\(\int \frac {-36+360 x-900 x^2+(-18+192 x-570 x^2+300 x^3) \log (2 x)+(5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)) \log ^3(2 x)}{(x^2-10 x^3+25 x^4) \log ^3(2 x)} \, dx\) [10321]

Optimal result

Integrand size = 88, antiderivative size = 34 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=1-\frac {\log (x)}{-\frac {x}{5}+x^2}+\frac {2 \left (x-\frac {3}{\log (2 x)}\right )^2}{x} \]

[Out]

2/x*(x-3/ln(2*x))^2-ln(x)/(x^2-1/5*x)+1

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35, number of steps used = 31, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.193, Rules used = {1608, 27, 6820, 14, 6874, 46, 45, 2404, 2341, 2351, 31, 2343, 2346, 2209, 2395, 2339, 30} \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2 x+\frac {18}{x \log ^2(2 x)}+\frac {125 x \log (x)}{1-5 x}+25 \log (x)-\frac {12}{\log (2 x)}+\frac {5 \log (x)}{x} \]

[In]

Int[(-36 + 360*x - 900*x^2 + (-18 + 192*x - 570*x^2 + 300*x^3)*Log[2*x] + (5 - 25*x + 2*x^2 - 20*x^3 + 50*x^4
+ (-5 + 50*x)*Log[x])*Log[2*x]^3)/((x^2 - 10*x^3 + 25*x^4)*Log[2*x]^3),x]

[Out]

2*x + 25*Log[x] + (5*Log[x])/x + (125*x*Log[x])/(1 - 5*x) + 18/(x*Log[2*x]^2) - 12/Log[2*x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 30

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

Rule 31

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

Rule 45

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 1608

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

Rule 2209

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

Rule 2339

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 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2343

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 2346

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 2351

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 2395

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 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{x^2 \left (1-10 x+25 x^2\right ) \log ^3(2 x)} \, dx \\ & = \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{x^2 (-1+5 x)^2 \log ^3(2 x)} \, dx \\ & = \int \frac {\frac {5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)}{(1-5 x)^2}-\frac {36}{\log ^3(2 x)}+\frac {6 (-3+2 x)}{\log ^2(2 x)}}{x^2} \, dx \\ & = \int \left (\frac {5-25 x+2 x^2-20 x^3+50 x^4-5 \log (x)+50 x \log (x)}{x^2 (-1+5 x)^2}-\frac {36}{x^2 \log ^3(2 x)}+\frac {6 (-3+2 x)}{x^2 \log ^2(2 x)}\right ) \, dx \\ & = 6 \int \frac {-3+2 x}{x^2 \log ^2(2 x)} \, dx-36 \int \frac {1}{x^2 \log ^3(2 x)} \, dx+\int \frac {5-25 x+2 x^2-20 x^3+50 x^4-5 \log (x)+50 x \log (x)}{x^2 (-1+5 x)^2} \, dx \\ & = \frac {18}{x \log ^2(2 x)}+6 \int \left (-\frac {3}{x^2 \log ^2(2 x)}+\frac {2}{x \log ^2(2 x)}\right ) \, dx+18 \int \frac {1}{x^2 \log ^2(2 x)} \, dx+\int \left (\frac {2}{(-1+5 x)^2}+\frac {5}{x^2 (-1+5 x)^2}-\frac {25}{x (-1+5 x)^2}-\frac {20 x}{(-1+5 x)^2}+\frac {50 x^2}{(-1+5 x)^2}+\frac {5 (-1+10 x) \log (x)}{x^2 (-1+5 x)^2}\right ) \, dx \\ & = \frac {2}{5 (1-5 x)}+\frac {18}{x \log ^2(2 x)}-\frac {18}{x \log (2 x)}+5 \int \frac {1}{x^2 (-1+5 x)^2} \, dx+5 \int \frac {(-1+10 x) \log (x)}{x^2 (-1+5 x)^2} \, dx+12 \int \frac {1}{x \log ^2(2 x)} \, dx-18 \int \frac {1}{x^2 \log ^2(2 x)} \, dx-18 \int \frac {1}{x^2 \log (2 x)} \, dx-20 \int \frac {x}{(-1+5 x)^2} \, dx-25 \int \frac {1}{x (-1+5 x)^2} \, dx+50 \int \frac {x^2}{(-1+5 x)^2} \, dx \\ & = \frac {2}{5 (1-5 x)}+\frac {18}{x \log ^2(2 x)}+5 \int \left (\frac {1}{x^2}+\frac {10}{x}+\frac {25}{(-1+5 x)^2}-\frac {50}{-1+5 x}\right ) \, dx+5 \int \left (-\frac {\log (x)}{x^2}+\frac {25 \log (x)}{(-1+5 x)^2}\right ) \, dx+12 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (2 x)\right )+18 \int \frac {1}{x^2 \log (2 x)} \, dx-20 \int \left (\frac {1}{5 (-1+5 x)^2}+\frac {1}{5 (-1+5 x)}\right ) \, dx-25 \int \left (\frac {1}{x}+\frac {5}{(-1+5 x)^2}-\frac {5}{-1+5 x}\right ) \, dx-36 \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (2 x)\right )+50 \int \left (\frac {1}{25}+\frac {1}{25 (-1+5 x)^2}+\frac {2}{25 (-1+5 x)}\right ) \, dx \\ & = -\frac {5}{x}+2 x-36 \operatorname {ExpIntegralEi}(-\log (2 x))-25 \log (1-5 x)+25 \log (x)+\frac {18}{x \log ^2(2 x)}-\frac {12}{\log (2 x)}-5 \int \frac {\log (x)}{x^2} \, dx+36 \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (2 x)\right )+125 \int \frac {\log (x)}{(-1+5 x)^2} \, dx \\ & = 2 x-25 \log (1-5 x)+25 \log (x)+\frac {5 \log (x)}{x}+\frac {125 x \log (x)}{1-5 x}+\frac {18}{x \log ^2(2 x)}-\frac {12}{\log (2 x)}+125 \int \frac {1}{-1+5 x} \, dx \\ & = 2 x+25 \log (x)+\frac {5 \log (x)}{x}+\frac {125 x \log (x)}{1-5 x}+\frac {18}{x \log ^2(2 x)}-\frac {12}{\log (2 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2 x-\frac {5 \log (x)}{x (-1+5 x)}+\frac {18}{x \log ^2(2 x)}-\frac {12}{\log (2 x)} \]

[In]

Integrate[(-36 + 360*x - 900*x^2 + (-18 + 192*x - 570*x^2 + 300*x^3)*Log[2*x] + (5 - 25*x + 2*x^2 - 20*x^3 + 5
0*x^4 + (-5 + 50*x)*Log[x])*Log[2*x]^3)/((x^2 - 10*x^3 + 25*x^4)*Log[2*x]^3),x]

[Out]

2*x - (5*Log[x])/(x*(-1 + 5*x)) + 18/(x*Log[2*x]^2) - 12/Log[2*x]

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38

method	result	size
risch	\(-\frac {5 \ln \left (x \right )}{x \left (5 x -1\right )}+2 x -\frac {24 \left (-3+2 x \ln \left (2\right )+2 x \ln \left (x \right )\right )}{x \left (2 \ln \left (2\right )+2 \ln \left (x \right )\right )^{2}}\)	\(47\)
parts	\(2 x +25 \ln \left (x \right )+\frac {18}{x \ln \left (2 x \right )^{2}}-\frac {12}{\ln \left (2 x \right )}+\frac {5 \ln \left (x \right )}{x}-\frac {125 \ln \left (x \right ) x}{5 x -1}\)	\(47\)
parallelrisch	\(\frac {-540+2700 x -12 x \ln \left (2 x \right )^{2}-1800 x^{2} \ln \left (2 x \right )+300 x^{3} \ln \left (2 x \right )^{2}+360 x \ln \left (2 x \right )-150 \ln \left (2 x \right )^{2} \ln \left (x \right )}{30 x \ln \left (2 x \right )^{2} \left (5 x -1\right )}\)	\(70\)
default	\(\frac {-18-60 x^{2} \ln \left (2\right )+\left (90-\frac {2 \ln \left (2\right )^{2}}{5}+12 \ln \left (2\right )\right ) x -\frac {2 x \ln \left (x \right )^{2}}{5}-60 x^{2} \ln \left (x \right )+\left (-\frac {4 \ln \left (2\right )}{5}+12\right ) \ln \left (x \right ) x -5 \ln \left (x \right )^{3}+10 x^{3} \ln \left (x \right )^{2}+10 x^{3} \ln \left (2\right )^{2}-10 \ln \left (2\right ) \ln \left (x \right )^{2}-5 \ln \left (2\right )^{2} \ln \left (x \right )+20 x^{3} \ln \left (2\right ) \ln \left (x \right )}{x \left (5 x -1\right ) \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}\)	\(115\)

[In]

int((((50*x-5)*ln(x)+50*x^4-20*x^3+2*x^2-25*x+5)*ln(2*x)^3+(300*x^3-570*x^2+192*x-18)*ln(2*x)-900*x^2+360*x-36
)/(25*x^4-10*x^3+x^2)/ln(2*x)^3,x,method=_RETURNVERBOSE)

[Out]

-5/x/(5*x-1)*ln(x)+2*x-24*(-3+2*x*ln(2)+2*x*ln(x))/x/(2*ln(2)+2*ln(x))^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (34) = 68\).

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 4.21 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=\frac {2 \, {\left (5 \, x^{3} - x^{2}\right )} \log \left (2\right )^{2} + 2 \, {\left (5 \, x^{3} - x^{2} - 5 \, \log \left (2\right )\right )} \log \left (x\right )^{2} - 5 \, \log \left (x\right )^{3} - 12 \, {\left (5 \, x^{2} - x\right )} \log \left (2\right ) - {\left (60 \, x^{2} - 4 \, {\left (5 \, x^{3} - x^{2}\right )} \log \left (2\right ) + 5 \, \log \left (2\right )^{2} - 12 \, x\right )} \log \left (x\right ) + 90 \, x - 18}{{\left (5 \, x^{2} - x\right )} \log \left (2\right )^{2} + 2 \, {\left (5 \, x^{2} - x\right )} \log \left (2\right ) \log \left (x\right ) + {\left (5 \, x^{2} - x\right )} \log \left (x\right )^{2}} \]

[In]

integrate((((50*x-5)*log(x)+50*x^4-20*x^3+2*x^2-25*x+5)*log(2*x)^3+(300*x^3-570*x^2+192*x-18)*log(2*x)-900*x^2
+360*x-36)/(25*x^4-10*x^3+x^2)/log(2*x)^3,x, algorithm="fricas")

[Out]

(2*(5*x^3 - x^2)*log(2)^2 + 2*(5*x^3 - x^2 - 5*log(2))*log(x)^2 - 5*log(x)^3 - 12*(5*x^2 - x)*log(2) - (60*x^2
- 4*(5*x^3 - x^2)*log(2) + 5*log(2)^2 - 12*x)*log(x) + 90*x - 18)/((5*x^2 - x)*log(2)^2 + 2*(5*x^2 - x)*log(2
)*log(x) + (5*x^2 - x)*log(x)^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2 x + \frac {- 12 x \log {\left (x \right )} - 12 x \log {\left (2 \right )} + 18}{x \log {\left (x \right )}^{2} + 2 x \log {\left (2 \right )} \log {\left (x \right )} + x \log {\left (2 \right )}^{2}} - \frac {5 \log {\left (x \right )}}{5 x^{2} - x} \]

[In]

integrate((((50*x-5)*ln(x)+50*x**4-20*x**3+2*x**2-25*x+5)*ln(2*x)**3+(300*x**3-570*x**2+192*x-18)*ln(2*x)-900*
x**2+360*x-36)/(25*x**4-10*x**3+x**2)/ln(2*x)**3,x)

[Out]

2*x + (-12*x*log(x) - 12*x*log(2) + 18)/(x*log(x)**2 + 2*x*log(2)*log(x) + x*log(2)**2) - 5*log(x)/(5*x**2 - x
)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (34) = 68\).

Time = 0.31 (sec) , antiderivative size = 141, normalized size of antiderivative = 4.15 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=\frac {10 \, x^{3} \log \left (2\right )^{2} - 2 \, {\left (\log \left (2\right )^{2} + 30 \, \log \left (2\right )\right )} x^{2} + 2 \, {\left (5 \, x^{3} - x^{2} - 5 \, \log \left (2\right )\right )} \log \left (x\right )^{2} - 5 \, \log \left (x\right )^{3} + 6 \, x {\left (2 \, \log \left (2\right ) + 15\right )} + {\left (20 \, x^{3} \log \left (2\right ) - 4 \, x^{2} {\left (\log \left (2\right ) + 15\right )} - 5 \, \log \left (2\right )^{2} + 12 \, x\right )} \log \left (x\right ) - 18}{5 \, x^{2} \log \left (2\right )^{2} - x \log \left (2\right )^{2} + {\left (5 \, x^{2} - x\right )} \log \left (x\right )^{2} + 2 \, {\left (5 \, x^{2} \log \left (2\right ) - x \log \left (2\right )\right )} \log \left (x\right )} \]

[In]

integrate((((50*x-5)*log(x)+50*x^4-20*x^3+2*x^2-25*x+5)*log(2*x)^3+(300*x^3-570*x^2+192*x-18)*log(2*x)-900*x^2
+360*x-36)/(25*x^4-10*x^3+x^2)/log(2*x)^3,x, algorithm="maxima")

[Out]

(10*x^3*log(2)^2 - 2*(log(2)^2 + 30*log(2))*x^2 + 2*(5*x^3 - x^2 - 5*log(2))*log(x)^2 - 5*log(x)^3 + 6*x*(2*lo
g(2) + 15) + (20*x^3*log(2) - 4*x^2*(log(2) + 15) - 5*log(2)^2 + 12*x)*log(x) - 18)/(5*x^2*log(2)^2 - x*log(2)
^2 + (5*x^2 - x)*log(x)^2 + 2*(5*x^2*log(2) - x*log(2))*log(x))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=-5 \, {\left (\frac {5}{5 \, x - 1} - \frac {1}{x}\right )} \log \left (x\right ) + 2 \, x - \frac {6 \, {\left (2 \, x \log \left (2\right ) + 2 \, x \log \left (x\right ) - 3\right )}}{x \log \left (2\right )^{2} + 2 \, x \log \left (2\right ) \log \left (x\right ) + x \log \left (x\right )^{2}} \]

[In]

integrate((((50*x-5)*log(x)+50*x^4-20*x^3+2*x^2-25*x+5)*log(2*x)^3+(300*x^3-570*x^2+192*x-18)*log(2*x)-900*x^2
+360*x-36)/(25*x^4-10*x^3+x^2)/log(2*x)^3,x, algorithm="giac")

[Out]

-5*(5/(5*x - 1) - 1/x)*log(x) + 2*x - 6*(2*x*log(2) + 2*x*log(x) - 3)/(x*log(2)^2 + 2*x*log(2)*log(x) + x*log(
x)^2)

Mupad [B] (verification not implemented)

Time = 17.97 (sec) , antiderivative size = 133, normalized size of antiderivative = 3.91 \[ \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2\,x+\frac {\ln \left (x\right )}{\frac {x}{5}-x^2}+\frac {\frac {3\,\left (3\,\ln \left (2\,x\right )-3\,\ln \left (x\right )-2\,x\,\left (\ln \left (2\,x\right )-\ln \left (x\right )\right )+6\right )}{x}-\frac {3\,\ln \left (x\right )\,\left (2\,x-3\right )}{x}}{2\,\ln \left (x\right )\,\left (\ln \left (2\,x\right )-\ln \left (x\right )\right )+{\ln \left (x\right )}^2+{\left (\ln \left (2\,x\right )-\ln \left (x\right )\right )}^2}-\frac {\frac {9\,\ln \left (x\right )}{x}+\frac {3\,\left (2\,x+3\,\ln \left (2\,x\right )-3\,\ln \left (x\right )+3\right )}{x}}{\ln \left (2\,x\right )}+\frac {9}{x} \]

[In]

int((360*x + log(2*x)*(192*x - 570*x^2 + 300*x^3 - 18) + log(2*x)^3*(log(x)*(50*x - 5) - 25*x + 2*x^2 - 20*x^3
+ 50*x^4 + 5) - 900*x^2 - 36)/(log(2*x)^3*(x^2 - 10*x^3 + 25*x^4)),x)

[Out]

2*x + log(x)/(x/5 - x^2) + ((3*(3*log(2*x) - 3*log(x) - 2*x*(log(2*x) - log(x)) + 6))/x - (3*log(x)*(2*x - 3))
/x)/(2*log(x)*(log(2*x) - log(x)) + log(x)^2 + (log(2*x) - log(x))^2) - ((9*log(x))/x + (3*(2*x + 3*log(2*x) -
3*log(x) + 3))/x)/log(2*x) + 9/x