\(\int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx\) [6515]

Optimal result

Integrand size = 29, antiderivative size = 17 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=5+\frac {5 e^{14+2 x}}{x}-x \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2228} \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=\frac {5 e^{2 x+14}}{x}-x \]

[In]

Int[(-(E^2*x^2) + E^(16 + 2*x)*(-5 + 10*x))/(E^2*x^2),x]

[Out]

(5*E^(14 + 2*x))/x - 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 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{x^2} \, dx}{e^2} \\ & = \frac {\int \left (-e^2+\frac {5 e^{16+2 x} (-1+2 x)}{x^2}\right ) \, dx}{e^2} \\ & = -x+\frac {5 \int \frac {e^{16+2 x} (-1+2 x)}{x^2} \, dx}{e^2} \\ & = \frac {5 e^{14+2 x}}{x}-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=\frac {5 e^{2 (7+x)}}{x}-x \]

[In]

Integrate[(-(E^2*x^2) + E^(16 + 2*x)*(-5 + 10*x))/(E^2*x^2),x]

[Out]

(5*E^(2*(7 + x)))/x - x

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
risch	\(-x +\frac {5 \,{\mathrm e}^{14+2 x}}{x}\)	\(16\)
parts	\(-x +\frac {5 \,{\mathrm e}^{2 x} {\mathrm e}^{-2} {\mathrm e}^{16}}{x}\)	\(20\)
norman	\(\frac {-x^{2}+5 \,{\mathrm e}^{2 x} {\mathrm e}^{-2} {\mathrm e}^{16}}{x}\)	\(23\)
parallelrisch	\(\frac {{\mathrm e}^{-2} \left (-x^{2} {\mathrm e}^{2}+5 \,{\mathrm e}^{16} {\mathrm e}^{2 x}\right )}{x}\)	\(25\)
default	\({\mathrm e}^{-2} \left (-10 \,{\mathrm e}^{16} \left (-\frac {{\mathrm e}^{2 x}}{2 x}-\operatorname {Ei}_{1}\left (-2 x \right )\right )-10 \,{\mathrm e}^{16} \operatorname {Ei}_{1}\left (-2 x \right )-{\mathrm e}^{2} x \right )\)	\(42\)
derivativedivides	\(\frac {{\mathrm e}^{-2} \left (-20 \,{\mathrm e}^{16} \left (-\frac {{\mathrm e}^{2 x}}{2 x}-\operatorname {Ei}_{1}\left (-2 x \right )\right )-20 \,{\mathrm e}^{16} \operatorname {Ei}_{1}\left (-2 x \right )-2 \,{\mathrm e}^{2} x \right )}{2}\)	\(43\)

[In]

int(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=-\frac {{\left (x^{2} e^{2} - 5 \, e^{\left (2 \, x + 16\right )}\right )} e^{\left (-2\right )}}{x} \]

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x, algorithm="fricas")

[Out]

-(x^2*e^2 - 5*e^(2*x + 16))*e^(-2)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=- x + \frac {5 e^{14} e^{2 x}}{x} \]

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x**2*exp(2))/x**2/exp(2),x)

[Out]

-x + 5*exp(14)*exp(2*x)/x

Maxima [C] (verification not implemented)

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

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx={\left (10 \, {\rm Ei}\left (2 \, x\right ) e^{16} - x e^{2} - 10 \, e^{16} \Gamma \left (-1, -2 \, x\right )\right )} e^{\left (-2\right )} \]

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x, algorithm="maxima")

[Out]

(10*Ei(2*x)*e^16 - x*e^2 - 10*e^16*gamma(-1, -2*x))*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=-\frac {{\left (x^{2} e^{2} - 5 \, e^{\left (2 \, x + 16\right )}\right )} e^{\left (-2\right )}}{x} \]

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x, algorithm="giac")

[Out]

-(x^2*e^2 - 5*e^(2*x + 16))*e^(-2)/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx=\frac {5\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{14}}{x}-x \]

[In]

int(-(exp(-2)*(x^2*exp(2) - exp(2*x)*exp(16)*(10*x - 5)))/x^2,x)

[Out]

(5*exp(2*x)*exp(14))/x - x