\(\int \frac {(-1-240 x-240 e^{16} x) \log (5 e^3)}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} (240 x^3+28800 x^4)} \, dx\) [3265]

Optimal result

Integrand size = 57, antiderivative size = 24 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {\log \left (5 e^3\right )}{x \left (1+24 \left (5+5 e^{16}\right ) x\right )} \]

[Out]

ln(5*exp(3))/(24*(5+5*exp(16))*x+1)/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 1694, 267} \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {3+\log (5)}{x \left (120 \left (1+e^{16}\right ) x+1\right )} \]

[In]

Int[((-1 - 240*x - 240*E^16*x)*Log[5*E^3])/(x^2 + 240*x^3 + 14400*x^4 + 14400*E^32*x^4 + E^16*(240*x^3 + 28800
*x^4)),x]

[Out]

(3 + Log[5])/(x*(1 + 120*(1 + E^16)*x))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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}& = \int \frac {\left (-1+\left (-240-240 e^{16}\right ) x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx \\ & = \int \frac {\left (-1+\left (-240-240 e^{16}\right ) x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+\left (14400+14400 e^{32}\right ) x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx \\ & = (3+\log (5)) \int \frac {-1+\left (-240-240 e^{16}\right ) x}{x^2+240 x^3+\left (14400+14400 e^{32}\right ) x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx \\ & = (3+\log (5)) \text {Subst}\left (\int \frac {55296000 \left (-1-e^{16}\right )^3 x}{\left (1-57600 \left (1+e^{16}\right )^2 x^2\right )^2} \, dx,x,\frac {240+240 e^{16}}{4 \left (14400+28800 e^{16}+14400 e^{32}\right )}+x\right ) \\ & = -\left (\left (55296000 \left (1+e^{16}\right )^3 (3+\log (5))\right ) \text {Subst}\left (\int \frac {x}{\left (1-57600 \left (1+e^{16}\right )^2 x^2\right )^2} \, dx,x,\frac {240+240 e^{16}}{4 \left (14400+28800 e^{16}+14400 e^{32}\right )}+x\right )\right ) \\ & = \frac {3+\log (5)}{x \left (1+120 \left (1+e^{16}\right ) x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {3+\log (5)}{x \left (1+120 \left (1+e^{16}\right ) x\right )} \]

[In]

Integrate[((-1 - 240*x - 240*E^16*x)*Log[5*E^3])/(x^2 + 240*x^3 + 14400*x^4 + 14400*E^32*x^4 + E^16*(240*x^3 +
28800*x^4)),x]

[Out]

(3 + Log[5])/(x*(1 + 120*(1 + E^16)*x))

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
norman	\(\frac {\ln \left (5\right )+3}{x \left (120 x \,{\mathrm e}^{16}+120 x +1\right )}\)	\(21\)
risch	\(\frac {\ln \left (5\right )+3}{x \left (120 x \,{\mathrm e}^{16}+120 x +1\right )}\)	\(21\)
gosper	\(\frac {\ln \left (5 \,{\mathrm e}^{3}\right )}{x \left (120 x \,{\mathrm e}^{16}+120 x +1\right )}\)	\(22\)
parallelrisch	\(\frac {\ln \left (5 \,{\mathrm e}^{3}\right )}{x \left (120 x \,{\mathrm e}^{16}+120 x +1\right )}\)	\(22\)

[In]

int((-240*x*exp(16)-240*x-1)*ln(5*exp(3))/(14400*x^4*exp(16)^2+(28800*x^4+240*x^3)*exp(16)+14400*x^4+240*x^3+x
^2),x,method=_RETURNVERBOSE)

[Out]

(ln(5)+3)/x/(120*x*exp(16)+120*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {\log \left (5\right ) + 3}{120 \, x^{2} e^{16} + 120 \, x^{2} + x} \]

[In]

integrate((-240*x*exp(16)-240*x-1)*log(5*exp(3))/(14400*x^4*exp(16)^2+(28800*x^4+240*x^3)*exp(16)+14400*x^4+24
0*x^3+x^2),x, algorithm="fricas")

[Out]

(log(5) + 3)/(120*x^2*e^16 + 120*x^2 + x)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=- \frac {-3 - \log {\left (5 \right )}}{x^{2} \cdot \left (120 + 120 e^{16}\right ) + x} \]

[In]

integrate((-240*x*exp(16)-240*x-1)*ln(5*exp(3))/(14400*x**4*exp(16)**2+(28800*x**4+240*x**3)*exp(16)+14400*x**
4+240*x**3+x**2),x)

[Out]

-(-3 - log(5))/(x**2*(120 + 120*exp(16)) + x)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {\log \left (5 \, e^{3}\right )}{120 \, x^{2} {\left (e^{16} + 1\right )} + x} \]

[In]

integrate((-240*x*exp(16)-240*x-1)*log(5*exp(3))/(14400*x^4*exp(16)^2+(28800*x^4+240*x^3)*exp(16)+14400*x^4+24
0*x^3+x^2),x, algorithm="maxima")

[Out]

log(5*e^3)/(120*x^2*(e^16 + 1) + x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {\log \left (5 \, e^{3}\right )}{120 \, x^{2} e^{16} + 120 \, x^{2} + x} \]

[In]

integrate((-240*x*exp(16)-240*x-1)*log(5*exp(3))/(14400*x^4*exp(16)^2+(28800*x^4+240*x^3)*exp(16)+14400*x^4+24
0*x^3+x^2),x, algorithm="giac")

[Out]

log(5*e^3)/(120*x^2*e^16 + 120*x^2 + x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-1-240 x-240 e^{16} x\right ) \log \left (5 e^3\right )}{x^2+240 x^3+14400 x^4+14400 e^{32} x^4+e^{16} \left (240 x^3+28800 x^4\right )} \, dx=\frac {\ln \left (5\right )+3}{\left (120\,{\mathrm {e}}^{16}+120\right )\,x^2+x} \]

[In]

int(-(log(5*exp(3))*(240*x + 240*x*exp(16) + 1))/(exp(16)*(240*x^3 + 28800*x^4) + 14400*x^4*exp(32) + x^2 + 24
0*x^3 + 14400*x^4),x)

[Out]

(log(5) + 3)/(x + x^2*(120*exp(16) + 120))