\(\int \frac {e^{-2 x} (2 e^{2 x} x^2+e^{\frac {e^{-2 x} (-1+e^{2 x} x \log ^2(4))}{2 x}} (1+2 x))}{2 x^2} \, dx\) [3984]

Optimal result

Integrand size = 56, antiderivative size = 23 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}+\log ^2(4)\right )}+x \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=\int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx \]

[In]

Int[(2*E^(2*x)*x^2 + E^((-1 + E^(2*x)*x*Log[4]^2)/(2*E^(2*x)*x))*(1 + 2*x))/(2*E^(2*x)*x^2),x]

[Out]

x + Defer[Int][E^((-(1/(E^(2*x)*x)) - 4*x + Log[4]^2)/2)/x^2, x]/2 + Defer[Int][E^((-(1/(E^(2*x)*x)) - 4*x + L
og[4]^2)/2)/x, x]

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=e^{-\frac {e^{-2 x}}{2 x}+\frac {\log ^2(4)}{2}}+x \]

[In]

Integrate[(2*E^(2*x)*x^2 + E^((-1 + E^(2*x)*x*Log[4]^2)/(2*E^(2*x)*x))*(1 + 2*x))/(2*E^(2*x)*x^2),x]

[Out]

E^(-1/2*1/(E^(2*x)*x) + Log[4]^2/2) + x

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
risch	\(x +{\mathrm e}^{\frac {4 x \ln \left (2\right )^{2}-{\mathrm e}^{-2 x}}{2 x}}\)	\(23\)
parts	\(x +{\mathrm e}^{\frac {\left (4 x \ln \left (2\right )^{2} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-2 x}}{2 x}}\)	\(28\)
parallelrisch	\({\mathrm e}^{\frac {\left (4 x \ln \left (2\right )^{2} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-2 x}}{2 x}}+\frac {\ln \left ({\mathrm e}^{2 x}\right )}{2}\)	\(34\)
norman	\(\frac {\left ({\mathrm e}^{2 x} x^{2}+x \,{\mathrm e}^{2 x} {\mathrm e}^{\frac {\left (4 x \ln \left (2\right )^{2} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-2 x}}{2 x}}\right ) {\mathrm e}^{-2 x}}{x}\)	\(51\)

[In]

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

[Out]

x+exp(1/2*(4*x*ln(2)^2-exp(-2*x))/x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x + e^{\left (\frac {{\left (4 \, x e^{\left (2 \, x\right )} \log \left (2\right )^{2} - 1\right )} e^{\left (-2 \, x\right )}}{2 \, x}\right )} \]

[In]

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

[Out]

x + e^(1/2*(4*x*e^(2*x)*log(2)^2 - 1)*e^(-2*x)/x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x + e^{\frac {\left (2 x e^{2 x} \log {\left (2 \right )}^{2} - \frac {1}{2}\right ) e^{- 2 x}}{x}} \]

[In]

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

[Out]

x + exp((2*x*exp(2*x)*log(2)**2 - 1/2)*exp(-2*x)/x)

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x + e^{\left (2 \, \log \left (2\right )^{2} - \frac {e^{\left (-2 \, x\right )}}{2 \, x}\right )} \]

[In]

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

[Out]

x + e^(2*log(2)^2 - 1/2*e^(-2*x)/x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx={\left (x e^{\left (-2 \, x\right )} + e^{\left (\frac {4 \, x \log \left (2\right )^{2} - 4 \, x^{2} - e^{\left (-2 \, x\right )}}{2 \, x}\right )}\right )} e^{\left (2 \, x\right )} \]

[In]

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

[Out]

(x*e^(-2*x) + e^(1/2*(4*x*log(2)^2 - 4*x^2 - e^(-2*x))/x))*e^(2*x)

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 x} \left (2 e^{2 x} x^2+e^{\frac {e^{-2 x} \left (-1+e^{2 x} x \log ^2(4)\right )}{2 x}} (1+2 x)\right )}{2 x^2} \, dx=x+{\mathrm {e}}^{2\,{\ln \left (2\right )}^2-\frac {{\mathrm {e}}^{-2\,x}}{2\,x}} \]

[In]

int((exp(-2*x)*(x^2*exp(2*x) + (exp((exp(-2*x)*(2*x*exp(2*x)*log(2)^2 - 1/2))/x)*(2*x + 1))/2))/x^2,x)

[Out]

x + exp(2*log(2)^2 - exp(-2*x)/(2*x))