3.72.17 \(\int \frac {e^x+x^2-2 x^3+x^4+(-10 x+32 x^2-26 x^3+4 x^4+e^x (-5+2 e^2+x)+e^2 (4 x-12 x^2+8 x^3)) \log (\frac {1}{2} (-5+2 e^2+x))}{-5+2 e^2+x} \, dx\)

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

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Rubi [B]  time = 1.90, antiderivative size = 1057, normalized size of antiderivative = 40.65, number of steps used = 42, number of rules used = 13, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6742, 43, 2411, 12, 2346, 2301, 2295, 2418, 2389, 2395, 2390, 2334, 2288} \begin {gather*} -\frac {1}{4} \left (-x-2 e^2+5\right )^4+\frac {16}{9} \left (5-2 e^2\right ) \left (-x-2 e^2+5\right )^3-\frac {26}{9} \left (-x-2 e^2+5\right )^3-6 \left (5-2 e^2\right )^2 \left (-x-2 e^2+5\right )^2+\frac {39}{2} \left (5-2 e^2\right ) \left (-x-2 e^2+5\right )^2-8 \left (-x-2 e^2+5\right )^2-8 e^2 \left (18-17 e^2+4 e^4\right ) \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right ) \left (-x-2 e^2+5\right )+10 \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right ) \left (-x-2 e^2+5\right )+\frac {x^4}{4}+\frac {1}{3} \left (5-2 e^2\right ) x^3-\frac {8 e^2 x^3}{9}-\frac {2 x^3}{3}+\frac {1}{2} \left (5-2 e^2\right )^2 x^2-\frac {4}{3} e^2 \left (5-2 e^2\right ) x^2-\left (5-2 e^2\right ) x^2-e^2 \left (7-4 e^2\right ) x^2+\frac {x^2}{2}+4 e^2 \left (90-121 e^2+54 e^4-8 e^6\right ) \log ^2\left (\frac {1}{2} \left (x+2 e^2-5\right )\right )-2 \left (5-2 e^2\right )^4 \log ^2\left (\frac {1}{2} \left (x+2 e^2-5\right )\right )+13 \left (5-2 e^2\right )^3 \log ^2\left (\frac {1}{2} \left (x+2 e^2-5\right )\right )-16 \left (5-2 e^2\right )^2 \log ^2\left (\frac {1}{2} \left (x+2 e^2-5\right )\right )-5 \left (5-2 e^2\right ) \log ^2\left (\frac {1}{2} \left (x+2 e^2-5\right )\right )-8 e^2 \left (18-17 e^2+4 e^4\right ) x-15 \left (5-2 e^2\right )^3 x-\frac {8}{3} e^2 \left (5-2 e^2\right )^2 x+76 \left (5-2 e^2\right )^2 x-2 e^2 \left (7-4 e^2\right ) \left (5-2 e^2\right ) x-63 \left (5-2 e^2\right ) x+10 x+\left (5-2 e^2\right )^4 \log \left (-x-2 e^2+5\right )-\frac {8}{3} e^2 \left (5-2 e^2\right )^3 \log \left (-x-2 e^2+5\right )-2 \left (5-2 e^2\right )^3 \log \left (-x-2 e^2+5\right )-2 e^2 \left (7-4 e^2\right ) \left (5-2 e^2\right )^2 \log \left (-x-2 e^2+5\right )+\left (5-2 e^2\right )^2 \log \left (-x-2 e^2+5\right )+\frac {8}{3} e^2 x^3 \log \left (\frac {x}{2}+\frac {1}{2} \left (-5+2 e^2\right )\right )+2 e^2 \left (7-4 e^2\right ) x^2 \log \left (\frac {x}{2}+\frac {1}{2} \left (-5+2 e^2\right )\right )-16 \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right ) \left (-\left (-x-2 e^2+5\right )^2+4 \left (5-2 e^2\right ) \left (-x-2 e^2+5\right )-2 \left (5-2 e^2\right )^2 \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right )\right )+\frac {13}{3} \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right ) \left (2 \left (-x-2 e^2+5\right )^3-9 \left (5-2 e^2\right ) \left (-x-2 e^2+5\right )^2+18 \left (5-2 e^2\right )^2 \left (-x-2 e^2+5\right )-6 \left (5-2 e^2\right )^3 \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right )\right )-\frac {1}{3} \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right ) \left (-3 \left (-x-2 e^2+5\right )^4+16 \left (5-2 e^2\right ) \left (-x-2 e^2+5\right )^3-36 \left (5-2 e^2\right )^2 \left (-x-2 e^2+5\right )^2+48 \left (5-2 e^2\right )^3 \left (-x-2 e^2+5\right )-12 \left (5-2 e^2\right )^4 \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right )\right )-\frac {e^x \left (x \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right )+2 e^2 \log \left (\frac {1}{2} \left (x+2 e^2-5\right )\right )-5 \log \left (x+2 e^2-5\right )+\log (32)\right )}{-x-2 e^2+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x + x^2 - 2*x^3 + x^4 + (-10*x + 32*x^2 - 26*x^3 + 4*x^4 + E^x*(-5 + 2*E^2 + x) + E^2*(4*x - 12*x^2 + 8
*x^3))*Log[(-5 + 2*E^2 + x)/2])/(-5 + 2*E^2 + x),x]

[Out]

-8*(5 - 2*E^2 - x)^2 + (39*(5 - 2*E^2)*(5 - 2*E^2 - x)^2)/2 - 6*(5 - 2*E^2)^2*(5 - 2*E^2 - x)^2 - (26*(5 - 2*E
^2 - x)^3)/9 + (16*(5 - 2*E^2)*(5 - 2*E^2 - x)^3)/9 - (5 - 2*E^2 - x)^4/4 + 10*x - 63*(5 - 2*E^2)*x - 2*E^2*(7
- 4*E^2)*(5 - 2*E^2)*x + 76*(5 - 2*E^2)^2*x - (8*E^2*(5 - 2*E^2)^2*x)/3 - 15*(5 - 2*E^2)^3*x - 8*E^2*(18 - 17
*E^2 + 4*E^4)*x + x^2/2 - E^2*(7 - 4*E^2)*x^2 - (5 - 2*E^2)*x^2 - (4*E^2*(5 - 2*E^2)*x^2)/3 + ((5 - 2*E^2)^2*x
^2)/2 - (2*x^3)/3 - (8*E^2*x^3)/9 + ((5 - 2*E^2)*x^3)/3 + x^4/4 + (5 - 2*E^2)^2*Log[5 - 2*E^2 - x] - 2*E^2*(7
- 4*E^2)*(5 - 2*E^2)^2*Log[5 - 2*E^2 - x] - 2*(5 - 2*E^2)^3*Log[5 - 2*E^2 - x] - (8*E^2*(5 - 2*E^2)^3*Log[5 -
2*E^2 - x])/3 + (5 - 2*E^2)^4*Log[5 - 2*E^2 - x] + 2*E^2*(7 - 4*E^2)*x^2*Log[(-5 + 2*E^2)/2 + x/2] + (8*E^2*x^
3*Log[(-5 + 2*E^2)/2 + x/2])/3 + 10*(5 - 2*E^2 - x)*Log[(-5 + 2*E^2 + x)/2] - 8*E^2*(18 - 17*E^2 + 4*E^4)*(5 -
2*E^2 - x)*Log[(-5 + 2*E^2 + x)/2] - 5*(5 - 2*E^2)*Log[(-5 + 2*E^2 + x)/2]^2 - 16*(5 - 2*E^2)^2*Log[(-5 + 2*E
^2 + x)/2]^2 + 13*(5 - 2*E^2)^3*Log[(-5 + 2*E^2 + x)/2]^2 - 2*(5 - 2*E^2)^4*Log[(-5 + 2*E^2 + x)/2]^2 + 4*E^2*
(90 - 121*E^2 + 54*E^4 - 8*E^6)*Log[(-5 + 2*E^2 + x)/2]^2 - 16*Log[(-5 + 2*E^2 + x)/2]*(4*(5 - 2*E^2)*(5 - 2*E
^2 - x) - (5 - 2*E^2 - x)^2 - 2*(5 - 2*E^2)^2*Log[(-5 + 2*E^2 + x)/2]) + (13*Log[(-5 + 2*E^2 + x)/2]*(18*(5 -
2*E^2)^2*(5 - 2*E^2 - x) - 9*(5 - 2*E^2)*(5 - 2*E^2 - x)^2 + 2*(5 - 2*E^2 - x)^3 - 6*(5 - 2*E^2)^3*Log[(-5 + 2
*E^2 + x)/2]))/3 - (Log[(-5 + 2*E^2 + x)/2]*(48*(5 - 2*E^2)^3*(5 - 2*E^2 - x) - 36*(5 - 2*E^2)^2*(5 - 2*E^2 -
x)^2 + 16*(5 - 2*E^2)*(5 - 2*E^2 - x)^3 - 3*(5 - 2*E^2 - x)^4 - 12*(5 - 2*E^2)^4*Log[(-5 + 2*E^2 + x)/2]))/3 -
(E^x*(Log[32] + 2*E^2*Log[(-5 + 2*E^2 + x)/2] + x*Log[(-5 + 2*E^2 + x)/2] - 5*Log[-5 + 2*E^2 + x]))/(5 - 2*E^
2 - 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 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

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 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 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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {x^2}{-5+2 e^2+x}-\frac {2 x^3}{-5+2 e^2+x}+\frac {x^4}{-5+2 e^2+x}+\frac {10 x \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{5-2 e^2-x}+\frac {4 e^2 (1-2 x) (-1+x) x \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{5-2 e^2-x}+\frac {26 x^3 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{5-2 e^2-x}+\frac {32 x^2 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{-5+2 e^2+x}+\frac {4 x^4 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{-5+2 e^2+x}+\frac {e^x \left (1+\log (32)+2 e^2 \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )+x \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )-5 \log \left (-5+2 e^2+x\right )\right )}{-5+2 e^2+x}\right ) \, dx\\ &=-\left (2 \int \frac {x^3}{-5+2 e^2+x} \, dx\right )+4 \int \frac {x^4 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{-5+2 e^2+x} \, dx+10 \int \frac {x \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{5-2 e^2-x} \, dx+26 \int \frac {x^3 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{5-2 e^2-x} \, dx+32 \int \frac {x^2 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{-5+2 e^2+x} \, dx+\left (4 e^2\right ) \int \frac {(1-2 x) (-1+x) x \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{5-2 e^2-x} \, dx+\int \frac {x^2}{-5+2 e^2+x} \, dx+\int \frac {x^4}{-5+2 e^2+x} \, dx+\int \frac {e^x \left (1+\log (32)+2 e^2 \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )+x \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )-5 \log \left (-5+2 e^2+x\right )\right )}{-5+2 e^2+x} \, dx\\ &=-\frac {e^x \left (\log (32)+2 e^2 \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )+x \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )-5 \log \left (-5+2 e^2+x\right )\right )}{5-2 e^2-x}-2 \int \left (\left (-5+2 e^2\right )^2-\left (-5+2 e^2\right ) x+x^2-\frac {\left (-5+2 e^2\right )^3}{-5+2 e^2+x}\right ) \, dx+8 \operatorname {Subst}\left (\int \frac {\left (5-2 e^2+2 x\right )^4 \log (x)}{2 x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+20 \operatorname {Subst}\left (\int -\frac {\left (5-2 e^2+2 x\right ) \log (x)}{2 x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+52 \operatorname {Subst}\left (\int -\frac {\left (5-2 e^2+2 x\right )^3 \log (x)}{2 x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+64 \operatorname {Subst}\left (\int \frac {\left (5-2 e^2+2 x\right )^2 \log (x)}{2 x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+\left (4 e^2\right ) \int \left (2 \left (18-17 e^2+4 e^4\right ) \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )-\left (-7+4 e^2\right ) x \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+2 x^2 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )-\frac {2 \left (-90+121 e^2-54 e^4+8 e^6\right ) \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{-5+2 e^2+x}\right ) \, dx+\int \left (5 \left (1-\frac {2 e^2}{5}\right )+x+\frac {\left (-5+2 e^2\right )^2}{-5+2 e^2+x}\right ) \, dx+\int \left (-\left (-5+2 e^2\right )^3+\left (-5+2 e^2\right )^2 x-\left (-5+2 e^2\right ) x^2+x^3+\frac {\left (-5+2 e^2\right )^4}{-5+2 e^2+x}\right ) \, dx\\ &=\left (5-2 e^2\right ) x-2 \left (5-2 e^2\right )^2 x+\left (5-2 e^2\right )^3 x+\frac {x^2}{2}-\left (5-2 e^2\right ) x^2+\frac {1}{2} \left (5-2 e^2\right )^2 x^2-\frac {2 x^3}{3}+\frac {1}{3} \left (5-2 e^2\right ) x^3+\frac {x^4}{4}+\left (5-2 e^2\right )^2 \log \left (5-2 e^2-x\right )-2 \left (5-2 e^2\right )^3 \log \left (5-2 e^2-x\right )+\left (5-2 e^2\right )^4 \log \left (5-2 e^2-x\right )-\frac {e^x \left (\log (32)+2 e^2 \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )+x \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )-5 \log \left (-5+2 e^2+x\right )\right )}{5-2 e^2-x}+4 \operatorname {Subst}\left (\int \frac {\left (5-2 e^2+2 x\right )^4 \log (x)}{x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )-10 \operatorname {Subst}\left (\int \frac {\left (5-2 e^2+2 x\right ) \log (x)}{x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )-26 \operatorname {Subst}\left (\int \frac {\left (5-2 e^2+2 x\right )^3 \log (x)}{x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+32 \operatorname {Subst}\left (\int \frac {\left (5-2 e^2+2 x\right )^2 \log (x)}{x} \, dx,x,\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )+\left (8 e^2\right ) \int x^2 \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right ) \, dx+\left (4 e^2 \left (7-4 e^2\right )\right ) \int x \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right ) \, dx+\left (8 e^2 \left (18-17 e^2+4 e^4\right )\right ) \int \log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right ) \, dx+\left (8 e^2 \left (90-121 e^2+54 e^4-8 e^6\right )\right ) \int \frac {\log \left (\frac {1}{2} \left (-5+2 e^2\right )+\frac {x}{2}\right )}{-5+2 e^2+x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.51, size = 82, normalized size = 3.15 \begin {gather*} \frac {\left (2 e^{2+x}+e^x x+2 e^2 (-1+x)^2 x^2+(-5+x) (-1+x)^2 x^2\right ) \log \left (\frac {1}{2} \left (-5+2 e^2+x\right )\right )+e^x \left (\log (32)-5 \log \left (-5+2 e^2+x\right )\right )}{-5+2 e^2+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x + x^2 - 2*x^3 + x^4 + (-10*x + 32*x^2 - 26*x^3 + 4*x^4 + E^x*(-5 + 2*E^2 + x) + E^2*(4*x - 12*x
^2 + 8*x^3))*Log[(-5 + 2*E^2 + x)/2])/(-5 + 2*E^2 + x),x]

[Out]

((2*E^(2 + x) + E^x*x + 2*E^2*(-1 + x)^2*x^2 + (-5 + x)*(-1 + x)^2*x^2)*Log[(-5 + 2*E^2 + x)/2] + E^x*(Log[32]
- 5*Log[-5 + 2*E^2 + x]))/(-5 + 2*E^2 + x)

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fricas [A]  time = 1.33, size = 23, normalized size = 0.88 \begin {gather*} {\left (x^{4} - 2 \, x^{3} + x^{2} + e^{x}\right )} \log \left (\frac {1}{2} \, x + e^{2} - \frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(2)+x-5)*exp(x)+(8*x^3-12*x^2+4*x)*exp(2)+4*x^4-26*x^3+32*x^2-10*x)*log(exp(2)+1/2*x-5/2)+ex
p(x)+x^4-2*x^3+x^2)/(2*exp(2)+x-5),x, algorithm="fricas")

[Out]

(x^4 - 2*x^3 + x^2 + e^x)*log(1/2*x + e^2 - 5/2)

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giac [B]  time = 0.18, size = 49, normalized size = 1.88 \begin {gather*} x^{4} \log \left (\frac {1}{2} \, x + e^{2} - \frac {5}{2}\right ) - 2 \, x^{3} \log \left (\frac {1}{2} \, x + e^{2} - \frac {5}{2}\right ) + x^{2} \log \left (\frac {1}{2} \, x + e^{2} - \frac {5}{2}\right ) + e^{x} \log \left (\frac {1}{2} \, x + e^{2} - \frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(2)+x-5)*exp(x)+(8*x^3-12*x^2+4*x)*exp(2)+4*x^4-26*x^3+32*x^2-10*x)*log(exp(2)+1/2*x-5/2)+ex
p(x)+x^4-2*x^3+x^2)/(2*exp(2)+x-5),x, algorithm="giac")

[Out]

x^4*log(1/2*x + e^2 - 5/2) - 2*x^3*log(1/2*x + e^2 - 5/2) + x^2*log(1/2*x + e^2 - 5/2) + e^x*log(1/2*x + e^2 -
5/2)

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maple [A]  time = 0.15, size = 24, normalized size = 0.92

method	result	size
risch	\(\left (x^{4}-2 x^{3}+x^{2}+{\mathrm e}^{x}\right ) \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )\)	\(24\)
norman	\({\mathrm e}^{x} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+\ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x^{2}+\ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x^{4}-2 \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x^{3}\)	\(50\)
default	\(\frac {100 \,{\mathrm e}^{8}}{3}-304 \,{\mathrm e}^{6}+1036 \,{\mathrm e}^{4}-\frac {4690 \,{\mathrm e}^{2}}{3}+720 \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+144 \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )^{3} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+16 \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )^{4} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+484 \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )^{2}+720 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+{\mathrm e}^{x} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+160 \,{\mathrm e}^{6} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )-488 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )-64 \,{\mathrm e}^{6} \ln \left (2 \,{\mathrm e}^{2}+x -5\right )-32 \,{\mathrm e}^{8} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right )+16 \left ({\mathrm e}^{4}\right )^{2} \ln \left (2 \,{\mathrm e}^{2}+x -5\right )-32 \,{\mathrm e}^{6} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x -24 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x^{2}+24 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x -720 \ln \left (2 \,{\mathrm e}^{2}+x -5\right ) {\mathrm e}^{2}+484 \,{\mathrm e}^{4} \ln \left (2 \,{\mathrm e}^{2}+x -5\right )-80 \,{\mathrm e}^{2} {\mathrm e}^{4} \ln \left (2 \,{\mathrm e}^{2}+x -5\right )+\frac {3525}{4}-8 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x^{3}+12 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x^{2}-4 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{2}+\frac {x}{2}-\frac {5}{2}\right ) x +400 \ln \left (2 \,{\mathrm e}^{2}+x -5\right )\)	\(346\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*exp(2)+x-5)*exp(x)+(8*x^3-12*x^2+4*x)*exp(2)+4*x^4-26*x^3+32*x^2-10*x)*ln(exp(2)+1/2*x-5/2)+exp(x)+x^
4-2*x^3+x^2)/(2*exp(2)+x-5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(2)+x-5)*exp(x)+(8*x^3-12*x^2+4*x)*exp(2)+4*x^4-26*x^3+32*x^2-10*x)*log(exp(2)+1/2*x-5/2)+ex
p(x)+x^4-2*x^3+x^2)/(2*exp(2)+x-5),x, algorithm="maxima")

[Out]

4/9*x^3*(2*e^2 - 5) + 20/9*x^3 - 5/3*x^2*(4*e^4 - 20*e^2 + 25) - 59/6*x^2*(2*e^2 - 5) - 2*(16*e^8 - 160*e^6 +
600*e^4 - 1000*e^2 + 625)*log(x + 2*e^2 - 5)^2 - 13*(8*e^6 - 60*e^4 + 150*e^2 - 125)*log(x + 2*e^2 - 5)^2 - 16
*(4*e^4 - 20*e^2 + 25)*log(x + 2*e^2 - 5)^2 - 5*(2*e^2 - 5)*log(x + 2*e^2 - 5)^2 + 4/3*(2*x^3 - 3*x^2*(2*e^2 -
5) + 6*x*(4*e^4 - 20*e^2 + 25) - 6*(8*e^6 - 60*e^4 + 150*e^2 - 125)*log(x + 2*e^2 - 5))*e^2*log(1/2*x + e^2 -
5/2) - 6*(x^2 - 2*x*(2*e^2 - 5) + 2*(4*e^4 - 20*e^2 + 25)*log(x + 2*e^2 - 5))*e^2*log(1/2*x + e^2 - 5/2) - 4*
((2*e^2 - 5)*log(x + 2*e^2 - 5) - x)*e^2*log(1/2*x + e^2 - 5/2) - 15/2*x^2 + 22/3*x*(8*e^6 - 60*e^4 + 150*e^2
- 125) + 137/3*x*(4*e^4 - 20*e^2 + 25) + 47*x*(2*e^2 - 5) - 2/9*(4*x^3 - 15*x^2*(2*e^2 - 5) - 18*(8*e^6 - 60*e
^4 + 150*e^2 - 125)*log(x + 2*e^2 - 5)^2 + 66*x*(4*e^4 - 20*e^2 + 25) - 66*(8*e^6 - 60*e^4 + 150*e^2 - 125)*lo
g(x + 2*e^2 - 5))*e^2 + 3*(2*(4*e^4 - 20*e^2 + 25)*log(x + 2*e^2 - 5)^2 + x^2 - 6*x*(2*e^2 - 5) + 6*(4*e^4 - 2
0*e^2 + 25)*log(x + 2*e^2 - 5))*e^2 + 2*((2*e^2 - 5)*log(x + 2*e^2 - 5)^2 + 2*(2*e^2 - 5)*log(x + 2*e^2 - 5) -
2*x)*e^2 - e^(-2*e^2 + 5)*exp_integral_e(1, -x - 2*e^2 + 5) - 22/3*(16*e^8 - 160*e^6 + 600*e^4 - 1000*e^2 + 6
25)*log(x + 2*e^2 - 5) - 137/3*(8*e^6 - 60*e^4 + 150*e^2 - 125)*log(x + 2*e^2 - 5) - 47*(4*e^4 - 20*e^2 + 25)*
log(x + 2*e^2 - 5) - 10*(2*e^2 - 5)*log(x + 2*e^2 - 5) + 1/3*(3*x^4 - 4*x^3*(2*e^2 - 5) + 6*x^2*(4*e^4 - 20*e^
2 + 25) - 12*x*(8*e^6 - 60*e^4 + 150*e^2 - 125) + 12*(16*e^8 - 160*e^6 + 600*e^4 - 1000*e^2 + 625)*log(x + 2*e
^2 - 5))*log(1/2*x + e^2 - 5/2) - 13/3*(2*x^3 - 3*x^2*(2*e^2 - 5) + 6*x*(4*e^4 - 20*e^2 + 25) - 6*(8*e^6 - 60*
e^4 + 150*e^2 - 125)*log(x + 2*e^2 - 5))*log(1/2*x + e^2 - 5/2) + 16*(x^2 - 2*x*(2*e^2 - 5) + 2*(4*e^4 - 20*e^
2 + 25)*log(x + 2*e^2 - 5))*log(1/2*x + e^2 - 5/2) + 10*((2*e^2 - 5)*log(x + 2*e^2 - 5) - x)*log(1/2*x + e^2 -
5/2) + 10*x - (x*e^x*log(2) - (x + 2*e^2 - 5)*e^x*log(x + 2*e^2 - 5))/(x + 2*e^2 - 5) - integrate(((2*e^2*log
(2) - 5*log(2) + 1)*x - 2*(11*log(2) - 1)*e^2 + 4*e^4*log(2) + 30*log(2) - 5)*e^x/(x^2 + 2*x*(2*e^2 - 5) + 4*e
^4 - 20*e^2 + 25), x)

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mupad [B]  time = 4.57, size = 23, normalized size = 0.88 \begin {gather*} \ln \left (\frac {x}{2}+{\mathrm {e}}^2-\frac {5}{2}\right )\,\left ({\mathrm {e}}^x+x^2-2\,x^3+x^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x) + log(x/2 + exp(2) - 5/2)*(exp(2)*(4*x - 12*x^2 + 8*x^3) - 10*x + 32*x^2 - 26*x^3 + 4*x^4 + exp(x)
*(x + 2*exp(2) - 5)) + x^2 - 2*x^3 + x^4)/(x + 2*exp(2) - 5),x)

[Out]

log(x/2 + exp(2) - 5/2)*(exp(x) + x^2 - 2*x^3 + x^4)

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sympy [A]  time = 0.56, size = 37, normalized size = 1.42 \begin {gather*} \left (x^{4} - 2 x^{3} + x^{2}\right ) \log {\left (\frac {x}{2} - \frac {5}{2} + e^{2} \right )} + e^{x} \log {\left (\frac {x}{2} - \frac {5}{2} + e^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(2)+x-5)*exp(x)+(8*x**3-12*x**2+4*x)*exp(2)+4*x**4-26*x**3+32*x**2-10*x)*ln(exp(2)+1/2*x-5/2
)+exp(x)+x**4-2*x**3+x**2)/(2*exp(2)+x-5),x)

[Out]

(x**4 - 2*x**3 + x**2)*log(x/2 - 5/2 + exp(2)) + exp(x)*log(x/2 - 5/2 + exp(2))

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