3.85.96 \(\int \frac {(-10+2 x^2) \log (\frac {5+x^2}{x})+(5+x^2) \log ^2(\frac {5+x^2}{x})}{20+4 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{4} x \left (-\frac {\log (8)}{x}+\log ^2\left (\frac {5}{x}+x\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.88, antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 35, number of rules used = 15, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.319, Rules used = {6725, 2528, 2523, 388, 203, 2526, 12, 446, 72, 4848, 2391, 4920, 4854, 2402, 2315} \begin {gather*} \frac {1}{4} x \log ^2\left (\frac {x^2+5}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-10 + 2*x^2)*Log[(5 + x^2)/x] + (5 + x^2)*Log[(5 + x^2)/x]^2)/(20 + 4*x^2),x]

[Out]

(x*Log[(5 + x^2)/x]^2)/4

Rule 12

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 446

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 2526

Int[Log[(c_.)*(RFx_)^(n_.)]/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2), x]}, Simp[u*L
og[c*RFx^n], x] - Dist[n, Int[SimplifyIntegrand[(u*D[RFx, x])/RFx, x], x], x]] /; FreeQ[{c, d, e, n}, x] && Ra
tionalFunctionQ[RFx, x] &&  !PolynomialQ[RFx, x]

Rule 2528

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

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {\left (-5+x^2\right ) \log \left (\frac {5+x^2}{x}\right )}{2 \left (5+x^2\right )}+\frac {1}{4} \log ^2\left (\frac {5+x^2}{x}\right )\right ) \, dx\\ &=\frac {1}{4} \int \log ^2\left (\frac {5+x^2}{x}\right ) \, dx+\frac {1}{2} \int \frac {\left (-5+x^2\right ) \log \left (\frac {5+x^2}{x}\right )}{5+x^2} \, dx\\ &=\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-\frac {1}{2} \int \frac {\left (-5+x^2\right ) \log \left (\frac {5+x^2}{x}\right )}{5+x^2} \, dx+\frac {1}{2} \int \left (\log \left (\frac {5+x^2}{x}\right )-\frac {10 \log \left (\frac {5+x^2}{x}\right )}{5+x^2}\right ) \, dx\\ &=\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )+\frac {1}{2} \int \log \left (\frac {5+x^2}{x}\right ) \, dx-\frac {1}{2} \int \left (\log \left (\frac {5+x^2}{x}\right )-\frac {10 \log \left (\frac {5+x^2}{x}\right )}{5+x^2}\right ) \, dx-5 \int \frac {\log \left (\frac {5+x^2}{x}\right )}{5+x^2} \, dx\\ &=\frac {1}{2} x \log \left (\frac {5+x^2}{x}\right )-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \log \left (\frac {5+x^2}{x}\right )+\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-\frac {1}{2} \int \frac {-5+x^2}{5+x^2} \, dx-\frac {1}{2} \int \log \left (\frac {5+x^2}{x}\right ) \, dx+5 \int \frac {\left (-5+x^2\right ) \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5} x \left (5+x^2\right )} \, dx+5 \int \frac {\log \left (\frac {5+x^2}{x}\right )}{5+x^2} \, dx\\ &=-\frac {x}{2}+\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )+\frac {1}{2} \int \frac {-5+x^2}{5+x^2} \, dx+5 \int \frac {1}{5+x^2} \, dx-5 \int \frac {\left (-5+x^2\right ) \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5} x \left (5+x^2\right )} \, dx+\sqrt {5} \int \frac {\left (-5+x^2\right ) \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x \left (5+x^2\right )} \, dx\\ &=\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-5 \int \frac {1}{5+x^2} \, dx-\sqrt {5} \int \frac {\left (-5+x^2\right ) \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x \left (5+x^2\right )} \, dx+\sqrt {5} \int \left (-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x}+\frac {2 x \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{5+x^2}\right ) \, dx\\ &=\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-\sqrt {5} \int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x} \, dx-\sqrt {5} \int \left (-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x}+\frac {2 x \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{5+x^2}\right ) \, dx+\left (2 \sqrt {5}\right ) \int \frac {x \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{5+x^2} \, dx\\ &=-i \sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )^2+\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-2 \int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{i-\frac {x}{\sqrt {5}}} \, dx-\frac {1}{2} \left (i \sqrt {5}\right ) \int \frac {\log \left (1-\frac {i x}{\sqrt {5}}\right )}{x} \, dx+\frac {1}{2} \left (i \sqrt {5}\right ) \int \frac {\log \left (1+\frac {i x}{\sqrt {5}}\right )}{x} \, dx+\sqrt {5} \int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x} \, dx-\left (2 \sqrt {5}\right ) \int \frac {x \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{5+x^2} \, dx\\ &=-2 \sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \log \left (\frac {2 \sqrt {5}}{\sqrt {5}+i x}\right )+\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-\frac {1}{2} i \sqrt {5} \text {Li}_2\left (-\frac {i x}{\sqrt {5}}\right )+\frac {1}{2} i \sqrt {5} \text {Li}_2\left (\frac {i x}{\sqrt {5}}\right )+2 \int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{i-\frac {x}{\sqrt {5}}} \, dx+2 \int \frac {\log \left (\frac {2}{1+\frac {i x}{\sqrt {5}}}\right )}{1+\frac {x^2}{5}} \, dx+\frac {1}{2} \left (i \sqrt {5}\right ) \int \frac {\log \left (1-\frac {i x}{\sqrt {5}}\right )}{x} \, dx-\frac {1}{2} \left (i \sqrt {5}\right ) \int \frac {\log \left (1+\frac {i x}{\sqrt {5}}\right )}{x} \, dx\\ &=\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-2 \int \frac {\log \left (\frac {2}{1+\frac {i x}{\sqrt {5}}}\right )}{1+\frac {x^2}{5}} \, dx-\left (2 i \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+\frac {i x}{\sqrt {5}}}\right )\\ &=\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )-i \sqrt {5} \text {Li}_2\left (1-\frac {2 \sqrt {5}}{\sqrt {5}+i x}\right )+\left (2 i \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+\frac {i x}{\sqrt {5}}}\right )\\ &=\frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{4} x \log ^2\left (\frac {5+x^2}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-10 + 2*x^2)*Log[(5 + x^2)/x] + (5 + x^2)*Log[(5 + x^2)/x]^2)/(20 + 4*x^2),x]

[Out]

(x*Log[(5 + x^2)/x]^2)/4

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 15, normalized size = 0.65 \begin {gather*} \frac {1}{4} \, x \log \left (\frac {x^{2} + 5}{x}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5)*log((x^2+5)/x)^2+(2*x^2-10)*log((x^2+5)/x))/(4*x^2+20),x, algorithm="fricas")

[Out]

1/4*x*log((x^2 + 5)/x)^2

________________________________________________________________________________________

giac [A]  time = 0.18, size = 15, normalized size = 0.65 \begin {gather*} \frac {1}{4} \, x \log \left (\frac {x^{2} + 5}{x}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5)*log((x^2+5)/x)^2+(2*x^2-10)*log((x^2+5)/x))/(4*x^2+20),x, algorithm="giac")

[Out]

1/4*x*log((x^2 + 5)/x)^2

________________________________________________________________________________________

maple [A]  time = 0.36, size = 16, normalized size = 0.70

method	result	size
norman	\(\frac {x \ln \left (\frac {x^{2}+5}{x}\right )^{2}}{4}\)	\(16\)
risch	\(\frac {x \ln \left (\frac {x^{2}+5}{x}\right )^{2}}{4}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+5)*ln((x^2+5)/x)^2+(2*x^2-10)*ln((x^2+5)/x))/(4*x^2+20),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.70, size = 30, normalized size = 1.30 \begin {gather*} \frac {1}{4} \, x \log \left (x^{2} + 5\right )^{2} - \frac {1}{2} \, x \log \left (x^{2} + 5\right ) \log \relax (x) + \frac {1}{4} \, x \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5)*log((x^2+5)/x)^2+(2*x^2-10)*log((x^2+5)/x))/(4*x^2+20),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.96, size = 15, normalized size = 0.65 \begin {gather*} \frac {x\,{\ln \left (\frac {x^2+5}{x}\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((x^2 + 5)/x)*(2*x^2 - 10) + log((x^2 + 5)/x)^2*(x^2 + 5))/(4*x^2 + 20),x)

[Out]

(x*log((x^2 + 5)/x)^2)/4

________________________________________________________________________________________

sympy [A]  time = 0.18, size = 12, normalized size = 0.52 \begin {gather*} \frac {x \log {\left (\frac {x^{2} + 5}{x} \right )}^{2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+5)*ln((x**2+5)/x)**2+(2*x**2-10)*ln((x**2+5)/x))/(4*x**2+20),x)

[Out]

x*log((x**2 + 5)/x)**2/4

________________________________________________________________________________________