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

Optimal. Leaf size=27 \[ \frac {4 x \log (x) \left (7-x-\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \]

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Rubi [A]  time = 0.34, antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps used = 28, number of rules used = 12, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {12, 6688, 2357, 2295, 2304, 203, 2324, 4848, 2391, 2523, 388, 2556} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[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 203

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

Rule 388

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 2295

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

Rule 2304

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 2324

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 2357

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 2391

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 2523

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 2556

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 4848

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 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} &=\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} \tan ^{-1}\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 {\tan ^{-1}\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 {\tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2} x} \, dx}{\log (25)}+\frac {\left (8 \sqrt {2}\right ) \int \frac {\tan ^{-1}\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 {\tan ^{-1}\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} \text {Li}_2\left (-\frac {i x}{\sqrt {2}}\right )}{\log (25)}-\frac {4 i \sqrt {2} \text {Li}_2\left (\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 23, normalized size = 0.85 \begin {gather*} -\frac {4 x \log (x) \left (-7+x+\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.54, size = 28, normalized size = 1.04 \begin {gather*} -\frac {2 \, {\left (x^{2} + x \log \left (\frac {2 \, x}{x^{2} + 2}\right ) - 7 \, x\right )} \log \relax (x)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.23, size = 38, normalized size = 1.41 \begin {gather*} \frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \relax (x) - x \log \relax (x)^{2} - {\left (x^{2} + x {\left (\log \relax (2) - 7\right )}\right )} \log \relax (x)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 0.22, size = 189, normalized size = 7.00

method	result	size
risch	\(\frac {2 \ln \relax (x ) x \ln \left (x^{2}+2\right )}{\ln \relax (5)}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}+2}\right ) \mathrm {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \relax (x )}{\ln \relax (5)}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}+2}\right ) \mathrm {csgn}\left (\frac {i x}{x^{2}+2}\right ) \mathrm {csgn}\left (i x \right ) \ln \relax (x )}{\ln \relax (5)}-\frac {i \pi x \mathrm {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \mathrm {csgn}\left (i x \right ) \ln \relax (x )}{\ln \relax (5)}+\frac {i \pi x \mathrm {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \relax (x )}{\ln \relax (5)}-\frac {2 x \ln \relax (2) \ln \relax (x )}{\ln \relax (5)}-\frac {2 x^{2} \ln \relax (x )}{\ln \relax (5)}-\frac {2 x \ln \relax (x )^{2}}{\ln \relax (5)}+\frac {14 x \ln \relax (x )}{\ln \relax (5)}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[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]

2/ln(5)*ln(x)*x*ln(x^2+2)-I/ln(5)*Pi*x*csgn(I/(x^2+2))*csgn(I*x/(x^2+2))^2*ln(x)+I/ln(5)*Pi*x*csgn(I/(x^2+2))*
csgn(I*x/(x^2+2))*csgn(I*x)*ln(x)-I/ln(5)*Pi*x*csgn(I*x/(x^2+2))^2*csgn(I*x)*ln(x)+I/ln(5)*Pi*x*csgn(I*x/(x^2+
2))^3*ln(x)-2/ln(5)*x*ln(2)*ln(x)-2/ln(5)*x^2*ln(x)-2/ln(5)*x*ln(x)^2+14/ln(5)*x*ln(x)

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maxima [A]  time = 0.61, size = 38, normalized size = 1.41 \begin {gather*} \frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \relax (x) - x \log \relax (x)^{2} - {\left (x^{2} + x {\left (\log \relax (2) - 7\right )}\right )} \log \relax (x)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {4\,x+\frac {\ln \relax (x)\,\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 \relax (5)\,\left (x^2+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.68, size = 34, normalized size = 1.26 \begin {gather*} - \frac {2 x \log {\relax (x )} \log {\left (\frac {2 x}{x^{2} + 2} \right )}}{\log {\relax (5 )}} + \frac {\left (- 2 x^{2} + 14 x\right ) \log {\relax (x )}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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