\(\int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} (-2 x-2 e^5 x+2 e^x x)}{x} \, dx\) [2044]

Optimal result

Integrand size = 44, antiderivative size = 23 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=e^{2 e^x-2 x-2 e^5 x}+2 x^2 \]

[Out]

2*x^2+exp(-exp(5+ln(x))+exp(x)-x)^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {14, 6838} \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=2 x^2+e^{2 \left (e^x-\left (1+e^5\right ) x\right )} \]

[In]

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

[Out]

E^(2*(E^x - (1 + E^5)*x)) + 2*x^2

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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^{2 \left (e^x-\left (1+e^5\right ) x\right )} \left (-1-e^5+e^x\right )+4 x\right ) \, dx \\ & = 2 x^2+2 \int e^{2 \left (e^x-\left (1+e^5\right ) x\right )} \left (-1-e^5+e^x\right ) \, dx \\ & = e^{2 \left (e^x-\left (1+e^5\right ) x\right )}+2 x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=e^{2 e^x-2 \left (1+e^5\right ) x}+2 x^2 \]

[In]

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

[Out]

E^(2*E^x - 2*(1 + E^5)*x) + 2*x^2

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

method	result	size
norman	\({\mathrm e}^{-2 x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{x}-2 x}+2 x^{2}\)	\(21\)
risch	\({\mathrm e}^{-2 x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{x}-2 x}+2 x^{2}\)	\(21\)
parallelrisch	\(\frac {2 x^{4}+{\mathrm e}^{-2 x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{x}-2 x} x^{2}}{x^{2}}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=2 \, x^{2} + e^{\left (-2 \, x e^{5} - 2 \, x + 2 \, e^{x}\right )} \]

[In]

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

[Out]

2*x^2 + e^(-2*x*e^5 - 2*x + 2*e^x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=2 x^{2} + e^{- 2 x e^{5} - 2 x + 2 e^{x}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=2 \, x^{2} + e^{\left (-2 \, x e^{5} - 2 \, x + 2 \, e^{x}\right )} \]

[In]

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

[Out]

2*x^2 + e^(-2*x*e^5 - 2*x + 2*e^x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx={\left (2 \, x^{2} e^{x} + e^{\left (-2 \, x e^{5} - x + 2 \, e^{x}\right )}\right )} e^{\left (-x\right )} \]

[In]

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

[Out]

(2*x^2*e^x + e^(-2*x*e^5 - x + 2*e^x))*e^(-x)

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4 x^2+e^{2 e^x-2 x-2 e^5 x} \left (-2 x-2 e^5 x+2 e^x x\right )}{x} \, dx=2\,x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^5} \]

[In]

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

[Out]

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