3.10.76 \(\int \frac {(5-26 x+10 x^2-x^3+(-3 x+30 x^2-6 x^3) \log (2)-9 x^3 \log ^2(2)) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+(-30 x^2+6 x^3) \log (2)+9 x^3 \log ^2(2)} \, dx\)

Optimal. Leaf size=27 \[ 2+\log (3) \left (1-x+\frac {\log (20 x)}{5-x-3 x \log (2)}\right ) \]

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Rubi [B]  time = 0.89, antiderivative size = 100, normalized size of antiderivative = 3.70, number of steps used = 13, number of rules used = 8, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6, 6688, 12, 6742, 44, 43, 2314, 31} \begin {gather*} \frac {x \log (3) (1+\log (8)) \log (20 x)}{5 (5-x (1+\log (8)))}-x \log (3)+\frac {1}{5} \log (3) \log (x)-\frac {\log (3) (26+\log (8))}{(1+\log (8)) (5-x (1+\log (8)))}+\frac {25 \log (3)}{(1+\log (8)) (5-x (1+\log (8)))}+\frac {\log (3)}{5-x (1+\log (8))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((5 - 26*x + 10*x^2 - x^3 + (-3*x + 30*x^2 - 6*x^3)*Log[2] - 9*x^3*Log[2]^2)*Log[3] + (x + 3*x*Log[2])*Log
[3]*Log[20*x])/(25*x - 10*x^2 + x^3 + (-30*x^2 + 6*x^3)*Log[2] + 9*x^3*Log[2]^2),x]

[Out]

-(x*Log[3]) + Log[3]/(5 - x*(1 + Log[8])) + (25*Log[3])/((1 + Log[8])*(5 - x*(1 + Log[8]))) - (Log[3]*(26 + Lo
g[8]))/((1 + Log[8])*(5 - x*(1 + Log[8]))) + (Log[3]*Log[x])/5 + (x*Log[3]*(1 + Log[8])*Log[20*x])/(5*(5 - x*(
1 + Log[8])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 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 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 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 {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+\left (-30 x^2+6 x^3\right ) \log (2)+x^3 \left (1+9 \log ^2(2)\right )} \, dx\\ &=\int \frac {\log (3) \left (5+10 x^2 (1+\log (8))-x^3 (1+\log (8))^2-x (26+\log (8))+x (1+\log (8)) \log (20 x)\right )}{x (5-x (1+\log (8)))^2} \, dx\\ &=\log (3) \int \frac {5+10 x^2 (1+\log (8))-x^3 (1+\log (8))^2-x (26+\log (8))+x (1+\log (8)) \log (20 x)}{x (5-x (1+\log (8)))^2} \, dx\\ &=\log (3) \int \left (\frac {5}{x (5-x (1+\log (8)))^2}+\frac {-26-\log (8)}{(5-x (1+\log (8)))^2}+\frac {10 x (1+\log (8))}{(5-x (1+\log (8)))^2}-\frac {x^2 (1+\log (8))^2}{(5-x (1+\log (8)))^2}+\frac {(1+\log (8)) \log (20 x)}{(5-x (1+\log (8)))^2}\right ) \, dx\\ &=-\frac {\log (3) (26+\log (8))}{(1+\log (8)) (5-x (1+\log (8)))}+(5 \log (3)) \int \frac {1}{x (5-x (1+\log (8)))^2} \, dx+(\log (3) (1+\log (8))) \int \frac {\log (20 x)}{(5+x (-1-\log (8)))^2} \, dx+(10 \log (3) (1+\log (8))) \int \frac {x}{(5-x (1+\log (8)))^2} \, dx-\left (\log (3) (1+\log (8))^2\right ) \int \frac {x^2}{(5-x (1+\log (8)))^2} \, dx\\ &=-\frac {\log (3) (26+\log (8))}{(1+\log (8)) (5-x (1+\log (8)))}+\frac {x \log (3) (1+\log (8)) \log (20 x)}{5 (5-x (1+\log (8)))}+(5 \log (3)) \int \left (\frac {1}{25 x}+\frac {1+\log (8)}{5 (5-x (1+\log (8)))^2}+\frac {1+\log (8)}{25 (5-x (1+\log (8)))}\right ) \, dx-\frac {1}{5} (\log (3) (1+\log (8))) \int \frac {1}{5+x (-1-\log (8))} \, dx+(10 \log (3) (1+\log (8))) \int \left (\frac {5}{(1+\log (8)) (5-x (1+\log (8)))^2}+\frac {1}{(-1-\log (8)) (5-x (1+\log (8)))}\right ) \, dx-\left (\log (3) (1+\log (8))^2\right ) \int \left (\frac {1}{(1+\log (8))^2}+\frac {25}{(1+\log (8))^2 (5-x (1+\log (8)))^2}+\frac {10}{(1+\log (8))^2 (-5+x (1+\log (8)))}\right ) \, dx\\ &=-x \log (3)+\frac {\log (3)}{5-x (1+\log (8))}+\frac {25 \log (3)}{(1+\log (8)) (5-x (1+\log (8)))}-\frac {\log (3) (26+\log (8))}{(1+\log (8)) (5-x (1+\log (8)))}+\frac {1}{5} \log (3) \log (x)+\frac {x \log (3) (1+\log (8)) \log (20 x)}{5 (5-x (1+\log (8)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 23, normalized size = 0.85 \begin {gather*} \log (3) \left (-x+\frac {\log (20 x)}{5-x (1+\log (8))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((5 - 26*x + 10*x^2 - x^3 + (-3*x + 30*x^2 - 6*x^3)*Log[2] - 9*x^3*Log[2]^2)*Log[3] + (x + 3*x*Log[2
])*Log[3]*Log[20*x])/(25*x - 10*x^2 + x^3 + (-30*x^2 + 6*x^3)*Log[2] + 9*x^3*Log[2]^2),x]

[Out]

Log[3]*(-x + Log[20*x]/(5 - x*(1 + Log[8])))

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fricas [A]  time = 0.85, size = 37, normalized size = 1.37 \begin {gather*} -\frac {{\left (3 \, x^{2} \log \relax (2) + x^{2} - 5 \, x\right )} \log \relax (3) + \log \relax (3) \log \left (20 \, x\right )}{3 \, x \log \relax (2) + x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x*log(2)+x)*log(3)*log(20*x)+(-9*x^3*log(2)^2+(-6*x^3+30*x^2-3*x)*log(2)-x^3+10*x^2-26*x+5)*log(
3))/(9*x^3*log(2)^2+(6*x^3-30*x^2)*log(2)+x^3-10*x^2+25*x),x, algorithm="fricas")

[Out]

-((3*x^2*log(2) + x^2 - 5*x)*log(3) + log(3)*log(20*x))/(3*x*log(2) + x - 5)

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giac [A]  time = 0.32, size = 40, normalized size = 1.48 \begin {gather*} -x \log \relax (3) - \frac {2 \, \log \relax (3) \log \relax (2)}{3 \, x \log \relax (2) + x - 5} - \frac {\log \relax (3) \log \left (5 \, x\right )}{3 \, x \log \relax (2) + x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x*log(2)+x)*log(3)*log(20*x)+(-9*x^3*log(2)^2+(-6*x^3+30*x^2-3*x)*log(2)-x^3+10*x^2-26*x+5)*log(
3))/(9*x^3*log(2)^2+(6*x^3-30*x^2)*log(2)+x^3-10*x^2+25*x),x, algorithm="giac")

[Out]

-x*log(3) - 2*log(3)*log(2)/(3*x*log(2) + x - 5) - log(3)*log(5*x)/(3*x*log(2) + x - 5)

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maple [A]  time = 0.11, size = 25, normalized size = 0.93

method	result	size
risch	\(-\frac {\ln \relax (3) \ln \left (20 x \right )}{3 x \ln \relax (2)+x -5}-x \ln \relax (3)\)	\(25\)
norman	\(\frac {-\ln \relax (3) \ln \left (20 x \right )+\left (-3 \ln \relax (2) \ln \relax (3)-\ln \relax (3)\right ) x^{2}+5 x \ln \relax (3)}{3 x \ln \relax (2)+x -5}\)	\(41\)
derivativedivides	\(\frac {\ln \relax (3) \left (\frac {12 \ln \relax (2) \ln \left (-100+20 \left (3 \ln \relax (2)+1\right ) x \right )}{3 \ln \relax (2)+1}-\frac {240 \ln \relax (2) \ln \left (20 x \right ) x}{60 x \ln \relax (2)+20 x -100}+\frac {4 \ln \left (-100+20 \left (3 \ln \relax (2)+1\right ) x \right )}{3 \ln \relax (2)+1}-\frac {80 \ln \left (20 x \right ) x}{60 x \ln \relax (2)+20 x -100}-20 x -\frac {12 \ln \left (60 x \ln \relax (2)+20 x -100\right ) \ln \relax (2)}{3 \ln \relax (2)+1}-\frac {4 \ln \left (60 x \ln \relax (2)+20 x -100\right )}{3 \ln \relax (2)+1}+4 \ln \left (20 x \right )\right )}{20}\)	\(145\)
default	\(\frac {\ln \relax (3) \left (\frac {12 \ln \relax (2) \ln \left (-100+20 \left (3 \ln \relax (2)+1\right ) x \right )}{3 \ln \relax (2)+1}-\frac {240 \ln \relax (2) \ln \left (20 x \right ) x}{60 x \ln \relax (2)+20 x -100}+\frac {4 \ln \left (-100+20 \left (3 \ln \relax (2)+1\right ) x \right )}{3 \ln \relax (2)+1}-\frac {80 \ln \left (20 x \right ) x}{60 x \ln \relax (2)+20 x -100}-20 x -\frac {12 \ln \left (60 x \ln \relax (2)+20 x -100\right ) \ln \relax (2)}{3 \ln \relax (2)+1}-\frac {4 \ln \left (60 x \ln \relax (2)+20 x -100\right )}{3 \ln \relax (2)+1}+4 \ln \left (20 x \right )\right )}{20}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x*ln(2)+x)*ln(3)*ln(20*x)+(-9*x^3*ln(2)^2+(-6*x^3+30*x^2-3*x)*ln(2)-x^3+10*x^2-26*x+5)*ln(3))/(9*x^3*l
n(2)^2+(6*x^3-30*x^2)*ln(2)+x^3-10*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

-ln(3)/(3*x*ln(2)+x-5)*ln(20*x)-x*ln(3)

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maxima [B]  time = 0.39, size = 678, normalized size = 25.11 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x*log(2)+x)*log(3)*log(20*x)+(-9*x^3*log(2)^2+(-6*x^3+30*x^2-3*x)*log(2)-x^3+10*x^2-26*x+5)*log(
3))/(9*x^3*log(2)^2+(6*x^3-30*x^2)*log(2)+x^3-10*x^2+25*x),x, algorithm="maxima")

[Out]

-9*(x/(9*log(2)^2 + 6*log(2) + 1) + 10*log(x*(3*log(2) + 1) - 5)/(27*log(2)^3 + 27*log(2)^2 + 9*log(2) + 1) +
25/(135*log(2)^3 - (81*log(2)^4 + 108*log(2)^3 + 54*log(2)^2 + 12*log(2) + 1)*x + 135*log(2)^2 + 45*log(2) + 5
))*log(3)*log(2)^2 - 6*(x/(9*log(2)^2 + 6*log(2) + 1) + 10*log(x*(3*log(2) + 1) - 5)/(27*log(2)^3 + 27*log(2)^
2 + 9*log(2) + 1) + 25/(135*log(2)^3 - (81*log(2)^4 + 108*log(2)^3 + 54*log(2)^2 + 12*log(2) + 1)*x + 135*log(
2)^2 + 45*log(2) + 5))*log(3)*log(2) + 30*(log(x*(3*log(2) + 1) - 5)/(9*log(2)^2 + 6*log(2) + 1) - 5/((27*log(
2)^3 + 27*log(2)^2 + 9*log(2) + 1)*x - 45*log(2)^2 - 30*log(2) - 5))*log(3)*log(2) + 3/5*(log(x*(3*log(2) + 1)
- 5)/(3*log(2) + 1) - log(x)/(3*log(2) + 1))*log(3)*log(2) - (x/(9*log(2)^2 + 6*log(2) + 1) + 10*log(x*(3*log
(2) + 1) - 5)/(27*log(2)^3 + 27*log(2)^2 + 9*log(2) + 1) + 25/(135*log(2)^3 - (81*log(2)^4 + 108*log(2)^3 + 54
*log(2)^2 + 12*log(2) + 1)*x + 135*log(2)^2 + 45*log(2) + 5))*log(3) + 10*(log(x*(3*log(2) + 1) - 5)/(9*log(2)
^2 + 6*log(2) + 1) - 5/((27*log(2)^3 + 27*log(2)^2 + 9*log(2) + 1)*x - 45*log(2)^2 - 30*log(2) - 5))*log(3) +
1/5*(log(x*(3*log(2) + 1) - 5)/(3*log(2) + 1) - log(x)/(3*log(2) + 1))*log(3) - 1/5*(5/(x*(3*log(2) + 1) - 5)
+ log(x*(3*log(2) + 1) - 5) - log(x))*log(3) - 3*log(3)*log(2)*log(20*x)/((9*log(2)^2 + 6*log(2) + 1)*x - 15*l
og(2) - 5) + 3*log(3)*log(2)/((9*log(2)^2 + 6*log(2) + 1)*x - 15*log(2) - 5) - log(3)*log(20*x)/((9*log(2)^2 +
6*log(2) + 1)*x - 15*log(2) - 5) + 26*log(3)/((9*log(2)^2 + 6*log(2) + 1)*x - 15*log(2) - 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {\ln \relax (3)\,\left (26\,x+9\,x^3\,{\ln \relax (2)}^2+\ln \relax (2)\,\left (6\,x^3-30\,x^2+3\,x\right )-10\,x^2+x^3-5\right )-\ln \left (20\,x\right )\,\ln \relax (3)\,\left (x+3\,x\,\ln \relax (2)\right )}{25\,x+9\,x^3\,{\ln \relax (2)}^2-\ln \relax (2)\,\left (30\,x^2-6\,x^3\right )-10\,x^2+x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(26*x + 9*x^3*log(2)^2 + log(2)*(3*x - 30*x^2 + 6*x^3) - 10*x^2 + x^3 - 5) - log(20*x)*log(3)*(x
+ 3*x*log(2)))/(25*x + 9*x^3*log(2)^2 - log(2)*(30*x^2 - 6*x^3) - 10*x^2 + x^3),x)

[Out]

-int((log(3)*(26*x + 9*x^3*log(2)^2 + log(2)*(3*x - 30*x^2 + 6*x^3) - 10*x^2 + x^3 - 5) - log(20*x)*log(3)*(x
+ 3*x*log(2)))/(25*x + 9*x^3*log(2)^2 - log(2)*(30*x^2 - 6*x^3) - 10*x^2 + x^3), x)

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sympy [A]  time = 0.24, size = 24, normalized size = 0.89 \begin {gather*} - x \log {\relax (3 )} - \frac {\log {\relax (3 )} \log {\left (20 x \right )}}{x + 3 x \log {\relax (2 )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x*ln(2)+x)*ln(3)*ln(20*x)+(-9*x**3*ln(2)**2+(-6*x**3+30*x**2-3*x)*ln(2)-x**3+10*x**2-26*x+5)*ln(
3))/(9*x**3*ln(2)**2+(6*x**3-30*x**2)*ln(2)+x**3-10*x**2+25*x),x)

[Out]

-x*log(3) - log(3)*log(20*x)/(x + 3*x*log(2) - 5)

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