\(\int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} (-80-32 x+4 x^9)}{25 x^9} \, dx\) [944]

Optimal result

Integrand size = 43, antiderivative size = 18 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {1}{25} \left (-5-e^{\frac {1}{x^8}}-2 x\right )^2 \]

[Out]

1/25*(-5-2*x-exp(1/x^8))^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(37\) vs. \(2(18)=36\).

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 14, 2240, 2326} \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {2}{25} e^{\frac {1}{x^8}} (2 x+5)+\frac {e^{\frac {2}{x^8}}}{25}+\frac {1}{25} (2 x+5)^2 \]

[In]

Int[(-16*E^(2/x^8) + 20*x^9 + 8*x^10 + E^x^(-8)*(-80 - 32*x + 4*x^9))/(25*x^9),x]

[Out]

E^(2/x^8)/25 + (2*E^x^(-8)*(5 + 2*x))/25 + (5 + 2*x)^2/25

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 2240

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]

Rule 2326

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{align*} \text {integral}& = \frac {1}{25} \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{x^9} \, dx \\ & = \frac {1}{25} \int \left (-\frac {16 e^{\frac {2}{x^8}}}{x^9}+4 (5+2 x)+\frac {4 e^{\frac {1}{x^8}} \left (-20-8 x+x^9\right )}{x^9}\right ) \, dx \\ & = \frac {1}{25} (5+2 x)^2+\frac {4}{25} \int \frac {e^{\frac {1}{x^8}} \left (-20-8 x+x^9\right )}{x^9} \, dx-\frac {16}{25} \int \frac {e^{\frac {2}{x^8}}}{x^9} \, dx \\ & = \frac {e^{\frac {2}{x^8}}}{25}+\frac {2}{25} e^{\frac {1}{x^8}} (5+2 x)+\frac {1}{25} (5+2 x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {4}{25} \left (\frac {e^{\frac {2}{x^8}}}{4}+e^{\frac {1}{x^8}} \left (\frac {5}{2}+x\right )+x (5+x)\right ) \]

[In]

Integrate[(-16*E^(2/x^8) + 20*x^9 + 8*x^10 + E^x^(-8)*(-80 - 32*x + 4*x^9))/(25*x^9),x]

[Out]

(4*(E^(2/x^8)/4 + E^x^(-8)*(5/2 + x) + x*(5 + x)))/25

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61

method	result	size
risch	\(\frac {4 x^{2}}{25}+\frac {{\mathrm e}^{\frac {2}{x^{8}}}}{25}+\frac {4 x}{5}+\frac {\left (4 x +10\right ) {\mathrm e}^{\frac {1}{x^{8}}}}{25}\)	\(29\)
parallelrisch	\(\frac {4 x^{2}}{25}+\frac {{\mathrm e}^{\frac {2}{x^{8}}}}{25}+\frac {4 x}{5}+\frac {4 \,{\mathrm e}^{\frac {1}{x^{8}}} x}{25}+\frac {2 \,{\mathrm e}^{\frac {1}{x^{8}}}}{5}\)	\(31\)
parts	\(\frac {4 x}{5}+\frac {4 x^{2}}{25}+\frac {{\mathrm e}^{\frac {2}{x^{8}}}}{25}+\frac {\frac {2 \,{\mathrm e}^{\frac {1}{x^{8}}} x^{8}}{5}+\frac {4 \,{\mathrm e}^{\frac {1}{x^{8}}} x^{9}}{25}}{x^{8}}\)	\(41\)
norman	\(\frac {\frac {4 x^{9}}{5}+\frac {4 x^{10}}{25}+\frac {2 \,{\mathrm e}^{\frac {1}{x^{8}}} x^{8}}{5}+\frac {4 \,{\mathrm e}^{\frac {1}{x^{8}}} x^{9}}{25}+\frac {{\mathrm e}^{\frac {2}{x^{8}}} x^{8}}{25}}{x^{8}}\)	\(45\)

[In]

int(1/25*(-16*exp(1/x^8)^2+(4*x^9-32*x-80)*exp(1/x^8)+8*x^10+20*x^9)/x^9,x,method=_RETURNVERBOSE)

[Out]

4/25*x^2+1/25*exp(2/x^8)+4/5*x+1/25*(4*x+10)*exp(1/x^8)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (13) = 26\).

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {4}{25} \, x^{2} + \frac {2}{25} \, {\left (2 \, x + 5\right )} e^{\left (\frac {1}{x^{8}}\right )} + \frac {4}{5} \, x + \frac {1}{25} \, e^{\left (\frac {2}{x^{8}}\right )} \]

[In]

integrate(1/25*(-16*exp(1/x^8)^2+(4*x^9-32*x-80)*exp(1/x^8)+8*x^10+20*x^9)/x^9,x, algorithm="fricas")

[Out]

4/25*x^2 + 2/25*(2*x + 5)*e^(x^(-8)) + 4/5*x + 1/25*e^(2/x^8)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (15) = 30\).

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {4 x^{2}}{25} + \frac {4 x}{5} + \frac {\left (100 x + 250\right ) e^{\frac {1}{x^{8}}}}{625} + \frac {e^{\frac {2}{x^{8}}}}{25} \]

[In]

integrate(1/25*(-16*exp(1/x**8)**2+(4*x**9-32*x-80)*exp(1/x**8)+8*x**10+20*x**9)/x**9,x)

[Out]

4*x**2/25 + 4*x/5 + (100*x + 250)*exp(x**(-8))/625 + exp(2/x**8)/25

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.28 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {1}{50} \, x \left (-\frac {1}{x^{8}}\right )^{\frac {1}{8}} \Gamma \left (-\frac {1}{8}, -\frac {1}{x^{8}}\right ) + \frac {4}{25} \, x^{2} + \frac {4}{5} \, x - \frac {4 \, \Gamma \left (\frac {7}{8}, -\frac {1}{x^{8}}\right )}{25 \, x^{7} \left (-\frac {1}{x^{8}}\right )^{\frac {7}{8}}} + \frac {1}{25} \, e^{\left (\frac {2}{x^{8}}\right )} + \frac {2}{5} \, e^{\left (\frac {1}{x^{8}}\right )} \]

[In]

integrate(1/25*(-16*exp(1/x^8)^2+(4*x^9-32*x-80)*exp(1/x^8)+8*x^10+20*x^9)/x^9,x, algorithm="maxima")

[Out]

1/50*x*(-1/x^8)^(1/8)*gamma(-1/8, -1/x^8) + 4/25*x^2 + 4/5*x - 4/25*gamma(7/8, -1/x^8)/(x^7*(-1/x^8)^(7/8)) +
1/25*e^(2/x^8) + 2/5*e^(x^(-8))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (13) = 26\).

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.67 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {4}{25} \, x^{2} + \frac {4}{25} \, x e^{\left (\frac {1}{x^{8}}\right )} + \frac {4}{5} \, x + \frac {1}{25} \, e^{\left (\frac {2}{x^{8}}\right )} + \frac {2}{5} \, e^{\left (\frac {1}{x^{8}}\right )} \]

[In]

integrate(1/25*(-16*exp(1/x^8)^2+(4*x^9-32*x-80)*exp(1/x^8)+8*x^10+20*x^9)/x^9,x, algorithm="giac")

[Out]

4/25*x^2 + 4/25*x*e^(x^(-8)) + 4/5*x + 1/25*e^(2/x^8) + 2/5*e^(x^(-8))

Mupad [B] (verification not implemented)

Time = 9.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.67 \[ \int \frac {-16 e^{\frac {2}{x^8}}+20 x^9+8 x^{10}+e^{\frac {1}{x^8}} \left (-80-32 x+4 x^9\right )}{25 x^9} \, dx=\frac {4\,x}{5}+\frac {2\,{\mathrm {e}}^{\frac {1}{x^8}}}{5}+\frac {{\mathrm {e}}^{\frac {2}{x^8}}}{25}+\frac {4\,x\,{\mathrm {e}}^{\frac {1}{x^8}}}{25}+\frac {4\,x^2}{25} \]

[In]

int(-((16*exp(2/x^8))/25 + (exp(1/x^8)*(32*x - 4*x^9 + 80))/25 - (4*x^9)/5 - (8*x^10)/25)/x^9,x)

[Out]

(4*x)/5 + (2*exp(1/x^8))/5 + exp(2/x^8)/25 + (4*x*exp(1/x^8))/25 + (4*x^2)/25