3.42.72 \(\int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx\)

Optimal. Leaf size=20 \[ 2 \left (1+\frac {5 \log ^2(x)}{\left (9+\frac {1}{x}\right )^2 x}\right ) \]

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Rubi [A]  time = 0.41, antiderivative size = 38, normalized size of antiderivative = 1.90, number of steps used = 19, number of rules used = 14, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {6688, 12, 6742, 2357, 2319, 44, 2314, 31, 2347, 2344, 2301, 2317, 2391, 2318} \begin {gather*} -\frac {10 x \log ^2(x)}{9 x+1}-\frac {10 \log ^2(x)}{9 (9 x+1)^2}+\frac {10 \log ^2(x)}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((20 + 180*x)*Log[x] + (10 - 90*x)*Log[x]^2)/(1 + 27*x + 243*x^2 + 729*x^3),x]

[Out]

(10*Log[x]^2)/9 - (10*Log[x]^2)/(9*(1 + 9*x)^2) - (10*x*Log[x]^2)/(1 + 9*x)

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2317

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

Rule 2318

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

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2344

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

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 \log (x) (2+18 x+\log (x)-9 x \log (x))}{(1+9 x)^3} \, dx\\ &=10 \int \frac {\log (x) (2+18 x+\log (x)-9 x \log (x))}{(1+9 x)^3} \, dx\\ &=10 \int \left (\frac {2 \log (x)}{(1+9 x)^2}-\frac {(-1+9 x) \log ^2(x)}{(1+9 x)^3}\right ) \, dx\\ &=-\left (10 \int \frac {(-1+9 x) \log ^2(x)}{(1+9 x)^3} \, dx\right )+20 \int \frac {\log (x)}{(1+9 x)^2} \, dx\\ &=\frac {20 x \log (x)}{1+9 x}-10 \int \left (-\frac {2 \log ^2(x)}{(1+9 x)^3}+\frac {\log ^2(x)}{(1+9 x)^2}\right ) \, dx-20 \int \frac {1}{1+9 x} \, dx\\ &=\frac {20 x \log (x)}{1+9 x}-\frac {20}{9} \log (1+9 x)-10 \int \frac {\log ^2(x)}{(1+9 x)^2} \, dx+20 \int \frac {\log ^2(x)}{(1+9 x)^3} \, dx\\ &=\frac {20 x \log (x)}{1+9 x}-\frac {10 \log ^2(x)}{9 (1+9 x)^2}-\frac {10 x \log ^2(x)}{1+9 x}-\frac {20}{9} \log (1+9 x)+\frac {20}{9} \int \frac {\log (x)}{x (1+9 x)^2} \, dx+20 \int \frac {\log (x)}{1+9 x} \, dx\\ &=\frac {20 x \log (x)}{1+9 x}-\frac {10 \log ^2(x)}{9 (1+9 x)^2}-\frac {10 x \log ^2(x)}{1+9 x}-\frac {20}{9} \log (1+9 x)+\frac {20}{9} \log (x) \log (1+9 x)+\frac {20}{9} \int \frac {\log (x)}{x (1+9 x)} \, dx-\frac {20}{9} \int \frac {\log (1+9 x)}{x} \, dx-20 \int \frac {\log (x)}{(1+9 x)^2} \, dx\\ &=-\frac {10 \log ^2(x)}{9 (1+9 x)^2}-\frac {10 x \log ^2(x)}{1+9 x}-\frac {20}{9} \log (1+9 x)+\frac {20}{9} \log (x) \log (1+9 x)+\frac {20 \text {Li}_2(-9 x)}{9}+\frac {20}{9} \int \frac {\log (x)}{x} \, dx+20 \int \frac {1}{1+9 x} \, dx-20 \int \frac {\log (x)}{1+9 x} \, dx\\ &=\frac {10 \log ^2(x)}{9}-\frac {10 \log ^2(x)}{9 (1+9 x)^2}-\frac {10 x \log ^2(x)}{1+9 x}+\frac {20 \text {Li}_2(-9 x)}{9}+\frac {20}{9} \int \frac {\log (1+9 x)}{x} \, dx\\ &=\frac {10 \log ^2(x)}{9}-\frac {10 \log ^2(x)}{9 (1+9 x)^2}-\frac {10 x \log ^2(x)}{1+9 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 14, normalized size = 0.70 \begin {gather*} \frac {10 x \log ^2(x)}{(1+9 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((20 + 180*x)*Log[x] + (10 - 90*x)*Log[x]^2)/(1 + 27*x + 243*x^2 + 729*x^3),x]

[Out]

(10*x*Log[x]^2)/(1 + 9*x)^2

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fricas [A]  time = 1.30, size = 19, normalized size = 0.95 \begin {gather*} \frac {10 \, x \log \relax (x)^{2}}{81 \, x^{2} + 18 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x+10)*log(x)^2+(180*x+20)*log(x))/(729*x^3+243*x^2+27*x+1),x, algorithm="fricas")

[Out]

10*x*log(x)^2/(81*x^2 + 18*x + 1)

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giac [A]  time = 0.23, size = 19, normalized size = 0.95 \begin {gather*} \frac {10 \, x \log \relax (x)^{2}}{81 \, x^{2} + 18 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x+10)*log(x)^2+(180*x+20)*log(x))/(729*x^3+243*x^2+27*x+1),x, algorithm="giac")

[Out]

10*x*log(x)^2/(81*x^2 + 18*x + 1)

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maple [A]  time = 0.05, size = 15, normalized size = 0.75

method	result	size
norman	\(\frac {10 x \ln \relax (x )^{2}}{\left (9 x +1\right )^{2}}\)	\(15\)
risch	\(\frac {10 x \ln \relax (x )^{2}}{81 x^{2}+18 x +1}\)	\(20\)
default	\(\frac {20 \ln \relax (x ) x}{9 x +1}+\frac {-20 x \ln \relax (x )-180 x^{2} \ln \relax (x )+10 x \ln \relax (x )^{2}}{\left (9 x +1\right )^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-90*x+10)*ln(x)^2+(180*x+20)*ln(x))/(729*x^3+243*x^2+27*x+1),x,method=_RETURNVERBOSE)

[Out]

10*x*ln(x)^2/(9*x+1)^2

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maxima [B]  time = 0.54, size = 69, normalized size = 3.45 \begin {gather*} -\frac {10 \, {\left (18 \, x + 1\right )} \log \relax (x)}{9 \, {\left (81 \, x^{2} + 18 \, x + 1\right )}} + \frac {10 \, {\left (9 \, x \log \relax (x)^{2} + 2 \, {\left (9 \, x + 1\right )} \log \relax (x)\right )}}{9 \, {\left (81 \, x^{2} + 18 \, x + 1\right )}} - \frac {10 \, \log \relax (x)}{9 \, {\left (81 \, x^{2} + 18 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x+10)*log(x)^2+(180*x+20)*log(x))/(729*x^3+243*x^2+27*x+1),x, algorithm="maxima")

[Out]

-10/9*(18*x + 1)*log(x)/(81*x^2 + 18*x + 1) + 10/9*(9*x*log(x)^2 + 2*(9*x + 1)*log(x))/(81*x^2 + 18*x + 1) - 1
0/9*log(x)/(81*x^2 + 18*x + 1)

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mupad [B]  time = 2.93, size = 14, normalized size = 0.70 \begin {gather*} \frac {10\,x\,{\ln \relax (x)}^2}{{\left (9\,x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(180*x + 20) - log(x)^2*(90*x - 10))/(27*x + 243*x^2 + 729*x^3 + 1),x)

[Out]

(10*x*log(x)^2)/(9*x + 1)^2

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sympy [A]  time = 0.14, size = 17, normalized size = 0.85 \begin {gather*} \frac {10 x \log {\relax (x )}^{2}}{81 x^{2} + 18 x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x+10)*ln(x)**2+(180*x+20)*ln(x))/(729*x**3+243*x**2+27*x+1),x)

[Out]

10*x*log(x)**2/(81*x**2 + 18*x + 1)

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