3.75.29 \(\int \frac {1+e^{-1+e^5} (-2+e^2 x^4)}{-e^2 x^4+2 e^{1+e^5} x^4} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{3 e^2 x^3}+\frac {3+x}{2-e^{1-e^5}} \]

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Rubi [A]  time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 14} \begin {gather*} \frac {1}{3 e^2 x^3}-\frac {e^{e^5} x}{e-2 e^{e^5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^(-1 + E^5)*(-2 + E^2*x^4))/(-(E^2*x^4) + 2*E^(1 + E^5)*x^4),x]

[Out]

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

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]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.47 \begin {gather*} -\frac {\frac {-e+2 e^{e^5}}{3 x^3}+e^{2+e^5} x}{e^2 \left (e-2 e^{e^5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(-1 + E^5)*(-2 + E^2*x^4))/(-(E^2*x^4) + 2*E^(1 + E^5)*x^4),x]

[Out]

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

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fricas [A]  time = 0.85, size = 41, normalized size = 1.37 \begin {gather*} -\frac {{\left (3 \, x^{4} e^{2} + 2\right )} e^{\left (e^{5} + 1\right )} - e^{2}}{3 \, {\left (x^{3} e^{4} - 2 \, x^{3} e^{\left (e^{5} + 3\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4*exp(1)^2-2)*exp(exp(5)-1)+1)/(2*x^4*exp(1)^2*exp(exp(5)-1)-x^4*exp(1)^2),x, algorithm="fricas"
)

[Out]

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

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giac [A]  time = 0.22, size = 47, normalized size = 1.57 \begin {gather*} -\frac {x e^{\left (e^{5} + 1\right )}}{e^{2} - 2 \, e^{\left (e^{5} + 1\right )}} - \frac {2 \, e^{\left (e^{5} - 1\right )} - 1}{3 \, x^{3} {\left (e^{2} - 2 \, e^{\left (e^{5} + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4*exp(1)^2-2)*exp(exp(5)-1)+1)/(2*x^4*exp(1)^2*exp(exp(5)-1)-x^4*exp(1)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 38, normalized size = 1.27

method	result	size
norman	\(\frac {\left (\frac {{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{5}} {\mathrm e}^{-1} x^{4}}{2 \,{\mathrm e}^{{\mathrm e}^{5}} {\mathrm e}^{-1}-1}+\frac {{\mathrm e}^{-1}}{3}\right ) {\mathrm e}^{-1}}{x^{3}}\)	\(38\)
default	\(\frac {{\mathrm e}^{-2} \left ({\mathrm e}^{{\mathrm e}^{5}+1} x -\frac {-2 \,{\mathrm e}^{{\mathrm e}^{5}-1}+1}{3 x^{3}}\right )}{2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1}\)	\(39\)
gosper	\(\frac {\left (3 x^{4} {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{5}-1}+2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1\right ) {\mathrm e}^{-2}}{3 x^{3} \left (2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1\right )}\)	\(44\)
risch	\(\frac {x \,{\mathrm e}^{{\mathrm e}^{5}-1}}{2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{5}-4}}{3 \left (2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1\right )^{2} x^{3}}-\frac {4 \,{\mathrm e}^{{\mathrm e}^{5}-3}}{3 \left (2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1\right )^{2} x^{3}}+\frac {{\mathrm e}^{-2}}{3 \left (2 \,{\mathrm e}^{{\mathrm e}^{5}-1}-1\right )^{2} x^{3}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4*exp(1)^2-2)*exp(exp(5)-1)+1)/(2*x^4*exp(1)^2*exp(exp(5)-1)-x^4*exp(1)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 24, normalized size = 0.80 \begin {gather*} -\frac {x e^{\left (e^{5}\right )}}{e - 2 \, e^{\left (e^{5}\right )}} + \frac {e^{\left (-2\right )}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4*exp(1)^2-2)*exp(exp(5)-1)+1)/(2*x^4*exp(1)^2*exp(exp(5)-1)-x^4*exp(1)^2),x, algorithm="maxima"
)

[Out]

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

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mupad [B]  time = 6.53, size = 28, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^{-2}}{3\,x^3}+\frac {x\,{\mathrm {e}}^2}{2\,{\mathrm {e}}^2-{\mathrm {e}}^{3-{\mathrm {e}}^5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(5) - 1)*(x^4*exp(2) - 2) + 1)/(x^4*exp(2) - 2*x^4*exp(exp(5) - 1)*exp(2)),x)

[Out]

exp(-2)/(3*x^3) + (x*exp(2))/(2*exp(2) - exp(3 - exp(5)))

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sympy [A]  time = 0.15, size = 39, normalized size = 1.30 \begin {gather*} \frac {- x e^{2} e^{e^{5}} - \frac {- e + 2 e^{e^{5}}}{3 x^{3}}}{- 2 e^{2} e^{e^{5}} + e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**4*exp(1)**2-2)*exp(exp(5)-1)+1)/(2*x**4*exp(1)**2*exp(exp(5)-1)-x**4*exp(1)**2),x)

[Out]

(-x*exp(2)*exp(exp(5)) - (-E + 2*exp(exp(5)))/(3*x**3))/(-2*exp(2)*exp(exp(5)) + exp(3))

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