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

Optimal. Leaf size=34 \[ 1-\frac {\log (x)}{-\frac {x}{5}+x^2}+\frac {2 \left (x-\frac {3}{\log (2 x)}\right )^2}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 1.15, antiderivative size = 46, normalized size of antiderivative = 1.35, number of steps used = 31, number of rules used = 17, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.193, Rules used = {1594, 27, 6688, 14, 6742, 44, 43, 2357, 2304, 2314, 31, 2306, 2309, 2178, 2353, 2302, 30} \begin {gather*} 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} \end {gather*}

Antiderivative was successfully verified.

[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 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 1594

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 2178

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

Rule 2302

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 2304

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 2306

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 2309

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 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 2353

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 2357

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 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 {-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 \operatorname {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 \operatorname {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 \text {Ei}(-\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 \operatorname {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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 37, normalized size = 1.09 \begin {gather*} 2 x-\frac {5 \log (x)}{x (-1+5 x)}+\frac {18}{x \log ^2(2 x)}-\frac {12}{\log (2 x)} \end {gather*}

Antiderivative was successfully verified.

[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]

________________________________________________________________________________________

fricas [B]  time = 1.15, size = 143, normalized size = 4.21 \begin {gather*} \frac {2 \, {\left (5 \, x^{3} - x^{2}\right )} \log \relax (2)^{2} + 2 \, {\left (5 \, x^{3} - x^{2} - 5 \, \log \relax (2)\right )} \log \relax (x)^{2} - 5 \, \log \relax (x)^{3} - 12 \, {\left (5 \, x^{2} - x\right )} \log \relax (2) - {\left (60 \, x^{2} - 4 \, {\left (5 \, x^{3} - x^{2}\right )} \log \relax (2) + 5 \, \log \relax (2)^{2} - 12 \, x\right )} \log \relax (x) + 90 \, x - 18}{{\left (5 \, x^{2} - x\right )} \log \relax (2)^{2} + 2 \, {\left (5 \, x^{2} - x\right )} \log \relax (2) \log \relax (x) + {\left (5 \, x^{2} - x\right )} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 59, normalized size = 1.74 \begin {gather*} -5 \, {\left (\frac {5}{5 \, x - 1} - \frac {1}{x}\right )} \log \relax (x) + 2 \, x - \frac {6 \, {\left (2 \, x \log \relax (2) + 2 \, x \log \relax (x) - 3\right )}}{x \log \relax (2)^{2} + 2 \, x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 47, normalized size = 1.38

method	result	size
risch	\(-\frac {5 \ln \relax (x )}{x \left (5 x -1\right )}+2 x -\frac {24 \left (-3+2 x \ln \relax (2)+2 x \ln \relax (x )\right )}{x \left (2 \ln \relax (2)+2 \ln \relax (x )\right )^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maxima [B]  time = 0.48, size = 141, normalized size = 4.15 \begin {gather*} \frac {10 \, x^{3} \log \relax (2)^{2} - 2 \, {\left (\log \relax (2)^{2} + 30 \, \log \relax (2)\right )} x^{2} + 2 \, {\left (5 \, x^{3} - x^{2} - 5 \, \log \relax (2)\right )} \log \relax (x)^{2} - 5 \, \log \relax (x)^{3} + 6 \, x {\left (2 \, \log \relax (2) + 15\right )} + {\left (20 \, x^{3} \log \relax (2) - 4 \, x^{2} {\left (\log \relax (2) + 15\right )} - 5 \, \log \relax (2)^{2} + 12 \, x\right )} \log \relax (x) - 18}{5 \, x^{2} \log \relax (2)^{2} - x \log \relax (2)^{2} + {\left (5 \, x^{2} - x\right )} \log \relax (x)^{2} + 2 \, {\left (5 \, x^{2} \log \relax (2) - x \log \relax (2)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 6.98, size = 133, normalized size = 3.91 \begin {gather*} 2\,x+\frac {\ln \relax (x)}{\frac {x}{5}-x^2}+\frac {\frac {3\,\left (3\,\ln \left (2\,x\right )-3\,\ln \relax (x)-2\,x\,\left (\ln \left (2\,x\right )-\ln \relax (x)\right )+6\right )}{x}-\frac {3\,\ln \relax (x)\,\left (2\,x-3\right )}{x}}{2\,\ln \relax (x)\,\left (\ln \left (2\,x\right )-\ln \relax (x)\right )+{\ln \relax (x)}^2+{\left (\ln \left (2\,x\right )-\ln \relax (x)\right )}^2}-\frac {\frac {9\,\ln \relax (x)}{x}+\frac {3\,\left (2\,x+3\,\ln \left (2\,x\right )-3\,\ln \relax (x)+3\right )}{x}}{\ln \left (2\,x\right )}+\frac {9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [B]  time = 0.44, size = 53, normalized size = 1.56 \begin {gather*} 2 x + \frac {- 12 x \log {\relax (x )} - 12 x \log {\relax (2 )} + 18}{x \log {\relax (x )}^{2} + 2 x \log {\relax (2 )} \log {\relax (x )} + x \log {\relax (2 )}^{2}} - \frac {5 \log {\relax (x )}}{5 x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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
)

________________________________________________________________________________________