3.52.3 \(\int \frac {20880+41760 x+20880 x^2+(41760 x+41760 x^2) \log (x)+(1600 x+1600 x^2) \log ^2(x)}{68121 x+5220 x \log (x)+100 x \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {(4+4 x)^2 \log (x)}{\frac {1}{5}+2 (26+\log (x))} \]

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Rubi [A]  time = 0.51, antiderivative size = 42, normalized size of antiderivative = 1.91, number of steps used = 20, number of rules used = 12, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.218, Rules used = {6688, 12, 6742, 2353, 2297, 2299, 2178, 2302, 30, 2306, 2309, 2330} \begin {gather*} -\frac {2088 x^2}{10 \log (x)+261}+8 (x+1)^2-\frac {4176 x}{10 \log (x)+261}-\frac {2088}{10 \log (x)+261} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20880 + 41760*x + 20880*x^2 + (41760*x + 41760*x^2)*Log[x] + (1600*x + 1600*x^2)*Log[x]^2)/(68121*x + 522
0*x*Log[x] + 100*x*Log[x]^2),x]

[Out]

8*(1 + x)^2 - 2088/(261 + 10*Log[x]) - (4176*x)/(261 + 10*Log[x]) - (2088*x^2)/(261 + 10*Log[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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

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

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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 {80 (1+x) \left (261 (1+x)+522 x \log (x)+20 x \log ^2(x)\right )}{x (261+10 \log (x))^2} \, dx\\ &=80 \int \frac {(1+x) \left (261 (1+x)+522 x \log (x)+20 x \log ^2(x)\right )}{x (261+10 \log (x))^2} \, dx\\ &=80 \int \left (\frac {1+x}{5}+\frac {261 (1+x)^2}{x (261+10 \log (x))^2}-\frac {261 (1+x)}{5 (261+10 \log (x))}\right ) \, dx\\ &=8 (1+x)^2-4176 \int \frac {1+x}{261+10 \log (x)} \, dx+20880 \int \frac {(1+x)^2}{x (261+10 \log (x))^2} \, dx\\ &=8 (1+x)^2-4176 \int \left (\frac {1}{261+10 \log (x)}+\frac {x}{261+10 \log (x)}\right ) \, dx+20880 \int \left (\frac {2}{(261+10 \log (x))^2}+\frac {1}{x (261+10 \log (x))^2}+\frac {x}{(261+10 \log (x))^2}\right ) \, dx\\ &=8 (1+x)^2-4176 \int \frac {1}{261+10 \log (x)} \, dx-4176 \int \frac {x}{261+10 \log (x)} \, dx+20880 \int \frac {1}{x (261+10 \log (x))^2} \, dx+20880 \int \frac {x}{(261+10 \log (x))^2} \, dx+41760 \int \frac {1}{(261+10 \log (x))^2} \, dx\\ &=8 (1+x)^2-\frac {4176 x}{261+10 \log (x)}-\frac {2088 x^2}{261+10 \log (x)}+2088 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,261+10 \log (x)\right )+4176 \int \frac {1}{261+10 \log (x)} \, dx+4176 \int \frac {x}{261+10 \log (x)} \, dx-4176 \operatorname {Subst}\left (\int \frac {e^x}{261+10 x} \, dx,x,\log (x)\right )-4176 \operatorname {Subst}\left (\int \frac {e^{2 x}}{261+10 x} \, dx,x,\log (x)\right )\\ &=8 (1+x)^2-\frac {2088 \text {Ei}\left (\frac {1}{10} (261+10 \log (x))\right )}{5 e^{261/10}}-\frac {2088 \text {Ei}\left (\frac {1}{5} (261+10 \log (x))\right )}{5 e^{261/5}}-\frac {2088}{261+10 \log (x)}-\frac {4176 x}{261+10 \log (x)}-\frac {2088 x^2}{261+10 \log (x)}+4176 \operatorname {Subst}\left (\int \frac {e^x}{261+10 x} \, dx,x,\log (x)\right )+4176 \operatorname {Subst}\left (\int \frac {e^{2 x}}{261+10 x} \, dx,x,\log (x)\right )\\ &=8 (1+x)^2-\frac {2088}{261+10 \log (x)}-\frac {4176 x}{261+10 \log (x)}-\frac {2088 x^2}{261+10 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 20, normalized size = 0.91 \begin {gather*} \frac {80 (-261+10 x (2+x) \log (x))}{2610+100 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20880 + 41760*x + 20880*x^2 + (41760*x + 41760*x^2)*Log[x] + (1600*x + 1600*x^2)*Log[x]^2)/(68121*x
+ 5220*x*Log[x] + 100*x*Log[x]^2),x]

[Out]

(80*(-261 + 10*x*(2 + x)*Log[x]))/(2610 + 100*Log[x])

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fricas [A]  time = 0.67, size = 23, normalized size = 1.05 \begin {gather*} \frac {8 \, {\left (10 \, {\left (x^{2} + 2 \, x\right )} \log \relax (x) - 261\right )}}{10 \, \log \relax (x) + 261} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1600*x^2+1600*x)*log(x)^2+(41760*x^2+41760*x)*log(x)+20880*x^2+41760*x+20880)/(100*x*log(x)^2+5220
*x*log(x)+68121*x),x, algorithm="fricas")

[Out]

8*(10*(x^2 + 2*x)*log(x) - 261)/(10*log(x) + 261)

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giac [A]  time = 0.15, size = 27, normalized size = 1.23 \begin {gather*} 8 \, x^{2} + 16 \, x - \frac {2088 \, {\left (x^{2} + 2 \, x + 1\right )}}{10 \, \log \relax (x) + 261} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1600*x^2+1600*x)*log(x)^2+(41760*x^2+41760*x)*log(x)+20880*x^2+41760*x+20880)/(100*x*log(x)^2+5220
*x*log(x)+68121*x),x, algorithm="giac")

[Out]

8*x^2 + 16*x - 2088*(x^2 + 2*x + 1)/(10*log(x) + 261)

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maple [A]  time = 0.03, size = 24, normalized size = 1.09

method	result	size
norman	\(\frac {160 x \ln \relax (x )+80 x^{2} \ln \relax (x )-2088}{10 \ln \relax (x )+261}\)	\(24\)
risch	\(8 x^{2}+16 x -\frac {2088 \left (x^{2}+2 x +1\right )}{10 \ln \relax (x )+261}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1600*x^2+1600*x)*ln(x)^2+(41760*x^2+41760*x)*ln(x)+20880*x^2+41760*x+20880)/(100*x*ln(x)^2+5220*x*ln(x)+
68121*x),x,method=_RETURNVERBOSE)

[Out]

(160*x*ln(x)+80*x^2*ln(x)-2088)/(10*ln(x)+261)

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maxima [A]  time = 0.36, size = 30, normalized size = 1.36 \begin {gather*} \frac {80 \, {\left (x^{2} + 2 \, x\right )} \log \relax (x)}{10 \, \log \relax (x) + 261} - \frac {2088}{10 \, \log \relax (x) + 261} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1600*x^2+1600*x)*log(x)^2+(41760*x^2+41760*x)*log(x)+20880*x^2+41760*x+20880)/(100*x*log(x)^2+5220
*x*log(x)+68121*x),x, algorithm="maxima")

[Out]

80*(x^2 + 2*x)*log(x)/(10*log(x) + 261) - 2088/(10*log(x) + 261)

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mupad [B]  time = 3.37, size = 23, normalized size = 1.05 \begin {gather*} 8\,{\left (x+1\right )}^2-\frac {2088\,{\left (x+1\right )}^2}{10\,\ln \relax (x)+261} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((41760*x + log(x)^2*(1600*x + 1600*x^2) + log(x)*(41760*x + 41760*x^2) + 20880*x^2 + 20880)/(68121*x + 100
*x*log(x)^2 + 5220*x*log(x)),x)

[Out]

8*(x + 1)^2 - (2088*(x + 1)^2)/(10*log(x) + 261)

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sympy [A]  time = 0.11, size = 26, normalized size = 1.18 \begin {gather*} 8 x^{2} + 16 x + \frac {- 2088 x^{2} - 4176 x - 2088}{10 \log {\relax (x )} + 261} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1600*x**2+1600*x)*ln(x)**2+(41760*x**2+41760*x)*ln(x)+20880*x**2+41760*x+20880)/(100*x*ln(x)**2+52
20*x*ln(x)+68121*x),x)

[Out]

8*x**2 + 16*x + (-2088*x**2 - 4176*x - 2088)/(10*log(x) + 261)

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