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

Optimal. Leaf size=21 \[ \frac {e^4 \left (4+e^{4+e^3}-\frac {x}{2}\right )}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 4, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 1586, 30} \begin {gather*} \frac {e^4 \left (4+e^{4+e^3}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^4*(4 + E^(4 + E^3)))/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 \left (-8-2 e^{4+e^3}\right ) \left (8+2 e^{4+e^3}-x\right )}{2 \left (\left (8+2 e^{4+e^3}\right ) x^2-x^3\right )} \, dx\\ &=-\left (\left (e^4 \left (4+e^{4+e^3}\right )\right ) \int \frac {8+2 e^{4+e^3}-x}{\left (8+2 e^{4+e^3}\right ) x^2-x^3} \, dx\right )\\ &=-\left (\left (e^4 \left (4+e^{4+e^3}\right )\right ) \int \frac {1}{x^2} \, dx\right )\\ &=\frac {e^4 \left (4+e^{4+e^3}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.76 \begin {gather*} \frac {e^4 \left (4+e^{4+e^3}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^4*(4 + E^(4 + E^3)))/x

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fricas [A]  time = 1.16, size = 14, normalized size = 0.67 \begin {gather*} \frac {4 \, e^{4} + e^{\left (e^{3} + 8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(4+exp(3))-8)*exp(log(exp(4+exp(3))+4-1/2*x)+4)/(2*x^2*exp(4+exp(3))-x^3+8*x^2),x, algorithm=
"fricas")

[Out]

(4*e^4 + e^(e^3 + 8))/x

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giac [A]  time = 0.16, size = 13, normalized size = 0.62 \begin {gather*} \frac {{\left (e^{\left (e^{3} + 4\right )} + 4\right )} e^{4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(4+exp(3))-8)*exp(log(exp(4+exp(3))+4-1/2*x)+4)/(2*x^2*exp(4+exp(3))-x^3+8*x^2),x, algorithm=
"giac")

[Out]

(e^(e^3 + 4) + 4)*e^4/x

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maple [A]  time = 0.23, size = 17, normalized size = 0.81

method	result	size
default	\(-\frac {\left (-2 \,{\mathrm e}^{4+{\mathrm e}^{3}}-8\right ) {\mathrm e}^{4}}{2 x}\)	\(17\)
norman	\(\frac {{\mathrm e}^{8} {\mathrm e}^{{\mathrm e}^{3}}+4 \,{\mathrm e}^{4}}{x}\)	\(18\)
risch	\(\frac {{\mathrm e}^{4} {\mathrm e}^{4+{\mathrm e}^{3}}}{x}+\frac {4 \,{\mathrm e}^{4}}{x}\)	\(20\)
gosper	\(\frac {2 \left ({\mathrm e}^{4+{\mathrm e}^{3}}+4\right ) {\mathrm e}^{\ln \left ({\mathrm e}^{4+{\mathrm e}^{3}}+4-\frac {x}{2}\right )+4}}{x \left (2 \,{\mathrm e}^{4+{\mathrm e}^{3}}+8-x \right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(4+exp(3))-8)*exp(ln(exp(4+exp(3))+4-1/2*x)+4)/(2*x^2*exp(4+exp(3))-x^3+8*x^2),x,method=_RETURNVERB
OSE)

[Out]

-1/2*(-2*exp(4+exp(3))-8)*exp(4)/x

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maxima [A]  time = 0.45, size = 13, normalized size = 0.62 \begin {gather*} \frac {{\left (e^{\left (e^{3} + 4\right )} + 4\right )} e^{4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(4+exp(3))-8)*exp(log(exp(4+exp(3))+4-1/2*x)+4)/(2*x^2*exp(4+exp(3))-x^3+8*x^2),x, algorithm=
"maxima")

[Out]

(e^(e^3 + 4) + 4)*e^4/x

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mupad [B]  time = 0.09, size = 14, normalized size = 0.67 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^3+8}+4\,{\mathrm {e}}^4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(exp(3) + 8) + 4*exp(4))/x

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sympy [A]  time = 0.08, size = 17, normalized size = 0.81 \begin {gather*} - \frac {- e^{8} e^{e^{3}} - 4 e^{4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(4+exp(3))-8)*exp(ln(exp(4+exp(3))+4-1/2*x)+4)/(2*x**2*exp(4+exp(3))-x**3+8*x**2),x)

[Out]

-(-exp(8)*exp(exp(3)) - 4*exp(4))/x

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