\(\int \frac {x-x^3+(-2 x+4 x^3) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx\) [9171]

Optimal result

Integrand size = 40, antiderivative size = 23 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=\frac {x \left (-x+x^3\right )}{\log (x)}+(x+10 \log (\log (2)))^2 \]

[Out]

(x+10*ln(ln(2)))^2+x*(x^3-x)/ln(x)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 16, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6874, 2395, 2343, 2346, 2209} \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=\frac {x^4}{\log (x)}-\frac {x^2}{\log (x)}+(x+10 \log (\log (2)))^2 \]

[In]

Int[(x - x^3 + (-2*x + 4*x^3)*Log[x] + 2*x*Log[x]^2 + 20*Log[x]^2*Log[Log[2]])/Log[x]^2,x]

[Out]

-(x^2/Log[x]) + x^4/Log[x] + (x + 10*Log[Log[2]])^2

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log
[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2346

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 2395

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x \left (1-x^2\right )}{\log ^2(x)}+\frac {2 x \left (-1+2 x^2\right )}{\log (x)}+2 (x+10 \log (\log (2)))\right ) \, dx \\ & = (x+10 \log (\log (2)))^2+2 \int \frac {x \left (-1+2 x^2\right )}{\log (x)} \, dx+\int \frac {x \left (1-x^2\right )}{\log ^2(x)} \, dx \\ & = (x+10 \log (\log (2)))^2+2 \int \left (-\frac {x}{\log (x)}+\frac {2 x^3}{\log (x)}\right ) \, dx+\int \left (\frac {x}{\log ^2(x)}-\frac {x^3}{\log ^2(x)}\right ) \, dx \\ & = (x+10 \log (\log (2)))^2-2 \int \frac {x}{\log (x)} \, dx+4 \int \frac {x^3}{\log (x)} \, dx+\int \frac {x}{\log ^2(x)} \, dx-\int \frac {x^3}{\log ^2(x)} \, dx \\ & = -\frac {x^2}{\log (x)}+\frac {x^4}{\log (x)}+(x+10 \log (\log (2)))^2+2 \int \frac {x}{\log (x)} \, dx-2 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-4 \int \frac {x^3}{\log (x)} \, dx+4 \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right ) \\ & = -2 \operatorname {ExpIntegralEi}(2 \log (x))+4 \operatorname {ExpIntegralEi}(4 \log (x))-\frac {x^2}{\log (x)}+\frac {x^4}{\log (x)}+(x+10 \log (\log (2)))^2+2 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-4 \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {x^2}{\log (x)}+\frac {x^4}{\log (x)}+(x+10 \log (\log (2)))^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=x^2-\frac {x^2}{\log (x)}+\frac {x^4}{\log (x)}+20 x \log (\log (2)) \]

[In]

Integrate[(x - x^3 + (-2*x + 4*x^3)*Log[x] + 2*x*Log[x]^2 + 20*Log[x]^2*Log[Log[2]])/Log[x]^2,x]

[Out]

x^2 - x^2/Log[x] + x^4/Log[x] + 20*x*Log[Log[2]]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
risch	\(20 x \ln \left (\ln \left (2\right )\right )+x^{2}+\frac {x^{2} \left (x^{2}-1\right )}{\ln \left (x \right )}\)	\(24\)
default	\(20 x \ln \left (\ln \left (2\right )\right )+x^{2}+\frac {x^{4}}{\ln \left (x \right )}-\frac {x^{2}}{\ln \left (x \right )}\)	\(28\)
parts	\(20 x \ln \left (\ln \left (2\right )\right )+x^{2}+\frac {x^{4}}{\ln \left (x \right )}-\frac {x^{2}}{\ln \left (x \right )}\)	\(28\)
norman	\(\frac {x^{4}+x^{2} \ln \left (x \right )-x^{2}+20 x \ln \left (\ln \left (2\right )\right ) \ln \left (x \right )}{\ln \left (x \right )}\)	\(29\)
parallelrisch	\(\frac {x^{4}+x^{2} \ln \left (x \right )-x^{2}+20 x \ln \left (\ln \left (2\right )\right ) \ln \left (x \right )}{\ln \left (x \right )}\)	\(29\)

[In]

int((20*ln(x)^2*ln(ln(2))+2*x*ln(x)^2+(4*x^3-2*x)*ln(x)-x^3+x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

20*x*ln(ln(2))+x^2+x^2*(x^2-1)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=\frac {x^{4} + x^{2} \log \left (x\right ) + 20 \, x \log \left (x\right ) \log \left (\log \left (2\right )\right ) - x^{2}}{\log \left (x\right )} \]

[In]

integrate((20*log(x)^2*log(log(2))+2*x*log(x)^2+(4*x^3-2*x)*log(x)-x^3+x)/log(x)^2,x, algorithm="fricas")

[Out]

(x^4 + x^2*log(x) + 20*x*log(x)*log(log(2)) - x^2)/log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=x^{2} + 20 x \log {\left (\log {\left (2 \right )} \right )} + \frac {x^{4} - x^{2}}{\log {\left (x \right )}} \]

[In]

integrate((20*ln(x)**2*ln(ln(2))+2*x*ln(x)**2+(4*x**3-2*x)*ln(x)-x**3+x)/ln(x)**2,x)

[Out]

x**2 + 20*x*log(log(2)) + (x**4 - x**2)/log(x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=x^{2} + 20 \, x \log \left (\log \left (2\right )\right ) + 4 \, {\rm Ei}\left (4 \, \log \left (x\right )\right ) - 2 \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) + 2 \, \Gamma \left (-1, -2 \, \log \left (x\right )\right ) - 4 \, \Gamma \left (-1, -4 \, \log \left (x\right )\right ) \]

[In]

integrate((20*log(x)^2*log(log(2))+2*x*log(x)^2+(4*x^3-2*x)*log(x)-x^3+x)/log(x)^2,x, algorithm="maxima")

[Out]

x^2 + 20*x*log(log(2)) + 4*Ei(4*log(x)) - 2*Ei(2*log(x)) + 2*gamma(-1, -2*log(x)) - 4*gamma(-1, -4*log(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=\frac {x^{4}}{\log \left (x\right )} + x^{2} + 20 \, x \log \left (\log \left (2\right )\right ) - \frac {x^{2}}{\log \left (x\right )} \]

[In]

integrate((20*log(x)^2*log(log(2))+2*x*log(x)^2+(4*x^3-2*x)*log(x)-x^3+x)/log(x)^2,x, algorithm="giac")

[Out]

x^4/log(x) + x^2 + 20*x*log(log(2)) - x^2/log(x)

Mupad [B] (verification not implemented)

Time = 12.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {x-x^3+\left (-2 x+4 x^3\right ) \log (x)+2 x \log ^2(x)+20 \log ^2(x) \log (\log (2))}{\log ^2(x)} \, dx=x\,\left (x+20\,\ln \left (\ln \left (2\right )\right )\right )-\frac {x\,\left (x-x^3\right )}{\ln \left (x\right )} \]

[In]

int((x + 2*x*log(x)^2 + 20*log(log(2))*log(x)^2 - log(x)*(2*x - 4*x^3) - x^3)/log(x)^2,x)

[Out]

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