3.100.31 \(\int \frac {e^{\frac {16}{x^4}} (-51200-6400 e^5)+2 x^5}{200 x^5+25 e^5 x^5} \, dx\)

Optimal. Leaf size=22 \[ 4 \left (e^{\frac {16}{x^4}}+\frac {x}{50 \left (8+e^5\right )}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6, 12, 14, 2209} \begin {gather*} 4 e^{\frac {16}{x^4}}+\frac {2 x}{25 \left (8+e^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(16/x^4)*(-51200 - 6400*E^5) + 2*x^5)/(200*x^5 + 25*E^5*x^5),x]

[Out]

4*E^(16/x^4) + (2*x)/(25*(8 + 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]]

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 {e^{\frac {16}{x^4}} \left (-51200-6400 e^5\right )+2 x^5}{\left (200+25 e^5\right ) x^5} \, dx\\ &=\frac {\int \frac {e^{\frac {16}{x^4}} \left (-51200-6400 e^5\right )+2 x^5}{x^5} \, dx}{25 \left (8+e^5\right )}\\ &=\frac {\int \left (2-\frac {6400 e^{\frac {16}{x^4}} \left (8+e^5\right )}{x^5}\right ) \, dx}{25 \left (8+e^5\right )}\\ &=\frac {2 x}{25 \left (8+e^5\right )}-256 \int \frac {e^{\frac {16}{x^4}}}{x^5} \, dx\\ &=4 e^{\frac {16}{x^4}}+\frac {2 x}{25 \left (8+e^5\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} 4 e^{\frac {16}{x^4}}+\frac {2 x}{25 \left (8+e^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(16/x^4)*(-51200 - 6400*E^5) + 2*x^5)/(200*x^5 + 25*E^5*x^5),x]

[Out]

4*E^(16/x^4) + (2*x)/(25*(8 + E^5))

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fricas [A]  time = 0.60, size = 22, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (50 \, {\left (e^{5} + 8\right )} e^{\left (\frac {16}{x^{4}}\right )} + x\right )}}{25 \, {\left (e^{5} + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6400*exp(5)-51200)*exp(16/x^4)+2*x^5)/(25*x^5*exp(5)+200*x^5),x, algorithm="fricas")

[Out]

2/25*(50*(e^5 + 8)*e^(16/x^4) + x)/(e^5 + 8)

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giac [A]  time = 0.20, size = 18, normalized size = 0.82 \begin {gather*} \frac {2 \, x}{25 \, {\left (e^{5} + 8\right )}} + 4 \, e^{\left (\frac {16}{x^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6400*exp(5)-51200)*exp(16/x^4)+2*x^5)/(25*x^5*exp(5)+200*x^5),x, algorithm="giac")

[Out]

2/25*x/(e^5 + 8) + 4*e^(16/x^4)

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maple [A]  time = 0.15, size = 21, normalized size = 0.95

method	result	size
risch	\(\frac {2 x}{25 \,{\mathrm e}^{5}+200}+4 \,{\mathrm e}^{\frac {16}{x^{4}}}\)	\(21\)
norman	\(\frac {4 x^{4} {\mathrm e}^{\frac {16}{x^{4}}}+\frac {2 x^{5}}{25 \left ({\mathrm e}^{5}+8\right )}}{x^{4}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6400*exp(5)-51200)*exp(16/x^4)+2*x^5)/(25*x^5*exp(5)+200*x^5),x,method=_RETURNVERBOSE)

[Out]

2*x/(25*exp(5)+200)+4*exp(16/x^4)

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maxima [B]  time = 0.35, size = 40, normalized size = 1.82 \begin {gather*} \frac {2 \, x}{25 \, {\left (e^{5} + 8\right )}} + \frac {32 \, e^{\left (\frac {16}{x^{4}}\right )}}{e^{5} + 8} + \frac {4 \, e^{\left (\frac {16}{x^{4}} + 5\right )}}{e^{5} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6400*exp(5)-51200)*exp(16/x^4)+2*x^5)/(25*x^5*exp(5)+200*x^5),x, algorithm="maxima")

[Out]

2/25*x/(e^5 + 8) + 32*e^(16/x^4)/(e^5 + 8) + 4*e^(16/x^4 + 5)/(e^5 + 8)

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mupad [B]  time = 7.31, size = 33, normalized size = 1.50 \begin {gather*} \frac {2\,x}{25\,\left ({\mathrm {e}}^5+8\right )}+\frac {{\mathrm {e}}^{\frac {16}{x^4}}\,\left (100\,{\mathrm {e}}^5+800\right )}{25\,\left ({\mathrm {e}}^5+8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(16/x^4)*(6400*exp(5) + 51200) - 2*x^5)/(25*x^5*exp(5) + 200*x^5),x)

[Out]

(2*x)/(25*(exp(5) + 8)) + (exp(16/x^4)*(100*exp(5) + 800))/(25*(exp(5) + 8))

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sympy [A]  time = 0.17, size = 17, normalized size = 0.77 \begin {gather*} \frac {2 x}{200 + 25 e^{5}} + 4 e^{\frac {16}{x^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6400*exp(5)-51200)*exp(16/x**4)+2*x**5)/(25*x**5*exp(5)+200*x**5),x)

[Out]

2*x/(200 + 25*exp(5)) + 4*exp(16/x**4)

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