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

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

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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} $$\begin{array}{c}\frac{1}{4}x{\log }^2\left(\frac{x^2+5}{x}\right)\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{(-5+x^2)\log \left(\frac{5+x^2}{x}\right)}{2(5+x^2)}+\frac{1}{4}{\log }^2\left(\frac{5+x^2}{x}\right))dx\\ & =\frac{1}{4}\int {\log }^2\left(\frac{5+x^2}{x}\right)dx+\frac{1}{2}\int \frac{(-5+x^2)\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{(-5+x^2)\log \left(\frac{5+x^2}{x}\right)}{5+x^2}dx+\frac{1}{2}\int (\log \left(\frac{5+x^2}{x}\right)-\frac{10\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 \log \left(\frac{5+x^2}{x}\right)dx-\frac{1}{2}\int (\log \left(\frac{5+x^2}{x}\right)-\frac{10\log \left(\frac{5+x^2}{x}\right)}{5+x^2})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{(-5+x^2){\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}x(5+x^2)}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{(-5+x^2){\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}x(5+x^2)}dx+\sqrt{5}\int \frac{(-5+x^2){\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{x(5+x^2)}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{(-5+x^2){\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{x(5+x^2)}dx+\sqrt{5}\int (-\frac{{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{x}+\frac{2x{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{5+x^2})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 (-\frac{{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{x}+\frac{2x{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)}{5+x^2})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 (1-\frac{ix}{\sqrt{5}})}{x}dx+\frac{1}{2}\left(i\sqrt{5}\right)\int \frac{\log (1+\frac{ix}{\sqrt{5}})}{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}+ix}\right)+\frac{1}{4}x{\log }^2\left(\frac{5+x^2}{x}\right)-\frac{1}{2}i\sqrt{5}{\text{Li}}_2(-\frac{ix}{\sqrt{5}})+\frac{1}{2}i\sqrt{5}{\text{Li}}_2\left(\frac{ix}{\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{ix}{\sqrt{5}}}\right)}{1+\frac{x^2}{5}}dx+\frac{1}{2}\left(i\sqrt{5}\right)\int \frac{\log (1-\frac{ix}{\sqrt{5}})}{x}dx-\frac{1}{2}\left(i\sqrt{5}\right)\int \frac{\log (1+\frac{ix}{\sqrt{5}})}{x}dx\\ & =\frac{1}{4}x{\log }^2\left(\frac{5+x^2}{x}\right)-2\int \frac{\log \left(\frac{2}{1+\frac{ix}{\sqrt{5}}}\right)}{1+\frac{x^2}{5}}dx-\left(2i\sqrt{5}\right)\mathrm{Subst}(\int \frac{\log (2x)}{1-2x}dx,x,\frac{1}{1+\frac{ix}{\sqrt{5}}})\\ & =\frac{1}{4}x{\log }^2\left(\frac{5+x^2}{x}\right)-i\sqrt{5}{\text{Li}}_2(1-\frac{2\sqrt{5}}{\sqrt{5}+ix})+\left(2i\sqrt{5}\right)\mathrm{Subst}(\int \frac{\log (2x)}{1-2x}dx,x,\frac{1}{1+\frac{ix}{\sqrt{5}}})\\ & =\frac{1}{4}x{\log }^2\left(\frac{5+x^2}{x}\right)\end{array}\end{array}$$

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

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

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

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

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

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

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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

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maxima [A]  time = 0.70, size = 30, normalized size = 1.30 $$\begin{array}{c}\frac{1}{4}x\log {(x^2+5)}^2-\frac{1}{2}x\log (x^2+5)\log (x)+\frac{1}{4}x\log (x{)}^2\end{array}$$

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

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

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

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

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

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