\(\int \frac {2 e^2+e^{16} (-x^2+e^5 x^2)}{e^{16} x^2} \, dx\) [1747]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx=-\frac {2}{e^{14} x}-x+e^5 x \]

[Out]

x*exp(5)-2/x*exp(2)/exp(16)-x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 14} \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx=-\left (\left (1-e^5\right ) x\right )-\frac {2}{e^{14} x} \]

[In]

Int[(2*E^2 + E^16*(-x^2 + E^5*x^2))/(E^16*x^2),x]

[Out]

-2/(E^14*x) - (1 - E^5)*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{align*} \text {integral}& = \frac {\int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{x^2} \, dx}{e^{16}} \\ & = \frac {\int \left (e^{16} \left (-1+e^5\right )+\frac {2 e^2}{x^2}\right ) \, dx}{e^{16}} \\ & = -\frac {2}{e^{14} x}-\left (1-e^5\right ) x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx=-\frac {2}{e^{14} x}-x+e^5 x \]

[In]

Integrate[(2*E^2 + E^16*(-x^2 + E^5*x^2))/(E^16*x^2),x]

[Out]

-2/(E^14*x) - x + E^5*x

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{5}-x^{2}-2 \,{\mathrm e}^{-14}}{x}\)	\(21\)
norman	\(\frac {\left ({\mathrm e}^{5}-1\right ) x^{2}-2 \,{\mathrm e}^{-16} {\mathrm e}^{2}}{x}\)	\(22\)
default	\({\mathrm e}^{-16} \left (x \,{\mathrm e}^{21}-x \,{\mathrm e}^{16}-\frac {2 \,{\mathrm e}^{2}}{x}\right )\)	\(23\)
gosper	\(-\frac {\left (-{\mathrm e}^{5} {\mathrm e}^{16} x^{2}+x^{2} {\mathrm e}^{16}+2 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-16}}{x}\)	\(30\)
parallelrisch	\(-\frac {\left (-{\mathrm e}^{5} {\mathrm e}^{16} x^{2}+x^{2} {\mathrm e}^{16}+2 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-16}}{x}\)	\(30\)

[In]

int(((x^2*exp(5)-x^2)*exp(16)+2*exp(2))/x^2/exp(16),x,method=_RETURNVERBOSE)

[Out]

(x^2*exp(5)-x^2-2*exp(-14))/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx=\frac {{\left (x^{2} e^{19} - x^{2} e^{14} - 2\right )} e^{\left (-14\right )}}{x} \]

[In]

integrate(((x^2*exp(5)-x^2)*exp(16)+2*exp(2))/x^2/exp(16),x, algorithm="fricas")

[Out]

(x^2*e^19 - x^2*e^14 - 2)*e^(-14)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx=\frac {- x \left (- e^{19} + e^{14}\right ) - \frac {2}{x}}{e^{14}} \]

[In]

integrate(((x**2*exp(5)-x**2)*exp(16)+2*exp(2))/x**2/exp(16),x)

[Out]

(-x*(-exp(19) + exp(14)) - 2/x)*exp(-14)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx={\left (x {\left (e^{21} - e^{16}\right )} - \frac {2 \, e^{2}}{x}\right )} e^{\left (-16\right )} \]

[In]

integrate(((x^2*exp(5)-x^2)*exp(16)+2*exp(2))/x^2/exp(16),x, algorithm="maxima")

[Out]

(x*(e^21 - e^16) - 2*e^2/x)*e^(-16)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx={\left (x e^{21} - x e^{16} - \frac {2 \, e^{2}}{x}\right )} e^{\left (-16\right )} \]

[In]

integrate(((x^2*exp(5)-x^2)*exp(16)+2*exp(2))/x^2/exp(16),x, algorithm="giac")

[Out]

(x*e^21 - x*e^16 - 2*e^2/x)*e^(-16)

Mupad [B] (verification not implemented)

Time = 8.69 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2 e^2+e^{16} \left (-x^2+e^5 x^2\right )}{e^{16} x^2} \, dx=x\,\left ({\mathrm {e}}^5-1\right )-\frac {2\,{\mathrm {e}}^{-14}}{x} \]

[In]

int((exp(-16)*(2*exp(2) + exp(16)*(x^2*exp(5) - x^2)))/x^2,x)

[Out]

x*(exp(5) - 1) - (2*exp(-14))/x