3.97.11 \(\int \frac {20 x+(50+60 x+16 x^2+(25+40 x+16 x^2) \log (8)) \log (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)})}{50+60 x+16 x^2+(25+40 x+16 x^2) \log (8)} \, dx\)

Optimal. Leaf size=22 \[ x \log \left (\frac {3}{-1-\frac {5}{5+4 x}-\log (8)}\right ) \]

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Rubi [B]  time = 0.18, antiderivative size = 98, normalized size of antiderivative = 4.45, number of steps used = 6, number of rules used = 4, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6688, 72, 2486, 31} \begin {gather*} -\frac {5}{4} \log (4 x+5)+\frac {1}{4} (4 x+5) \log \left (-\frac {3 (4 x+5)}{4 x (1+\log (8))+5 (2+\log (8))}\right )+\frac {5 (2+\log (8)) \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))}-\frac {5 \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*x + (50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8])*Log[(-15 - 12*x)/(10 + 4*x + (5 + 4*x)*Log[8])]
)/(50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8]),x]

[Out]

(-5*Log[5 + 4*x])/4 + ((5 + 4*x)*Log[(-3*(5 + 4*x))/(4*x*(1 + Log[8]) + 5*(2 + Log[8]))])/4 - (5*Log[4*x*(1 +
Log[8]) + 5*(2 + Log[8])])/(4*(1 + Log[8])) + (5*(2 + Log[8])*Log[4*x*(1 + Log[8]) + 5*(2 + Log[8])])/(4*(1 +
Log[8]))

Rule 31

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 2486

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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {20 x}{(5+4 x) (4 x (1+\log (8))+5 (2+\log (8)))}+\log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )\right ) \, dx\\ &=20 \int \frac {x}{(5+4 x) (4 x (1+\log (8))+5 (2+\log (8)))} \, dx+\int \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right ) \, dx\\ &=\frac {1}{4} (5+4 x) \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )-5 \int \frac {1}{4 x (1+\log (8))+5 (2+\log (8))} \, dx+20 \int \left (-\frac {1}{4 (5+4 x)}+\frac {2+\log (8)}{4 (4 x (1+\log (8))+5 (2+\log (8)))}\right ) \, dx\\ &=-\frac {5}{4} \log (5+4 x)+\frac {1}{4} (5+4 x) \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )-\frac {5 \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))}+\frac {5 (2+\log (8)) \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.05, size = 60, normalized size = 2.73 \begin {gather*} \frac {1}{4} \left (-5 \log (5+4 x)+(5+4 x) \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )+5 \log (4 x (1+\log (8))+5 (2+\log (8)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x + (50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8])*Log[(-15 - 12*x)/(10 + 4*x + (5 + 4*x)*Lo
g[8])])/(50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8]),x]

[Out]

(-5*Log[5 + 4*x] + (5 + 4*x)*Log[(-3*(5 + 4*x))/(4*x*(1 + Log[8]) + 5*(2 + Log[8]))] + 5*Log[4*x*(1 + Log[8])
+ 5*(2 + Log[8])])/4

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fricas [A]  time = 0.68, size = 26, normalized size = 1.18 \begin {gather*} x \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{3 \, {\left (4 \, x + 5\right )} \log \relax (2) + 4 \, x + 10}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(4*x+5)*log(2)+4*x+10))+20*x)/(3*(16*x
^2+40*x+25)*log(2)+16*x^2+60*x+50),x, algorithm="fricas")

[Out]

x*log(-3*(4*x + 5)/(3*(4*x + 5)*log(2) + 4*x + 10))

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giac [B]  time = 0.30, size = 149, normalized size = 6.77 \begin {gather*} -\frac {5 \, {\left (3 \, \log \relax (2) + 2\right )} \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10}\right )}{4 \, {\left (3 \, \log \relax (2) + 1\right )}} - \frac {5 \, \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10}\right )}{4 \, {\left (\frac {9 \, {\left (4 \, x + 5\right )} \log \relax (2)^{2}}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10} + \frac {6 \, {\left (4 \, x + 5\right )} \log \relax (2)}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10} + \frac {4 \, x + 5}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10} - 3 \, \log \relax (2) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(4*x+5)*log(2)+4*x+10))+20*x)/(3*(16*x
^2+40*x+25)*log(2)+16*x^2+60*x+50),x, algorithm="giac")

[Out]

-5/4*(3*log(2) + 2)*log(-3*(4*x + 5)/(12*x*log(2) + 4*x + 15*log(2) + 10))/(3*log(2) + 1) - 5/4*log(-3*(4*x +
5)/(12*x*log(2) + 4*x + 15*log(2) + 10))/(9*(4*x + 5)*log(2)^2/(12*x*log(2) + 4*x + 15*log(2) + 10) + 6*(4*x +
5)*log(2)/(12*x*log(2) + 4*x + 15*log(2) + 10) + (4*x + 5)/(12*x*log(2) + 4*x + 15*log(2) + 10) - 3*log(2) -
1)

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maple [A]  time = 0.23, size = 26, normalized size = 1.18

method	result	size
norman	\(x \ln \left (\frac {-12 x -15}{3 \left (4 x +5\right ) \ln \relax (2)+4 x +10}\right )\)	\(26\)
risch	\(x \ln \left (\frac {-12 x -15}{3 \left (4 x +5\right ) \ln \relax (2)+4 x +10}\right )\)	\(26\)
derivativedivides	\(-\frac {15 \left (-\frac {3 \ln \left (3+\left (3 \ln \relax (2)+1\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )\right ) \ln \relax (2)^{2}}{3 \ln \relax (2)+1}-\frac {2 \ln \left (3+\left (3 \ln \relax (2)+1\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )\right ) \ln \relax (2)}{3 \ln \relax (2)+1}-\frac {\ln \left (3+\left (3 \ln \relax (2)+1\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )\right )}{3 \left (3 \ln \relax (2)+1\right )}+\frac {3 \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)^{2}}{3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3}+\frac {2 \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)}{3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3}+\frac {\ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {45}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {9}{3 \ln \relax (2)+1}+9}+\frac {3 \ln \relax (2)^{2} \ln \left (3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3\right )}{3 \ln \relax (2)+1}+3 \ln \relax (2)^{2} \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )+\frac {2 \ln \relax (2) \ln \left (3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3\right )}{3 \ln \relax (2)+1}+2 \ln \relax (2) \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )+\frac {\ln \left (3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3\right )}{9 \ln \relax (2)+3}+\frac {\ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )}{3}\right )}{4 \left (3 \ln \relax (2)+1\right )^{2}}\)	\(1071\)
default	\(-\frac {15 \left (-\frac {3 \ln \left (3+\left (3 \ln \relax (2)+1\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )\right ) \ln \relax (2)^{2}}{3 \ln \relax (2)+1}-\frac {2 \ln \left (3+\left (3 \ln \relax (2)+1\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )\right ) \ln \relax (2)}{3 \ln \relax (2)+1}-\frac {\ln \left (3+\left (3 \ln \relax (2)+1\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )\right )}{3 \left (3 \ln \relax (2)+1\right )}+\frac {3 \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)^{2}}{3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3}+\frac {2 \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)}{3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3}+\frac {\ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {45}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {9}{3 \ln \relax (2)+1}+9}+\frac {3 \ln \relax (2)^{2} \ln \left (3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3\right )}{3 \ln \relax (2)+1}+3 \ln \relax (2)^{2} \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )+\frac {2 \ln \relax (2) \ln \left (3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3\right )}{3 \ln \relax (2)+1}+2 \ln \relax (2) \ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )+\frac {\ln \left (3 \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right ) \ln \relax (2)+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}-\frac {3}{3 \ln \relax (2)+1}+3\right )}{9 \ln \relax (2)+3}+\frac {\ln \left (-\frac {3}{3 \ln \relax (2)+1}+\frac {15}{\left (3 \ln \relax (2)+1\right ) \left (12 x \ln \relax (2)+15 \ln \relax (2)+4 x +10\right )}\right )}{3}\right )}{4 \left (3 \ln \relax (2)+1\right )^{2}}\)	\(1071\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*(16*x^2+40*x+25)*ln(2)+16*x^2+60*x+50)*ln((-12*x-15)/(3*(4*x+5)*ln(2)+4*x+10))+20*x)/(3*(16*x^2+40*x+2
5)*ln(2)+16*x^2+60*x+50),x,method=_RETURNVERBOSE)

[Out]

x*ln((-12*x-15)/(3*(4*x+5)*ln(2)+4*x+10))

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maxima [C]  time = 0.53, size = 512, normalized size = 23.27 \begin {gather*} -\frac {15}{4} \, {\left (\log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right ) - \log \left (4 \, x + 5\right )\right )} \log \relax (2) \log \left (-\frac {12 \, x}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10} - \frac {15}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10}\right ) - \frac {15}{8} \, {\left (\log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right )^{2} - 2 \, \log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right ) \log \left (4 \, x + 5\right ) + \log \left (4 \, x + 5\right )^{2}\right )} \log \relax (2) - \frac {5}{4} \, \log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right )^{2} + \frac {5}{2} \, \log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right ) \log \left (4 \, x + 5\right ) - \frac {5}{4} \, \log \left (4 \, x + 5\right )^{2} - \frac {5}{2} \, {\left (\log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right ) - \log \left (4 \, x + 5\right )\right )} \log \left (-\frac {12 \, x}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10} - \frac {15}{12 \, x \log \relax (2) + 4 \, x + 15 \, \log \relax (2) + 10}\right ) + \frac {5 \, {\left (3 \, \log \relax (2) + 2\right )} \log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right )}{4 \, {\left (3 \, \log \relax (2) + 1\right )}} - \frac {5 \, {\left (9 \, \log \relax (2)^{2} + 9 \, \log \relax (2) + 2\right )} \log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right )^{2} + 5 \, {\left (9 \, \log \relax (2)^{2} + 9 \, \log \relax (2) + 2\right )} \log \left (4 \, x + 5\right )^{2} - 8 \, {\left (i \, \pi {\left (3 \, \log \relax (2) + 1\right )} + 3 \, \log \relax (3) \log \relax (2) + \log \relax (3)\right )} x - 2 \, {\left (45 \, \log \relax (3) \log \relax (2)^{2} + 5 i \, \pi {\left (9 \, \log \relax (2)^{2} + 9 \, \log \relax (2) + 2\right )} - 4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, {\left (3 \, \log \relax (3) - 1\right )} \log \relax (2) + 5 \, {\left (9 \, \log \relax (2)^{2} + 9 \, \log \relax (2) + 2\right )} \log \left (4 \, x + 5\right ) + 10 \, \log \relax (3) - 10\right )} \log \left (4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, \log \relax (2) + 10\right ) + 2 \, {\left (45 \, \log \relax (3) \log \relax (2)^{2} + 5 i \, \pi {\left (9 \, \log \relax (2)^{2} + 9 \, \log \relax (2) + 2\right )} - 4 \, x {\left (3 \, \log \relax (2) + 1\right )} + 15 \, {\left (3 \, \log \relax (3) - 1\right )} \log \relax (2) + 10 \, \log \relax (3) - 5\right )} \log \left (4 \, x + 5\right )}{8 \, {\left (3 \, \log \relax (2) + 1\right )}} - \frac {5}{4} \, \log \left (4 \, x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(4*x+5)*log(2)+4*x+10))+20*x)/(3*(16*x
^2+40*x+25)*log(2)+16*x^2+60*x+50),x, algorithm="maxima")

[Out]

-15/4*(log(4*x*(3*log(2) + 1) + 15*log(2) + 10) - log(4*x + 5))*log(2)*log(-12*x/(12*x*log(2) + 4*x + 15*log(2
) + 10) - 15/(12*x*log(2) + 4*x + 15*log(2) + 10)) - 15/8*(log(4*x*(3*log(2) + 1) + 15*log(2) + 10)^2 - 2*log(
4*x*(3*log(2) + 1) + 15*log(2) + 10)*log(4*x + 5) + log(4*x + 5)^2)*log(2) - 5/4*log(4*x*(3*log(2) + 1) + 15*l
og(2) + 10)^2 + 5/2*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)*log(4*x + 5) - 5/4*log(4*x + 5)^2 - 5/2*(log(4*x*
(3*log(2) + 1) + 15*log(2) + 10) - log(4*x + 5))*log(-12*x/(12*x*log(2) + 4*x + 15*log(2) + 10) - 15/(12*x*log
(2) + 4*x + 15*log(2) + 10)) + 5/4*(3*log(2) + 2)*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)/(3*log(2) + 1) - 1/
8*(5*(9*log(2)^2 + 9*log(2) + 2)*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)^2 + 5*(9*log(2)^2 + 9*log(2) + 2)*lo
g(4*x + 5)^2 - 8*(I*pi*(3*log(2) + 1) + 3*log(3)*log(2) + log(3))*x - 2*(45*log(3)*log(2)^2 + 5*I*pi*(9*log(2)
^2 + 9*log(2) + 2) - 4*x*(3*log(2) + 1) + 15*(3*log(3) - 1)*log(2) + 5*(9*log(2)^2 + 9*log(2) + 2)*log(4*x + 5
) + 10*log(3) - 10)*log(4*x*(3*log(2) + 1) + 15*log(2) + 10) + 2*(45*log(3)*log(2)^2 + 5*I*pi*(9*log(2)^2 + 9*
log(2) + 2) - 4*x*(3*log(2) + 1) + 15*(3*log(3) - 1)*log(2) + 10*log(3) - 5)*log(4*x + 5))/(3*log(2) + 1) - 5/
4*log(4*x + 5)

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mupad [B]  time = 7.01, size = 26, normalized size = 1.18 \begin {gather*} x\,\ln \left (-\frac {12\,x+15}{4\,x+3\,\ln \relax (2)\,\left (4\,x+5\right )+10}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x + log(-(12*x + 15)/(4*x + 3*log(2)*(4*x + 5) + 10))*(60*x + 3*log(2)*(40*x + 16*x^2 + 25) + 16*x^2 +
50))/(60*x + 3*log(2)*(40*x + 16*x^2 + 25) + 16*x^2 + 50),x)

[Out]

x*log(-(12*x + 15)/(4*x + 3*log(2)*(4*x + 5) + 10))

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sympy [A]  time = 0.22, size = 22, normalized size = 1.00 \begin {gather*} x \log {\left (\frac {- 12 x - 15}{4 x + \left (12 x + 15\right ) \log {\relax (2 )} + 10} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(16*x**2+40*x+25)*ln(2)+16*x**2+60*x+50)*ln((-12*x-15)/(3*(4*x+5)*ln(2)+4*x+10))+20*x)/(3*(16*x*
*2+40*x+25)*ln(2)+16*x**2+60*x+50),x)

[Out]

x*log((-12*x - 15)/(4*x + (12*x + 15)*log(2) + 10))

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