3.70.62 \(\int \frac {2 x^2+e^{4 x^x} (-2+x^x (8 x+8 x \log (x)))}{x^2} \, dx\)

Optimal. Leaf size=18 \[ \frac {2 e^{4 x^x}}{x}+2 (9+x) \]

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Rubi [F]  time = 0.27, 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 {2 x^2+e^{4 x^x} \left (-2+x^x (8 x+8 x \log (x))\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x^2 + E^(4*x^x)*(-2 + x^x*(8*x + 8*x*Log[x])))/x^2,x]

[Out]

2*x - 2*Defer[Int][E^(4*x^x)/x^2, x] + 8*Defer[Int][E^(4*x^x)*x^(-1 + x), x] + 8*Log[x]*Defer[Int][E^(4*x^x)*x
^(-1 + x), x] - 8*Defer[Int][Defer[Int][E^(4*x^x)*x^(-1 + x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-e^{4 x^x}+x^2\right )}{x^2}+8 e^{4 x^x} x^{-1+x} (1+\log (x))\right ) \, dx\\ &=2 \int \frac {-e^{4 x^x}+x^2}{x^2} \, dx+8 \int e^{4 x^x} x^{-1+x} (1+\log (x)) \, dx\\ &=2 \int \left (1-\frac {e^{4 x^x}}{x^2}\right ) \, dx+8 \int \left (e^{4 x^x} x^{-1+x}+e^{4 x^x} x^{-1+x} \log (x)\right ) \, dx\\ &=2 x-2 \int \frac {e^{4 x^x}}{x^2} \, dx+8 \int e^{4 x^x} x^{-1+x} \, dx+8 \int e^{4 x^x} x^{-1+x} \log (x) \, dx\\ &=2 x-2 \int \frac {e^{4 x^x}}{x^2} \, dx+8 \int e^{4 x^x} x^{-1+x} \, dx-8 \int \frac {\int e^{4 x^x} x^{-1+x} \, dx}{x} \, dx+(8 \log (x)) \int e^{4 x^x} x^{-1+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 16, normalized size = 0.89 \begin {gather*} \frac {2 e^{4 x^x}}{x}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + E^(4*x^x)*(-2 + x^x*(8*x + 8*x*Log[x])))/x^2,x]

[Out]

(2*E^(4*x^x))/x + 2*x

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fricas [A]  time = 0.96, size = 15, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (x^{2} + e^{\left (4 \, x^{x}\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*log(x)+8*x)*exp(x*log(x))-2)*exp(exp(x*log(x)))^4+2*x^2)/x^2,x, algorithm="fricas")

[Out]

2*(x^2 + e^(4*x^x))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{2} + {\left (4 \, {\left (x \log \relax (x) + x\right )} x^{x} - 1\right )} e^{\left (4 \, x^{x}\right )}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*log(x)+8*x)*exp(x*log(x))-2)*exp(exp(x*log(x)))^4+2*x^2)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 16, normalized size = 0.89

method	result	size
risch	\(2 x +\frac {2 \,{\mathrm e}^{4 x^{x}}}{x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x*ln(x)+8*x)*exp(x*ln(x))-2)*exp(exp(x*ln(x)))^4+2*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

2*x+2*exp(4*x^x)/x

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maxima [A]  time = 0.41, size = 15, normalized size = 0.83 \begin {gather*} 2 \, x + \frac {2 \, e^{\left (4 \, x^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*log(x)+8*x)*exp(x*log(x))-2)*exp(exp(x*log(x)))^4+2*x^2)/x^2,x, algorithm="maxima")

[Out]

2*x + 2*e^(4*x^x)/x

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mupad [B]  time = 4.20, size = 15, normalized size = 0.83 \begin {gather*} \frac {2\,\left ({\mathrm {e}}^{4\,x^x}+x^2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*exp(x*log(x)))*(exp(x*log(x))*(8*x + 8*x*log(x)) - 2) + 2*x^2)/x^2,x)

[Out]

(2*(exp(4*x^x) + x^2))/x

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sympy [A]  time = 0.36, size = 15, normalized size = 0.83 \begin {gather*} 2 x + \frac {2 e^{4 e^{x \log {\relax (x )}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*ln(x)+8*x)*exp(x*ln(x))-2)*exp(exp(x*ln(x)))**4+2*x**2)/x**2,x)

[Out]

2*x + 2*exp(4*exp(x*log(x)))/x

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