3.69.22 $\int \frac{-384e^{\frac{4(16-x)}{x}}-10x^3+10e^3{x}^3}{-x^2+e^3{x}^2}dx$

Optimal. Leaf size=29 $$\frac{6e^{\frac{4(16-x)}{x}}}{-1+e^3}+5(-5+x^2)$$

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 4, integrand size = 44, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.091, Rules used = {6, 12, 14, 2209} $$\begin{array}{c}5x^2-\frac{6e^{\frac{64}{x}-4}}{1-e^3}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-384*E^((4*(16 - x))/x) - 10*x^3 + 10*E^3*x^3)/(-x^2 + E^3*x^2),x]

[Out]

(-6*E^(-4 + 64/x))/(1 - E^3) + 5*x^2

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-384e^{\frac{4(16-x)}{x}}-10x^3+10e^3{x}^3}{(-1+e^3)x^2}dx\\ & =\int \frac{-384e^{\frac{4(16-x)}{x}}+(-10+10e^3)x^3}{(-1+e^3)x^2}dx\\ & =\frac{\int \frac{-384e^{\frac{4(16-x)}{x}}+(-10+10e^3)x^3}{x^2}dx}{-1+e^3}\\ & =\frac{\int (-\frac{384e^{-4+\frac{64}{x}}}{x^2}+10(-1+e^3)x)dx}{-1+e^3}\\ & =5x^2+\frac{384\int \frac{e^{-4+\frac{64}{x}}}{x^2}dx}{1-e^3}\\ & =-\frac{6e^{-4+\frac{64}{x}}}{1-e^3}+5x^2\end{array}\end{array}$$

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Mathematica [A]  time = 0.02, size = 39, normalized size = 1.34 $$\begin{array}{c}\frac{2(3e^{64/x}-\frac{5}{2}{e}^4(1-e^3)x^2)}{e^4(-1+e^3)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-384*E^((4*(16 - x))/x) - 10*x^3 + 10*E^3*x^3)/(-x^2 + E^3*x^2),x]

[Out]

(2*(3*E^(64/x) - (5*E^4*(1 - E^3)*x^2)/2))/(E^4*(-1 + E^3))

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fricas [A]  time = 0.45, size = 31, normalized size = 1.07 $$\begin{array}{c}\frac{5x^2{e}^3-5x^2+6e^{(-\frac{4(x-16)}{x})}}{e^3-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-384*exp((16-x)/x)^4+10*x^3*exp(3)-10*x^3)/(x^2*exp(3)-x^2),x, algorithm="fricas")

[Out]

(5*x^2*e^3 - 5*x^2 + 6*e^(-4*(x - 16)/x))/(e^3 - 1)

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giac [B]  time = 0.15, size = 100, normalized size = 3.45 $$\begin{array}{c}\frac{2(\frac{3{(x-16)}^2{e}^{(-\frac{4(x-16)}{x})}}{x^2}-\frac{6(x-16)e^{(-\frac{4(x-16)}{x})}}{x}+640e^3+3e^{(-\frac{4(x-16)}{x})}-640)}{\frac{{(x-16)}^2{e}^3}{x^2}-\frac{2(x-16)e^3}{x}-\frac{{(x-16)}^2}{x^2}+\frac{2(x-16)}{x}+e^3-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-384*exp((16-x)/x)^4+10*x^3*exp(3)-10*x^3)/(x^2*exp(3)-x^2),x, algorithm="giac")

[Out]

2*(3*(x - 16)^2*e^(-4*(x - 16)/x)/x^2 - 6*(x - 16)*e^(-4*(x - 16)/x)/x + 640*e^3 + 3*e^(-4*(x - 16)/x) - 640)/
((x - 16)^2*e^3/x^2 - 2*(x - 16)*e^3/x - (x - 16)^2/x^2 + 2*(x - 16)/x + e^3 - 1)

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maple [A]  time = 0.10, size = 24, normalized size = 0.83

method	result	size
risch	$5x^2+\frac{6{\mathrm{e}}^{-\frac{4(x-16)}{x}}}{{\mathrm{e}}^3-1}$	$24$
norman	$\frac{5x^3+\frac{6x{\mathrm{e}}^{\frac{-4x+64}{x}}}{{\mathrm{e}}^3-1}}{x}$	$32$
derivativedivides	$-\frac{1280x^2}{256{\mathrm{e}}^3-256}+\frac{6{\mathrm{e}}^{-4+\frac{64}{x}}}{{\mathrm{e}}^3-1}+\frac{1280{\mathrm{e}}^3{x}^2}{256{\mathrm{e}}^3-256}$	$48$
default	$-\frac{1280x^2}{256{\mathrm{e}}^3-256}+\frac{6{\mathrm{e}}^{-4+\frac{64}{x}}}{{\mathrm{e}}^3-1}+\frac{1280{\mathrm{e}}^3{x}^2}{256{\mathrm{e}}^3-256}$	$48$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-384*exp((16-x)/x)^4+10*x^3*exp(3)-10*x^3)/(x^2*exp(3)-x^2),x,method=_RETURNVERBOSE)

[Out]

5*x^2+6*exp(-4*(x-16)/x)/(exp(3)-1)

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maxima [A]  time = 0.37, size = 42, normalized size = 1.45 $$\begin{array}{c}\frac{5x^2{e}^3}{e^3-1}-\frac{5x^2}{e^3-1}+\frac{6e^{\frac{64}{x}}}{e^7-e^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-384*exp((16-x)/x)^4+10*x^3*exp(3)-10*x^3)/(x^2*exp(3)-x^2),x, algorithm="maxima")

[Out]

5*x^2*e^3/(e^3 - 1) - 5*x^2/(e^3 - 1) + 6*e^(64/x)/(e^7 - e^4)

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mupad [B]  time = 4.11, size = 33, normalized size = 1.14 $$\begin{array}{c}\frac{6{\mathrm{e}}^{\frac{64}{x}-4}}{{\mathrm{e}}^3-1}+\frac{x^2(5{\mathrm{e}}^3-5)}{{\mathrm{e}}^3-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(384*exp(-(4*(x - 16))/x) - 10*x^3*exp(3) + 10*x^3)/(x^2*exp(3) - x^2),x)

[Out]

(6*exp(64/x - 4))/(exp(3) - 1) + (x^2*(5*exp(3) - 5))/(exp(3) - 1)

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sympy [A]  time = 0.20, size = 19, normalized size = 0.66 $$\begin{array}{c}5x^2+\frac{6e^{\frac{4(16-x)}{x}}}{-1+e^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-384*exp((16-x)/x)**4+10*x**3*exp(3)-10*x**3)/(x**2*exp(3)-x**2),x)

[Out]

5*x**2 + 6*exp(4*(16 - x)/x)/(-1 + exp(3))

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