3.48.12 \(\int \frac {-3 x^3+e^3 (10 x^2-x^3-x^4)+e^x (-6 x^3+e^3 (-10 x^2-2 x^3-2 x^4))+e^{2 x} (6 x^3+e^3 (2 x^3+2 x^4))+(-30 x+e^3 (40 x-10 x^2)+e^x (30 x-30 x^2+e^3 (10 x-10 x^3))) \log (3+e^3 (1+x))+(-150+e^3 (-50-50 x)) \log ^2(3+e^3 (1+x))}{3 x^3+e^3 (x^3+x^4)} \, dx\)

Optimal. Leaf size=26 \[ -x+\left (-1+e^x-\frac {5 \log \left (3+e^3 (1+x)\right )}{x}\right )^2 \]

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Rubi [B]  time = 4.52, antiderivative size = 218, normalized size of antiderivative = 8.38, number of steps used = 41, number of rules used = 24, integrand size = 181, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6741, 6742, 2194, 6688, 2178, 2197, 2554, 893, 2418, 2395, 36, 29, 31, 2392, 2391, 2390, 2301, 2398, 2411, 2347, 2344, 2316, 2315, 2314} \begin {gather*} \frac {25 \log ^2\left (e^3 x+e^3+3\right )}{x^2}-2 e^x+e^{2 x}-x-\frac {10 e^x \log \left (e^3 x+e^3+3\right )}{x}+\frac {10 \left (3-4 e^3\right ) \log \left (e^3 x+e^3+3\right )}{\left (3+e^3\right ) x}-\frac {10 e^3 \log \left (e^3 x+e^3+3\right )}{3+e^3}+\frac {10 e^3 \left (3-4 e^3\right ) \log \left (e^3 x+e^3+3\right )}{\left (3+e^3\right )^2}+\frac {10 e^3 \log (x)}{3+e^3}-\frac {10 e^3 \left (3-4 e^3\right ) \log (x)}{\left (3+e^3\right )^2}-\frac {50 e^6 \log (x)}{\left (3+e^3\right )^2}+\frac {50 e^3 \left (e^3 x+e^3+3\right ) \log \left (e^3 (x+1)+3\right )}{\left (3+e^3\right )^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x^3 + E^3*(10*x^2 - x^3 - x^4) + E^x*(-6*x^3 + E^3*(-10*x^2 - 2*x^3 - 2*x^4)) + E^(2*x)*(6*x^3 + E^3*(
2*x^3 + 2*x^4)) + (-30*x + E^3*(40*x - 10*x^2) + E^x*(30*x - 30*x^2 + E^3*(10*x - 10*x^3)))*Log[3 + E^3*(1 + x
)] + (-150 + E^3*(-50 - 50*x))*Log[3 + E^3*(1 + x)]^2)/(3*x^3 + E^3*(x^3 + x^4)),x]

[Out]

-2*E^x + E^(2*x) - x - (50*E^6*Log[x])/(3 + E^3)^2 - (10*E^3*(3 - 4*E^3)*Log[x])/(3 + E^3)^2 + (10*E^3*Log[x])
/(3 + E^3) + (10*E^3*(3 - 4*E^3)*Log[3 + E^3 + E^3*x])/(3 + E^3)^2 - (10*E^3*Log[3 + E^3 + E^3*x])/(3 + E^3) -
(10*E^x*Log[3 + E^3 + E^3*x])/x + (10*(3 - 4*E^3)*Log[3 + E^3 + E^3*x])/((3 + E^3)*x) + (25*Log[3 + E^3 + E^3
*x]^2)/x^2 + (50*E^3*(3 + E^3 + E^3*x)*Log[3 + E^3*(1 + x)])/((3 + E^3)^2*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 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 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 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 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 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 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 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 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 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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{x^3 \left (3+e^3+e^3 x\right )} \, dx\\ &=\int \left (2 e^{2 x}+\frac {2 e^x \left (-5 e^3 x-3 \left (1+\frac {e^3}{3}\right ) x^2-e^3 x^3+15 \left (1+\frac {e^3}{3}\right ) \log \left (3+e^3 (1+x)\right )-15 x \log \left (3+e^3 (1+x)\right )-5 e^3 x^2 \log \left (3+e^3 (1+x)\right )\right )}{x^2 \left (3+e^3+e^3 x\right )}+\frac {10 e^3 x^2-3 \left (1+\frac {e^3}{3}\right ) x^3-e^3 x^4-30 \left (1-\frac {4 e^3}{3}\right ) x \log \left (3+e^3 (1+x)\right )-10 e^3 x^2 \log \left (3+e^3 (1+x)\right )-150 \left (1+\frac {e^3}{3}\right ) \log ^2\left (3+e^3 (1+x)\right )-50 e^3 x \log ^2\left (3+e^3 (1+x)\right )}{x^3 \left (3+e^3+e^3 x\right )}\right ) \, dx\\ &=2 \int e^{2 x} \, dx+2 \int \frac {e^x \left (-5 e^3 x-3 \left (1+\frac {e^3}{3}\right ) x^2-e^3 x^3+15 \left (1+\frac {e^3}{3}\right ) \log \left (3+e^3 (1+x)\right )-15 x \log \left (3+e^3 (1+x)\right )-5 e^3 x^2 \log \left (3+e^3 (1+x)\right )\right )}{x^2 \left (3+e^3+e^3 x\right )} \, dx+\int \frac {10 e^3 x^2-3 \left (1+\frac {e^3}{3}\right ) x^3-e^3 x^4-30 \left (1-\frac {4 e^3}{3}\right ) x \log \left (3+e^3 (1+x)\right )-10 e^3 x^2 \log \left (3+e^3 (1+x)\right )-150 \left (1+\frac {e^3}{3}\right ) \log ^2\left (3+e^3 (1+x)\right )-50 e^3 x \log ^2\left (3+e^3 (1+x)\right )}{x^3 \left (3+e^3+e^3 x\right )} \, dx\\ &=e^{2 x}+2 \int \frac {e^x \left (-x \left (3 x+e^3 \left (5+x+x^2\right )\right )-5 \left (3 (-1+x)+e^3 \left (-1+x^2\right )\right ) \log \left (3+e^3 (1+x)\right )\right )}{x^2 \left (3+e^3+e^3 x\right )} \, dx+\int \frac {-x^2 \left (3 x+e^3 \left (-10+x+x^2\right )\right )-10 \left (3+e^3 (-4+x)\right ) x \log \left (3+e^3 (1+x)\right )-50 \left (3+e^3 (1+x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{x^3 \left (3+e^3+e^3 x\right )} \, dx\\ &=e^{2 x}+2 \int \left (\frac {e^x \left (-5 e^3-\left (3+e^3\right ) x-e^3 x^2\right )}{x \left (3+e^3+e^3 x\right )}+\frac {5 e^x (1-x) \log \left (3+e^3+e^3 x\right )}{x^2}\right ) \, dx+\int \left (\frac {10 e^3-\left (3+e^3\right ) x-e^3 x^2}{x \left (3+e^3+e^3 x\right )}+\frac {10 \left (-3+4 e^3-e^3 x\right ) \log \left (3+e^3+e^3 x\right )}{x^2 \left (3+e^3+e^3 x\right )}-\frac {50 \log ^2\left (3+e^3+e^3 x\right )}{x^3}\right ) \, dx\\ &=e^{2 x}+2 \int \frac {e^x \left (-5 e^3-\left (3+e^3\right ) x-e^3 x^2\right )}{x \left (3+e^3+e^3 x\right )} \, dx+10 \int \frac {e^x (1-x) \log \left (3+e^3+e^3 x\right )}{x^2} \, dx+10 \int \frac {\left (-3+4 e^3-e^3 x\right ) \log \left (3+e^3+e^3 x\right )}{x^2 \left (3+e^3+e^3 x\right )} \, dx-50 \int \frac {\log ^2\left (3+e^3+e^3 x\right )}{x^3} \, dx+\int \frac {10 e^3-\left (3+e^3\right ) x-e^3 x^2}{x \left (3+e^3+e^3 x\right )} \, dx\\ &=e^{2 x}-\frac {10 e^x \log \left (3+e^3+e^3 x\right )}{x}+\frac {25 \log ^2\left (3+e^3+e^3 x\right )}{x^2}+2 \int \left (-e^x-\frac {5 e^{3+x}}{\left (3+e^3\right ) x}+\frac {5 e^{6+x}}{\left (3+e^3\right ) \left (3+e^3+e^3 x\right )}\right ) \, dx+10 \int \frac {e^{3+x}}{x \left (3+e^3+e^3 x\right )} \, dx+10 \int \left (\frac {\left (-3+4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right ) x^2}-\frac {5 e^6 \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right )^2 x}+\frac {5 e^9 \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right )^2 \left (3+e^3+e^3 x\right )}\right ) \, dx-\left (50 e^3\right ) \int \frac {\log \left (3+e^3+e^3 x\right )}{x^2 \left (3+e^3+e^3 x\right )} \, dx+\int \left (-1+\frac {10 e^3}{\left (3+e^3\right ) x}-\frac {10 e^6}{\left (3+e^3\right ) \left (3+e^3+e^3 x\right )}\right ) \, dx\\ &=e^{2 x}-x+\frac {10 e^3 \log (x)}{3+e^3}-\frac {10 e^3 \log \left (3+e^3+e^3 x\right )}{3+e^3}-\frac {10 e^x \log \left (3+e^3+e^3 x\right )}{x}+\frac {25 \log ^2\left (3+e^3+e^3 x\right )}{x^2}-2 \int e^x \, dx+10 \int \left (\frac {e^{3+x}}{\left (3+e^3\right ) x}-\frac {e^{6+x}}{\left (3+e^3\right ) \left (3+e^3+e^3 x\right )}\right ) \, dx-50 \operatorname {Subst}\left (\int \frac {\log (x)}{x \left (\frac {-3-e^3}{e^3}+\frac {x}{e^3}\right )^2} \, dx,x,3+e^3+e^3 x\right )-\frac {\left (50 e^6\right ) \int \frac {\log \left (3+e^3+e^3 x\right )}{x} \, dx}{\left (3+e^3\right )^2}+\frac {\left (50 e^9\right ) \int \frac {\log \left (3+e^3+e^3 x\right )}{3+e^3+e^3 x} \, dx}{\left (3+e^3\right )^2}-\frac {10 \int \frac {e^{3+x}}{x} \, dx}{3+e^3}+\frac {10 \int \frac {e^{6+x}}{3+e^3+e^3 x} \, dx}{3+e^3}-\frac {\left (10 \left (3-4 e^3\right )\right ) \int \frac {\log \left (3+e^3+e^3 x\right )}{x^2} \, dx}{3+e^3}\\ &=-2 e^x+e^{2 x}-x-\frac {10 e^3 \text {Ei}(x)}{3+e^3}+\frac {10 e^{2-\frac {3}{e^3}} \text {Ei}\left (\frac {3+e^3+e^3 x}{e^3}\right )}{3+e^3}+\frac {10 e^3 \log (x)}{3+e^3}-\frac {50 e^6 \log \left (3+e^3\right ) \log (x)}{\left (3+e^3\right )^2}-\frac {10 e^3 \log \left (3+e^3+e^3 x\right )}{3+e^3}-\frac {10 e^x \log \left (3+e^3+e^3 x\right )}{x}+\frac {10 \left (3-4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right ) x}+\frac {25 \log ^2\left (3+e^3+e^3 x\right )}{x^2}-\frac {\left (50 e^6\right ) \int \frac {\log \left (1+\frac {e^3 x}{3+e^3}\right )}{x} \, dx}{\left (3+e^3\right )^2}+\frac {\left (50 e^6\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}+\frac {10 \int \frac {e^{3+x}}{x} \, dx}{3+e^3}-\frac {10 \int \frac {e^{6+x}}{3+e^3+e^3 x} \, dx}{3+e^3}-\frac {50 \operatorname {Subst}\left (\int \frac {\log (x)}{\left (\frac {-3-e^3}{e^3}+\frac {x}{e^3}\right )^2} \, dx,x,3+e^3+e^3 x\right )}{3+e^3}+\frac {\left (50 e^3\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x \left (\frac {-3-e^3}{e^3}+\frac {x}{e^3}\right )} \, dx,x,3+e^3+e^3 x\right )}{3+e^3}-\frac {\left (10 e^3 \left (3-4 e^3\right )\right ) \int \frac {1}{x \left (3+e^3+e^3 x\right )} \, dx}{3+e^3}\\ &=-2 e^x+e^{2 x}-x+\frac {10 e^3 \log (x)}{3+e^3}-\frac {50 e^6 \log \left (3+e^3\right ) \log (x)}{\left (3+e^3\right )^2}-\frac {10 e^3 \log \left (3+e^3+e^3 x\right )}{3+e^3}-\frac {10 e^x \log \left (3+e^3+e^3 x\right )}{x}+\frac {10 \left (3-4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right ) x}+\frac {25 \log ^2\left (3+e^3+e^3 x\right )}{x^2}+\frac {50 e^3 \left (3+e^3+e^3 x\right ) \log \left (3+e^3 (1+x)\right )}{\left (3+e^3\right )^2 x}+\frac {25 e^6 \log ^2\left (3+e^3 (1+x)\right )}{\left (3+e^3\right )^2}+\frac {50 e^6 \text {Li}_2\left (-\frac {e^3 x}{3+e^3}\right )}{\left (3+e^3\right )^2}-\frac {\left (50 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {-3-e^3}{e^3}+\frac {x}{e^3}} \, dx,x,3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}+\frac {\left (50 e^3\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{\frac {-3-e^3}{e^3}+\frac {x}{e^3}} \, dx,x,3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}-\frac {\left (50 e^6\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}-\frac {\left (10 e^3 \left (3-4 e^3\right )\right ) \int \frac {1}{x} \, dx}{\left (3+e^3\right )^2}+\frac {\left (10 e^6 \left (3-4 e^3\right )\right ) \int \frac {1}{3+e^3+e^3 x} \, dx}{\left (3+e^3\right )^2}\\ &=-2 e^x+e^{2 x}-x-\frac {50 e^6 \log (x)}{\left (3+e^3\right )^2}-\frac {10 e^3 \left (3-4 e^3\right ) \log (x)}{\left (3+e^3\right )^2}+\frac {10 e^3 \log (x)}{3+e^3}+\frac {10 e^3 \left (3-4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}-\frac {10 e^3 \log \left (3+e^3+e^3 x\right )}{3+e^3}-\frac {10 e^x \log \left (3+e^3+e^3 x\right )}{x}+\frac {10 \left (3-4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right ) x}+\frac {25 \log ^2\left (3+e^3+e^3 x\right )}{x^2}+\frac {50 e^3 \left (3+e^3+e^3 x\right ) \log \left (3+e^3 (1+x)\right )}{\left (3+e^3\right )^2 x}+\frac {50 e^6 \text {Li}_2\left (-\frac {e^3 x}{3+e^3}\right )}{\left (3+e^3\right )^2}+\frac {\left (50 e^3\right ) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {x}{-3-e^3}\right )}{\frac {-3-e^3}{e^3}+\frac {x}{e^3}} \, dx,x,3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}\\ &=-2 e^x+e^{2 x}-x-\frac {50 e^6 \log (x)}{\left (3+e^3\right )^2}-\frac {10 e^3 \left (3-4 e^3\right ) \log (x)}{\left (3+e^3\right )^2}+\frac {10 e^3 \log (x)}{3+e^3}+\frac {10 e^3 \left (3-4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right )^2}-\frac {10 e^3 \log \left (3+e^3+e^3 x\right )}{3+e^3}-\frac {10 e^x \log \left (3+e^3+e^3 x\right )}{x}+\frac {10 \left (3-4 e^3\right ) \log \left (3+e^3+e^3 x\right )}{\left (3+e^3\right ) x}+\frac {25 \log ^2\left (3+e^3+e^3 x\right )}{x^2}+\frac {50 e^3 \left (3+e^3+e^3 x\right ) \log \left (3+e^3 (1+x)\right )}{\left (3+e^3\right )^2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 51, normalized size = 1.96 \begin {gather*} -2 e^x+e^{2 x}-x-\frac {10 \left (-1+e^x\right ) \log \left (3+e^3 (1+x)\right )}{x}+\frac {25 \log ^2\left (3+e^3 (1+x)\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^3 + E^3*(10*x^2 - x^3 - x^4) + E^x*(-6*x^3 + E^3*(-10*x^2 - 2*x^3 - 2*x^4)) + E^(2*x)*(6*x^3 +
E^3*(2*x^3 + 2*x^4)) + (-30*x + E^3*(40*x - 10*x^2) + E^x*(30*x - 30*x^2 + E^3*(10*x - 10*x^3)))*Log[3 + E^3*
(1 + x)] + (-150 + E^3*(-50 - 50*x))*Log[3 + E^3*(1 + x)]^2)/(3*x^3 + E^3*(x^3 + x^4)),x]

[Out]

-2*E^x + E^(2*x) - x - (10*(-1 + E^x)*Log[3 + E^3*(1 + x)])/x + (25*Log[3 + E^3*(1 + x)]^2)/x^2

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fricas [B]  time = 0.62, size = 57, normalized size = 2.19 \begin {gather*} -\frac {x^{3} - x^{2} e^{\left (2 \, x\right )} + 2 \, x^{2} e^{x} + 10 \, {\left (x e^{x} - x\right )} \log \left ({\left (x + 1\right )} e^{3} + 3\right ) - 25 \, \log \left ({\left (x + 1\right )} e^{3} + 3\right )^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*x-50)*exp(3)-150)*log((x+1)*exp(3)+3)^2+(((-10*x^3+10*x)*exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+
40*x)*exp(3)-30*x)*log((x+1)*exp(3)+3)+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x
^3)*exp(x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x, algorithm="fricas")

[Out]

-(x^3 - x^2*e^(2*x) + 2*x^2*e^x + 10*(x*e^x - x)*log((x + 1)*e^3 + 3) - 25*log((x + 1)*e^3 + 3)^2)/x^2

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giac [B]  time = 0.26, size = 64, normalized size = 2.46 \begin {gather*} -\frac {x^{3} - x^{2} e^{\left (2 \, x\right )} + 2 \, x^{2} e^{x} + 10 \, x e^{x} \log \left (x e^{3} + e^{3} + 3\right ) - 10 \, x \log \left (x e^{3} + e^{3} + 3\right ) - 25 \, \log \left (x e^{3} + e^{3} + 3\right )^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*x-50)*exp(3)-150)*log((x+1)*exp(3)+3)^2+(((-10*x^3+10*x)*exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+
40*x)*exp(3)-30*x)*log((x+1)*exp(3)+3)+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x
^3)*exp(x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x, algorithm="giac")

[Out]

-(x^3 - x^2*e^(2*x) + 2*x^2*e^x + 10*x*e^x*log(x*e^3 + e^3 + 3) - 10*x*log(x*e^3 + e^3 + 3) - 25*log(x*e^3 + e
^3 + 3)^2)/x^2

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maple [A]  time = 0.40, size = 47, normalized size = 1.81

method	result	size
risch	\(\frac {25 \ln \left (\left (x +1\right ) {\mathrm e}^{3}+3\right )^{2}}{x^{2}}-\frac {10 \left ({\mathrm e}^{x}-1\right ) \ln \left (\left (x +1\right ) {\mathrm e}^{3}+3\right )}{x}+{\mathrm e}^{2 x}-x -2 \,{\mathrm e}^{x}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-50*x-50)*exp(3)-150)*ln((x+1)*exp(3)+3)^2+(((-10*x^3+10*x)*exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+40*x)*e
xp(3)-30*x)*ln((x+1)*exp(3)+3)+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x^3)*exp(
x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x,method=_RETURNVERBOSE)

[Out]

25/x^2*ln((x+1)*exp(3)+3)^2-10*(exp(x)-1)/x*ln((x+1)*exp(3)+3)+exp(2*x)-x-2*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} {\left ({\left (e^{3} + 3\right )} e^{\left (-6\right )} \log \left (x e^{3} + e^{3} + 3\right ) - x e^{\left (-3\right )}\right )} e^{3} - 10 \, {\left (\frac {\log \left (x e^{3} + e^{3} + 3\right )}{e^{3} + 3} - \frac {\log \relax (x)}{e^{3} + 3}\right )} e^{3} + 6 \, e^{\left (-{\left (e^{3} + 3\right )} e^{\left (-3\right )} - 3\right )} E_{1}\left (-{\left (x e^{3} + e^{3} + 3\right )} e^{\left (-3\right )}\right ) - 6 \, e^{\left (-2 \, {\left (e^{3} + 3\right )} e^{\left (-3\right )} - 3\right )} E_{1}\left (-2 \, {\left (x e^{3} + e^{3} + 3\right )} e^{\left (-3\right )}\right ) - 3 \, e^{\left (-3\right )} \log \left (x e^{3} + e^{3} + 3\right ) - \frac {10 \, e^{3} \log \relax (x)}{e^{3} + 3} + \frac {x^{3} {\left (e^{6} + 3 \, e^{3}\right )} e^{\left (2 \, x\right )} - 2 \, x^{3} {\left (e^{6} + 3 \, e^{3}\right )} e^{x} + 25 \, {\left (x {\left (e^{6} + 3 \, e^{3}\right )} + e^{6} + 6 \, e^{3} + 9\right )} \log \left (x e^{3} + e^{3} + 3\right )^{2} + 10 \, {\left (x^{3} e^{6} + 2 \, x^{2} {\left (e^{6} + 3 \, e^{3}\right )} + x {\left (e^{6} + 6 \, e^{3} + 9\right )} - {\left (x^{2} {\left (e^{6} + 3 \, e^{3}\right )} + x {\left (e^{6} + 6 \, e^{3} + 9\right )}\right )} e^{x}\right )} \log \left (x e^{3} + e^{3} + 3\right )}{x^{3} {\left (e^{6} + 3 \, e^{3}\right )} + x^{2} {\left (e^{6} + 6 \, e^{3} + 9\right )}} + \int \frac {{\left (2 \, x e^{6} + e^{6} + 3 \, e^{3}\right )} e^{\left (2 \, x\right )}}{x^{2} e^{6} + 2 \, x {\left (e^{6} + 3 \, e^{3}\right )} + e^{6} + 6 \, e^{3} + 9}\,{d x} - 2 \, \int \frac {x e^{\left (x + 6\right )}}{x^{2} e^{6} + 2 \, x {\left (e^{6} + 3 \, e^{3}\right )} + e^{6} + 6 \, e^{3} + 9}\,{d x} - \log \left (x e^{3} + e^{3} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*x-50)*exp(3)-150)*log((x+1)*exp(3)+3)^2+(((-10*x^3+10*x)*exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+
40*x)*exp(3)-30*x)*log((x+1)*exp(3)+3)+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x
^3)*exp(x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x, algorithm="maxima")

[Out]

((e^3 + 3)*e^(-6)*log(x*e^3 + e^3 + 3) - x*e^(-3))*e^3 - 10*(log(x*e^3 + e^3 + 3)/(e^3 + 3) - log(x)/(e^3 + 3)
)*e^3 + 6*e^(-(e^3 + 3)*e^(-3) - 3)*exp_integral_e(1, -(x*e^3 + e^3 + 3)*e^(-3)) - 6*e^(-2*(e^3 + 3)*e^(-3) -
3)*exp_integral_e(1, -2*(x*e^3 + e^3 + 3)*e^(-3)) - 3*e^(-3)*log(x*e^3 + e^3 + 3) - 10*e^3*log(x)/(e^3 + 3) +
(x^3*(e^6 + 3*e^3)*e^(2*x) - 2*x^3*(e^6 + 3*e^3)*e^x + 25*(x*(e^6 + 3*e^3) + e^6 + 6*e^3 + 9)*log(x*e^3 + e^3
+ 3)^2 + 10*(x^3*e^6 + 2*x^2*(e^6 + 3*e^3) + x*(e^6 + 6*e^3 + 9) - (x^2*(e^6 + 3*e^3) + x*(e^6 + 6*e^3 + 9))*e
^x)*log(x*e^3 + e^3 + 3))/(x^3*(e^6 + 3*e^3) + x^2*(e^6 + 6*e^3 + 9)) + integrate((2*x*e^6 + e^6 + 3*e^3)*e^(2
*x)/(x^2*e^6 + 2*x*(e^6 + 3*e^3) + e^6 + 6*e^3 + 9), x) - 2*integrate(x*e^(x + 6)/(x^2*e^6 + 2*x*(e^6 + 3*e^3)
+ e^6 + 6*e^3 + 9), x) - log(x*e^3 + e^3 + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^3\,\left (x^4+x^3-10\,x^2\right )-\ln \left ({\mathrm {e}}^3\,\left (x+1\right )+3\right )\,\left ({\mathrm {e}}^3\,\left (40\,x-10\,x^2\right )-30\,x+{\mathrm {e}}^x\,\left (30\,x+{\mathrm {e}}^3\,\left (10\,x-10\,x^3\right )-30\,x^2\right )\right )+{\ln \left ({\mathrm {e}}^3\,\left (x+1\right )+3\right )}^2\,\left ({\mathrm {e}}^3\,\left (50\,x+50\right )+150\right )+{\mathrm {e}}^x\,\left ({\mathrm {e}}^3\,\left (2\,x^4+2\,x^3+10\,x^2\right )+6\,x^3\right )-{\mathrm {e}}^{2\,x}\,\left ({\mathrm {e}}^3\,\left (2\,x^4+2\,x^3\right )+6\,x^3\right )+3\,x^3}{{\mathrm {e}}^3\,\left (x^4+x^3\right )+3\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*(x^3 - 10*x^2 + x^4) - log(exp(3)*(x + 1) + 3)*(exp(3)*(40*x - 10*x^2) - 30*x + exp(x)*(30*x + ex
p(3)*(10*x - 10*x^3) - 30*x^2)) + log(exp(3)*(x + 1) + 3)^2*(exp(3)*(50*x + 50) + 150) + exp(x)*(exp(3)*(10*x^
2 + 2*x^3 + 2*x^4) + 6*x^3) - exp(2*x)*(exp(3)*(2*x^3 + 2*x^4) + 6*x^3) + 3*x^3)/(exp(3)*(x^3 + x^4) + 3*x^3),
x)

[Out]

int(-(exp(3)*(x^3 - 10*x^2 + x^4) - log(exp(3)*(x + 1) + 3)*(exp(3)*(40*x - 10*x^2) - 30*x + exp(x)*(30*x + ex
p(3)*(10*x - 10*x^3) - 30*x^2)) + log(exp(3)*(x + 1) + 3)^2*(exp(3)*(50*x + 50) + 150) + exp(x)*(exp(3)*(10*x^
2 + 2*x^3 + 2*x^4) + 6*x^3) - exp(2*x)*(exp(3)*(2*x^3 + 2*x^4) + 6*x^3) + 3*x^3)/(exp(3)*(x^3 + x^4) + 3*x^3),
x)

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sympy [B]  time = 0.60, size = 60, normalized size = 2.31 \begin {gather*} - x + \frac {x e^{2 x} + \left (- 2 x - 10 \log {\left (\left (x + 1\right ) e^{3} + 3 \right )}\right ) e^{x}}{x} + \frac {10 \log {\left (\left (x + 1\right ) e^{3} + 3 \right )}}{x} + \frac {25 \log {\left (\left (x + 1\right ) e^{3} + 3 \right )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-50*x-50)*exp(3)-150)*ln((x+1)*exp(3)+3)**2+(((-10*x**3+10*x)*exp(3)-30*x**2+30*x)*exp(x)+(-10*x*
*2+40*x)*exp(3)-30*x)*ln((x+1)*exp(3)+3)+((2*x**4+2*x**3)*exp(3)+6*x**3)*exp(x)**2+((-2*x**4-2*x**3-10*x**2)*e
xp(3)-6*x**3)*exp(x)+(-x**4-x**3+10*x**2)*exp(3)-3*x**3)/((x**4+x**3)*exp(3)+3*x**3),x)

[Out]

-x + (x*exp(2*x) + (-2*x - 10*log((x + 1)*exp(3) + 3))*exp(x))/x + 10*log((x + 1)*exp(3) + 3)/x + 25*log((x +
1)*exp(3) + 3)**2/x**2

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