\(\int \frac {1-4 e^{2 x} x}{x} \, dx\) [350]

Optimal result

Integrand size = 14, antiderivative size = 22 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=\log \left (\frac {1}{5} e^{-2-2 e^{2 x}} (1+e)^2 x\right ) \]

[Out]

ln(1/5*x/exp(2)/exp(exp(x)^2)^2*(1+exp(1))^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2225} \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=\log (x)-2 e^{2 x} \]

[In]

Int[(1 - 4*E^(2*x)*x)/x,x]

[Out]

-2*E^(2*x) + Log[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 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=-2 e^{2 x}+\log (x) \]

[In]

Integrate[(1 - 4*E^(2*x)*x)/x,x]

[Out]

-2*E^(2*x) + Log[x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45

method	result	size
default	\(\ln \left (x \right )-2 \,{\mathrm e}^{2 x}\)	\(10\)
norman	\(\ln \left (x \right )-2 \,{\mathrm e}^{2 x}\)	\(10\)
risch	\(\ln \left (x \right )-2 \,{\mathrm e}^{2 x}\)	\(10\)
parallelrisch	\(\ln \left (x \right )-2 \,{\mathrm e}^{2 x}\)	\(10\)
parts	\(\ln \left (x \right )-2 \,{\mathrm e}^{2 x}\)	\(10\)

[In]

int((-4*x*exp(x)^2+1)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.41 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=-2 \, e^{\left (2 \, x\right )} + \log \left (x\right ) \]

[In]

integrate((-4*x*exp(x)^2+1)/x,x, algorithm="fricas")

[Out]

-2*e^(2*x) + log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.36 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=- 2 e^{2 x} + \log {\left (x \right )} \]

[In]

integrate((-4*x*exp(x)**2+1)/x,x)

[Out]

-2*exp(2*x) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.41 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=-2 \, e^{\left (2 \, x\right )} + \log \left (x\right ) \]

[In]

integrate((-4*x*exp(x)^2+1)/x,x, algorithm="maxima")

[Out]

-2*e^(2*x) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.41 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=-2 \, e^{\left (2 \, x\right )} + \log \left (x\right ) \]

[In]

integrate((-4*x*exp(x)^2+1)/x,x, algorithm="giac")

[Out]

-2*e^(2*x) + log(x)

Mupad [B] (verification not implemented)

Time = 9.09 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.41 \[ \int \frac {1-4 e^{2 x} x}{x} \, dx=\ln \left (x\right )-2\,{\mathrm {e}}^{2\,x} \]

[In]

int(-(4*x*exp(2*x) - 1)/x,x)

[Out]

log(x) - 2*exp(2*x)