3.53.36 \(\int \frac {10+x+(-20-2 x) \log (x)-2 x \log (\frac {1}{3} (20+2 x))+(20+2 x) \log ^2(\frac {1}{3} (20+2 x))}{450 x^3+45 x^4} \, dx\)

Optimal. Leaf size=24 \[ \frac {\log (x)-\log ^2\left (\frac {1}{3} (20+2 x)\right )}{45 x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.55, antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 22, number of rules used = 13, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.232, Rules used = {1593, 6742, 2411, 12, 2347, 2344, 2301, 2316, 2315, 2314, 31, 2398, 2303} \begin {gather*} \frac {\log (x)}{45 x^2}-\frac {\log ^2\left (\frac {2 x}{3}+\frac {20}{3}\right )}{45 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + x + (-20 - 2*x)*Log[x] - 2*x*Log[(20 + 2*x)/3] + (20 + 2*x)*Log[(20 + 2*x)/3]^2)/(450*x^3 + 45*x^4),
x]

[Out]

-1/45*Log[20/3 + (2*x)/3]^2/x^2 + Log[x]/(45*x^2)

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

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 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 2316

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

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 2398

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

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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+x+(-20-2 x) \log (x)-2 x \log \left (\frac {1}{3} (20+2 x)\right )+(20+2 x) \log ^2\left (\frac {1}{3} (20+2 x)\right )}{x^3 (450+45 x)} \, dx\\ &=\int \left (\frac {2 \log \left (\frac {20}{3}+\frac {2 x}{3}\right )}{45 (-10-x) x^2}+\frac {2 \log ^2\left (\frac {20}{3}+\frac {2 x}{3}\right )}{45 x^3}+\frac {1-2 \log (x)}{45 x^3}\right ) \, dx\\ &=\frac {1}{45} \int \frac {1-2 \log (x)}{x^3} \, dx+\frac {2}{45} \int \frac {\log \left (\frac {20}{3}+\frac {2 x}{3}\right )}{(-10-x) x^2} \, dx+\frac {2}{45} \int \frac {\log ^2\left (\frac {20}{3}+\frac {2 x}{3}\right )}{x^3} \, dx\\ &=-\frac {\log ^2\left (\frac {20}{3}+\frac {2 x}{3}\right )}{45 x^2}+\frac {\log (x)}{45 x^2}+\frac {4}{135} \int \frac {\log \left (\frac {20}{3}+\frac {2 x}{3}\right )}{\left (\frac {20}{3}+\frac {2 x}{3}\right ) x^2} \, dx+\frac {1}{15} \operatorname {Subst}\left (\int -\frac {2 \log (x)}{3 x \left (-10+\frac {3 x}{2}\right )^2} \, dx,x,\frac {20}{3}+\frac {2 x}{3}\right )\\ &=-\frac {\log ^2\left (\frac {20}{3}+\frac {2 x}{3}\right )}{45 x^2}+\frac {\log (x)}{45 x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 28, normalized size = 1.17 \begin {gather*} \frac {1}{45} \left (-\frac {\log ^2\left (\frac {20}{3}+\frac {2 x}{3}\right )}{x^2}+\frac {\log (x)}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + x + (-20 - 2*x)*Log[x] - 2*x*Log[(20 + 2*x)/3] + (20 + 2*x)*Log[(20 + 2*x)/3]^2)/(450*x^3 + 45
*x^4),x]

[Out]

(-(Log[20/3 + (2*x)/3]^2/x^2) + Log[x]/x^2)/45

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 18, normalized size = 0.75 \begin {gather*} -\frac {\log \left (\frac {2}{3} \, x + \frac {20}{3}\right )^{2} - \log \relax (x)}{45 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-20)*log(x)+(2*x+20)*log(2/3*x+20/3)^2-2*x*log(2/3*x+20/3)+x+10)/(45*x^4+450*x^3),x, algorithm
="fricas")

[Out]

-1/45*(log(2/3*x + 20/3)^2 - log(x))/x^2

________________________________________________________________________________________

giac [B]  time = 2.52, size = 43, normalized size = 1.79 \begin {gather*} -\frac {\log \relax (3)^{2}}{45 \, x^{2}} + \frac {2 \, \log \relax (3) \log \left (2 \, x + 20\right )}{45 \, x^{2}} - \frac {\log \left (2 \, x + 20\right )^{2}}{45 \, x^{2}} + \frac {\log \relax (x)}{45 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-20)*log(x)+(2*x+20)*log(2/3*x+20/3)^2-2*x*log(2/3*x+20/3)+x+10)/(45*x^4+450*x^3),x, algorithm
="giac")

[Out]

-1/45*log(3)^2/x^2 + 2/45*log(3)*log(2*x + 20)/x^2 - 1/45*log(2*x + 20)^2/x^2 + 1/45*log(x)/x^2

________________________________________________________________________________________

maple [A]  time = 0.19, size = 22, normalized size = 0.92

method	result	size
risch	\(-\frac {\ln \left (\frac {2 x}{3}+\frac {20}{3}\right )^{2}}{45 x^{2}}+\frac {\ln \relax (x )}{45 x^{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-20)*ln(x)+(2*x+20)*ln(2/3*x+20/3)^2-2*x*ln(2/3*x+20/3)+x+10)/(45*x^4+450*x^3),x,method=_RETURNVERBO
SE)

[Out]

-1/45/x^2*ln(2/3*x+20/3)^2+1/45*ln(x)/x^2

________________________________________________________________________________________

maxima [B]  time = 0.50, size = 81, normalized size = 3.38 \begin {gather*} -\frac {100 \, \log \relax (3)^{2} - 200 \, \log \relax (3) \log \relax (2) + 100 \, \log \relax (2)^{2} - {\left (x^{2} + 200 \, \log \relax (3) - 200 \, \log \relax (2)\right )} \log \left (x + 10\right ) + 100 \, \log \left (x + 10\right )^{2} + {\left (x^{2} - 100\right )} \log \relax (x) + 10 \, x - 50}{4500 \, x^{2}} + \frac {x - 5}{450 \, x^{2}} - \frac {1}{4500} \, \log \left (x + 10\right ) + \frac {1}{4500} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-20)*log(x)+(2*x+20)*log(2/3*x+20/3)^2-2*x*log(2/3*x+20/3)+x+10)/(45*x^4+450*x^3),x, algorithm
="maxima")

[Out]

-1/4500*(100*log(3)^2 - 200*log(3)*log(2) + 100*log(2)^2 - (x^2 + 200*log(3) - 200*log(2))*log(x + 10) + 100*l
og(x + 10)^2 + (x^2 - 100)*log(x) + 10*x - 50)/x^2 + 1/450*(x - 5)/x^2 - 1/4500*log(x + 10) + 1/4500*log(x)

________________________________________________________________________________________

mupad [B]  time = 3.50, size = 18, normalized size = 0.75 \begin {gather*} \frac {\ln \relax (x)-{\ln \left (\frac {2\,x}{3}+\frac {20}{3}\right )}^2}{45\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(x)*(2*x + 20) - 2*x*log((2*x)/3 + 20/3) + log((2*x)/3 + 20/3)^2*(2*x + 20) + 10)/(450*x^3 + 45*x^
4),x)

[Out]

(log(x) - log((2*x)/3 + 20/3)^2)/(45*x^2)

________________________________________________________________________________________

sympy [A]  time = 0.34, size = 24, normalized size = 1.00 \begin {gather*} \frac {\log {\relax (x )}}{45 x^{2}} - \frac {\log {\left (\frac {2 x}{3} + \frac {20}{3} \right )}^{2}}{45 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-20)*ln(x)+(2*x+20)*ln(2/3*x+20/3)**2-2*x*ln(2/3*x+20/3)+x+10)/(45*x**4+450*x**3),x)

[Out]

log(x)/(45*x**2) - log(2*x/3 + 20/3)**2/(45*x**2)

________________________________________________________________________________________