3.17.37 \(\int \frac {-64 x-16 x^2-8 x^3+e^{e^x+x} (-8 x^2-4 x^3)+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} (2 x^2+x^3)+(256 x+256 x^2+64 x^3) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} (16 x^2+16 x^3+4 x^4+(64 x+32 x^2) \log (2+x))} \, dx\)

Optimal. Leaf size=23 \[ \frac {4}{4+e^{e^x}+2 x+\frac {16 \log (2+x)}{x}} \]

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Rubi [F]  time = 4.41, 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 {-64 x-16 x^2-8 x^3+e^{e^x+x} \left (-8 x^2-4 x^3\right )+(128+64 x) \log (2+x)}{32 x^2+48 x^3+24 x^4+4 x^5+e^{2 e^x} \left (2 x^2+x^3\right )+\left (256 x+256 x^2+64 x^3\right ) \log (2+x)+(512+256 x) \log ^2(2+x)+e^{e^x} \left (16 x^2+16 x^3+4 x^4+\left (64 x+32 x^2\right ) \log (2+x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-64*x - 16*x^2 - 8*x^3 + E^(E^x + x)*(-8*x^2 - 4*x^3) + (128 + 64*x)*Log[2 + x])/(32*x^2 + 48*x^3 + 24*x^
4 + 4*x^5 + E^(2*E^x)*(2*x^2 + x^3) + (256*x + 256*x^2 + 64*x^3)*Log[2 + x] + (512 + 256*x)*Log[2 + x]^2 + E^E
^x*(16*x^2 + 16*x^3 + 4*x^4 + (64*x + 32*x^2)*Log[2 + x])),x]

[Out]

-64*Defer[Int][(4*x + E^E^x*x + 2*x^2 + 16*Log[2 + x])^(-2), x] - 8*Defer[Int][x^2/(4*x + E^E^x*x + 2*x^2 + 16
*Log[2 + x])^2, x] - 4*Defer[Int][(E^(E^x + x)*x^2)/(4*x + E^E^x*x + 2*x^2 + 16*Log[2 + x])^2, x] + 128*Defer[
Int][1/((2 + x)*(4*x + E^E^x*x + 2*x^2 + 16*Log[2 + x])^2), x] + 64*Defer[Int][Log[2 + x]/(4*x + E^E^x*x + 2*x
^2 + 16*Log[2 + x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x \left (16+2 \left (2+e^{e^x+x}\right ) x+\left (2+e^{e^x+x}\right ) x^2\right )+64 (2+x) \log (2+x)}{(2+x) \left (x \left (4+e^{e^x}+2 x\right )+16 \log (2+x)\right )^2} \, dx\\ &=\int \left (-\frac {4 e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {64 x}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {16 x^2}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {8 x^3}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {64 \log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\right )-8 \int \frac {x^3}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-16 \int \frac {x^2}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-64 \int \frac {x}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx+64 \int \frac {\log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\right )-8 \int \left (\frac {4}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {2 x}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {8}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx-16 \int \left (-\frac {2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {x}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}+\frac {4}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx+64 \int \frac {\log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-64 \int \left (\frac {1}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}-\frac {2}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\right )-8 \int \frac {x^2}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx-64 \int \frac {1}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx+64 \int \frac {\log (2+x)}{\left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx+128 \int \frac {1}{(2+x) \left (4 x+e^{e^x} x+2 x^2+16 \log (2+x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.02, size = 24, normalized size = 1.04 \begin {gather*} \frac {4 x}{x \left (4+e^{e^x}+2 x\right )+16 \log (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64*x - 16*x^2 - 8*x^3 + E^(E^x + x)*(-8*x^2 - 4*x^3) + (128 + 64*x)*Log[2 + x])/(32*x^2 + 48*x^3 +
24*x^4 + 4*x^5 + E^(2*E^x)*(2*x^2 + x^3) + (256*x + 256*x^2 + 64*x^3)*Log[2 + x] + (512 + 256*x)*Log[2 + x]^2
+ E^E^x*(16*x^2 + 16*x^3 + 4*x^4 + (64*x + 32*x^2)*Log[2 + x])),x]

[Out]

(4*x)/(x*(4 + E^E^x + 2*x) + 16*Log[2 + x])

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fricas [A]  time = 0.83, size = 34, normalized size = 1.48 \begin {gather*} \frac {4 \, x e^{x}}{x e^{\left (x + e^{x}\right )} + 2 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 16 \, e^{x} \log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*log(2+x)-8*x^3-16*x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2
+((32*x^2+64*x)*log(2+x)+4*x^4+16*x^3+16*x^2)*exp(exp(x))+(256*x+512)*log(2+x)^2+(64*x^3+256*x^2+256*x)*log(2+
x)+4*x^5+24*x^4+48*x^3+32*x^2),x, algorithm="fricas")

[Out]

4*x*e^x/(x*e^(x + e^x) + 2*(x^2 + 2*x)*e^x + 16*e^x*log(x + 2))

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giac [A]  time = 0.42, size = 25, normalized size = 1.09 \begin {gather*} \frac {4 \, x}{2 \, x^{2} + x e^{\left (e^{x}\right )} + 4 \, x + 16 \, \log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*log(2+x)-8*x^3-16*x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2
+((32*x^2+64*x)*log(2+x)+4*x^4+16*x^3+16*x^2)*exp(exp(x))+(256*x+512)*log(2+x)^2+(64*x^3+256*x^2+256*x)*log(2+
x)+4*x^5+24*x^4+48*x^3+32*x^2),x, algorithm="giac")

[Out]

4*x/(2*x^2 + x*e^(e^x) + 4*x + 16*log(x + 2))

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maple [A]  time = 0.08, size = 26, normalized size = 1.13

method	result	size
risch	\(\frac {4 x}{x \,{\mathrm e}^{{\mathrm e}^{x}}+2 x^{2}+16 \ln \left (2+x \right )+4 x}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*ln(2+x)-8*x^3-16*x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2+((32*x
^2+64*x)*ln(2+x)+4*x^4+16*x^3+16*x^2)*exp(exp(x))+(256*x+512)*ln(2+x)^2+(64*x^3+256*x^2+256*x)*ln(2+x)+4*x^5+2
4*x^4+48*x^3+32*x^2),x,method=_RETURNVERBOSE)

[Out]

4*x/(x*exp(exp(x))+2*x^2+16*ln(2+x)+4*x)

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maxima [A]  time = 0.76, size = 25, normalized size = 1.09 \begin {gather*} \frac {4 \, x}{2 \, x^{2} + x e^{\left (e^{x}\right )} + 4 \, x + 16 \, \log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-8*x^2)*exp(x)*exp(exp(x))+(64*x+128)*log(2+x)-8*x^3-16*x^2-64*x)/((x^3+2*x^2)*exp(exp(x))^2
+((32*x^2+64*x)*log(2+x)+4*x^4+16*x^3+16*x^2)*exp(exp(x))+(256*x+512)*log(2+x)^2+(64*x^3+256*x^2+256*x)*log(2+
x)+4*x^5+24*x^4+48*x^3+32*x^2),x, algorithm="maxima")

[Out]

4*x/(2*x^2 + x*e^(e^x) + 4*x + 16*log(x + 2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {64\,x+16\,x^2+8\,x^3-\ln \left (x+2\right )\,\left (64\,x+128\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\left (4\,x^3+8\,x^2\right )}{{\ln \left (x+2\right )}^2\,\left (256\,x+512\right )+\ln \left (x+2\right )\,\left (64\,x^3+256\,x^2+256\,x\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (x^3+2\,x^2\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (\ln \left (x+2\right )\,\left (32\,x^2+64\,x\right )+16\,x^2+16\,x^3+4\,x^4\right )+32\,x^2+48\,x^3+24\,x^4+4\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(64*x + 16*x^2 + 8*x^3 - log(x + 2)*(64*x + 128) + exp(exp(x))*exp(x)*(8*x^2 + 4*x^3))/(log(x + 2)^2*(256
*x + 512) + log(x + 2)*(256*x + 256*x^2 + 64*x^3) + exp(2*exp(x))*(2*x^2 + x^3) + exp(exp(x))*(log(x + 2)*(64*
x + 32*x^2) + 16*x^2 + 16*x^3 + 4*x^4) + 32*x^2 + 48*x^3 + 24*x^4 + 4*x^5),x)

[Out]

int(-(64*x + 16*x^2 + 8*x^3 - log(x + 2)*(64*x + 128) + exp(exp(x))*exp(x)*(8*x^2 + 4*x^3))/(log(x + 2)^2*(256
*x + 512) + log(x + 2)*(256*x + 256*x^2 + 64*x^3) + exp(2*exp(x))*(2*x^2 + x^3) + exp(exp(x))*(log(x + 2)*(64*
x + 32*x^2) + 16*x^2 + 16*x^3 + 4*x^4) + 32*x^2 + 48*x^3 + 24*x^4 + 4*x^5), x)

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sympy [A]  time = 0.46, size = 24, normalized size = 1.04 \begin {gather*} \frac {4 x}{2 x^{2} + x e^{e^{x}} + 4 x + 16 \log {\left (x + 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**3-8*x**2)*exp(x)*exp(exp(x))+(64*x+128)*ln(2+x)-8*x**3-16*x**2-64*x)/((x**3+2*x**2)*exp(exp(
x))**2+((32*x**2+64*x)*ln(2+x)+4*x**4+16*x**3+16*x**2)*exp(exp(x))+(256*x+512)*ln(2+x)**2+(64*x**3+256*x**2+25
6*x)*ln(2+x)+4*x**5+24*x**4+48*x**3+32*x**2),x)

[Out]

4*x/(2*x**2 + x*exp(exp(x)) + 4*x + 16*log(x + 2))

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