3.40.32 \(\int \frac {5 x^2-4 x^3-x^4+(5 x+x^2) \log (\frac {15}{x})+(-5-6 x-7 x^2-6 x^3+(-5-7 x-2 x^2) \log (\frac {15}{x})) \log (\frac {1+x}{4})}{400 x^2+560 x^3+176 x^4+16 x^5} \, dx\)

Optimal. Leaf size=33 \[ \frac {\left (x-x^2+\log \left (\frac {15}{x}\right )\right ) \log \left (\frac {1+x}{4}\right )}{16 x (5+x)} \]

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Rubi [B]  time = 2.04, antiderivative size = 132, normalized size of antiderivative = 4.00, number of steps used = 53, number of rules used = 24, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.253, Rules used = {6741, 12, 6742, 44, 77, 88, 2357, 2301, 2317, 2391, 893, 2304, 2314, 31, 2418, 2395, 36, 29, 2392, 2394, 2393, 74, 2557, 2416} \begin {gather*} \frac {\log \left (\frac {15}{x}\right ) \log \left (\frac {x}{4}+\frac {1}{4}\right )}{16 x (x+5)}-\frac {1}{400} \log \left (\frac {x+5}{4}\right ) \log \left (\frac {x}{4}+\frac {1}{4}\right )+\frac {\log \left (\frac {x}{4}+\frac {1}{4}\right )}{80 x}+\frac {3 \log \left (\frac {x}{4}+\frac {1}{4}\right )}{8 (x+5)}+\frac {\log (4)-\log (x+1)}{80 x}-\frac {1}{16} \log (x+1)-\frac {1}{400} (\log (4)-\log (x+1)) \log \left (\frac {x+5}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^2 - 4*x^3 - x^4 + (5*x + x^2)*Log[15/x] + (-5 - 6*x - 7*x^2 - 6*x^3 + (-5 - 7*x - 2*x^2)*Log[15/x])*L
og[(1 + x)/4])/(400*x^2 + 560*x^3 + 176*x^4 + 16*x^5),x]

[Out]

Log[1/4 + x/4]/(80*x) + (3*Log[1/4 + x/4])/(8*(5 + x)) + (Log[1/4 + x/4]*Log[15/x])/(16*x*(5 + x)) + (Log[4] -
Log[1 + x])/(80*x) - Log[1 + x]/16 - (Log[1/4 + x/4]*Log[(5 + x)/4])/400 - ((Log[4] - Log[1 + x])*Log[(5 + x)
/4])/400

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, 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 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 2317

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

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 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 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 6741

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

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 {5 x^2-4 x^3-x^4+\left (5 x+x^2\right ) \log \left (\frac {15}{x}\right )+\left (-5-6 x-7 x^2-6 x^3+\left (-5-7 x-2 x^2\right ) \log \left (\frac {15}{x}\right )\right ) \log \left (\frac {1+x}{4}\right )}{16 x^2 (1+x) (5+x)^2} \, dx\\ &=\frac {1}{16} \int \frac {5 x^2-4 x^3-x^4+\left (5 x+x^2\right ) \log \left (\frac {15}{x}\right )+\left (-5-6 x-7 x^2-6 x^3+\left (-5-7 x-2 x^2\right ) \log \left (\frac {15}{x}\right )\right ) \log \left (\frac {1+x}{4}\right )}{x^2 (1+x) (5+x)^2} \, dx\\ &=\frac {1}{16} \int \left (\frac {5}{(1+x) (5+x)^2}-\frac {4 x}{(1+x) (5+x)^2}-\frac {x^2}{(1+x) (5+x)^2}+\frac {\log \left (\frac {15}{x}\right )}{x \left (5+6 x+x^2\right )}+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \left (-5-x-6 x^2-5 \log \left (\frac {15}{x}\right )-2 x \log \left (\frac {15}{x}\right )\right )}{x^2 (5+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{16} \int \frac {x^2}{(1+x) (5+x)^2} \, dx\right )+\frac {1}{16} \int \frac {\log \left (\frac {15}{x}\right )}{x \left (5+6 x+x^2\right )} \, dx+\frac {1}{16} \int \frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \left (-5-x-6 x^2-5 \log \left (\frac {15}{x}\right )-2 x \log \left (\frac {15}{x}\right )\right )}{x^2 (5+x)^2} \, dx-\frac {1}{4} \int \frac {x}{(1+x) (5+x)^2} \, dx+\frac {5}{16} \int \frac {1}{(1+x) (5+x)^2} \, dx\\ &=-\left (\frac {1}{16} \int \left (\frac {1}{16 (1+x)}-\frac {25}{4 (5+x)^2}+\frac {15}{16 (5+x)}\right ) \, dx\right )+\frac {1}{16} \int \left (\frac {\log \left (\frac {15}{x}\right )}{5 x}-\frac {\log \left (\frac {15}{x}\right )}{4 (1+x)}+\frac {\log \left (\frac {15}{x}\right )}{20 (5+x)}\right ) \, dx+\frac {1}{16} \int \left (\frac {\left (-5-x-6 x^2\right ) \log \left (\frac {1}{4}+\frac {x}{4}\right )}{x^2 (5+x)^2}+\frac {(-5-2 x) \log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{x^2 (5+x)^2}\right ) \, dx-\frac {1}{4} \int \left (-\frac {1}{16 (1+x)}+\frac {5}{4 (5+x)^2}+\frac {1}{16 (5+x)}\right ) \, dx+\frac {5}{16} \int \left (\frac {1}{16 (1+x)}-\frac {1}{4 (5+x)^2}-\frac {1}{16 (5+x)}\right ) \, dx\\ &=\frac {1}{32} \log (1+x)-\frac {3}{32} \log (5+x)+\frac {1}{320} \int \frac {\log \left (\frac {15}{x}\right )}{5+x} \, dx+\frac {1}{80} \int \frac {\log \left (\frac {15}{x}\right )}{x} \, dx-\frac {1}{64} \int \frac {\log \left (\frac {15}{x}\right )}{1+x} \, dx+\frac {1}{16} \int \frac {\left (-5-x-6 x^2\right ) \log \left (\frac {1}{4}+\frac {x}{4}\right )}{x^2 (5+x)^2} \, dx+\frac {1}{16} \int \frac {(-5-2 x) \log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{x^2 (5+x)^2} \, dx\\ &=\frac {1}{320} \log \left (1+\frac {x}{5}\right ) \log \left (\frac {15}{x}\right )+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{16 x (5+x)}-\frac {1}{160} \log ^2\left (\frac {15}{x}\right )+\frac {1}{32} \log (1+x)-\frac {1}{64} \log \left (\frac {15}{x}\right ) \log (1+x)-\frac {3}{32} \log (5+x)+\frac {1}{320} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx-\frac {1}{64} \int \frac {\log (1+x)}{x} \, dx+\frac {1}{16} \int \left (-\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{5 x^2}+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{25 x}-\frac {6 \log \left (\frac {1}{4}+\frac {x}{4}\right )}{(5+x)^2}-\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{25 (5+x)}\right ) \, dx-\frac {1}{16} \int \frac {\log \left (\frac {15}{x}\right )}{x \left (5+6 x+x^2\right )} \, dx-\frac {1}{16} \int \frac {\log (4)-\log (1+x)}{x^2 (5+x)} \, dx\\ &=\frac {1}{320} \log \left (1+\frac {x}{5}\right ) \log \left (\frac {15}{x}\right )+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{16 x (5+x)}-\frac {1}{160} \log ^2\left (\frac {15}{x}\right )+\frac {1}{32} \log (1+x)-\frac {1}{64} \log \left (\frac {15}{x}\right ) \log (1+x)-\frac {3}{32} \log (5+x)+\frac {\text {Li}_2(-x)}{64}-\frac {\text {Li}_2\left (-\frac {x}{5}\right )}{320}+\frac {1}{400} \int \frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{x} \, dx-\frac {1}{400} \int \frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{5+x} \, dx-\frac {1}{80} \int \frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{x^2} \, dx-\frac {1}{16} \int \left (\frac {\log \left (\frac {15}{x}\right )}{5 x}-\frac {\log \left (\frac {15}{x}\right )}{4 (1+x)}+\frac {\log \left (\frac {15}{x}\right )}{20 (5+x)}\right ) \, dx-\frac {1}{16} \int \left (\frac {\log (4)-\log (1+x)}{5 x^2}-\frac {\log (4)-\log (1+x)}{25 x}+\frac {\log (4)-\log (1+x)}{25 (5+x)}\right ) \, dx-\frac {3}{8} \int \frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{(5+x)^2} \, dx\\ &=\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{80 x}+\frac {3 \log \left (\frac {1}{4}+\frac {x}{4}\right )}{8 (5+x)}+\frac {1}{320} \log \left (1+\frac {x}{5}\right ) \log \left (\frac {15}{x}\right )+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{16 x (5+x)}-\frac {1}{160} \log ^2\left (\frac {15}{x}\right )-\frac {1}{400} \log (4) \log (x)+\frac {1}{32} \log (1+x)-\frac {1}{64} \log \left (\frac {15}{x}\right ) \log (1+x)-\frac {1}{400} \log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {5+x}{4}\right )-\frac {3}{32} \log (5+x)+\frac {\text {Li}_2(-x)}{64}-\frac {\text {Li}_2\left (-\frac {x}{5}\right )}{320}+\frac {\int \frac {\log \left (\frac {5+x}{4}\right )}{\frac {1}{4}+\frac {x}{4}} \, dx}{1600}+\frac {1}{400} \int \frac {\log (4)-\log (1+x)}{x} \, dx-\frac {1}{400} \int \frac {\log (4)-\log (1+x)}{5+x} \, dx+\frac {1}{400} \int \frac {\log (1+x)}{x} \, dx-\frac {1}{320} \int \frac {1}{\left (\frac {1}{4}+\frac {x}{4}\right ) x} \, dx-\frac {1}{320} \int \frac {\log \left (\frac {15}{x}\right )}{5+x} \, dx-\frac {1}{80} \int \frac {\log \left (\frac {15}{x}\right )}{x} \, dx-\frac {1}{80} \int \frac {\log (4)-\log (1+x)}{x^2} \, dx+\frac {1}{64} \int \frac {\log \left (\frac {15}{x}\right )}{1+x} \, dx-\frac {3}{32} \int \frac {1}{\left (\frac {1}{4}+\frac {x}{4}\right ) (5+x)} \, dx\\ &=\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{80 x}+\frac {3 \log \left (\frac {1}{4}+\frac {x}{4}\right )}{8 (5+x)}+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{16 x (5+x)}+\frac {\log (4)-\log (1+x)}{80 x}+\frac {1}{32} \log (1+x)-\frac {1}{400} \log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {5+x}{4}\right )-\frac {1}{400} (\log (4)-\log (1+x)) \log \left (\frac {5+x}{4}\right )-\frac {3}{32} \log (5+x)+\frac {21 \text {Li}_2(-x)}{1600}-\frac {\text {Li}_2\left (-\frac {x}{5}\right )}{320}-\frac {1}{400} \int \frac {\log (1+x)}{x} \, dx-\frac {1}{400} \int \frac {\log \left (\frac {5+x}{4}\right )}{1+x} \, dx+\frac {1}{400} \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,\frac {1}{4}+\frac {x}{4}\right )+\frac {1}{320} \int \frac {1}{\frac {1}{4}+\frac {x}{4}} \, dx-\frac {1}{320} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx-\frac {1}{80} \int \frac {1}{x} \, dx+\frac {1}{80} \int \frac {1}{x (1+x)} \, dx+\frac {1}{64} \int \frac {\log (1+x)}{x} \, dx-\frac {3}{128} \int \frac {1}{\frac {1}{4}+\frac {x}{4}} \, dx+\frac {3}{32} \int \frac {1}{5+x} \, dx\\ &=\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{80 x}+\frac {3 \log \left (\frac {1}{4}+\frac {x}{4}\right )}{8 (5+x)}+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{16 x (5+x)}-\frac {\log (x)}{80}+\frac {\log (4)-\log (1+x)}{80 x}-\frac {1}{20} \log (1+x)-\frac {1}{400} \log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {5+x}{4}\right )-\frac {1}{400} (\log (4)-\log (1+x)) \log \left (\frac {5+x}{4}\right )-\frac {1}{400} \text {Li}_2\left (\frac {1}{4} (-1-x)\right )-\frac {1}{400} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx,x,1+x\right )+\frac {1}{80} \int \frac {1}{x} \, dx-\frac {1}{80} \int \frac {1}{1+x} \, dx\\ &=\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right )}{80 x}+\frac {3 \log \left (\frac {1}{4}+\frac {x}{4}\right )}{8 (5+x)}+\frac {\log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {15}{x}\right )}{16 x (5+x)}+\frac {\log (4)-\log (1+x)}{80 x}-\frac {1}{16} \log (1+x)-\frac {1}{400} \log \left (\frac {1}{4}+\frac {x}{4}\right ) \log \left (\frac {5+x}{4}\right )-\frac {1}{400} (\log (4)-\log (1+x)) \log \left (\frac {5+x}{4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 38, normalized size = 1.15 \begin {gather*} \frac {1}{16} \left (\frac {\left (6 x+\log \left (\frac {15}{x}\right )\right ) \log \left (\frac {1+x}{4}\right )}{x (5+x)}-\log (1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 - 4*x^3 - x^4 + (5*x + x^2)*Log[15/x] + (-5 - 6*x - 7*x^2 - 6*x^3 + (-5 - 7*x - 2*x^2)*Log[15
/x])*Log[(1 + x)/4])/(400*x^2 + 560*x^3 + 176*x^4 + 16*x^5),x]

[Out]

(((6*x + Log[15/x])*Log[(1 + x)/4])/(x*(5 + x)) - Log[1 + x])/16

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fricas [A]  time = 0.61, size = 32, normalized size = 0.97 \begin {gather*} -\frac {{\left (x^{2} - x - \log \left (\frac {15}{x}\right )\right )} \log \left (\frac {1}{4} \, x + \frac {1}{4}\right )}{16 \, {\left (x^{2} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-7*x-5)*log(15/x)-6*x^3-7*x^2-6*x-5)*log(1/4*x+1/4)+(x^2+5*x)*log(15/x)-x^4-4*x^3+5*x^2)/(1
6*x^5+176*x^4+560*x^3+400*x^2),x, algorithm="fricas")

[Out]

-1/16*(x^2 - x - log(15/x))*log(1/4*x + 1/4)/(x^2 + 5*x)

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giac [B]  time = 0.17, size = 91, normalized size = 2.76 \begin {gather*} \frac {1}{80} \, {\left ({\left (\frac {1}{x + 5} - \frac {1}{x}\right )} \log \relax (x) - \frac {\log \left (15\right ) - 30}{x + 5} + \frac {\log \left (15\right )}{x}\right )} \log \left (x + 1\right ) - \frac {1}{40} \, {\left (\frac {\log \relax (2)}{x + 5} - \frac {\log \relax (2)}{x}\right )} \log \relax (x) - \frac {\log \left (15\right ) \log \relax (2)}{40 \, x} + \frac {\log \left (15\right ) \log \relax (2) - 30 \, \log \relax (2)}{40 \, {\left (x + 5\right )}} - \frac {1}{16} \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-7*x-5)*log(15/x)-6*x^3-7*x^2-6*x-5)*log(1/4*x+1/4)+(x^2+5*x)*log(15/x)-x^4-4*x^3+5*x^2)/(1
6*x^5+176*x^4+560*x^3+400*x^2),x, algorithm="giac")

[Out]

1/80*((1/(x + 5) - 1/x)*log(x) - (log(15) - 30)/(x + 5) + log(15)/x)*log(x + 1) - 1/40*(log(2)/(x + 5) - log(2
)/x)*log(x) - 1/40*log(15)*log(2)/x + 1/40*(log(15)*log(2) - 30*log(2))/(x + 5) - 1/16*log(x + 1)

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maple [A]  time = 0.22, size = 40, normalized size = 1.21

method	result	size
risch	\(\frac {\left (2 \ln \relax (3)+2 \ln \relax (5)+12 x -2 \ln \relax (x )\right ) \ln \left (\frac {x}{4}+\frac {1}{4}\right )}{32 \left (5+x \right ) x}-\frac {\ln \left (x +1\right )}{16}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-7*x-5)*ln(15/x)-6*x^3-7*x^2-6*x-5)*ln(1/4*x+1/4)+(x^2+5*x)*ln(15/x)-x^4-4*x^3+5*x^2)/(16*x^5+176
*x^4+560*x^3+400*x^2),x,method=_RETURNVERBOSE)

[Out]

1/32*(2*ln(3)+2*ln(5)+12*x-2*ln(x))/(5+x)/x*ln(1/4*x+1/4)-1/16*ln(x+1)

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maxima [A]  time = 0.48, size = 54, normalized size = 1.64 \begin {gather*} -\frac {12 \, x \log \relax (2) + 2 \, {\left (\log \relax (5) + \log \relax (3)\right )} \log \relax (2) + {\left (x^{2} - x - \log \relax (5) - \log \relax (3) + \log \relax (x)\right )} \log \left (x + 1\right ) - 2 \, \log \relax (2) \log \relax (x)}{16 \, {\left (x^{2} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-7*x-5)*log(15/x)-6*x^3-7*x^2-6*x-5)*log(1/4*x+1/4)+(x^2+5*x)*log(15/x)-x^4-4*x^3+5*x^2)/(1
6*x^5+176*x^4+560*x^3+400*x^2),x, algorithm="maxima")

[Out]

-1/16*(12*x*log(2) + 2*(log(5) + log(3))*log(2) + (x^2 - x - log(5) - log(3) + log(x))*log(x + 1) - 2*log(2)*l
og(x))/(x^2 + 5*x)

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mupad [B]  time = 2.66, size = 46, normalized size = 1.39 \begin {gather*} \frac {3\,\ln \left (\frac {x}{4}+\frac {1}{4}\right )}{8\,\left (x+5\right )}-\frac {\ln \left (x+1\right )}{16}+\frac {\ln \left (\frac {x}{4}+\frac {1}{4}\right )\,\ln \left (\frac {15}{x}\right )}{16\,\left (x^2+5\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x/4 + 1/4)*(6*x + log(15/x)*(7*x + 2*x^2 + 5) + 7*x^2 + 6*x^3 + 5) - log(15/x)*(5*x + x^2) - 5*x^2 +
4*x^3 + x^4)/(400*x^2 + 560*x^3 + 176*x^4 + 16*x^5),x)

[Out]

(3*log(x/4 + 1/4))/(8*(x + 5)) - log(x + 1)/16 + (log(x/4 + 1/4)*log(15/x))/(16*(5*x + x^2))

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sympy [A]  time = 0.64, size = 32, normalized size = 0.97 \begin {gather*} \frac {\left (6 x + \log {\left (\frac {15}{x} \right )}\right ) \log {\left (\frac {x}{4} + \frac {1}{4} \right )}}{16 x^{2} + 80 x} - \frac {\log {\left (16 x + 16 \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-7*x-5)*ln(15/x)-6*x**3-7*x**2-6*x-5)*ln(1/4*x+1/4)+(x**2+5*x)*ln(15/x)-x**4-4*x**3+5*x**2
)/(16*x**5+176*x**4+560*x**3+400*x**2),x)

[Out]

(6*x + log(15/x))*log(x/4 + 1/4)/(16*x**2 + 80*x) - log(16*x + 16)/16

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