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

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

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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 {gather*} 5 x^2-\frac {6 e^{\frac {64}{x}-4}}{1-e^3} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{\left (-1+e^3\right ) x^2} \, dx\\ &=\int \frac {-384 e^{\frac {4 (16-x)}{x}}+\left (-10+10 e^3\right ) x^3}{\left (-1+e^3\right ) x^2} \, dx\\ &=\frac {\int \frac {-384 e^{\frac {4 (16-x)}{x}}+\left (-10+10 e^3\right ) x^3}{x^2} \, dx}{-1+e^3}\\ &=\frac {\int \left (-\frac {384 e^{-4+\frac {64}{x}}}{x^2}+10 \left (-1+e^3\right ) x\right ) \, dx}{-1+e^3}\\ &=5 x^2+\frac {384 \int \frac {e^{-4+\frac {64}{x}}}{x^2} \, dx}{1-e^3}\\ &=-\frac {6 e^{-4+\frac {64}{x}}}{1-e^3}+5 x^2\\ \end {aligned} \end {gather*}

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

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 {gather*} \frac {5 \, x^{2} e^{3} - 5 \, x^{2} + 6 \, e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )}}{e^{3} - 1} \end {gather*}

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 {gather*} \frac {2 \, {\left (\frac {3 \, {\left (x - 16\right )}^{2} e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )}}{x^{2}} - \frac {6 \, {\left (x - 16\right )} e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )}}{x} + 640 \, e^{3} + 3 \, e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )} - 640\right )}}{\frac {{\left (x - 16\right )}^{2} e^{3}}{x^{2}} - \frac {2 \, {\left (x - 16\right )} e^{3}}{x} - \frac {{\left (x - 16\right )}^{2}}{x^{2}} + \frac {2 \, {\left (x - 16\right )}}{x} + e^{3} - 1} \end {gather*}

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	\(5 x^{2}+\frac {6 \,{\mathrm e}^{-\frac {4 \left (x -16\right )}{x}}}{{\mathrm e}^{3}-1}\)	\(24\)
norman	\(\frac {5 x^{3}+\frac {6 x \,{\mathrm e}^{\frac {-4 x +64}{x}}}{{\mathrm e}^{3}-1}}{x}\)	\(32\)
derivativedivides	\(-\frac {1280 x^{2}}{256 \,{\mathrm e}^{3}-256}+\frac {6 \,{\mathrm e}^{-4+\frac {64}{x}}}{{\mathrm e}^{3}-1}+\frac {1280 \,{\mathrm e}^{3} x^{2}}{256 \,{\mathrm e}^{3}-256}\)	\(48\)
default	\(-\frac {1280 x^{2}}{256 \,{\mathrm e}^{3}-256}+\frac {6 \,{\mathrm e}^{-4+\frac {64}{x}}}{{\mathrm e}^{3}-1}+\frac {1280 \,{\mathrm e}^{3} x^{2}}{256 \,{\mathrm e}^{3}-256}\)	\(48\)

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 {gather*} \frac {5 \, x^{2} e^{3}}{e^{3} - 1} - \frac {5 \, x^{2}}{e^{3} - 1} + \frac {6 \, e^{\frac {64}{x}}}{e^{7} - e^{4}} \end {gather*}

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 {gather*} \frac {6\,{\mathrm {e}}^{\frac {64}{x}-4}}{{\mathrm {e}}^3-1}+\frac {x^2\,\left (5\,{\mathrm {e}}^3-5\right )}{{\mathrm {e}}^3-1} \end {gather*}

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 {gather*} 5 x^{2} + \frac {6 e^{\frac {4 \left (16 - x\right )}{x}}}{-1 + e^{3}} \end {gather*}

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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