\(\int \frac {-1+64 x+8 x^2+e^5 (-96 x^2-12 x^3)+(-16 x-2 x^2+e^5 (24 x^2+3 x^3)) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx\) [6268]

Optimal result

Integrand size = 70, antiderivative size = 25 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=-2-x^2+e^5 x^3-\log (4-\log (8+x)) \]

[Out]

x^3*exp(5)-ln(4-ln(x+8))-2-x^2

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6873, 6874, 45, 2437, 2339, 29} \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=e^5 x^3-x^2-\log (4-\log (x+8)) \]

[In]

Int[(-1 + 64*x + 8*x^2 + E^5*(-96*x^2 - 12*x^3) + (-16*x - 2*x^2 + E^5*(24*x^2 + 3*x^3))*Log[8 + x])/(-32 - 4*
x + (8 + x)*Log[8 + x]),x]

[Out]

-x^2 + E^5*x^3 - Log[4 - Log[8 + x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 45

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 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

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 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 {1-64 x-8 x^2-e^5 \left (-96 x^2-12 x^3\right )-\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{(8+x) (4-\log (8+x))} \, dx \\ & = \int \left (x \left (-2+3 e^5 x\right )-\frac {1}{(8+x) (-4+\log (8+x))}\right ) \, dx \\ & = \int x \left (-2+3 e^5 x\right ) \, dx-\int \frac {1}{(8+x) (-4+\log (8+x))} \, dx \\ & = \int \left (-2 x+3 e^5 x^2\right ) \, dx-\text {Subst}\left (\int \frac {1}{x (-4+\log (x))} \, dx,x,8+x\right ) \\ & = -x^2+e^5 x^3-\text {Subst}\left (\int \frac {1}{x} \, dx,x,-4+\log (8+x)\right ) \\ & = -x^2+e^5 x^3-\log (4-\log (8+x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=64-x^2+e^5 \left (512+x^3\right )-\log (4-\log (8+x)) \]

[In]

Integrate[(-1 + 64*x + 8*x^2 + E^5*(-96*x^2 - 12*x^3) + (-16*x - 2*x^2 + E^5*(24*x^2 + 3*x^3))*Log[8 + x])/(-3
2 - 4*x + (8 + x)*Log[8 + x]),x]

[Out]

64 - x^2 + E^5*(512 + x^3) - Log[4 - Log[8 + x]]

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
norman	\(x^{3} {\mathrm e}^{5}-x^{2}-\ln \left (\ln \left (x +8\right )-4\right )\)	\(22\)
risch	\(x^{3} {\mathrm e}^{5}-x^{2}-\ln \left (\ln \left (x +8\right )-4\right )\)	\(22\)
parallelrisch	\(x^{3} {\mathrm e}^{5}-x^{2}-\ln \left (\ln \left (x +8\right )-4\right )+64\)	\(23\)
derivativedivides	\(-\ln \left (\ln \left (x +8\right )-4\right )+16 x +128-\left (x +8\right )^{2}+192 \,{\mathrm e}^{5} \left (x +8\right )-24 \,{\mathrm e}^{5} \left (x +8\right )^{2}+{\mathrm e}^{5} \left (x +8\right )^{3}\)	\(46\)
default	\(-\ln \left (\ln \left (x +8\right )-4\right )+16 x +128-\left (x +8\right )^{2}+192 \,{\mathrm e}^{5} \left (x +8\right )-24 \,{\mathrm e}^{5} \left (x +8\right )^{2}+{\mathrm e}^{5} \left (x +8\right )^{3}\)	\(46\)

[In]

int((((3*x^3+24*x^2)*exp(5)-2*x^2-16*x)*ln(x+8)+(-12*x^3-96*x^2)*exp(5)+8*x^2+64*x-1)/((x+8)*ln(x+8)-4*x-32),x
,method=_RETURNVERBOSE)

[Out]

x^3*exp(5)-x^2-ln(ln(x+8)-4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=x^{3} e^{5} - x^{2} - \log \left (\log \left (x + 8\right ) - 4\right ) \]

[In]

integrate((((3*x^3+24*x^2)*exp(5)-2*x^2-16*x)*log(x+8)+(-12*x^3-96*x^2)*exp(5)+8*x^2+64*x-1)/((x+8)*log(x+8)-4
*x-32),x, algorithm="fricas")

[Out]

x^3*e^5 - x^2 - log(log(x + 8) - 4)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=x^{3} e^{5} - x^{2} - \log {\left (\log {\left (x + 8 \right )} - 4 \right )} \]

[In]

integrate((((3*x**3+24*x**2)*exp(5)-2*x**2-16*x)*ln(x+8)+(-12*x**3-96*x**2)*exp(5)+8*x**2+64*x-1)/((x+8)*ln(x+
8)-4*x-32),x)

[Out]

x**3*exp(5) - x**2 - log(log(x + 8) - 4)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=x^{3} e^{5} - x^{2} - \log \left (\log \left (x + 8\right ) - 4\right ) \]

[In]

integrate((((3*x^3+24*x^2)*exp(5)-2*x^2-16*x)*log(x+8)+(-12*x^3-96*x^2)*exp(5)+8*x^2+64*x-1)/((x+8)*log(x+8)-4
*x-32),x, algorithm="maxima")

[Out]

x^3*e^5 - x^2 - log(log(x + 8) - 4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=x^{3} e^{5} - x^{2} - \log \left (\log \left (x + 8\right ) - 4\right ) \]

[In]

integrate((((3*x^3+24*x^2)*exp(5)-2*x^2-16*x)*log(x+8)+(-12*x^3-96*x^2)*exp(5)+8*x^2+64*x-1)/((x+8)*log(x+8)-4
*x-32),x, algorithm="giac")

[Out]

x^3*e^5 - x^2 - log(log(x + 8) - 4)

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-1+64 x+8 x^2+e^5 \left (-96 x^2-12 x^3\right )+\left (-16 x-2 x^2+e^5 \left (24 x^2+3 x^3\right )\right ) \log (8+x)}{-32-4 x+(8+x) \log (8+x)} \, dx=x^3\,{\mathrm {e}}^5-\ln \left (\ln \left (x+8\right )-4\right )-x^2 \]

[In]

int((exp(5)*(96*x^2 + 12*x^3) - 64*x + log(x + 8)*(16*x - exp(5)*(24*x^2 + 3*x^3) + 2*x^2) - 8*x^2 + 1)/(4*x -
log(x + 8)*(x + 8) + 32),x)

[Out]

x^3*exp(5) - log(log(x + 8) - 4) - x^2