3.26.22 \(\int \frac {-350 x+5 x^2+(-8-36 x+20 x^2) \log (2)+(8+36 x-20 x^2) \log (6-3 x)-4 x \log (x)}{-50 x+25 x^2} \, dx\)

Optimal. Leaf size=27 \[ x-\frac {4}{25} (-\log (2)+\log (3 (2-x))) (5 (15+x)+\log (x)) \]

________________________________________________________________________________________

Rubi [C]  time = 0.71, antiderivative size = 90, normalized size of antiderivative = 3.33, number of steps used = 15, number of rules used = 11, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.196, Rules used = {1593, 6742, 893, 43, 2416, 2389, 2295, 2392, 2391, 2316, 2315} \begin {gather*} \frac {4}{25} \text {Li}_2\left (1-\frac {x}{2}\right )+\frac {4 \text {Li}_2\left (\frac {x}{2}\right )}{25}+\frac {4 x}{5}+\frac {1}{5} x (1+\log (16))+\frac {4}{5} (2-x) \log (6-3 x)-\frac {68}{5} \log (2-x)-\frac {4}{25} \log (2) \log (x-2)+\frac {1}{25} \log (16) \log (x)-\frac {4}{25} \log (6) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-350*x + 5*x^2 + (-8 - 36*x + 20*x^2)*Log[2] + (8 + 36*x - 20*x^2)*Log[6 - 3*x] - 4*x*Log[x])/(-50*x + 25
*x^2),x]

[Out]

(4*x)/5 + (x*(1 + Log[16]))/5 + (4*(2 - x)*Log[6 - 3*x])/5 - (68*Log[2 - x])/5 - (4*Log[2]*Log[-2 + x])/25 - (
4*Log[6]*Log[x])/25 + (Log[16]*Log[x])/25 + (4*PolyLog[2, 1 - x/2])/25 + (4*PolyLog[2, x/2])/25

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2389

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

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 2392

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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 {-350 x+5 x^2+\left (-8-36 x+20 x^2\right ) \log (2)+\left (8+36 x-20 x^2\right ) \log (6-3 x)-4 x \log (x)}{x (-50+25 x)} \, dx\\ &=\int \left (\frac {350 x \left (1+\frac {18 \log (2)}{175}\right )+8 \log (2)-5 x^2 (1+\log (16))-8 \log (6-3 x)-36 x \log (6-3 x)+20 x^2 \log (6-3 x)}{25 (2-x) x}-\frac {4 \log (x)}{25 (-2+x)}\right ) \, dx\\ &=\frac {1}{25} \int \frac {350 x \left (1+\frac {18 \log (2)}{175}\right )+8 \log (2)-5 x^2 (1+\log (16))-8 \log (6-3 x)-36 x \log (6-3 x)+20 x^2 \log (6-3 x)}{(2-x) x} \, dx-\frac {4}{25} \int \frac {\log (x)}{-2+x} \, dx\\ &=-\frac {4}{25} \log (2) \log (-2+x)+\frac {1}{25} \int \left (\frac {8 \log (2)+2 x (175+18 \log (2))-5 x^2 (1+\log (16))}{(2-x) x}-\frac {4 (1+5 x) \log (6-3 x)}{x}\right ) \, dx-\frac {4}{25} \int \frac {\log \left (\frac {x}{2}\right )}{-2+x} \, dx\\ &=-\frac {4}{25} \log (2) \log (-2+x)+\frac {4}{25} \text {Li}_2\left (1-\frac {x}{2}\right )+\frac {1}{25} \int \frac {8 \log (2)+2 x (175+18 \log (2))-5 x^2 (1+\log (16))}{(2-x) x} \, dx-\frac {4}{25} \int \frac {(1+5 x) \log (6-3 x)}{x} \, dx\\ &=-\frac {4}{25} \log (2) \log (-2+x)+\frac {4}{25} \text {Li}_2\left (1-\frac {x}{2}\right )+\frac {1}{25} \int \left (-\frac {340}{-2+x}+\frac {\log (16)}{x}+5 (1+\log (16))\right ) \, dx-\frac {4}{25} \int \left (5 \log (6-3 x)+\frac {\log (6-3 x)}{x}\right ) \, dx\\ &=\frac {1}{5} x (1+\log (16))-\frac {68}{5} \log (2-x)-\frac {4}{25} \log (2) \log (-2+x)+\frac {1}{25} \log (16) \log (x)+\frac {4}{25} \text {Li}_2\left (1-\frac {x}{2}\right )-\frac {4}{25} \int \frac {\log (6-3 x)}{x} \, dx-\frac {4}{5} \int \log (6-3 x) \, dx\\ &=\frac {1}{5} x (1+\log (16))-\frac {68}{5} \log (2-x)-\frac {4}{25} \log (2) \log (-2+x)-\frac {4}{25} \log (6) \log (x)+\frac {1}{25} \log (16) \log (x)+\frac {4}{25} \text {Li}_2\left (1-\frac {x}{2}\right )-\frac {4}{25} \int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx+\frac {4}{15} \operatorname {Subst}(\int \log (x) \, dx,x,6-3 x)\\ &=\frac {4 x}{5}+\frac {1}{5} x (1+\log (16))+\frac {4}{5} (2-x) \log (6-3 x)-\frac {68}{5} \log (2-x)-\frac {4}{25} \log (2) \log (-2+x)-\frac {4}{25} \log (6) \log (x)+\frac {1}{25} \log (16) \log (x)+\frac {4}{25} \text {Li}_2\left (1-\frac {x}{2}\right )+\frac {4 \text {Li}_2\left (\frac {x}{2}\right )}{25}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [C]  time = 0.09, size = 62, normalized size = 2.30 \begin {gather*} \frac {1}{25} \left (20 x+5 x (1+\log (16))-20 (-2+x) \log (6-3 x)-4 (85+\log (2)) \log (-2+x)-4 \log (3) \log (x)+4 \text {Li}_2\left (1-\frac {x}{2}\right )+4 \text {Li}_2\left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-350*x + 5*x^2 + (-8 - 36*x + 20*x^2)*Log[2] + (8 + 36*x - 20*x^2)*Log[6 - 3*x] - 4*x*Log[x])/(-50*
x + 25*x^2),x]

[Out]

(20*x + 5*x*(1 + Log[16]) - 20*(-2 + x)*Log[6 - 3*x] - 4*(85 + Log[2])*Log[-2 + x] - 4*Log[3]*Log[x] + 4*PolyL
og[2, 1 - x/2] + 4*PolyLog[2, x/2])/25

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 33, normalized size = 1.22 \begin {gather*} \frac {4}{5} \, x \log \relax (2) + \frac {4}{25} \, {\left (\log \relax (2) - \log \left (-3 \, x + 6\right )\right )} \log \relax (x) - \frac {4}{5} \, {\left (x + 15\right )} \log \left (-3 \, x + 6\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(x)+(-20*x^2+36*x+8)*log(-3*x+6)+(20*x^2-36*x-8)*log(2)+5*x^2-350*x)/(25*x^2-50*x),x, algor
ithm="fricas")

[Out]

4/5*x*log(2) + 4/25*(log(2) - log(-3*x + 6))*log(x) - 4/5*(x + 15)*log(-3*x + 6) + x

________________________________________________________________________________________

giac [A]  time = 0.23, size = 36, normalized size = 1.33 \begin {gather*} \frac {1}{5} \, x {\left (4 \, \log \relax (2) + 5\right )} + \frac {4}{25} \, \log \relax (2) \log \relax (x) - \frac {4}{25} \, {\left (5 \, x + \log \relax (x)\right )} \log \left (-3 \, x + 6\right ) - 12 \, \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(x)+(-20*x^2+36*x+8)*log(-3*x+6)+(20*x^2-36*x-8)*log(2)+5*x^2-350*x)/(25*x^2-50*x),x, algor
ithm="giac")

[Out]

1/5*x*(4*log(2) + 5) + 4/25*log(2)*log(x) - 4/25*(5*x + log(x))*log(-3*x + 6) - 12*log(x - 2)

________________________________________________________________________________________

maple [A]  time = 0.40, size = 37, normalized size = 1.37

method	result	size
risch	\(\left (-\frac {4 x}{5}-\frac {4 \ln \relax (x )}{25}\right ) \ln \left (-3 x +6\right )+\frac {4 x \ln \relax (2)}{5}+x -12 \ln \left (x -2\right )+\frac {4 \ln \relax (2) \ln \left (-x \right )}{25}\)	\(37\)
norman	\(\left (1+\frac {4 \ln \relax (2)}{5}\right ) x +\frac {4 \ln \relax (2) \ln \relax (x )}{25}-12 \ln \left (-3 x +6\right )-\frac {4 x \ln \left (-3 x +6\right )}{5}-\frac {4 \ln \relax (x ) \ln \left (-3 x +6\right )}{25}\)	\(43\)
default	\(x -\frac {68 \ln \left (x -2\right )}{5}-\frac {4 \left (\ln \relax (x )-\ln \left (\frac {x}{2}\right )\right ) \ln \left (1-\frac {x}{2}\right )}{25}+\frac {4 \dilog \left (\frac {x}{2}\right )}{25}+\frac {4 \ln \relax (2) \ln \relax (x )}{25}+\frac {4 x \ln \relax (2)}{5}-\frac {4 \ln \relax (3) \ln \relax (x )}{25}-\frac {4 x \ln \relax (3)}{5}+\frac {4 \left (2-x \right ) \ln \left (2-x \right )}{5}-\frac {8}{5}-\frac {4 \left (\ln \left (2-x \right )-\ln \left (1-\frac {x}{2}\right )\right ) \ln \left (\frac {x}{2}\right )}{25}+\frac {4 \dilog \left (1-\frac {x}{2}\right )}{25}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x*ln(x)+(-20*x^2+36*x+8)*ln(-3*x+6)+(20*x^2-36*x-8)*ln(2)+5*x^2-350*x)/(25*x^2-50*x),x,method=_RETURNV
ERBOSE)

[Out]

(-4/5*x-4/25*ln(x))*ln(-3*x+6)+4/5*x*ln(2)+x-12*ln(x-2)+4/25*ln(2)*ln(-x)

________________________________________________________________________________________

maxima [C]  time = 0.56, size = 57, normalized size = 2.11 \begin {gather*} -\frac {1}{5} \, {\left (4 i \, \pi + 4 \, \log \relax (3) - 4 \, \log \relax (2) - 5\right )} x - \frac {4}{25} \, {\left (\log \left (x - 2\right ) - \log \relax (x)\right )} \log \relax (2) - \frac {4}{25} \, {\left (5 \, x - \log \relax (2) + \log \relax (x) + 75\right )} \log \left (x - 2\right ) - \frac {4}{25} \, {\left (i \, \pi + \log \relax (3)\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(x)+(-20*x^2+36*x+8)*log(-3*x+6)+(20*x^2-36*x-8)*log(2)+5*x^2-350*x)/(25*x^2-50*x),x, algor
ithm="maxima")

[Out]

-1/5*(4*I*pi + 4*log(3) - 4*log(2) - 5)*x - 4/25*(log(x - 2) - log(x))*log(2) - 4/25*(5*x - log(2) + log(x) +
75)*log(x - 2) - 4/25*(I*pi + log(3))*log(x)

________________________________________________________________________________________

mupad [B]  time = 1.70, size = 38, normalized size = 1.41 \begin {gather*} x-12\,\ln \left (x-2\right )+\frac {4\,x\,\ln \relax (2)}{5}-\frac {4\,x\,\ln \left (6-3\,x\right )}{5}+\frac {4\,\ln \relax (2)\,\ln \relax (x)}{25}-\frac {4\,\ln \left (6-3\,x\right )\,\ln \relax (x)}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((350*x + log(2)*(36*x - 20*x^2 + 8) - log(6 - 3*x)*(36*x - 20*x^2 + 8) + 4*x*log(x) - 5*x^2)/(50*x - 25*x^
2),x)

[Out]

x - 12*log(x - 2) + (4*x*log(2))/5 - (4*x*log(6 - 3*x))/5 + (4*log(2)*log(x))/25 - (4*log(6 - 3*x)*log(x))/25

________________________________________________________________________________________

sympy [B]  time = 0.72, size = 60, normalized size = 2.22 \begin {gather*} x \left (\frac {4 \log {\relax (2 )}}{5} + 1\right ) + \left (- \frac {4 x}{5} - \frac {4 \log {\relax (x )}}{25}\right ) \log {\left (6 - 3 x \right )} + \frac {4 \log {\relax (2 )} \log {\relax (x )}}{25} - 12 \log {\left (x + \frac {-300 - 4 \log {\relax (2 )}}{2 \log {\relax (2 )} + 150} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*ln(x)+(-20*x**2+36*x+8)*ln(-3*x+6)+(20*x**2-36*x-8)*ln(2)+5*x**2-350*x)/(25*x**2-50*x),x)

[Out]

x*(4*log(2)/5 + 1) + (-4*x/5 - 4*log(x)/25)*log(6 - 3*x) + 4*log(2)*log(x)/25 - 12*log(x + (-300 - 4*log(2))/(
2*log(2) + 150))

________________________________________________________________________________________