3.48.7 \(\int \frac {x^3+(2 x^2+x^3) \log (\frac {1}{2} (6+3 x))+e^{e^x} (-4-2 x-x^2+e^x (4 x+2 x^2)+(-4 x-2 x^2+e^x (2 x^2+x^3)) \log (\frac {1}{2} (6+3 x)))}{8 x^2+4 x^3+(8 x^3+4 x^4) \log (\frac {1}{2} (6+3 x))+(2 x^4+x^5) \log ^2(\frac {1}{2} (6+3 x))} \, dx\)

Optimal. Leaf size=30 \[ \frac {\log \left (e^{e^{e^x}-x}\right )}{x \left (2+x \log \left (3+\frac {3 x}{2}\right )\right )} \]

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Rubi [F]  time = 4.04, 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 {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^3 + (2*x^2 + x^3)*Log[(6 + 3*x)/2] + E^E^x*(-4 - 2*x - x^2 + E^x*(4*x + 2*x^2) + (-4*x - 2*x^2 + E^x*(2
*x^2 + x^3))*Log[(6 + 3*x)/2]))/(8*x^2 + 4*x^3 + (8*x^3 + 4*x^4)*Log[(6 + 3*x)/2] + (2*x^4 + x^5)*Log[(6 + 3*x
)/2]^2),x]

[Out]

Defer[Int][(2 + x*Log[(3*(2 + x))/2])^(-2), x] - 2*Defer[Int][E^E^x/(x^2*(2 + x*Log[(3*(2 + x))/2])^2), x] - 2
*Defer[Int][1/((2 + x)*(2 + x*Log[(3*(2 + x))/2])^2), x] - Defer[Int][E^E^x/((2 + x)*(2 + x*Log[(3*(2 + x))/2]
)^2), x] + Defer[Int][E^(E^x + x)/(x*(2 + x*Log[(3*(2 + x))/2])), x] + Defer[Int][Log[3 + (3*x)/2]/(2 + x*Log[
(3*(2 + x))/2])^2, x] - 2*Defer[Int][(E^E^x*Log[3 + (3*x)/2])/(x*(2 + x*Log[(3*(2 + x))/2])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^3+2 e^{e^x+x} x (2+x)-e^{e^x} \left (4+2 x+x^2\right )+x (2+x) \left (-2 e^{e^x}+x+e^{e^x+x} x\right ) \log \left (\frac {3 (2+x)}{2}\right )}{x^2 (2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ &=\int \left (\frac {x}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}-\frac {e^{e^x} \left (4+2 x+x^2\right )}{x^2 (2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}+\frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )}+\frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}-\frac {2 e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\right )+\int \frac {x}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx-\int \frac {e^{e^x} \left (4+2 x+x^2\right )}{x^2 (2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx+\int \frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \, dx+\int \frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ &=-\left (2 \int \frac {e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\right )+\int \frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \, dx+\int \left (\frac {1}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}-\frac {2}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}\right ) \, dx-\int \left (\frac {2 e^{e^x}}{x^2 \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}+\frac {e^{e^x}}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2}\right ) \, dx+\int \frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ &=-\left (2 \int \frac {e^{e^x}}{x^2 \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\right )-2 \int \frac {1}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx-2 \int \frac {e^{e^x} \log \left (3+\frac {3 x}{2}\right )}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx+\int \frac {1}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx-\int \frac {e^{e^x}}{(2+x) \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx+\int \frac {e^{e^x+x}}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \, dx+\int \frac {\log \left (3+\frac {3 x}{2}\right )}{\left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 27, normalized size = 0.90 \begin {gather*} \frac {e^{e^x}-x}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 + (2*x^2 + x^3)*Log[(6 + 3*x)/2] + E^E^x*(-4 - 2*x - x^2 + E^x*(4*x + 2*x^2) + (-4*x - 2*x^2 +
E^x*(2*x^2 + x^3))*Log[(6 + 3*x)/2]))/(8*x^2 + 4*x^3 + (8*x^3 + 4*x^4)*Log[(6 + 3*x)/2] + (2*x^4 + x^5)*Log[(6
+ 3*x)/2]^2),x]

[Out]

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

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fricas [A]  time = 0.61, size = 25, normalized size = 0.83 \begin {gather*} -\frac {x - e^{\left (e^{x}\right )}}{x^{2} \log \left (\frac {3}{2} \, x + 3\right ) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x)-x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*
log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3/2*x)^2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.31, size = 25, normalized size = 0.83 \begin {gather*} -\frac {x - e^{\left (e^{x}\right )}}{x^{2} \log \left (\frac {3}{2} \, x + 3\right ) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x)-x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*
log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3/2*x)^2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.71, size = 35, normalized size = 1.17

method	result	size
risch	\(-\frac {1}{2+x \ln \left (3+\frac {3 x}{2}\right )}+\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\left (2+x \ln \left (3+\frac {3 x}{2}\right )\right ) x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*ln(3+3/2*x)+(2*x^2+4*x)*exp(x)-x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*ln(3+3/
2*x)+x^3)/((x^5+2*x^4)*ln(3+3/2*x)^2+(4*x^4+8*x^3)*ln(3+3/2*x)+4*x^3+8*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 34, normalized size = 1.13 \begin {gather*} -\frac {x - e^{\left (e^{x}\right )}}{x^{2} {\left (\log \relax (3) - \log \relax (2)\right )} + x^{2} \log \left (x + 2\right ) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x)-x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*
log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3/2*x)^2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.49, size = 24, normalized size = 0.80 \begin {gather*} -\frac {x-{\mathrm {e}}^{{\mathrm {e}}^x}}{x\,\left (x\,\ln \left (\frac {3\,x}{2}+3\right )+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((3*x)/2 + 3)*(2*x^2 + x^3) - exp(exp(x))*(2*x + log((3*x)/2 + 3)*(4*x - exp(x)*(2*x^2 + x^3) + 2*x^2)
- exp(x)*(4*x + 2*x^2) + x^2 + 4) + x^3)/(log((3*x)/2 + 3)*(8*x^3 + 4*x^4) + log((3*x)/2 + 3)^2*(2*x^4 + x^5)
+ 8*x^2 + 4*x^3),x)

[Out]

-(x - exp(exp(x)))/(x*(x*log((3*x)/2 + 3) + 2))

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sympy [A]  time = 0.43, size = 32, normalized size = 1.07 \begin {gather*} \frac {e^{e^{x}}}{x^{2} \log {\left (\frac {3 x}{2} + 3 \right )} + 2 x} - \frac {1}{x \log {\left (\frac {3 x}{2} + 3 \right )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x**3+2*x**2)*exp(x)-2*x**2-4*x)*ln(3+3/2*x)+(2*x**2+4*x)*exp(x)-x**2-2*x-4)*exp(exp(x))+(x**3+2*
x**2)*ln(3+3/2*x)+x**3)/((x**5+2*x**4)*ln(3+3/2*x)**2+(4*x**4+8*x**3)*ln(3+3/2*x)+4*x**3+8*x**2),x)

[Out]

exp(exp(x))/(x**2*log(3*x/2 + 3) + 2*x) - 1/(x*log(3*x/2 + 3) + 2)

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