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

Optimal. Leaf size=23 \[ e^{\frac {1}{2} \left (-\frac {e^{-2 x}}{x}+\log ^2(4)\right )}+x \]

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Rubi [F]  time = 1.05, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 25, normalized size = 1.09 \begin {gather*} e^{-\frac {e^{-2 x}}{2 x}+\frac {\log ^2(4)}{2}}+x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.61, size = 25, normalized size = 1.09 \begin {gather*} x + e^{\left (\frac {{\left (4 \, x e^{\left (2 \, x\right )} \log \relax (2)^{2} - 1\right )} e^{\left (-2 \, x\right )}}{2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x+1)*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)

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giac [A]  time = 0.14, size = 37, normalized size = 1.61 \begin {gather*} {\left (x e^{\left (-2 \, x\right )} + e^{\left (\frac {4 \, x \log \relax (2)^{2} - 4 \, x^{2} - e^{\left (-2 \, x\right )}}{2 \, x}\right )}\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x+1)*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)

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maple [A]  time = 0.07, size = 23, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((2*x+1)*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)

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maxima [A]  time = 0.68, size = 19, normalized size = 0.83 \begin {gather*} x + e^{\left (2 \, \log \relax (2)^{2} - \frac {e^{\left (-2 \, x\right )}}{2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x+1)*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)

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mupad [B]  time = 2.67, size = 19, normalized size = 0.83 \begin {gather*} x+{\mathrm {e}}^{2\,{\ln \relax (2)}^2-\frac {{\mathrm {e}}^{-2\,x}}{2\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.19, size = 26, normalized size = 1.13 \begin {gather*} x + e^{\frac {\left (2 x e^{2 x} \log {\relax (2 )}^{2} - \frac {1}{2}\right ) e^{- 2 x}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x+1)*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)

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