3.20.68 \(\int (8+e^{\frac {5-4 x}{x}} (5 x^2-4 x^3)) \, dx\)

Optimal. Leaf size=21 \[ -5+8 x-e^{-5+\frac {5+x}{x}} x^4 \]

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Rubi [A]  time = 0.07, antiderivative size = 37, normalized size of antiderivative = 1.76, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1593, 2288} \begin {gather*} 8 x-\frac {5 e^{\frac {5-4 x}{x}} x^2}{\frac {5-4 x}{x^2}+\frac {4}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8 + E^((5 - 4*x)/x)*(5*x^2 - 4*x^3),x]

[Out]

8*x - (5*E^((5 - 4*x)/x)*x^2)/((5 - 4*x)/x^2 + 4/x)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=8 x+\int e^{\frac {5-4 x}{x}} \left (5 x^2-4 x^3\right ) \, dx\\ &=8 x+\int e^{\frac {5-4 x}{x}} (5-4 x) x^2 \, dx\\ &=8 x-\frac {5 e^{\frac {5-4 x}{x}} x^2}{\frac {5-4 x}{x^2}+\frac {4}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 0.86 \begin {gather*} 8 x-e^{-4+\frac {5}{x}} x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8 + E^((5 - 4*x)/x)*(5*x^2 - 4*x^3),x]

[Out]

8*x - E^(-4 + 5/x)*x^4

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fricas [A]  time = 0.82, size = 20, normalized size = 0.95 \begin {gather*} -x^{4} e^{\left (-\frac {4 \, x - 5}{x}\right )} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3+5*x^2)*exp((-4*x+5)/x)+8,x, algorithm="fricas")

[Out]

-x^4*e^(-(4*x - 5)/x) + 8*x

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giac [B]  time = 0.21, size = 66, normalized size = 3.14 \begin {gather*} 8 \, x - \frac {625 \, e^{\left (-\frac {4 \, x - 5}{x}\right )}}{\frac {{\left (4 \, x - 5\right )}^{4}}{x^{4}} - \frac {16 \, {\left (4 \, x - 5\right )}^{3}}{x^{3}} + \frac {96 \, {\left (4 \, x - 5\right )}^{2}}{x^{2}} - \frac {256 \, {\left (4 \, x - 5\right )}}{x} + 256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3+5*x^2)*exp((-4*x+5)/x)+8,x, algorithm="giac")

[Out]

8*x - 625*e^(-(4*x - 5)/x)/((4*x - 5)^4/x^4 - 16*(4*x - 5)^3/x^3 + 96*(4*x - 5)^2/x^2 - 256*(4*x - 5)/x + 256)

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maple [A]  time = 0.09, size = 18, normalized size = 0.86

method	result	size
derivativedivides	\(8 x -{\mathrm e}^{\frac {5}{x}-4} x^{4}\)	\(18\)
default	\(8 x -{\mathrm e}^{\frac {5}{x}-4} x^{4}\)	\(18\)
norman	\(8 x -x^{4} {\mathrm e}^{\frac {-4 x +5}{x}}\)	\(20\)
risch	\(8 x -x^{4} {\mathrm e}^{-\frac {4 x -5}{x}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^3+5*x^2)*exp((-4*x+5)/x)+8,x,method=_RETURNVERBOSE)

[Out]

8*x-exp(5/x-4)*x^4

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maxima [C]  time = 0.58, size = 26, normalized size = 1.24 \begin {gather*} -625 \, e^{\left (-4\right )} \Gamma \left (-3, -\frac {5}{x}\right ) - 2500 \, e^{\left (-4\right )} \Gamma \left (-4, -\frac {5}{x}\right ) + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3+5*x^2)*exp((-4*x+5)/x)+8,x, algorithm="maxima")

[Out]

-625*e^(-4)*gamma(-3, -5/x) - 2500*e^(-4)*gamma(-4, -5/x) + 8*x

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mupad [B]  time = 1.15, size = 17, normalized size = 0.81 \begin {gather*} 8\,x-x^4\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{5/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-(4*x - 5)/x)*(5*x^2 - 4*x^3) + 8,x)

[Out]

8*x - x^4*exp(-4)*exp(5/x)

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sympy [A]  time = 0.09, size = 14, normalized size = 0.67 \begin {gather*} - x^{4} e^{\frac {5 - 4 x}{x}} + 8 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**3+5*x**2)*exp((-4*x+5)/x)+8,x)

[Out]

-x**4*exp((5 - 4*x)/x) + 8*x

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