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

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

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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 {gather*} 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 \left (x \log ^2(x) \log (2 x)\right )}{x} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-15+x^2+\frac {10}{\log (x)}+\frac {5}{\log (2 x)}-5 \log \left (x \log ^2(x) \log (2 x)\right )}{x^2} \, dx\\ &=\int \left (\frac {5 \log (x)+10 \log (2 x)-15 \log (x) \log (2 x)+x^2 \log (x) \log (2 x)}{x^2 \log (x) \log (2 x)}-\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x^2}\right ) \, dx\\ &=-\left (5 \int \frac {\log \left (x \log ^2(x) \log (2 x)\right )}{x^2} \, dx\right )+\int \frac {5 \log (x)+10 \log (2 x)-15 \log (x) \log (2 x)+x^2 \log (x) \log (2 x)}{x^2 \log (x) \log (2 x)} \, dx\\ &=\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x}-5 \int \frac {1+\frac {2}{\log (x)}+\frac {1}{\log (2 x)}}{x^2} \, dx+\int \left (1-\frac {15}{x^2}+\frac {10}{x^2 \log (x)}+\frac {5}{x^2 \log (2 x)}\right ) \, dx\\ &=\frac {15}{x}+x+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x}-5 \int \left (\frac {2+\log (x)}{x^2 \log (x)}+\frac {1}{x^2 \log (2 x)}\right ) \, dx+5 \int \frac {1}{x^2 \log (2 x)} \, dx+10 \int \frac {1}{x^2 \log (x)} \, dx\\ &=\frac {15}{x}+x+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x}-5 \int \frac {2+\log (x)}{x^2 \log (x)} \, dx-5 \int \frac {1}{x^2 \log (2 x)} \, dx+10 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )+10 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (2 x)\right )\\ &=\frac {15}{x}+x+10 \text {Ei}(-\log (x))+10 \text {Ei}(-\log (2 x))-5 \text {Ei}(-\log (x)) (2+\log (x))+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x}+5 \int \frac {\text {Ei}(-\log (x))}{x} \, dx-10 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (2 x)\right )\\ &=\frac {15}{x}+x+10 \text {Ei}(-\log (x))-5 \text {Ei}(-\log (x)) (2+\log (x))+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x}+5 \operatorname {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 \left (x \log ^2(x) \log (2 x)\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 23, normalized size = 0.85 \begin {gather*} \frac {20}{x}+x+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x} \end {gather*}

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 {gather*} \frac {x^{2} + 5 \, \log \left (x \log \relax (2) \log \relax (x)^{2} + x \log \relax (x)^{3}\right ) + 20}{x} \end {gather*}

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 {gather*} x + \frac {5 \, \log \left (\log \relax (2) \log \relax (x)^{2} + \log \relax (x)^{3}\right )}{x} + \frac {5 \, \log \relax (x)}{x} + \frac {20}{x} \end {gather*}

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 \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )}{x}-\frac {5 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i x \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right ) \ln \relax (x )^{2}\right )+5 i \pi \,\mathrm {csgn}\left (i \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )-5 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right ) \ln \relax (x )^{2}\right )^{2}-10 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}+5 i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )-5 i \pi \mathrm {csgn}\left (x \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )^{2}+5 i \pi \,\mathrm {csgn}\left (i x \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right ) \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (x \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )-5 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i x \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right ) \ln \relax (x )^{2}\right )^{2}-5 i \pi \,\mathrm {csgn}\left (i x \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right ) \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (x \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )^{2}+5 i \pi \mathrm {csgn}\left (i x \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right ) \ln \relax (x )^{2}\right )^{3}+5 i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )^{3}-5 i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )+5 i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}+5 i \pi \mathrm {csgn}\left (x \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )^{3}-5 i \pi \,\mathrm {csgn}\left (i \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2} \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )\right )^{2}+5 i \pi -2 x^{2}+10 \ln \relax (2)-10 \ln \relax (x )-20 \ln \left (\ln \relax (x )\right )-40}{2 x}\)	\(530\)

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 {gather*} x + \frac {5 \, {\left (\log \relax (x) + \log \left (\log \relax (2) + \log \relax (x)\right ) + 2 \, \log \left (\log \relax (x)\right ) + 1\right )}}{x} + \frac {15}{x} + 10 \, {\rm Ei}\left (-\log \left (2 \, x\right )\right ) + 10 \, {\rm Ei}\left (-\log \relax (x)\right ) - 5 \, \int \frac {1}{x^{2} \log \relax (2) + x^{2} \log \relax (x)}\,{d x} - 10 \, \int \frac {1}{x^{2} \log \relax (x)}\,{d x} \end {gather*}

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 {gather*} x+\frac {5\,\ln \left (x\,\ln \left (2\,x\right )\,{\ln \relax (x)}^2\right )+20}{x} \end {gather*}

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 {gather*} x + \frac {5 \log {\left (x \left (\log {\relax (x )} + \log {\relax (2 )}\right ) \log {\relax (x )}^{2} \right )}}{x} + \frac {20}{x} \end {gather*}

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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