\(\int \frac {2 x+32 x^2+32 x^4+6 x^6+(5+e^2+2 x+16 x^2+8 x^4+x^6) \log (-5-e^2-2 x-16 x^2-8 x^4-x^6)}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx\) [1787]

Optimal result

Integrand size = 91, antiderivative size = 25 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log \left (-5-e^2-2 x-x^2 \left (4+x^2\right )^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6873, 6874, 2603, 12} \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log \left (-x^6-8 x^4-16 x^2-2 x-e^2-5\right ) \]

[In]

Int[(2*x + 32*x^2 + 32*x^4 + 6*x^6 + (5 + E^2 + 2*x + 16*x^2 + 8*x^4 + x^6)*Log[-5 - E^2 - 2*x - 16*x^2 - 8*x^
4 - x^6])/(5 + E^2 + 2*x + 16*x^2 + 8*x^4 + x^6),x]

[Out]

x*Log[-5 - E^2 - 2*x - 16*x^2 - 8*x^4 - x^6]

Rule 12

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

Rule 2603

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[x*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx \\ & = \int \left (\frac {2 x \left (1+16 x+16 x^3+3 x^5\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\log \left (-5 \left (1+\frac {e^2}{5}\right )-2 x-16 x^2-8 x^4-x^6\right )\right ) \, dx \\ & = 2 \int \frac {x \left (1+16 x+16 x^3+3 x^5\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx+\int \log \left (-5 \left (1+\frac {e^2}{5}\right )-2 x-16 x^2-8 x^4-x^6\right ) \, dx \\ & = x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )+2 \int \left (3-\frac {3 \left (5+e^2\right )+5 x+32 x^2+8 x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}\right ) \, dx-\int \frac {2 x \left (1+16 x+16 x^3+3 x^5\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx \\ & = 6 x+x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )-2 \int \frac {3 \left (5+e^2\right )+5 x+32 x^2+8 x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-2 \int \frac {x \left (1+16 x+16 x^3+3 x^5\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx \\ & = 6 x+x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )-2 \int \left (\frac {3 \left (5+e^2\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\frac {5 x}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\frac {32 x^2}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\frac {8 x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}\right ) \, dx-2 \int \left (3-\frac {3 \left (5+e^2\right )+5 x+32 x^2+8 x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}\right ) \, dx \\ & = x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )+2 \int \frac {3 \left (5+e^2\right )+5 x+32 x^2+8 x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-10 \int \frac {x}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-16 \int \frac {x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-64 \int \frac {x^2}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-\left (6 \left (5+e^2\right )\right ) \int \frac {1}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx \\ & = x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )+2 \int \left (\frac {3 \left (5+e^2\right )}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\frac {5 x}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\frac {32 x^2}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}+\frac {8 x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6}\right ) \, dx-10 \int \frac {x}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-16 \int \frac {x^4}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-64 \int \frac {x^2}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx-\left (6 \left (5+e^2\right )\right ) \int \frac {1}{5 \left (1+\frac {e^2}{5}\right )+2 x+16 x^2+8 x^4+x^6} \, dx \\ & = x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right ) \]

[In]

Integrate[(2*x + 32*x^2 + 32*x^4 + 6*x^6 + (5 + E^2 + 2*x + 16*x^2 + 8*x^4 + x^6)*Log[-5 - E^2 - 2*x - 16*x^2
- 8*x^4 - x^6])/(5 + E^2 + 2*x + 16*x^2 + 8*x^4 + x^6),x]

[Out]

x*Log[-5 - E^2 - 2*x - 16*x^2 - 8*x^4 - x^6]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12

method	result	size
default	\(\ln \left (-{\mathrm e}^{2}-x^{6}-8 x^{4}-16 x^{2}-2 x -5\right ) x\)	\(28\)
norman	\(\ln \left (-{\mathrm e}^{2}-x^{6}-8 x^{4}-16 x^{2}-2 x -5\right ) x\)	\(28\)
risch	\(\ln \left (-{\mathrm e}^{2}-x^{6}-8 x^{4}-16 x^{2}-2 x -5\right ) x\)	\(28\)
parallelrisch	\(\ln \left (-{\mathrm e}^{2}-x^{6}-8 x^{4}-16 x^{2}-2 x -5\right ) x\)	\(28\)
parts	\(\ln \left (-{\mathrm e}^{2}-x^{6}-8 x^{4}-16 x^{2}-2 x -5\right ) x\)	\(28\)

[In]

int(((exp(2)+x^6+8*x^4+16*x^2+2*x+5)*ln(-exp(2)-x^6-8*x^4-16*x^2-2*x-5)+6*x^6+32*x^4+32*x^2+2*x)/(exp(2)+x^6+8
*x^4+16*x^2+2*x+5),x,method=_RETURNVERBOSE)

[Out]

ln(-exp(2)-x^6-8*x^4-16*x^2-2*x-5)*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log \left (-x^{6} - 8 \, x^{4} - 16 \, x^{2} - 2 \, x - e^{2} - 5\right ) \]

[In]

integrate(((exp(2)+x^6+8*x^4+16*x^2+2*x+5)*log(-exp(2)-x^6-8*x^4-16*x^2-2*x-5)+6*x^6+32*x^4+32*x^2+2*x)/(exp(2
)+x^6+8*x^4+16*x^2+2*x+5),x, algorithm="fricas")

[Out]

x*log(-x^6 - 8*x^4 - 16*x^2 - 2*x - e^2 - 5)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log {\left (- x^{6} - 8 x^{4} - 16 x^{2} - 2 x - e^{2} - 5 \right )} \]

[In]

integrate(((exp(2)+x**6+8*x**4+16*x**2+2*x+5)*ln(-exp(2)-x**6-8*x**4-16*x**2-2*x-5)+6*x**6+32*x**4+32*x**2+2*x
)/(exp(2)+x**6+8*x**4+16*x**2+2*x+5),x)

[Out]

x*log(-x**6 - 8*x**4 - 16*x**2 - 2*x - exp(2) - 5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log \left (-x^{6} - 8 \, x^{4} - 16 \, x^{2} - 2 \, x - e^{2} - 5\right ) \]

[In]

integrate(((exp(2)+x^6+8*x^4+16*x^2+2*x+5)*log(-exp(2)-x^6-8*x^4-16*x^2-2*x-5)+6*x^6+32*x^4+32*x^2+2*x)/(exp(2
)+x^6+8*x^4+16*x^2+2*x+5),x, algorithm="maxima")

[Out]

x*log(-x^6 - 8*x^4 - 16*x^2 - 2*x - e^2 - 5)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x \log \left (-x^{6} - 8 \, x^{4} - 16 \, x^{2} - 2 \, x - e^{2} - 5\right ) \]

[In]

integrate(((exp(2)+x^6+8*x^4+16*x^2+2*x+5)*log(-exp(2)-x^6-8*x^4-16*x^2-2*x-5)+6*x^6+32*x^4+32*x^2+2*x)/(exp(2
)+x^6+8*x^4+16*x^2+2*x+5),x, algorithm="giac")

[Out]

x*log(-x^6 - 8*x^4 - 16*x^2 - 2*x - e^2 - 5)

Mupad [B] (verification not implemented)

Time = 9.69 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {2 x+32 x^2+32 x^4+6 x^6+\left (5+e^2+2 x+16 x^2+8 x^4+x^6\right ) \log \left (-5-e^2-2 x-16 x^2-8 x^4-x^6\right )}{5+e^2+2 x+16 x^2+8 x^4+x^6} \, dx=x\,\ln \left (-x^6-8\,x^4-16\,x^2-2\,x-{\mathrm {e}}^2-5\right ) \]

[In]

int((2*x + log(- 2*x - exp(2) - 16*x^2 - 8*x^4 - x^6 - 5)*(2*x + exp(2) + 16*x^2 + 8*x^4 + x^6 + 5) + 32*x^2 +
32*x^4 + 6*x^6)/(2*x + exp(2) + 16*x^2 + 8*x^4 + x^6 + 5),x)

[Out]

x*log(- 2*x - exp(2) - 16*x^2 - 8*x^4 - x^6 - 5)