3.57.81 \(\int \frac {-45 x^3+5 x^4+x^5+(65 x^2-10 x^3-x^4) \log (-4+x)+(-20 x+5 x^2) \log ^2(-4+x)}{e^8 (-4000-1400 x+120 x^2+88 x^3+8 x^4)} \, dx\)

Optimal. Leaf size=31 \[ \frac {x^2 \left (\frac {x}{2}-\frac {1}{2} \log (-4+x)\right )^2}{4 e^8 (5+x)^2} \]

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Rubi [B]  time = 0.66, antiderivative size = 179, normalized size of antiderivative = 5.77, number of steps used = 40, number of rules used = 23, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.299, Rules used = {12, 6742, 88, 2418, 2389, 2295, 2390, 2301, 2395, 44, 36, 31, 2394, 2393, 2391, 2416, 2398, 2411, 2347, 2344, 2317, 2314, 2397} \begin {gather*} \frac {x^2}{16 e^8}-\frac {5 x}{8 e^8}-\frac {125}{4 e^8 (x+5)}+\frac {625}{16 e^8 (x+5)^2}-\frac {5 (4-x) \log ^2(x-4)}{72 e^8 (x+5)}+\frac {25 \log ^2(x-4)}{16 e^8 (x+5)^2}-\frac {\log ^2(x-4)}{144 e^8}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (x-4)}{8 e^8}-\frac {25 (4-x) \log (x-4)}{648 e^8 (x+5)}-\frac {325 \log (x-4)}{36 e^8 (x+5)}+\frac {125 \log (x-4)}{8 e^8 (x+5)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45*x^3 + 5*x^4 + x^5 + (65*x^2 - 10*x^3 - x^4)*Log[-4 + x] + (-20*x + 5*x^2)*Log[-4 + x]^2)/(E^8*(-4000
- 1400*x + 120*x^2 + 88*x^3 + 8*x^4)),x]

[Out]

(-5*x)/(8*E^8) + x^2/(16*E^8) + 625/(16*E^8*(5 + x)^2) - 125/(4*E^8*(5 + x)) + (461*Log[4 - x])/(648*E^8) + ((
4 - x)*Log[-4 + x])/(8*E^8) + (125*Log[-4 + x])/(8*E^8*(5 + x)^2) - (325*Log[-4 + x])/(36*E^8*(5 + x)) - (25*(
4 - x)*Log[-4 + x])/(648*E^8*(5 + x)) - Log[-4 + x]^2/(144*E^8) + (25*Log[-4 + x]^2)/(16*E^8*(5 + x)^2) - (5*(
4 - x)*Log[-4 + x]^2)/(72*E^8*(5 + 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 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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 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 2344

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

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2389

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

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 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 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 2397

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

Rule 2398

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

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-45 x^3+5 x^4+x^5+\left (65 x^2-10 x^3-x^4\right ) \log (-4+x)+\left (-20 x+5 x^2\right ) \log ^2(-4+x)}{-4000-1400 x+120 x^2+88 x^3+8 x^4} \, dx}{e^8}\\ &=\frac {\int \left (-\frac {45 x^3}{8 (-4+x) (5+x)^3}+\frac {5 x^4}{8 (-4+x) (5+x)^3}+\frac {x^5}{8 (-4+x) (5+x)^3}-\frac {x^2 \left (-65+10 x+x^2\right ) \log (-4+x)}{8 (-4+x) (5+x)^3}+\frac {5 x \log ^2(-4+x)}{8 (5+x)^3}\right ) \, dx}{e^8}\\ &=\frac {\int \frac {x^5}{(-4+x) (5+x)^3} \, dx}{8 e^8}-\frac {\int \frac {x^2 \left (-65+10 x+x^2\right ) \log (-4+x)}{(-4+x) (5+x)^3} \, dx}{8 e^8}+\frac {5 \int \frac {x^4}{(-4+x) (5+x)^3} \, dx}{8 e^8}+\frac {5 \int \frac {x \log ^2(-4+x)}{(5+x)^3} \, dx}{8 e^8}-\frac {45 \int \frac {x^3}{(-4+x) (5+x)^3} \, dx}{8 e^8}\\ &=\frac {\int \left (-11+\frac {1024}{729 (-4+x)}+x+\frac {3125}{9 (5+x)^3}-\frac {25000}{81 (5+x)^2}+\frac {76250}{729 (5+x)}\right ) \, dx}{8 e^8}-\frac {\int \left (\log (-4+x)-\frac {16 \log (-4+x)}{81 (-4+x)}+\frac {250 \log (-4+x)}{(5+x)^3}-\frac {650 \log (-4+x)}{9 (5+x)^2}-\frac {65 \log (-4+x)}{81 (5+x)}\right ) \, dx}{8 e^8}+\frac {5 \int \left (1+\frac {256}{729 (-4+x)}-\frac {625}{9 (5+x)^3}+\frac {3875}{81 (5+x)^2}-\frac {8275}{729 (5+x)}\right ) \, dx}{8 e^8}+\frac {5 \int \left (-\frac {5 \log ^2(-4+x)}{(5+x)^3}+\frac {\log ^2(-4+x)}{(5+x)^2}\right ) \, dx}{8 e^8}-\frac {45 \int \left (\frac {64}{729 (-4+x)}+\frac {125}{9 (5+x)^3}-\frac {550}{81 (5+x)^2}+\frac {665}{729 (5+x)}\right ) \, dx}{8 e^8}\\ &=-\frac {3 x}{4 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {2125}{72 e^8 (5+x)}-\frac {8 \log (4-x)}{81 e^8}+\frac {275 \log (5+x)}{324 e^8}+\frac {2 \int \frac {\log (-4+x)}{-4+x} \, dx}{81 e^8}+\frac {65 \int \frac {\log (-4+x)}{5+x} \, dx}{648 e^8}-\frac {\int \log (-4+x) \, dx}{8 e^8}+\frac {5 \int \frac {\log ^2(-4+x)}{(5+x)^2} \, dx}{8 e^8}-\frac {25 \int \frac {\log ^2(-4+x)}{(5+x)^3} \, dx}{8 e^8}+\frac {325 \int \frac {\log (-4+x)}{(5+x)^2} \, dx}{36 e^8}-\frac {125 \int \frac {\log (-4+x)}{(5+x)^3} \, dx}{4 e^8}\\ &=-\frac {3 x}{4 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {2125}{72 e^8 (5+x)}-\frac {8 \log (4-x)}{81 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}+\frac {65 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {275 \log (5+x)}{324 e^8}+\frac {2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-4+x\right )}{81 e^8}-\frac {65 \int \frac {\log \left (\frac {5+x}{9}\right )}{-4+x} \, dx}{648 e^8}-\frac {\operatorname {Subst}(\int \log (x) \, dx,x,-4+x)}{8 e^8}-\frac {5 \int \frac {\log (-4+x)}{5+x} \, dx}{36 e^8}-\frac {25 \int \frac {\log (-4+x)}{(-4+x) (5+x)^2} \, dx}{8 e^8}+\frac {325 \int \frac {1}{(-4+x) (5+x)} \, dx}{36 e^8}-\frac {125 \int \frac {1}{(-4+x) (5+x)^2} \, dx}{8 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {2125}{72 e^8 (5+x)}-\frac {8 \log (4-x)}{81 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}+\frac {\log ^2(-4+x)}{81 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {275 \log (5+x)}{324 e^8}-\frac {65 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{9}\right )}{x} \, dx,x,-4+x\right )}{648 e^8}+\frac {5 \int \frac {\log \left (\frac {5+x}{9}\right )}{-4+x} \, dx}{36 e^8}+\frac {325 \int \frac {1}{-4+x} \, dx}{324 e^8}-\frac {325 \int \frac {1}{5+x} \, dx}{324 e^8}-\frac {25 \operatorname {Subst}\left (\int \frac {\log (x)}{x (9+x)^2} \, dx,x,-4+x\right )}{8 e^8}-\frac {125 \int \left (\frac {1}{81 (-4+x)}-\frac {1}{9 (5+x)^2}-\frac {1}{81 (5+x)}\right ) \, dx}{8 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}+\frac {\log ^2(-4+x)}{81 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {25 \log (5+x)}{648 e^8}+\frac {65 \text {Li}_2\left (\frac {4-x}{9}\right )}{648 e^8}+\frac {5 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{9}\right )}{x} \, dx,x,-4+x\right )}{36 e^8}+\frac {25 \operatorname {Subst}\left (\int \frac {\log (x)}{(9+x)^2} \, dx,x,-4+x\right )}{72 e^8}-\frac {25 \operatorname {Subst}\left (\int \frac {\log (x)}{x (9+x)} \, dx,x,-4+x\right )}{72 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}-\frac {25 (4-x) \log (-4+x)}{648 e^8 (5+x)}+\frac {\log ^2(-4+x)}{81 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {25 \log (5+x)}{648 e^8}-\frac {25 \text {Li}_2\left (\frac {4-x}{9}\right )}{648 e^8}-\frac {25 \operatorname {Subst}\left (\int \frac {1}{9+x} \, dx,x,-4+x\right )}{648 e^8}-\frac {25 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-4+x\right )}{648 e^8}+\frac {25 \operatorname {Subst}\left (\int \frac {\log (x)}{9+x} \, dx,x,-4+x\right )}{648 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}-\frac {25 (4-x) \log (-4+x)}{648 e^8 (5+x)}-\frac {\log ^2(-4+x)}{144 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \text {Li}_2\left (\frac {4-x}{9}\right )}{648 e^8}-\frac {25 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{9}\right )}{x} \, dx,x,-4+x\right )}{648 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}-\frac {25 (4-x) \log (-4+x)}{648 e^8 (5+x)}-\frac {\log ^2(-4+x)}{144 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.14, size = 128, normalized size = 4.13 \begin {gather*} \frac {-5625-2250 x-225 x^2+3 x^4+50 (5+x)^2 \log (4-x)-2 \left (625+250 x+25 x^2+3 x^3\right ) \log (-4+x)+3 x^2 \log ^2(-4+x)+1250 \log \left (\frac {5+x}{9}\right )+500 x \log \left (\frac {5+x}{9}\right )+50 x^2 \log \left (\frac {5+x}{9}\right )-1250 \log (5+x)-500 x \log (5+x)-50 x^2 \log (5+x)}{48 e^8 (5+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45*x^3 + 5*x^4 + x^5 + (65*x^2 - 10*x^3 - x^4)*Log[-4 + x] + (-20*x + 5*x^2)*Log[-4 + x]^2)/(E^8*(
-4000 - 1400*x + 120*x^2 + 88*x^3 + 8*x^4)),x]

[Out]

(-5625 - 2250*x - 225*x^2 + 3*x^4 + 50*(5 + x)^2*Log[4 - x] - 2*(625 + 250*x + 25*x^2 + 3*x^3)*Log[-4 + x] + 3
*x^2*Log[-4 + x]^2 + 1250*Log[(5 + x)/9] + 500*x*Log[(5 + x)/9] + 50*x^2*Log[(5 + x)/9] - 1250*Log[5 + x] - 50
0*x*Log[5 + x] - 50*x^2*Log[5 + x])/(48*E^8*(5 + x)^2)

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fricas [B]  time = 0.76, size = 46, normalized size = 1.48 \begin {gather*} \frac {{\left (x^{4} - 2 \, x^{3} \log \left (x - 4\right ) + x^{2} \log \left (x - 4\right )^{2} - 75 \, x^{2} - 750 \, x - 1875\right )} e^{\left (-8\right )}}{16 \, {\left (x^{2} + 10 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-20*x)*log(x-4)^2+(-x^4-10*x^3+65*x^2)*log(x-4)+x^5+5*x^4-45*x^3)/(8*x^4+88*x^3+120*x^2-1400*
x-4000)/exp(4)^2,x, algorithm="fricas")

[Out]

1/16*(x^4 - 2*x^3*log(x - 4) + x^2*log(x - 4)^2 - 75*x^2 - 750*x - 1875)*e^(-8)/(x^2 + 10*x + 25)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{5} + 5 \, x^{4} - 45 \, x^{3} + 5 \, {\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right )^{2} - {\left (x^{4} + 10 \, x^{3} - 65 \, x^{2}\right )} \log \left (x - 4\right )\right )} e^{\left (-8\right )}}{8 \, {\left (x^{4} + 11 \, x^{3} + 15 \, x^{2} - 175 \, x - 500\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-20*x)*log(x-4)^2+(-x^4-10*x^3+65*x^2)*log(x-4)+x^5+5*x^4-45*x^3)/(8*x^4+88*x^3+120*x^2-1400*
x-4000)/exp(4)^2,x, algorithm="giac")

[Out]

integrate(1/8*(x^5 + 5*x^4 - 45*x^3 + 5*(x^2 - 4*x)*log(x - 4)^2 - (x^4 + 10*x^3 - 65*x^2)*log(x - 4))*e^(-8)/
(x^4 + 11*x^3 + 15*x^2 - 175*x - 500), x)

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maple [A]  time = 0.26, size = 49, normalized size = 1.58

method	result	size
norman	\(\frac {\left (\frac {{\mathrm e}^{-4} x^{4}}{16}+\frac {x^{2} {\mathrm e}^{-4} \ln \left (x -4\right )^{2}}{16}-\frac {{\mathrm e}^{-4} x^{3} \ln \left (x -4\right )}{8}\right ) {\mathrm e}^{-4}}{\left (5+x \right )^{2}}\)	\(49\)
risch	\(\frac {{\mathrm e}^{-8} x^{2} \ln \left (x -4\right )^{2}}{16 x^{2}+160 x +400}-\frac {{\mathrm e}^{-8} \left (x^{3}+10 x^{2}+100 x +250\right ) \ln \left (x -4\right )}{8 \left (x^{2}+10 x +25\right )}+\frac {{\mathrm e}^{-8} \left (x^{4}+20 x^{2} \ln \left (x -4\right )+200 x \ln \left (x -4\right )-75 x^{2}+500 \ln \left (x -4\right )-750 x -1875\right )}{16 x^{2}+160 x +400}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2-20*x)*ln(x-4)^2+(-x^4-10*x^3+65*x^2)*ln(x-4)+x^5+5*x^4-45*x^3)/(8*x^4+88*x^3+120*x^2-1400*x-4000)/
exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

(1/16/exp(4)*x^4+1/16*x^2/exp(4)*ln(x-4)^2-1/8/exp(4)*x^3*ln(x-4))/(5+x)^2/exp(4)

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maxima [B]  time = 0.43, size = 46, normalized size = 1.48 \begin {gather*} \frac {{\left (x^{4} - 2 \, x^{3} \log \left (x - 4\right ) + x^{2} \log \left (x - 4\right )^{2} - 75 \, x^{2} - 750 \, x - 1875\right )} e^{\left (-8\right )}}{16 \, {\left (x^{2} + 10 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-20*x)*log(x-4)^2+(-x^4-10*x^3+65*x^2)*log(x-4)+x^5+5*x^4-45*x^3)/(8*x^4+88*x^3+120*x^2-1400*
x-4000)/exp(4)^2,x, algorithm="maxima")

[Out]

1/16*(x^4 - 2*x^3*log(x - 4) + x^2*log(x - 4)^2 - 75*x^2 - 750*x - 1875)*e^(-8)/(x^2 + 10*x + 25)

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mupad [B]  time = 0.37, size = 22, normalized size = 0.71 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-8}\,{\left (x-\ln \left (x-4\right )\right )}^2}{16\,{\left (x+5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(log(x - 4)*(10*x^3 - 65*x^2 + x^4) + log(x - 4)^2*(20*x - 5*x^2) + 45*x^3 - 5*x^4 - x^5))/(120*
x^2 - 1400*x + 88*x^3 + 8*x^4 - 4000),x)

[Out]

(x^2*exp(-8)*(x - log(x - 4))^2)/(16*(x + 5)^2)

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sympy [B]  time = 0.60, size = 126, normalized size = 4.06 \begin {gather*} \frac {x^{2}}{16 e^{8}} + \frac {x^{2} \log {\left (x - 4 \right )}^{2}}{16 x^{2} e^{8} + 160 x e^{8} + 400 e^{8}} - \frac {5 x}{8 e^{8}} + \frac {- 500 x - 1875}{16 x^{2} e^{8} + 160 x e^{8} + 400 e^{8}} + \frac {5 \log {\left (x - 4 \right )}}{4 e^{8}} + \frac {\left (- x^{3} - 10 x^{2} - 100 x - 250\right ) \log {\left (x - 4 \right )}}{8 x^{2} e^{8} + 80 x e^{8} + 200 e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2-20*x)*ln(x-4)**2+(-x**4-10*x**3+65*x**2)*ln(x-4)+x**5+5*x**4-45*x**3)/(8*x**4+88*x**3+120*x
**2-1400*x-4000)/exp(4)**2,x)

[Out]

x**2*exp(-8)/16 + x**2*log(x - 4)**2/(16*x**2*exp(8) + 160*x*exp(8) + 400*exp(8)) - 5*x*exp(-8)/8 + (-500*x -
1875)/(16*x**2*exp(8) + 160*x*exp(8) + 400*exp(8)) + 5*exp(-8)*log(x - 4)/4 + (-x**3 - 10*x**2 - 100*x - 250)*
log(x - 4)/(8*x**2*exp(8) + 80*x*exp(8) + 200*exp(8))

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