\(\int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 (8+32 x+32 x^2)} \, dx\) [9374]

Optimal result

Integrand size = 50, antiderivative size = 22 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=\frac {\log (5)}{2+\log \left (e^{\frac {2 e^8}{(2+4 x)^2}}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 1694, 267} \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^8 \log (5)}{2 \left (4 (2 x+1)^2+e^8\right )} \]

[In]

Int[(E^8*(8 + 16*x)*Log[5])/(16 + E^16 + 128*x + 384*x^2 + 512*x^3 + 256*x^4 + E^8*(8 + 32*x + 32*x^2)),x]

[Out]

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

Rule 12

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

Rule 267

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

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \left (e^8 \log (5)\right ) \int \frac {8+16 x}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx \\ & = \left (e^8 \log (5)\right ) \text {Subst}\left (\int \frac {16 x}{\left (e^8+16 x^2\right )^2} \, dx,x,\frac {1}{2}+x\right ) \\ & = \left (16 e^8 \log (5)\right ) \text {Subst}\left (\int \frac {x}{\left (e^8+16 x^2\right )^2} \, dx,x,\frac {1}{2}+x\right ) \\ & = -\frac {e^8 \log (5)}{2 \left (e^8+4 (1+2 x)^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^8 \log (5)}{2 \left (e^8+4 (1+2 x)^2\right )} \]

[In]

Integrate[(E^8*(8 + 16*x)*Log[5])/(16 + E^16 + 128*x + 384*x^2 + 512*x^3 + 256*x^4 + E^8*(8 + 32*x + 32*x^2)),
x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\)	\(21\)
gosper	\(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\)	\(25\)
norman	\(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\)	\(25\)
parallelrisch	\(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\)	\(25\)

[In]

int((16*x+8)*exp(4)^2*ln(5)/(exp(4)^4+(32*x^2+32*x+8)*exp(4)^2+256*x^4+512*x^3+384*x^2+128*x+16),x,method=_RET
URNVERBOSE)

[Out]

-1/2*exp(8)*ln(5)/(exp(8)+16*x^2+16*x+4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^{8} \log \left (5\right )}{2 \, {\left (16 \, x^{2} + 16 \, x + e^{8} + 4\right )}} \]

[In]

integrate((16*x+8)*exp(4)^2*log(5)/(exp(4)^4+(32*x^2+32*x+8)*exp(4)^2+256*x^4+512*x^3+384*x^2+128*x+16),x, alg
orithm="fricas")

[Out]

-1/2*e^8*log(5)/(16*x^2 + 16*x + e^8 + 4)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=- \frac {e^{8} \log {\left (5 \right )}}{32 x^{2} + 32 x + 8 + 2 e^{8}} \]

[In]

integrate((16*x+8)*exp(4)**2*ln(5)/(exp(4)**4+(32*x**2+32*x+8)*exp(4)**2+256*x**4+512*x**3+384*x**2+128*x+16),
x)

[Out]

-exp(8)*log(5)/(32*x**2 + 32*x + 8 + 2*exp(8))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^{8} \log \left (5\right )}{2 \, {\left (16 \, x^{2} + 16 \, x + e^{8} + 4\right )}} \]

[In]

integrate((16*x+8)*exp(4)^2*log(5)/(exp(4)^4+(32*x^2+32*x+8)*exp(4)^2+256*x^4+512*x^3+384*x^2+128*x+16),x, alg
orithm="maxima")

[Out]

-1/2*e^8*log(5)/(16*x^2 + 16*x + e^8 + 4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^{8} \log \left (5\right )}{2 \, {\left (16 \, x^{2} + 16 \, x + e^{8} + 4\right )}} \]

[In]

integrate((16*x+8)*exp(4)^2*log(5)/(exp(4)^4+(32*x^2+32*x+8)*exp(4)^2+256*x^4+512*x^3+384*x^2+128*x+16),x, alg
orithm="giac")

[Out]

-1/2*e^8*log(5)/(16*x^2 + 16*x + e^8 + 4)

Mupad [B] (verification not implemented)

Time = 13.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {{\mathrm {e}}^8\,\ln \left (5\right )}{2\,\left (16\,x^2+16\,x+{\mathrm {e}}^8+4\right )} \]

[In]

int((exp(8)*log(5)*(16*x + 8))/(128*x + exp(16) + exp(8)*(32*x + 32*x^2 + 8) + 384*x^2 + 512*x^3 + 256*x^4 + 1
6),x)

[Out]

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