3.33.65 $\int \frac{(-1-240x-240e^{16}x)\log (5e^3)}{x^2+240x^3+14400x^4+14400e^{32}{x}^4+e^{16}(240x^3+28800x^4)}dx$

Optimal. Leaf size=24 $$\frac{\log \left(5e^3\right)}{x(1+24(5+5e^{16})x)}$$

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Rubi [A]  time = 0.08, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 4, integrand size = 57, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.070, Rules used = {6, 12, 1680, 261} $$\begin{array}{c}\frac{3+\log (5)}{x(120(1+e^{16})x+1)}\end{array}$$

Antiderivative was successfully verified.

[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 261

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 1680

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{(-1+(-240-240e^{16})x)\log \left(5e^3\right)}{x^2+240x^3+14400x^4+14400e^{32}{x}^4+e^{16}(240x^3+28800x^4)}dx\\ & =\int \frac{(-1+(-240-240e^{16})x)\log \left(5e^3\right)}{x^2+240x^3+(14400+14400e^{32})x^4+e^{16}(240x^3+28800x^4)}dx\\ & =(3+\log (5))\int \frac{-1+(-240-240e^{16})x}{x^2+240x^3+(14400+14400e^{32})x^4+e^{16}(240x^3+28800x^4)}dx\\ & =(3+\log (5))\mathrm{Subst}(\int \frac{55296000{(-1-e^{16})}^{3}x}{{(1-57600{(1+e^{16})}^2{x}^2)}^2}dx,x,\frac{240+240e^{16}}{4(14400+28800e^{16}+14400e^{32})}+x)\\ & =-\left((55296000{(1+e^{16})}^3(3+\log (5)))\mathrm{Subst}(\int \frac{x}{{(1-57600{(1+e^{16})}^2{x}^2)}^2}dx,x,\frac{240+240e^{16}}{4(14400+28800e^{16}+14400e^{32})}+x)\right)\\ & =\frac{3+\log (5)}{x(1+120(1+e^{16})x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.83 $$\begin{array}{c}\frac{3+\log (5)}{x(1+120(1+e^{16})x)}\end{array}$$

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.65, size = 21, normalized size = 0.88 $$\begin{array}{c}\frac{\log (5)+3}{120x^2{e}^{16}+120x^2+x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Exception raised: NotImplementedError}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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]

Exception raised: NotImplementedError >> Unable to parse Giac output: -ln(5*exp(3))*(-1/sageVARx+(-14400*exp(3
2)-28800*exp(16)-14400)*1/240/sqrt(exp(16)^2-exp(32))*ln(sqrt((28800*sageVARx*exp(32)+57600*sageVARx*exp(16)+2
8800*sageVARx+240*exp

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maple [A]  time = 0.09, size = 21, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.39, size = 19, normalized size = 0.79 $$\begin{array}{c}\frac{\log \left(5e^3\right)}{120x^2(e^{16}+1)+x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.17, size = 19, normalized size = 0.79 $$\begin{array}{c}\frac{\ln (5)+3}{(120{\mathrm{e}}^{16}+120)x^2+x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.81, size = 19, normalized size = 0.79 $$\begin{array}{c}-\frac{-3-\log (5)}{x^2(120+120e^{16})+x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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