3.52.45 \(\int \frac {4 x+e^{2 x} x+x^2+x \log ^2(4 e^3 x)+(5 x+x^2+e^x (10+2 x)+e^{2 x} (10 x+2 x^2)) \log (5+x)+\log (4 e^3 x) (2 e^x x+(10+2 x+e^x (10 x+2 x^2)) \log (5+x))}{5 x+x^2} \, dx\)

Optimal. Leaf size=21 \[ \left (4+x+\left (e^x+\log \left (4 e^3 x\right )\right )^2\right ) \log (5+x) \]

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Rubi [C]  time = 5.03, antiderivative size = 122, normalized size of antiderivative = 5.81, number of steps used = 49, number of rules used = 23, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1593, 6742, 43, 2317, 2374, 6589, 2390, 2301, 2411, 2346, 2295, 2433, 2425, 2344, 2391, 2375, 2288, 6688, 2178, 2554, 12, 2194, 2557} \begin {gather*} \frac {6 \text {Ei}(x+5)}{e^5}-\frac {2 \text {Ei}(x+5) \log (x)}{e^5}+\frac {2 \text {Ei}(x+5) \log (4 x)}{e^5}-\frac {2 (3+\log (4)) \text {Ei}(x+5)}{e^5}+\log (x+5) (\log (x)+3+\log (4))^2+(x+5) \log (x+5)+2 e^x \log (x) \log (x+5)+2 e^x (3+\log (4)) \log (x+5)-\log (x+5)+\frac {e^{2 x} (x \log (x+5)+5 \log (x+5))}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + E^(2*x)*x + x^2 + x*Log[4*E^3*x]^2 + (5*x + x^2 + E^x*(10 + 2*x) + E^(2*x)*(10*x + 2*x^2))*Log[5 +
x] + Log[4*E^3*x]*(2*E^x*x + (10 + 2*x + E^x*(10*x + 2*x^2))*Log[5 + x]))/(5*x + x^2),x]

[Out]

(6*ExpIntegralEi[5 + x])/E^5 - (2*ExpIntegralEi[5 + x]*(3 + Log[4]))/E^5 - (2*ExpIntegralEi[5 + x]*Log[x])/E^5
+ (2*ExpIntegralEi[5 + x]*Log[4*x])/E^5 - Log[5 + x] + (5 + x)*Log[5 + x] + 2*E^x*(3 + Log[4])*Log[5 + x] + 2
*E^x*Log[x]*Log[5 + x] + (3 + Log[4] + Log[x])^2*Log[5 + x] + (E^(2*x)*(5*Log[5 + x] + x*Log[5 + x]))/(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 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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 2194

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 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 2346

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

Rule 2374

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

Rule 2375

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

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 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 2425

Int[(Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)))/(x_), x_Symbol] :> Simp[(Log[
f*x^m]^2*(a + b*Log[c*(d + e*x)^n]))/(2*m), x] - Dist[(b*e*n)/(2*m), Int[Log[f*x^m]^2/(d + e*x), x], x] /; Fre
eQ[{a, b, c, d, e, f, m, n}, x]

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

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 6589

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

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 {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{x (5+x)} \, dx\\ &=\int \left (\frac {4}{5+x}+\frac {x}{5+x}+\frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right )^2}{5+x}+\frac {5 \log (5+x)}{5+x}+\frac {x \log (5+x)}{5+x}+\frac {2 \left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right ) \log (5+x)}{5+x}+\frac {10 \left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right ) \log (5+x)}{x (5+x)}+\frac {e^{2 x} (1+10 \log (5+x)+2 x \log (5+x))}{5+x}+\frac {2 e^x \left (3 x \left (1+\frac {2 \log (2)}{3}\right )+x \log (x)+5 \log (5+x)+16 x \left (1+\frac {5 \log (2)}{8}\right ) \log (5+x)+3 x^2 \left (1+\frac {2 \log (2)}{3}\right ) \log (5+x)+5 x \log (x) \log (5+x)+x^2 \log (x) \log (5+x)\right )}{x (5+x)}\right ) \, dx\\ &=4 \log (5+x)+2 \int \frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right ) \log (5+x)}{5+x} \, dx+2 \int \frac {e^x \left (3 x \left (1+\frac {2 \log (2)}{3}\right )+x \log (x)+5 \log (5+x)+16 x \left (1+\frac {5 \log (2)}{8}\right ) \log (5+x)+3 x^2 \left (1+\frac {2 \log (2)}{3}\right ) \log (5+x)+5 x \log (x) \log (5+x)+x^2 \log (x) \log (5+x)\right )}{x (5+x)} \, dx+5 \int \frac {\log (5+x)}{5+x} \, dx+10 \int \frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right ) \log (5+x)}{x (5+x)} \, dx+\int \frac {x}{5+x} \, dx+\int \frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right )^2}{5+x} \, dx+\int \frac {x \log (5+x)}{5+x} \, dx+\int \frac {e^{2 x} (1+10 \log (5+x)+2 x \log (5+x))}{5+x} \, dx\\ &=\log \left (1+\frac {x}{5}\right ) (3+\log (4)+\log (x))^2+4 \log (5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}-2 \int \frac {\log \left (1+\frac {x}{5}\right ) \left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right )}{x} \, dx+2 \int \frac {e^x \left (x (3+\log (4))+\left (5+x^2 (3+\log (4))+2 x (8+\log (32))\right ) \log (5+x)+\log (x) (x+x (5+x) \log (5+x))\right )}{x (5+x)} \, dx+2 \operatorname {Subst}\left (\int \frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (-5+x)\right ) \log (x)}{x} \, dx,x,5+x\right )+5 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x\right )+10 \int \left (\frac {\left (-3 \left (1+\frac {2 \log (2)}{3}\right )-\log (x)\right ) \log (5+x)}{5 (5+x)}+\frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right ) \log (5+x)}{5 x}\right ) \, dx+\int \left (1-\frac {5}{5+x}\right ) \, dx+\operatorname {Subst}\left (\int \frac {(-5+x) \log (x)}{x} \, dx,x,5+x\right )\\ &=x+\log \left (1+\frac {x}{5}\right ) (3+\log (4)+\log (x))^2-\log (5+x)+\frac {5}{2} \log ^2(5+x)+(3+\log (4)+\log (x)) \log ^2(5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}+2 (3+\log (4)+\log (x)) \text {Li}_2\left (-\frac {x}{5}\right )+2 \int \frac {\left (-3 \left (1+\frac {2 \log (2)}{3}\right )-\log (x)\right ) \log (5+x)}{5+x} \, dx+2 \int \frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right ) \log (5+x)}{x} \, dx+2 \int \left (\frac {e^x (3+\log (4 x))}{5+x}+\frac {e^x \left (5+16 x \left (1+\frac {5 \log (2)}{8}\right )+3 x^2 \left (1+\frac {2 \log (2)}{3}\right )+5 x \log (x)+x^2 \log (x)\right ) \log (5+x)}{x (5+x)}\right ) \, dx-2 \int \frac {\text {Li}_2\left (-\frac {x}{5}\right )}{x} \, dx-5 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x\right )+\operatorname {Subst}(\int \log (x) \, dx,x,5+x)-\operatorname {Subst}\left (\int \frac {\log ^2(x)}{-5+x} \, dx,x,5+x\right )\\ &=\log \left (1+\frac {x}{5}\right ) (3+\log (4)+\log (x))^2-\log (5+x)+(5+x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)-\log \left (-\frac {x}{5}\right ) \log ^2(5+x)+(3+\log (4)+\log (x)) \log ^2(5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}+2 (3+\log (4)+\log (x)) \text {Li}_2\left (-\frac {x}{5}\right )-2 \text {Li}_3\left (-\frac {x}{5}\right )+2 \int \frac {e^x (3+\log (4 x))}{5+x} \, dx+2 \int \frac {e^x \left (5+16 x \left (1+\frac {5 \log (2)}{8}\right )+3 x^2 \left (1+\frac {2 \log (2)}{3}\right )+5 x \log (x)+x^2 \log (x)\right ) \log (5+x)}{x (5+x)} \, dx+2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right ) \log (x)}{x} \, dx,x,5+x\right )+2 \operatorname {Subst}\left (\int \frac {\left (-3 \left (1+\frac {2 \log (2)}{3}\right )-\log (-5+x)\right ) \log (x)}{x} \, dx,x,5+x\right )-\int \frac {\left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right )^2}{5+x} \, dx\\ &=-\log (5+x)+(5+x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)-\log \left (-\frac {x}{5}\right ) \log ^2(5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}+2 (3+\log (4)+\log (x)) \text {Li}_2\left (-\frac {x}{5}\right )-2 \log (5+x) \text {Li}_2\left (\frac {5+x}{5}\right )-2 \text {Li}_3\left (-\frac {x}{5}\right )+2 \int \frac {\log \left (1+\frac {x}{5}\right ) \left (3 \left (1+\frac {2 \log (2)}{3}\right )+\log (x)\right )}{x} \, dx+2 \int \left (\frac {3 e^x}{5+x}+\frac {e^x \log (4 x)}{5+x}\right ) \, dx+2 \int \frac {e^x (1+x (3+\log (4))+x \log (x)) \log (5+x)}{x} \, dx+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{5}\right )}{x} \, dx,x,5+x\right )+\operatorname {Subst}\left (\int \frac {\log ^2(x)}{-5+x} \, dx,x,5+x\right )\\ &=-\log (5+x)+(5+x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}-2 \log (5+x) \text {Li}_2\left (\frac {5+x}{5}\right )-2 \text {Li}_3\left (-\frac {x}{5}\right )+2 \text {Li}_3\left (\frac {5+x}{5}\right )+2 \int \frac {e^x \log (4 x)}{5+x} \, dx+2 \int \left (\frac {e^x \log (5+x)}{x}+e^x (3+\log (4)) \log (5+x)+e^x \log (x) \log (5+x)\right ) \, dx+2 \int \frac {\text {Li}_2\left (-\frac {x}{5}\right )}{x} \, dx-2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right ) \log (x)}{x} \, dx,x,5+x\right )+6 \int \frac {e^x}{5+x} \, dx\\ &=\frac {6 \text {Ei}(5+x)}{e^5}+\frac {2 \text {Ei}(5+x) \log (4 x)}{e^5}-\log (5+x)+(5+x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}+2 \text {Li}_3\left (\frac {5+x}{5}\right )-2 \int \frac {\text {Ei}(5+x)}{e^5 x} \, dx+2 \int \frac {e^x \log (5+x)}{x} \, dx+2 \int e^x \log (x) \log (5+x) \, dx-2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{5}\right )}{x} \, dx,x,5+x\right )+(2 (3+\log (4))) \int e^x \log (5+x) \, dx\\ &=\frac {6 \text {Ei}(5+x)}{e^5}+\frac {2 \text {Ei}(5+x) \log (4 x)}{e^5}-\log (5+x)+(5+x) \log (5+x)+2 \text {Ei}(x) \log (5+x)+2 e^x (3+\log (4)) \log (5+x)+2 e^x \log (x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}-2 \int \frac {\text {Ei}(x)}{5+x} \, dx-2 \int \frac {e^x \log (x)}{5+x} \, dx-2 \int \frac {e^x \log (5+x)}{x} \, dx-\frac {2 \int \frac {\text {Ei}(5+x)}{x} \, dx}{e^5}-(2 (3+\log (4))) \int \frac {e^x}{5+x} \, dx\\ &=\frac {6 \text {Ei}(5+x)}{e^5}-\frac {2 \text {Ei}(5+x) (3+\log (4))}{e^5}-\frac {2 \text {Ei}(5+x) \log (x)}{e^5}+\frac {2 \text {Ei}(5+x) \log (4 x)}{e^5}-\log (5+x)+(5+x) \log (5+x)+2 e^x (3+\log (4)) \log (5+x)+2 e^x \log (x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}+2 \int \frac {\text {Ei}(5+x)}{e^5 x} \, dx-\frac {2 \int \frac {\text {Ei}(5+x)}{x} \, dx}{e^5}\\ &=\frac {6 \text {Ei}(5+x)}{e^5}-\frac {2 \text {Ei}(5+x) (3+\log (4))}{e^5}-\frac {2 \text {Ei}(5+x) \log (x)}{e^5}+\frac {2 \text {Ei}(5+x) \log (4 x)}{e^5}-\log (5+x)+(5+x) \log (5+x)+2 e^x (3+\log (4)) \log (5+x)+2 e^x \log (x) \log (5+x)+(3+\log (4)+\log (x))^2 \log (5+x)+\frac {e^{2 x} (5 \log (5+x)+x \log (5+x))}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.22, size = 45, normalized size = 2.14 \begin {gather*} \left (13+e^{2 x}+x+6 \log (4)+\log ^2(4)+e^x (6+\log (16))+\left (6+2 e^x+\log (16)\right ) \log (x)+\log ^2(x)\right ) \log (5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + E^(2*x)*x + x^2 + x*Log[4*E^3*x]^2 + (5*x + x^2 + E^x*(10 + 2*x) + E^(2*x)*(10*x + 2*x^2))*Lo
g[5 + x] + Log[4*E^3*x]*(2*E^x*x + (10 + 2*x + E^x*(10*x + 2*x^2))*Log[5 + x]))/(5*x + x^2),x]

[Out]

(13 + E^(2*x) + x + 6*Log[4] + Log[4]^2 + E^x*(6 + Log[16]) + (6 + 2*E^x + Log[16])*Log[x] + Log[x]^2)*Log[5 +
x]

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fricas [B]  time = 0.53, size = 40, normalized size = 1.90 \begin {gather*} 2 \, e^{x} \log \left (4 \, x e^{3}\right ) \log \left (x + 5\right ) + \log \left (4 \, x e^{3}\right )^{2} \log \left (x + 5\right ) + {\left (x + e^{\left (2 \, x\right )} + 4\right )} \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*exp(x)*x)*log(4*x*exp(3))+((2*x^2+10*x
)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*log(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x, algorithm="fricas")

[Out]

2*e^x*log(4*x*e^3)*log(x + 5) + log(4*x*e^3)^2*log(x + 5) + (x + e^(2*x) + 4)*log(x + 5)

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giac [B]  time = 1.02, size = 95, normalized size = 4.52 \begin {gather*} 4 \, e^{x} \log \relax (2) \log \left (x + 5\right ) + 4 \, \log \relax (2)^{2} \log \left (x + 5\right ) + 2 \, e^{x} \log \left (x + 5\right ) \log \relax (x) + 4 \, \log \relax (2) \log \left (x + 5\right ) \log \relax (x) + \log \left (x + 5\right ) \log \relax (x)^{2} + x \log \left (x + 5\right ) + e^{\left (2 \, x\right )} \log \left (x + 5\right ) + 6 \, e^{x} \log \left (x + 5\right ) + 12 \, \log \relax (2) \log \left (x + 5\right ) + 6 \, \log \left (x + 5\right ) \log \relax (x) + 13 \, \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*exp(x)*x)*log(4*x*exp(3))+((2*x^2+10*x
)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*log(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x, algorithm="giac")

[Out]

4*e^x*log(2)*log(x + 5) + 4*log(2)^2*log(x + 5) + 2*e^x*log(x + 5)*log(x) + 4*log(2)*log(x + 5)*log(x) + log(x
+ 5)*log(x)^2 + x*log(x + 5) + e^(2*x)*log(x + 5) + 6*e^x*log(x + 5) + 12*log(2)*log(x + 5) + 6*log(x + 5)*lo
g(x) + 13*log(x + 5)

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maple [B]  time = 0.17, size = 50, normalized size = 2.38

method	result	size
risch	\(\ln \left (5+x \right ) \ln \left (4 x \,{\mathrm e}^{3}\right )^{2}+2 \,{\mathrm e}^{x} \ln \left (5+x \right ) \ln \left (4 x \,{\mathrm e}^{3}\right )+{\mathrm e}^{2 x} \ln \left (5+x \right )+x \ln \left (5+x \right )+4 \ln \left (5+x \right )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*ln(5+x)+2*exp(x)*x)*ln(4*x*exp(3))+((2*x^2+10*x)*exp(x)^
2+(2*x+10)*exp(x)+x^2+5*x)*ln(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

ln(5+x)*ln(4*x*exp(3))^2+2*exp(x)*ln(5+x)*ln(4*x*exp(3))+exp(2*x)*ln(5+x)+x*ln(5+x)+4*ln(5+x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -e^{\left (-10\right )} E_{1}\left (-2 \, x - 10\right ) + {\left (2 \, {\left (2 \, \log \relax (2) + \log \relax (x) + 3\right )} e^{x} + 4 \, \log \relax (2)^{2} + 2 \, {\left (2 \, \log \relax (2) + 3\right )} \log \relax (x) + \log \relax (x)^{2} + x + e^{\left (2 \, x\right )} + 12 \, \log \relax (2) + 14\right )} \log \left (x + 5\right ) - \int \frac {e^{\left (2 \, x\right )}}{x + 5}\,{d x} - \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*exp(x)*x)*log(4*x*exp(3))+((2*x^2+10*x
)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*log(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x, algorithm="maxima")

[Out]

-e^(-10)*exp_integral_e(1, -2*x - 10) + (2*(2*log(2) + log(x) + 3)*e^x + 4*log(2)^2 + 2*(2*log(2) + 3)*log(x)
+ log(x)^2 + x + e^(2*x) + 12*log(2) + 14)*log(x + 5) - integrate(e^(2*x)/(x + 5), x) - log(x + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {4\,x+x\,{\mathrm {e}}^{2\,x}+\ln \left (x+5\right )\,\left (5\,x+{\mathrm {e}}^{2\,x}\,\left (2\,x^2+10\,x\right )+{\mathrm {e}}^x\,\left (2\,x+10\right )+x^2\right )+\ln \left (4\,x\,{\mathrm {e}}^3\right )\,\left (\ln \left (x+5\right )\,\left (2\,x+{\mathrm {e}}^x\,\left (2\,x^2+10\,x\right )+10\right )+2\,x\,{\mathrm {e}}^x\right )+x^2+x\,{\ln \left (4\,x\,{\mathrm {e}}^3\right )}^2}{x^2+5\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + x*exp(2*x) + log(x + 5)*(5*x + exp(2*x)*(10*x + 2*x^2) + exp(x)*(2*x + 10) + x^2) + log(4*x*exp(3))
*(log(x + 5)*(2*x + exp(x)*(10*x + 2*x^2) + 10) + 2*x*exp(x)) + x^2 + x*log(4*x*exp(3))^2)/(5*x + x^2),x)

[Out]

int((4*x + x*exp(2*x) + log(x + 5)*(5*x + exp(2*x)*(10*x + 2*x^2) + exp(x)*(2*x + 10) + x^2) + log(4*x*exp(3))
*(log(x + 5)*(2*x + exp(x)*(10*x + 2*x^2) + 10) + 2*x*exp(x)) + x^2 + x*log(4*x*exp(3))^2)/(5*x + x^2), x)

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sympy [B]  time = 8.19, size = 56, normalized size = 2.67 \begin {gather*} x \log {\left (x + 5 \right )} + e^{2 x} \log {\left (x + 5 \right )} + 2 e^{x} \log {\left (4 x e^{3} \right )} \log {\left (x + 5 \right )} + \log {\left (4 x e^{3} \right )}^{2} \log {\left (x + 5 \right )} + 4 \log {\left (x + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(4*x*exp(3))**2+(((2*x**2+10*x)*exp(x)+2*x+10)*ln(5+x)+2*exp(x)*x)*ln(4*x*exp(3))+((2*x**2+10*x
)*exp(x)**2+(2*x+10)*exp(x)+x**2+5*x)*ln(5+x)+x*exp(x)**2+x**2+4*x)/(x**2+5*x),x)

[Out]

x*log(x + 5) + exp(2*x)*log(x + 5) + 2*exp(x)*log(4*x*exp(3))*log(x + 5) + log(4*x*exp(3))**2*log(x + 5) + 4*l
og(x + 5)

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