3.36.46 \(\int \frac {e^{3-e^{\log ^2(5)}} (10 x+4 x^2+2 x^3+(2+2 x-2 x^2-2 x^3) \log (1-x^2))}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{5+2 x+x^2} \]

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Rubi [C]  time = 1.30, antiderivative size = 172, normalized size of antiderivative = 5.55, number of steps used = 64, number of rules used = 17, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {12, 6688, 6725, 6728, 634, 618, 204, 628, 2528, 2463, 801, 2462, 260, 2416, 2394, 2393, 2391} \begin {gather*} \frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 (-2 x-(2-4 i))}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 (2 x+(2+4 i))}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (x^2+2 x+5\right )+\left (\frac {3}{16}-\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log (x+(1-2 i))+\left (\frac {3}{16}+\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log (x+(1+2 i))-\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {x+1}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 - E^Log[5]^2)*(10*x + 4*x^2 + 2*x^3 + (2 + 2*x - 2*x^2 - 2*x^3)*Log[1 - x^2]))/(-25 - 20*x + 11*x^2
+ 16*x^3 + 13*x^4 + 4*x^5 + x^6),x]

[Out]

-1/8*(E^(3 - E^Log[5]^2)*ArcTan[(1 + x)/2]) + (3/16 - I/16)*E^(3 - E^Log[5]^2)*Log[(1 - 2*I) + x] + (3/16 + I/
16)*E^(3 - E^Log[5]^2)*Log[(1 + 2*I) + x] + ((I/2)*E^(3 - E^Log[5]^2)*Log[1 - x^2])/((-2 + 4*I) - 2*x) + ((I/2
)*E^(3 - E^Log[5]^2)*Log[1 - x^2])/((2 + 4*I) + 2*x) - (3*E^(3 - E^Log[5]^2)*Log[5 + 2*x + x^2])/16

Rule 12

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 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 2462

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

Rule 2463

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

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rule 6688

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{3-e^{\log ^2(5)}} \int \frac {10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx\\ &=e^{3-e^{\log ^2(5)}} \int \frac {-2 x \left (5+2 x+x^2\right )+2 (-1+x) (1+x)^2 \log \left (1-x^2\right )}{\left (1-x^2\right ) \left (5+2 x+x^2\right )^2} \, dx\\ &=e^{3-e^{\log ^2(5)}} \int \left (\frac {2 x}{(-1+x) (1+x) \left (5+2 x+x^2\right )}-\frac {2 (1+x) \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2}\right ) \, dx\\ &=\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {x}{(-1+x) (1+x) \left (5+2 x+x^2\right )} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {(1+x) \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2} \, dx\\ &=\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {1}{16 (-1+x)}+\frac {1}{8 (1+x)}+\frac {-5-3 x}{16 \left (5+2 x+x^2\right )}\right ) \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {\log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2}+\frac {x \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2}\right ) \, dx\\ &=\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)+\frac {1}{8} e^{3-e^{\log ^2(5)}} \int \frac {-5-3 x}{5+2 x+x^2} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {\log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {x \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2} \, dx\\ &=\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)-\frac {1}{16} \left (3 e^{3-e^{\log ^2(5)}}\right ) \int \frac {2+2 x}{5+2 x+x^2} \, dx-\frac {1}{4} e^{3-e^{\log ^2(5)}} \int \frac {1}{5+2 x+x^2} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {\left (\frac {1}{4}-\frac {i}{2}\right ) \log \left (1-x^2\right )}{((-2+4 i)-2 x)^2}-\frac {i \log \left (1-x^2\right )}{16 ((-2+4 i)-2 x)}+\frac {\left (\frac {1}{4}+\frac {i}{2}\right ) \log \left (1-x^2\right )}{((2+4 i)+2 x)^2}-\frac {i \log \left (1-x^2\right )}{16 ((2+4 i)+2 x)}\right ) \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (-\frac {\log \left (1-x^2\right )}{4 ((-2+4 i)-2 x)^2}+\frac {i \log \left (1-x^2\right )}{16 ((-2+4 i)-2 x)}-\frac {\log \left (1-x^2\right )}{4 ((2+4 i)+2 x)^2}+\frac {i \log \left (1-x^2\right )}{16 ((2+4 i)+2 x)}\right ) \, dx\\ &=\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {\log \left (1-x^2\right )}{((-2+4 i)-2 x)^2} \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {\log \left (1-x^2\right )}{((2+4 i)+2 x)^2} \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \operatorname {Subst}\left (\int \frac {1}{-16-x^2} \, dx,x,2+2 x\right )-\left (\left (\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {\log \left (1-x^2\right )}{((-2+4 i)-2 x)^2} \, dx-\left (\left (\frac {1}{2}+i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {\log \left (1-x^2\right )}{((2+4 i)+2 x)^2} \, dx\\ &=-\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {1+x}{2}\right )+\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((-2+4 i)-2 x)}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((2+4 i)+2 x)}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )-\left (\left (-\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {x}{((2+4 i)+2 x) \left (1-x^2\right )} \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {x}{((-2+4 i)-2 x) \left (1-x^2\right )} \, dx-\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {x}{((2+4 i)+2 x) \left (1-x^2\right )} \, dx-\left (\left (\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {x}{((-2+4 i)-2 x) \left (1-x^2\right )} \, dx\\ &=-\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {1+x}{2}\right )+\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((-2+4 i)-2 x)}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((2+4 i)+2 x)}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )-\left (\left (-\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \left (-\frac {\frac {1}{16}-\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}+\frac {\frac {1}{16}-\frac {3 i}{16}}{(1+2 i)+x}\right ) \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \left (\frac {\frac {1}{16}+\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}-\frac {\frac {1}{16}+\frac {3 i}{16}}{(1-2 i)+x}\right ) \, dx-\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \left (-\frac {\frac {1}{16}-\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}+\frac {\frac {1}{16}-\frac {3 i}{16}}{(1+2 i)+x}\right ) \, dx-\left (\left (\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {\frac {1}{16}+\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}-\frac {\frac {1}{16}+\frac {3 i}{16}}{(1-2 i)+x}\right ) \, dx\\ &=-\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {1+x}{2}\right )+\left (\frac {3}{16}-\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log ((1-2 i)+x)+\left (\frac {3}{16}+\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log ((1+2 i)+x)+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((-2+4 i)-2 x)}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((2+4 i)+2 x)}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 31, normalized size = 1.00 \begin {gather*} \frac {e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{5+2 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 - E^Log[5]^2)*(10*x + 4*x^2 + 2*x^3 + (2 + 2*x - 2*x^2 - 2*x^3)*Log[1 - x^2]))/(-25 - 20*x + 1
1*x^2 + 16*x^3 + 13*x^4 + 4*x^5 + x^6),x]

[Out]

(E^(3 - E^Log[5]^2)*Log[1 - x^2])/(5 + 2*x + x^2)

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fricas [A]  time = 0.58, size = 29, normalized size = 0.94 \begin {gather*} \frac {e^{\left (-e^{\left (\log \relax (5)^{2}\right )} + 3\right )} \log \left (-x^{2} + 1\right )}{x^{2} + 2 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+13*x^4+16*x^3+11*x^2-20*x-25)/exp(exp
(log(5)^2)-3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 29, normalized size = 0.94 \begin {gather*} \frac {e^{\left (-e^{\left (\log \relax (5)^{2}\right )} + 3\right )} \log \left (-x^{2} + 1\right )}{x^{2} + 2 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+13*x^4+16*x^3+11*x^2-20*x-25)/exp(exp
(log(5)^2)-3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.23, size = 30, normalized size = 0.97

method	result	size
risch	\(\frac {\ln \left (-x^{2}+1\right ) {\mathrm e}^{3-{\mathrm e}^{\ln \relax (5)^{2}}}}{x^{2}+2 x +5}\)	\(30\)
norman	\(\frac {{\mathrm e}^{-{\mathrm e}^{\ln \relax (5)^{2}}} {\mathrm e}^{3} \ln \left (-x^{2}+1\right )}{x^{2}+2 x +5}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3-2*x^2+2*x+2)*ln(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+13*x^4+16*x^3+11*x^2-20*x-25)/exp(exp(ln(5)^
2)-3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 1.33, size = 115, normalized size = 3.71 \begin {gather*} -\frac {1}{32} \, {\left (\frac {4 \, {\left (2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + {\left (x^{2} + 2 \, x - 3\right )} \log \left (-x + 1\right )\right )}}{x^{2} + 2 \, x + 5} + \frac {9 \, x + 35}{x^{2} + 2 \, x + 5} - \frac {2 \, {\left (7 \, x + 5\right )}}{x^{2} + 2 \, x + 5} + \frac {5 \, {\left (x - 5\right )}}{x^{2} + 2 \, x + 5} - 8 \, \log \left (x + 1\right ) - 4 \, \log \left (x - 1\right )\right )} e^{\left (-e^{\left (\log \relax (5)^{2}\right )} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+13*x^4+16*x^3+11*x^2-20*x-25)/exp(exp
(log(5)^2)-3),x, algorithm="maxima")

[Out]

-1/32*(4*(2*(x^2 + 2*x + 1)*log(x + 1) + (x^2 + 2*x - 3)*log(-x + 1))/(x^2 + 2*x + 5) + (9*x + 35)/(x^2 + 2*x
+ 5) - 2*(7*x + 5)/(x^2 + 2*x + 5) + 5*(x - 5)/(x^2 + 2*x + 5) - 8*log(x + 1) - 4*log(x - 1))*e^(-e^(log(5)^2)
+ 3)

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mupad [B]  time = 2.50, size = 29, normalized size = 0.94 \begin {gather*} \frac {\ln \left (1-x^2\right )\,{\mathrm {e}}^{3-{\mathrm {e}}^{{\ln \relax (5)}^2}}}{x^2+2\,x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3 - exp(log(5)^2))*(10*x + log(1 - x^2)*(2*x - 2*x^2 - 2*x^3 + 2) + 4*x^2 + 2*x^3))/(11*x^2 - 20*x +
16*x^3 + 13*x^4 + 4*x^5 + x^6 - 25),x)

[Out]

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

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sympy [A]  time = 0.37, size = 42, normalized size = 1.35 \begin {gather*} \frac {e^{3} \log {\left (1 - x^{2} \right )}}{x^{2} e^{e^{\log {\relax (5 )}^{2}}} + 2 x e^{e^{\log {\relax (5 )}^{2}}} + 5 e^{e^{\log {\relax (5 )}^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3-2*x**2+2*x+2)*ln(-x**2+1)+2*x**3+4*x**2+10*x)/(x**6+4*x**5+13*x**4+16*x**3+11*x**2-20*x-25
)/exp(exp(ln(5)**2)-3),x)

[Out]

exp(3)*log(1 - x**2)/(x**2*exp(exp(log(5)**2)) + 2*x*exp(exp(log(5)**2)) + 5*exp(exp(log(5)**2)))

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