\(\int \frac {56-8 x+28 x^2-4 x^3+(-8-4 x^2) \log (\frac {2 x}{2+x^2})+\log (x) (48-16 x+32 x^2-8 x^3+(-8-4 x^2) \log (\frac {2 x}{2+x^2}))}{(2+x^2) \log (25)} \, dx\) [7460]

Optimal result

Integrand size = 83, antiderivative size = 27 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {4 x \log (x) \left (7-x-\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \]

[Out]

2*x*ln(x)*(7-ln(2*x/(x^2+2))-x)/ln(5)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps used = 28, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {12, 6820, 2404, 2332, 2341, 209, 2361, 4940, 2438, 2603, 396, 2636} \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {4 x^2 \log (x)}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{x^2+2}\right )}{\log (25)}+\frac {28 x \log (x)}{\log (25)} \]

[In]

Int[(56 - 8*x + 28*x^2 - 4*x^3 + (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)] + Log[x]*(48 - 16*x + 32*x^2 - 8*x^3 + (-8
- 4*x^2)*Log[(2*x)/(2 + x^2)]))/((2 + x^2)*Log[25]),x]

[Out]

(28*x*Log[x])/Log[25] - (4*x^2*Log[x])/Log[25] - (4*x*Log[x]*Log[(2*x)/(2 + x^2)])/Log[25]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 2332

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

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2603

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[x*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2636

Int[Log[v_]*Log[w_], x_Symbol] :> Simp[x*Log[v]*Log[w], x] + (-Int[SimplifyIntegrand[x*Log[w]*(D[v, x]/v), x],
x] - Int[SimplifyIntegrand[x*Log[v]*(D[w, x]/w), x], x]) /; InverseFunctionFreeQ[v, x] && InverseFunctionFree
Q[w, x]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{2+x^2} \, dx}{\log (25)} \\ & = \frac {\int \left (-\frac {8 \left (-6+2 x-4 x^2+x^3\right ) \log (x)}{2+x^2}-4 \log (x) \log \left (\frac {2 x}{2+x^2}\right )-4 \left (-7+x+\log \left (\frac {2 x}{2+x^2}\right )\right )\right ) \, dx}{\log (25)} \\ & = -\frac {4 \int \log (x) \log \left (\frac {2 x}{2+x^2}\right ) \, dx}{\log (25)}-\frac {4 \int \left (-7+x+\log \left (\frac {2 x}{2+x^2}\right )\right ) \, dx}{\log (25)}-\frac {8 \int \frac {\left (-6+2 x-4 x^2+x^3\right ) \log (x)}{2+x^2} \, dx}{\log (25)} \\ & = \frac {28 x}{\log (25)}-\frac {2 x^2}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)}+\frac {4 \int \frac {\left (2-x^2\right ) \log (x)}{2+x^2} \, dx}{\log (25)}-\frac {8 \int \left (-4 \log (x)+x \log (x)+\frac {2 \log (x)}{2+x^2}\right ) \, dx}{\log (25)} \\ & = \frac {28 x}{\log (25)}-\frac {2 x^2}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)}+\frac {4 \int \left (-\log (x)+\frac {4 \log (x)}{2+x^2}\right ) \, dx}{\log (25)}-\frac {8 \int x \log (x) \, dx}{\log (25)}-\frac {16 \int \frac {\log (x)}{2+x^2} \, dx}{\log (25)}+\frac {32 \int \log (x) \, dx}{\log (25)} \\ & = -\frac {4 x}{\log (25)}+\frac {32 x \log (x)}{\log (25)}-\frac {4 x^2 \log (x)}{\log (25)}-\frac {8 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right ) \log (x)}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)}-\frac {4 \int \log (x) \, dx}{\log (25)}+\frac {16 \int \frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2} x} \, dx}{\log (25)}+\frac {16 \int \frac {\log (x)}{2+x^2} \, dx}{\log (25)} \\ & = \frac {28 x \log (x)}{\log (25)}-\frac {4 x^2 \log (x)}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)}-\frac {16 \int \frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2} x} \, dx}{\log (25)}+\frac {\left (8 \sqrt {2}\right ) \int \frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{x} \, dx}{\log (25)} \\ & = \frac {28 x \log (x)}{\log (25)}-\frac {4 x^2 \log (x)}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)}+\frac {\left (4 i \sqrt {2}\right ) \int \frac {\log \left (1-\frac {i x}{\sqrt {2}}\right )}{x} \, dx}{\log (25)}-\frac {\left (4 i \sqrt {2}\right ) \int \frac {\log \left (1+\frac {i x}{\sqrt {2}}\right )}{x} \, dx}{\log (25)}-\frac {\left (8 \sqrt {2}\right ) \int \frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{x} \, dx}{\log (25)} \\ & = \frac {28 x \log (x)}{\log (25)}-\frac {4 x^2 \log (x)}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)}+\frac {4 i \sqrt {2} \operatorname {PolyLog}\left (2,-\frac {i x}{\sqrt {2}}\right )}{\log (25)}-\frac {4 i \sqrt {2} \operatorname {PolyLog}\left (2,\frac {i x}{\sqrt {2}}\right )}{\log (25)}-\frac {\left (4 i \sqrt {2}\right ) \int \frac {\log \left (1-\frac {i x}{\sqrt {2}}\right )}{x} \, dx}{\log (25)}+\frac {\left (4 i \sqrt {2}\right ) \int \frac {\log \left (1+\frac {i x}{\sqrt {2}}\right )}{x} \, dx}{\log (25)} \\ & = \frac {28 x \log (x)}{\log (25)}-\frac {4 x^2 \log (x)}{\log (25)}-\frac {4 x \log (x) \log \left (\frac {2 x}{2+x^2}\right )}{\log (25)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {4 x \log (x) \left (-7+x+\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \]

[In]

Integrate[(56 - 8*x + 28*x^2 - 4*x^3 + (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)] + Log[x]*(48 - 16*x + 32*x^2 - 8*x^3
+ (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)]))/((2 + x^2)*Log[25]),x]

[Out]

(-4*x*Log[x]*(-7 + x + Log[(2*x)/(2 + x^2)]))/Log[25]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(\frac {-4 x^{2} \ln \left (x \right )-4 x \ln \left (x \right ) \ln \left (\frac {2 x}{x^{2}+2}\right )+28 x \ln \left (x \right )}{2 \ln \left (5\right )}\)	\(36\)
risch	\(\frac {2 \ln \left (x \right ) x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x \ln \left (2\right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x^{2} \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x \ln \left (x \right )^{2}}{\ln \left (5\right )}+\frac {14 x \ln \left (x \right )}{\ln \left (5\right )}\)	\(189\)
default	\(\frac {-4 x \ln \left (2\right ) \ln \left (x \right )-4 \left (\ln \left (x \right )-1\right ) x \ln \left (x \right )+4 \ln \left (x \right ) x \ln \left (x^{2}+2\right )-4 x \ln \left (x^{2}+2\right )-4 x^{2} \ln \left (x \right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x +2 i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )+2 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x -2 i \pi \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} x +2 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \ln \left (x \right )-2 i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \operatorname {csgn}\left (i x \right ) \ln \left (x \right )-2 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) x -2 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )+28 x \ln \left (x \right )-4 x \ln \left (\frac {x}{x^{2}+2}\right )}{2 \ln \left (5\right )}\)	\(289\)
parts	\(-\frac {2 x \ln \left (x \right )^{2}}{\ln \left (5\right )}+\frac {16 x \ln \left (x \right )}{\ln \left (5\right )}+\frac {2 \ln \left (x \right ) x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}-\frac {2 x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}+\frac {x^{2}}{\ln \left (5\right )}-\frac {2 x^{2} \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x}{\ln \left (5\right )}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x}{\ln \left (5\right )}-\frac {i \pi \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} x}{\ln \left (5\right )}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) x}{\ln \left (5\right )}-\frac {4 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )}{\ln \left (5\right )}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {12 x}{\ln \left (5\right )}-\frac {2 \ln \left (2\right ) \left (x \ln \left (x \right )-x \right )}{\ln \left (5\right )}-\frac {2 \left (-7 x +\frac {1}{2} x^{2}\right )}{\ln \left (5\right )}-\frac {2 \left (x \ln \left (2\right )+x \ln \left (\frac {x}{x^{2}+2}\right )+x -2 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )\right )}{\ln \left (5\right )}\)	\(411\)

[In]

int(1/2*(((-4*x^2-8)*ln(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*ln(x)+(-4*x^2-8)*ln(2*x/(x^2+2))-4*x^3+28*x^2-8*x+5
6)/(x^2+2)/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/2/ln(5)*(-4*x^2*ln(x)-4*x*ln(x)*ln(2*x/(x^2+2))+28*x*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {2 \, {\left (x^{2} + x \log \left (\frac {2 \, x}{x^{2} + 2}\right ) - 7 \, x\right )} \log \left (x\right )}{\log \left (5\right )} \]

[In]

integrate(1/2*(((-4*x^2-8)*log(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*log(x)+(-4*x^2-8)*log(2*x/(x^2+2))-4*x^3+28*
x^2-8*x+56)/(x^2+2)/log(5),x, algorithm="fricas")

[Out]

-2*(x^2 + x*log(2*x/(x^2 + 2)) - 7*x)*log(x)/log(5)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=- \frac {2 x \log {\left (x \right )} \log {\left (\frac {2 x}{x^{2} + 2} \right )}}{\log {\left (5 \right )}} + \frac {\left (- 2 x^{2} + 14 x\right ) \log {\left (x \right )}}{\log {\left (5 \right )}} \]

[In]

integrate(1/2*(((-4*x**2-8)*ln(2*x/(x**2+2))-8*x**3+32*x**2-16*x+48)*ln(x)+(-4*x**2-8)*ln(2*x/(x**2+2))-4*x**3
+28*x**2-8*x+56)/(x**2+2)/ln(5),x)

[Out]

-2*x*log(x)*log(2*x/(x**2 + 2))/log(5) + (-2*x**2 + 14*x)*log(x)/log(5)

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \left (x\right ) - x \log \left (x\right )^{2} - {\left (x^{2} + x {\left (\log \left (2\right ) - 7\right )}\right )} \log \left (x\right )\right )}}{\log \left (5\right )} \]

[In]

integrate(1/2*(((-4*x^2-8)*log(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*log(x)+(-4*x^2-8)*log(2*x/(x^2+2))-4*x^3+28*
x^2-8*x+56)/(x^2+2)/log(5),x, algorithm="maxima")

[Out]

2*(x*log(x^2 + 2)*log(x) - x*log(x)^2 - (x^2 + x*(log(2) - 7))*log(x))/log(5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \left (x\right ) - x \log \left (x\right )^{2} - {\left (x^{2} + x {\left (\log \left (2\right ) - 7\right )}\right )} \log \left (x\right )\right )}}{\log \left (5\right )} \]

[In]

integrate(1/2*(((-4*x^2-8)*log(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*log(x)+(-4*x^2-8)*log(2*x/(x^2+2))-4*x^3+28*
x^2-8*x+56)/(x^2+2)/log(5),x, algorithm="giac")

[Out]

2*(x*log(x^2 + 2)*log(x) - x*log(x)^2 - (x^2 + x*(log(2) - 7))*log(x))/log(5)

Mupad [F(-1)]

Timed out. \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\int -\frac {4\,x+\frac {\ln \left (x\right )\,\left (16\,x+\ln \left (\frac {2\,x}{x^2+2}\right )\,\left (4\,x^2+8\right )-32\,x^2+8\,x^3-48\right )}{2}+\frac {\ln \left (\frac {2\,x}{x^2+2}\right )\,\left (4\,x^2+8\right )}{2}-14\,x^2+2\,x^3-28}{\ln \left (5\right )\,\left (x^2+2\right )} \,d x \]

[In]

int(-(4*x + (log(x)*(16*x + log((2*x)/(x^2 + 2))*(4*x^2 + 8) - 32*x^2 + 8*x^3 - 48))/2 + (log((2*x)/(x^2 + 2))
*(4*x^2 + 8))/2 - 14*x^2 + 2*x^3 - 28)/(log(5)*(x^2 + 2)),x)

[Out]

int(-(4*x + (log(x)*(16*x + log((2*x)/(x^2 + 2))*(4*x^2 + 8) - 32*x^2 + 8*x^3 - 48))/2 + (log((2*x)/(x^2 + 2))
*(4*x^2 + 8))/2 - 14*x^2 + 2*x^3 - 28)/(log(5)*(x^2 + 2)), x)