\(\int \frac {-15-12 x+3 x^2+e^4 (-4-2 x^2+10 x^3-2 x^4)+(-3+e^4 (-2 x+2 x^2)) \log (3+e^4 (2 x-2 x^2))}{-12+12 x-3 x^2+e^4 (-8 x+16 x^2-10 x^3+2 x^4)} \, dx\) [9160]

Optimal result

Integrand size = 99, antiderivative size = 30 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=-x+\frac {9+\log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{2-x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27, number of steps used = 18, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6873, 6820, 6860, 1642, 648, 632, 212, 642, 2605, 12, 814} \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=\frac {\log \left (-2 e^4 x^2+2 e^4 x+3\right )}{2-x}-x+\frac {9}{2-x} \]

[In]

Int[(-15 - 12*x + 3*x^2 + E^4*(-4 - 2*x^2 + 10*x^3 - 2*x^4) + (-3 + E^4*(-2*x + 2*x^2))*Log[3 + E^4*(2*x - 2*x
^2)])/(-12 + 12*x - 3*x^2 + E^4*(-8*x + 16*x^2 - 10*x^3 + 2*x^4)),x]

[Out]

9/(2 - x) - x + Log[3 + 2*E^4*x - 2*E^4*x^2]/(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 212

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

Rule 632

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 642

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 648

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 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2605

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m +
1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))), x] - Dist[b*n*(p/(e*(m + 1))), Int[SimplifyIntegrand[(d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6820

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

Rule 6860

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]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {15+12 x-3 x^2-e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )-\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{12-4 \left (3-2 e^4\right ) x+\left (3-16 e^4\right ) x^2+10 e^4 x^3-2 e^4 x^4} \, dx \\ & = \int \frac {-3 \left (-5-4 x+x^2\right )+2 e^4 \left (2+x^2-5 x^3+x^4\right )-\left (-3+2 e^4 (-1+x) x\right ) \log \left (3-2 e^4 (-1+x) x\right )}{(2-x)^2 \left (3+2 e^4 x-2 e^4 x^2\right )} \, dx \\ & = \int \left (\frac {15+4 e^4+12 x-\left (3-2 e^4\right ) x^2-10 e^4 x^3+2 e^4 x^4}{(2-x)^2 \left (3+2 e^4 x-2 e^4 x^2\right )}+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{(2-x)^2}\right ) \, dx \\ & = \int \frac {15+4 e^4+12 x-\left (3-2 e^4\right ) x^2-10 e^4 x^3+2 e^4 x^4}{(2-x)^2 \left (3+2 e^4 x-2 e^4 x^2\right )} \, dx+\int \frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{(2-x)^2} \, dx \\ & = \frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x}-\int \frac {2 e^4 (1-2 x)}{(2-x) \left (3+2 e^4 x-2 e^4 x^2\right )} \, dx+\int \left (-1+\frac {9}{(-2+x)^2}-\frac {6 e^4}{\left (-3+4 e^4\right ) (-2+x)}+\frac {4 e^4 \left (3-e^4+3 e^4 x\right )}{\left (-3+4 e^4\right ) \left (-3-2 e^4 x+2 e^4 x^2\right )}\right ) \, dx \\ & = \frac {9}{2-x}-x+\frac {6 e^4 \log (2-x)}{3-4 e^4}+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x}-\left (2 e^4\right ) \int \frac {1-2 x}{(2-x) \left (3+2 e^4 x-2 e^4 x^2\right )} \, dx-\frac {\left (4 e^4\right ) \int \frac {3-e^4+3 e^4 x}{-3-2 e^4 x+2 e^4 x^2} \, dx}{3-4 e^4} \\ & = \frac {9}{2-x}-x+\frac {6 e^4 \log (2-x)}{3-4 e^4}+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x}-\left (2 e^4\right ) \int \left (-\frac {3}{\left (-3+4 e^4\right ) (-2+x)}+\frac {2 \left (3-e^4+3 e^4 x\right )}{\left (-3+4 e^4\right ) \left (-3-2 e^4 x+2 e^4 x^2\right )}\right ) \, dx-\frac {\left (3 e^4\right ) \int \frac {-2 e^4+4 e^4 x}{-3-2 e^4 x+2 e^4 x^2} \, dx}{3-4 e^4}-\frac {\left (2 e^4 \left (6+e^4\right )\right ) \int \frac {1}{-3-2 e^4 x+2 e^4 x^2} \, dx}{3-4 e^4} \\ & = \frac {9}{2-x}-x-\frac {3 e^4 \log \left (3+2 e^4 x-2 e^4 x^2\right )}{3-4 e^4}+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x}+\frac {\left (4 e^4\right ) \int \frac {3-e^4+3 e^4 x}{-3-2 e^4 x+2 e^4 x^2} \, dx}{3-4 e^4}+\frac {\left (4 e^4 \left (6+e^4\right )\right ) \text {Subst}\left (\int \frac {1}{4 e^4 \left (6+e^4\right )-x^2} \, dx,x,-2 e^4+4 e^4 x\right )}{3-4 e^4} \\ & = \frac {9}{2-x}-x-\frac {2 e^2 \sqrt {6+e^4} \text {arctanh}\left (\frac {e^2 (1-2 x)}{\sqrt {6+e^4}}\right )}{3-4 e^4}-\frac {3 e^4 \log \left (3+2 e^4 x-2 e^4 x^2\right )}{3-4 e^4}+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x}+\frac {\left (3 e^4\right ) \int \frac {-2 e^4+4 e^4 x}{-3-2 e^4 x+2 e^4 x^2} \, dx}{3-4 e^4}+\frac {\left (2 e^4 \left (6+e^4\right )\right ) \int \frac {1}{-3-2 e^4 x+2 e^4 x^2} \, dx}{3-4 e^4} \\ & = \frac {9}{2-x}-x-\frac {2 e^2 \sqrt {6+e^4} \text {arctanh}\left (\frac {e^2 (1-2 x)}{\sqrt {6+e^4}}\right )}{3-4 e^4}+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x}-\frac {\left (4 e^4 \left (6+e^4\right )\right ) \text {Subst}\left (\int \frac {1}{4 e^4 \left (6+e^4\right )-x^2} \, dx,x,-2 e^4+4 e^4 x\right )}{3-4 e^4} \\ & = \frac {9}{2-x}-x+\frac {\log \left (3+2 e^4 x-2 e^4 x^2\right )}{2-x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=-x+\frac {-9-\log \left (3-2 e^4 (-1+x) x\right )}{-2+x} \]

[In]

Integrate[(-15 - 12*x + 3*x^2 + E^4*(-4 - 2*x^2 + 10*x^3 - 2*x^4) + (-3 + E^4*(-2*x + 2*x^2))*Log[3 + E^4*(2*x
- 2*x^2)])/(-12 + 12*x - 3*x^2 + E^4*(-8*x + 16*x^2 - 10*x^3 + 2*x^4)),x]

[Out]

-x + (-9 - Log[3 - 2*E^4*(-1 + x)*x])/(-2 + x)

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
norman	\(\frac {-\ln \left (\left (-2 x^{2}+2 x \right ) {\mathrm e}^{4}+3\right )-x^{2}-5}{-2+x}\)	\(31\)
parallelrisch	\(\frac {-18 x^{2}-45 x -18 \ln \left (\left (-2 x^{2}+2 x \right ) {\mathrm e}^{4}+3\right )}{-36+18 x}\)	\(34\)
risch	\(-\frac {\ln \left (\left (-2 x^{2}+2 x \right ) {\mathrm e}^{4}+3\right )}{-2+x}-\frac {x^{2}-2 x +9}{-2+x}\)	\(39\)

[In]

int((((2*x^2-2*x)*exp(4)-3)*ln((-2*x^2+2*x)*exp(4)+3)+(-2*x^4+10*x^3-2*x^2-4)*exp(4)+3*x^2-12*x-15)/((2*x^4-10
*x^3+16*x^2-8*x)*exp(4)-3*x^2+12*x-12),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=-\frac {x^{2} - 2 \, x + \log \left (-2 \, {\left (x^{2} - x\right )} e^{4} + 3\right ) + 9}{x - 2} \]

[In]

integrate((((2*x^2-2*x)*exp(4)-3)*log((-2*x^2+2*x)*exp(4)+3)+(-2*x^4+10*x^3-2*x^2-4)*exp(4)+3*x^2-12*x-15)/((2
*x^4-10*x^3+16*x^2-8*x)*exp(4)-3*x^2+12*x-12),x, algorithm="fricas")

[Out]

-(x^2 - 2*x + log(-2*(x^2 - x)*e^4 + 3) + 9)/(x - 2)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=- x - \frac {\log {\left (\left (- 2 x^{2} + 2 x\right ) e^{4} + 3 \right )}}{x - 2} - \frac {9}{x - 2} \]

[In]

integrate((((2*x**2-2*x)*exp(4)-3)*ln((-2*x**2+2*x)*exp(4)+3)+(-2*x**4+10*x**3-2*x**2-4)*exp(4)+3*x**2-12*x-15
)/((2*x**4-10*x**3+16*x**2-8*x)*exp(4)-3*x**2+12*x-12),x)

[Out]

-x - log((-2*x**2 + 2*x)*exp(4) + 3)/(x - 2) - 9/(x - 2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1175 vs. \(2 (27) = 54\).

Time = 0.35 (sec) , antiderivative size = 1175, normalized size of antiderivative = 39.17 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=\text {Too large to display} \]

[In]

integrate((((2*x^2-2*x)*exp(4)-3)*log((-2*x^2+2*x)*exp(4)+3)+(-2*x^4+10*x^3-2*x^2-4)*exp(4)+3*x^2-12*x-15)/((2
*x^4-10*x^3+16*x^2-8*x)*exp(4)-3*x^2+12*x-12),x, algorithm="maxima")

[Out]

-1/2*(2*x*e^(-4) + (16*e^12 + 120*e^8 + 189*e^4 + 27)*e^(-2)*log((2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4
+ sqrt(e^4 + 6)*e^2 - e^4))/((16*e^12 - 24*e^8 + 9*e^4)*sqrt(e^4 + 6)) + (16*e^8 + 72*e^4 + 45)*log(2*x^2*e^4
- 2*x*e^4 - 3)/(16*e^12 - 24*e^8 + 9*e^4) + 128*(e^4 - 3)*log(x - 2)/(16*e^8 - 24*e^4 + 9) - 64/(x*(4*e^4 - 3)
- 8*e^4 + 6))*e^4 + 5/2*((16*e^8 + 96*e^4 + 81)*e^(-2)*log((2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4 + sqr
t(e^4 + 6)*e^2 - e^4))/((16*e^8 - 24*e^4 + 9)*sqrt(e^4 + 6)) + (16*e^8 + 48*e^4 + 9)*log(2*x^2*e^4 - 2*x*e^4 -
3)/(16*e^12 - 24*e^8 + 9*e^4) - 144*log(x - 2)/(16*e^8 - 24*e^4 + 9) - 32/(x*(4*e^4 - 3) - 8*e^4 + 6))*e^4 -
((8*e^8 + 36*e^4 + 9)*e^(-2)*log((2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4 + sqrt(e^4 + 6)*e^2 - e^4))/((16
*e^8 - 24*e^4 + 9)*sqrt(e^4 + 6)) + 4*(2*e^4 + 3)*log(2*x^2*e^4 - 2*x*e^4 - 3)/(16*e^8 - 24*e^4 + 9) - 8*(2*e^
4 + 3)*log(x - 2)/(16*e^8 - 24*e^4 + 9) - 8/(x*(4*e^4 - 3) - 8*e^4 + 6))*e^4 - 4*((5*e^8 + 3*e^4)*e^(-2)*log((
2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4 + sqrt(e^4 + 6)*e^2 - e^4))/((16*e^8 - 24*e^4 + 9)*sqrt(e^4 + 6))
+ 3*e^4*log(2*x^2*e^4 - 2*x*e^4 - 3)/(16*e^8 - 24*e^4 + 9) - 6*e^4*log(x - 2)/(16*e^8 - 24*e^4 + 9) - 1/(x*(4*
e^4 - 3) - 8*e^4 + 6))*e^4 + 3/2*(8*e^8 + 36*e^4 + 9)*e^(-2)*log((2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4
+ sqrt(e^4 + 6)*e^2 - e^4))/((16*e^8 - 24*e^4 + 9)*sqrt(e^4 + 6)) - 6*(8*e^8 + 21*e^4)*e^(-2)*log((2*x*e^4 - s
qrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4 + sqrt(e^4 + 6)*e^2 - e^4))/((16*e^8 - 24*e^4 + 9)*sqrt(e^4 + 6)) - 15*(5*e^8
+ 3*e^4)*e^(-2)*log((2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4 + sqrt(e^4 + 6)*e^2 - e^4))/((16*e^8 - 24*e^
4 + 9)*sqrt(e^4 + 6)) - (e^8 + 6*e^4)*e^(-2)*log((2*x*e^4 - sqrt(e^4 + 6)*e^2 - e^4)/(2*x*e^4 + sqrt(e^4 + 6)*
e^2 - e^4))/((4*e^4 - 3)*sqrt(e^4 + 6)) - 6*(8*e^4 + 3)*log(2*x^2*e^4 - 2*x*e^4 - 3)/(16*e^8 - 24*e^4 + 9) + 6
*(2*e^4 + 3)*log(2*x^2*e^4 - 2*x*e^4 - 3)/(16*e^8 - 24*e^4 + 9) - 45*e^4*log(2*x^2*e^4 - 2*x*e^4 - 3)/(16*e^8
- 24*e^4 + 9) - 3*e^4*log(2*x^2*e^4 - 2*x*e^4 - 3)/(4*e^4 - 3) + 12*(8*e^4 + 3)*log(x - 2)/(16*e^8 - 24*e^4 +
9) - 12*(2*e^4 + 3)*log(x - 2)/(16*e^8 - 24*e^4 + 9) + 90*e^4*log(x - 2)/(16*e^8 - 24*e^4 + 9) + 6*e^4*log(x -
2)/(4*e^4 - 3) - log(-2*x^2*e^4 + 2*x*e^4 + 3)/(x - 2) + 27/(x*(4*e^4 - 3) - 8*e^4 + 6)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=-\frac {x^{2} - 2 \, x + \log \left (-2 \, x^{2} e^{4} + 2 \, x e^{4} + 3\right ) + 9}{x - 2} \]

[In]

integrate((((2*x^2-2*x)*exp(4)-3)*log((-2*x^2+2*x)*exp(4)+3)+(-2*x^4+10*x^3-2*x^2-4)*exp(4)+3*x^2-12*x-15)/((2
*x^4-10*x^3+16*x^2-8*x)*exp(4)-3*x^2+12*x-12),x, algorithm="giac")

[Out]

-(x^2 - 2*x + log(-2*x^2*e^4 + 2*x*e^4 + 3) + 9)/(x - 2)

Mupad [B] (verification not implemented)

Time = 14.63 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-15-12 x+3 x^2+e^4 \left (-4-2 x^2+10 x^3-2 x^4\right )+\left (-3+e^4 \left (-2 x+2 x^2\right )\right ) \log \left (3+e^4 \left (2 x-2 x^2\right )\right )}{-12+12 x-3 x^2+e^4 \left (-8 x+16 x^2-10 x^3+2 x^4\right )} \, dx=-\frac {\ln \left ({\mathrm {e}}^4\,\left (2\,x-2\,x^2\right )+3\right )-2\,x+x^2+9}{x-2} \]

[In]

int((12*x + exp(4)*(2*x^2 - 10*x^3 + 2*x^4 + 4) - 3*x^2 + log(exp(4)*(2*x - 2*x^2) + 3)*(exp(4)*(2*x - 2*x^2)
+ 3) + 15)/(exp(4)*(8*x - 16*x^2 + 10*x^3 - 2*x^4) - 12*x + 3*x^2 + 12),x)

[Out]

-(log(exp(4)*(2*x - 2*x^2) + 3) - 2*x + x^2 + 9)/(x - 2)