3.10.91 \(\int \frac {-3+3 x+e^{e^x} (2 x^2+e^x x^3)}{3 x} \, dx\)

Optimal. Leaf size=22 \[ x+\frac {1}{3} e^{e^x} x^2+\log (5)-\log (2 x) \]

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Rubi [F]  time = 0.10, 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 {-3+3 x+e^{e^x} \left (2 x^2+e^x x^3\right )}{3 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3 + 3*x + E^E^x*(2*x^2 + E^x*x^3))/(3*x),x]

[Out]

x - Log[x] + (2*Defer[Int][E^E^x*x, x])/3 + Defer[Int][E^(E^x + x)*x^2, x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-3+3 x+e^{e^x} \left (2 x^2+e^x x^3\right )}{x} \, dx\\ &=\frac {1}{3} \int \left (e^{e^x+x} x^2+\frac {-3+3 x+2 e^{e^x} x^2}{x}\right ) \, dx\\ &=\frac {1}{3} \int e^{e^x+x} x^2 \, dx+\frac {1}{3} \int \frac {-3+3 x+2 e^{e^x} x^2}{x} \, dx\\ &=\frac {1}{3} \int e^{e^x+x} x^2 \, dx+\frac {1}{3} \int \left (\frac {3 (-1+x)}{x}+2 e^{e^x} x\right ) \, dx\\ &=\frac {1}{3} \int e^{e^x+x} x^2 \, dx+\frac {2}{3} \int e^{e^x} x \, dx+\int \frac {-1+x}{x} \, dx\\ &=\frac {1}{3} \int e^{e^x+x} x^2 \, dx+\frac {2}{3} \int e^{e^x} x \, dx+\int \left (1-\frac {1}{x}\right ) \, dx\\ &=x-\log (x)+\frac {1}{3} \int e^{e^x+x} x^2 \, dx+\frac {2}{3} \int e^{e^x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.82 \begin {gather*} x+\frac {1}{3} e^{e^x} x^2-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + 3*x + E^E^x*(2*x^2 + E^x*x^3))/(3*x),x]

[Out]

x + (E^E^x*x^2)/3 - Log[x]

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fricas [A]  time = 0.70, size = 14, normalized size = 0.64 \begin {gather*} \frac {1}{3} \, x^{2} e^{\left (e^{x}\right )} + x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((exp(x)*x^3+2*x^2)*exp(exp(x))+3*x-3)/x,x, algorithm="fricas")

[Out]

1/3*x^2*e^(e^x) + x - log(x)

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giac [A]  time = 0.25, size = 27, normalized size = 1.23 \begin {gather*} \frac {1}{3} \, {\left (x^{2} e^{\left (x + e^{x}\right )} + 3 \, x e^{x} - 3 \, e^{x} \log \relax (x)\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((exp(x)*x^3+2*x^2)*exp(exp(x))+3*x-3)/x,x, algorithm="giac")

[Out]

1/3*(x^2*e^(x + e^x) + 3*x*e^x - 3*e^x*log(x))*e^(-x)

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maple [A]  time = 0.03, size = 15, normalized size = 0.68

method	result	size
norman	\(x +\frac {{\mathrm e}^{{\mathrm e}^{x}} x^{2}}{3}-\ln \relax (x )\)	\(15\)
risch	\(x +\frac {{\mathrm e}^{{\mathrm e}^{x}} x^{2}}{3}-\ln \relax (x )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((exp(x)*x^3+2*x^2)*exp(exp(x))+3*x-3)/x,x,method=_RETURNVERBOSE)

[Out]

x+1/3*exp(exp(x))*x^2-ln(x)

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maxima [A]  time = 0.54, size = 14, normalized size = 0.64 \begin {gather*} \frac {1}{3} \, x^{2} e^{\left (e^{x}\right )} + x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((exp(x)*x^3+2*x^2)*exp(exp(x))+3*x-3)/x,x, algorithm="maxima")

[Out]

1/3*x^2*e^(e^x) + x - log(x)

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mupad [B]  time = 0.70, size = 14, normalized size = 0.64 \begin {gather*} x-\ln \relax (x)+\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + (exp(exp(x))*(x^3*exp(x) + 2*x^2))/3 - 1)/x,x)

[Out]

x - log(x) + (x^2*exp(exp(x)))/3

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sympy [A]  time = 0.16, size = 14, normalized size = 0.64 \begin {gather*} \frac {x^{2} e^{e^{x}}}{3} + x - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((exp(x)*x**3+2*x**2)*exp(exp(x))+3*x-3)/x,x)

[Out]

x**2*exp(exp(x))/3 + x - log(x)

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