\log (x)}+\frac{1}{x^2\log (2x)})dx+5\int \frac{1}{x^2\log (2x)}dx+10\int \frac{1}{x^2\log (x)}dx\\ & =\frac{15}{x}+x+\frac{5\log (x{\log }^2(x)\log (2x))}{x}-5\int \frac{2+\log (x)}{x^2\log (x)}dx-5\int \frac{1}{x^2\log (2x)}dx+10\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (x))+10\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (2x))\\ & =\frac{15}{x}+x+10\text{Ei}(-\log (x))+10\text{Ei}(-\log (2x))-5\text{Ei}(-\log (x))(2+\log (x))+\frac{5\log (x{\log }^2(x)\log (2x))}{x}+5\int \frac{\text{Ei}(-\log (x))}{x}dx-10\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (2x))\\ & =\frac{15}{x}+x+10\text{Ei}(-\log (x))-5\text{Ei}(-\log (x))(2+\log (x))+\frac{5\log (x{\log }^2(x)\log (2x))}{x}+5\mathrm{Subst}(\int \text{Ei}(-x)dx,x,\log (x))\\ & =\frac{20}{x}+x+10\text{Ei}(-\log (x))+5\text{Ei}(-\log (x))\log (x)-5\text{Ei}(-\log (x))(2+\log (x))+\frac{5\log (x{\log }^2(x)\log (2x))}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.21, size = 23, normalized size = 0.85 $$\begin{array}{c}\frac{20}{x}+x+\frac{5\log (x{\log }^2(x)\log (2x))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(5*Log[x] + (10 + (-15 + x^2)*Log[x])*Log[2*x] - 5*Log[x]*Log[2*x]*Log[x*Log[x]^2*Log[2*x]])/(x^2*Lo
g[x]*Log[2*x]),x]

[Out]

20/x + x + (5*Log[x*Log[x]^2*Log[2*x]])/x

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fricas [A]  time = 0.50, size = 27, normalized size = 1.00 $$\begin{array}{c}\frac{x^2+5\log (x\log (2)\log (x{)}^2+x\log (x{)}^3)+20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10)*log(2*x)+5*log(x))/x^2/log(x)/log(
2*x),x, algorithm="fricas")

[Out]

(x^2 + 5*log(x*log(2)*log(x)^2 + x*log(x)^3) + 20)/x

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giac [A]  time = 0.20, size = 32, normalized size = 1.19 $$\begin{array}{c}x+\frac{5\log (\log (2)\log (x{)}^2+\log (x{)}^3)}{x}+\frac{5\log (x)}{x}+\frac{20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10)*log(2*x)+5*log(x))/x^2/log(x)/log(
2*x),x, algorithm="giac")

[Out]

x + 5*log(log(2)*log(x)^2 + log(x)^3)/x + 5*log(x)/x + 20/x

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maple [C]  time = 0.60, size = 530, normalized size = 19.63

method	result	size
risch	$\frac{5\ln (2i\ln (2)+2i\ln (x))}{x}-\frac{5i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}(i\ln (x{)}^2(2i\ln (2)+2i\ln (x)))\mathrm{csgn}(ix(2i\ln (2)+2i\ln (x))\ln (x{)}^2)+5i\pi \mathrm{csgn}\left(i(2i\ln (2)+2i\ln (x))\right)\mathrm{csgn}(i\ln (x{)}^2(2i\ln (2)+2i\ln (x)))\mathrm{csgn}(i\ln (x{)}^2)-5i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{(ix(2i\ln (2)+2i\ln (x))\ln (x{)}^2)}^2-10i\pi \mathrm{csgn}(i\ln (x))\mathrm{csgn}{(i\ln (x{)}^2)}^2+5i\pi \mathrm{csgn}{(i\ln (x))}^2\mathrm{csgn}(i\ln (x{)}^2)-5i\pi \mathrm{csgn}{(x\ln (x{)}^2(2i\ln (2)+2i\ln (x)))}^2+5i\pi \mathrm{csgn}(ix(2i\ln (2)+2i\ln (x))\ln (x{)}^2)\mathrm{csgn}(x\ln (x{)}^2(2i\ln (2)+2i\ln (x)))-5i\pi \mathrm{csgn}(i\ln (x{)}^2(2i\ln (2)+2i\ln (x)))\mathrm{csgn}{(ix(2i\ln (2)+2i\ln (x))\ln (x{)}^2)}^2-5i\pi \mathrm{csgn}(ix(2i\ln (2)+2i\ln (x))\ln (x{)}^2)\mathrm{csgn}{(x\ln (x{)}^2(2i\ln (2)+2i\ln (x)))}^2+5i\pi \mathrm{csgn}{(ix(2i\ln (2)+2i\ln (x))\ln (x{)}^2)}^3+5i\pi \mathrm{csgn}{(i\ln (x{)}^2(2i\ln (2)+2i\ln (x)))}^3-5i\pi \mathrm{csgn}{(i\ln (x{)}^2(2i\ln (2)+2i\ln (x)))}^2\mathrm{csgn}(i\ln (x{)}^2)+5i\pi \mathrm{csgn}{(i\ln (x{)}^2)}^3+5i\pi \mathrm{csgn}{(x\ln (x{)}^2(2i\ln (2)+2i\ln (x)))}^3-5i\pi \mathrm{csgn}\left(i(2i\ln (2)+2i\ln (x))\right)\mathrm{csgn}{(i\ln (x{)}^2(2i\ln (2)+2i\ln (x)))}^2+5i\pi -2x^2+10\ln (2)-10\ln (x)-20\ln (\ln (x))-40}{2x}$	$530$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(x)*ln(2*x)*ln(x*ln(x)^2*ln(2*x))+((x^2-15)*ln(x)+10)*ln(2*x)+5*ln(x))/x^2/ln(x)/ln(2*x),x,method=_R
ETURNVERBOSE)

[Out]

5/x*ln(2*I*ln(2)+2*I*ln(x))-1/2*(5*I*Pi*csgn(I*x)*csgn(I*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))*csgn(I*x*(2*I*ln(2)+2*
I*ln(x))*ln(x)^2)+5*I*Pi*csgn(I*(2*I*ln(2)+2*I*ln(x)))*csgn(I*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))*csgn(I*ln(x)^2)-5
*I*Pi*csgn(I*x)*csgn(I*x*(2*I*ln(2)+2*I*ln(x))*ln(x)^2)^2-10*I*Pi*csgn(I*ln(x))*csgn(I*ln(x)^2)^2+5*I*Pi*csgn(
I*ln(x))^2*csgn(I*ln(x)^2)-5*I*Pi*csgn(x*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))^2+5*I*Pi*csgn(I*x*(2*I*ln(2)+2*I*ln(x)
)*ln(x)^2)*csgn(x*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))-5*I*Pi*csgn(I*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))*csgn(I*x*(2*I*ln
(2)+2*I*ln(x))*ln(x)^2)^2-5*I*Pi*csgn(I*x*(2*I*ln(2)+2*I*ln(x))*ln(x)^2)*csgn(x*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))
^2+5*I*Pi*csgn(I*x*(2*I*ln(2)+2*I*ln(x))*ln(x)^2)^3+5*I*Pi*csgn(I*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))^3-5*I*Pi*csgn
(I*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))^2*csgn(I*ln(x)^2)+5*I*Pi*csgn(I*ln(x)^2)^3+5*I*Pi*csgn(x*ln(x)^2*(2*I*ln(2)+
2*I*ln(x)))^3-5*I*Pi*csgn(I*(2*I*ln(2)+2*I*ln(x)))*csgn(I*ln(x)^2*(2*I*ln(2)+2*I*ln(x)))^2+5*I*Pi-2*x^2+10*ln(
2)-10*ln(x)-20*ln(ln(x))-40)/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}x+\frac{5(\log (x)+\log (\log (2)+\log (x))+2\log (\log (x))+1)}{x}+\frac{15}{x}+10\mathrm{E}\mathrm{i}(-\log \left(2x\right))+10\mathrm{E}\mathrm{i}(-\log (x))-5\int \frac{1}{x^2\log (2)+x^2\log (x)}dx-10\int \frac{1}{x^2\log (x)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10)*log(2*x)+5*log(x))/x^2/log(x)/log(
2*x),x, algorithm="maxima")

[Out]

x + 5*(log(x) + log(log(2) + log(x)) + 2*log(log(x)) + 1)/x + 15/x + 10*Ei(-log(2*x)) + 10*Ei(-log(x)) - 5*int
egrate(1/(x^2*log(2) + x^2*log(x)), x) - 10*integrate(1/(x^2*log(x)), x)

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mupad [B]  time = 5.80, size = 21, normalized size = 0.78 $$\begin{array}{c}x+\frac{5\ln (x\ln \left(2x\right){\ln (x)}^2)+20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(x) + log(2*x)*(log(x)*(x^2 - 15) + 10) - 5*log(2*x)*log(x*log(2*x)*log(x)^2)*log(x))/(x^2*log(2*x)*
log(x)),x)

[Out]

x + (5*log(x*log(2*x)*log(x)^2) + 20)/x

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sympy [A]  time = 0.41, size = 22, normalized size = 0.81 $$\begin{array}{c}x+\frac{5\log (x(\log (x)+\log (2))\log {(x)}^2)}{x}+\frac{20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(x)*ln(2*x)*ln(x*ln(x)**2*ln(2*x))+((x**2-15)*ln(x)+10)*ln(2*x)+5*ln(x))/x**2/ln(x)/ln(2*x),x)

[Out]

x + 5*log(x*(log(x) + log(2))*log(x)**2)/x + 20/x

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