3.33.21 \(\int \frac {4-2 x^2-2 e^x x^2+(2+e^x x+x^2) \log (\frac {8+4 e^x x+4 x^2}{x})+(-2-e^x x-x^2+e^{2+x} (-2-e^x x-x^2)) \log ^3(\frac {8+4 e^x x+4 x^2}{x})}{(2+e^x x+x^2) \log ^3(\frac {8+4 e^x x+4 x^2}{x})} \, dx\)

Optimal. Leaf size=29 \[ 1-e^{2+x}-x+\frac {x}{\log ^2\left (4 \left (e^x+\frac {2}{x}+x\right )\right )} \]

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Rubi [F]  time = 3.21, 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 {4-2 x^2-2 e^x x^2+\left (2+e^x x+x^2\right ) \log \left (\frac {8+4 e^x x+4 x^2}{x}\right )+\left (-2-e^x x-x^2+e^{2+x} \left (-2-e^x x-x^2\right )\right ) \log ^3\left (\frac {8+4 e^x x+4 x^2}{x}\right )}{\left (2+e^x x+x^2\right ) \log ^3\left (\frac {8+4 e^x x+4 x^2}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-E^(2 + x) - x - 2*Defer[Int][x/Log[4*(E^x + 2/x + x)]^3, x] + 4*Defer[Int][1/((2 + E^x*x + x^2)*Log[4*(E^x +
2/x + x)]^3), x] + 4*Defer[Int][x/((2 + E^x*x + x^2)*Log[4*(E^x + 2/x + x)]^3), x] - 2*Defer[Int][x^2/((2 + E^
x*x + x^2)*Log[4*(E^x + 2/x + x)]^3), x] + 2*Defer[Int][x^3/((2 + E^x*x + x^2)*Log[4*(E^x + 2/x + x)]^3), x] +
Defer[Int][Log[4*(E^x + 2/x + x)]^(-2), x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 28, normalized size = 0.97 \begin {gather*} -e^{2+x}-x+\frac {x}{\log ^2\left (4 \left (e^x+\frac {2}{x}+x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 63, normalized size = 2.17 \begin {gather*} -\frac {{\left (x + e^{\left (x + 2\right )}\right )} \log \left (\frac {4 \, {\left ({\left (x^{2} + 2\right )} e^{2} + x e^{\left (x + 2\right )}\right )} e^{\left (-2\right )}}{x}\right )^{2} - x}{\log \left (\frac {4 \, {\left ({\left (x^{2} + 2\right )} e^{2} + x e^{\left (x + 2\right )}\right )} e^{\left (-2\right )}}{x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.47, size = 64, normalized size = 2.21 \begin {gather*} -\frac {x \log \left (\frac {4 \, {\left (x^{2} + x e^{x} + 2\right )}}{x}\right )^{2} + e^{\left (x + 2\right )} \log \left (\frac {4 \, {\left (x^{2} + x e^{x} + 2\right )}}{x}\right )^{2} - x}{\log \left (\frac {4 \, {\left (x^{2} + x e^{x} + 2\right )}}{x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.40, size = 160, normalized size = 5.52

method	result	size
risch	\(-{\mathrm e}^{2+x}-x -\frac {4 x}{\left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x +x^{2}+2\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +x^{2}+2\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +x^{2}+2\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x +x^{2}+2\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +x^{2}+2\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +x^{2}+2\right )}{x}\right )^{3}+4 i \ln \relax (2)-2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{x} x +x^{2}+2\right )\right )^{2}}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(2+x)-x-4*x/(Pi*csgn(I/x)*csgn(I*(exp(x)*x+x^2+2))*csgn(I/x*(exp(x)*x+x^2+2))-Pi*csgn(I/x)*csgn(I/x*(exp(x
)*x+x^2+2))^2-Pi*csgn(I*(exp(x)*x+x^2+2))*csgn(I/x*(exp(x)*x+x^2+2))^2+Pi*csgn(I/x*(exp(x)*x+x^2+2))^3+4*I*ln(
2)-2*I*ln(x)+2*I*ln(exp(x)*x+x^2+2))^2

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maxima [B]  time = 0.73, size = 163, normalized size = 5.62 \begin {gather*} -\frac {{\left (x + e^{\left (x + 2\right )}\right )} \log \left (x^{2} + x e^{x} + 2\right )^{2} - 4 \, x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} + {\left (4 \, \log \relax (2)^{2} - 1\right )} x + {\left (4 \, e^{2} \log \relax (2)^{2} - 4 \, e^{2} \log \relax (2) \log \relax (x) + e^{2} \log \relax (x)^{2}\right )} e^{x} + 2 \, {\left ({\left (2 \, e^{2} \log \relax (2) - e^{2} \log \relax (x)\right )} e^{x} + 2 \, x \log \relax (2) - x \log \relax (x)\right )} \log \left (x^{2} + x e^{x} + 2\right )}{4 \, \log \relax (2)^{2} + 2 \, {\left (2 \, \log \relax (2) - \log \relax (x)\right )} \log \left (x^{2} + x e^{x} + 2\right ) + \log \left (x^{2} + x e^{x} + 2\right )^{2} - 4 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.66, size = 483, normalized size = 16.66 \begin {gather*} \frac {\frac {x\,\left (x\,{\mathrm {e}}^x+x^2+2\right )}{2\,\left (x^2\,{\mathrm {e}}^x+x^2-2\right )}+\frac {x\,\ln \left (\frac {4\,x\,{\mathrm {e}}^x+4\,x^2+8}{x}\right )\,\left (x\,{\mathrm {e}}^x+x^2+2\right )\,\left (4\,x^2\,{\mathrm {e}}^x+2\,x^3\,{\mathrm {e}}^x-2\,x^4\,{\mathrm {e}}^x+x^5\,{\mathrm {e}}^x+4\,x\,{\mathrm {e}}^x+8\,x^2-x^4+4\right )}{2\,{\left (x^2\,{\mathrm {e}}^x+x^2-2\right )}^3}}{\ln \left (\frac {4\,x\,{\mathrm {e}}^x+4\,x^2+8}{x}\right )}-{\mathrm {e}}^{x+2}-x+\frac {x-\frac {x\,\ln \left (\frac {4\,x\,{\mathrm {e}}^x+4\,x^2+8}{x}\right )\,\left (x\,{\mathrm {e}}^x+x^2+2\right )}{2\,\left (x^2\,{\mathrm {e}}^x+x^2-2\right )}}{{\ln \left (\frac {4\,x\,{\mathrm {e}}^x+4\,x^2+8}{x}\right )}^2}-\frac {-x^7+2\,x^6-4\,x^4-8\,x^3+16\,x^2+24\,x+16}{2\,\left (x^2\,{\mathrm {e}}^x+x^2-2\right )\,\left (-x^4+2\,x^2+4\,x\right )}-\frac {-x^{12}+4\,x^{11}-5\,x^{10}-4\,x^9-2\,x^8+20\,x^7-28\,x^6+88\,x^4+144\,x^3+64\,x^2}{2\,x^2\,\left (-x^4+2\,x^2+4\,x\right )\,\left (x^4\,{\mathrm {e}}^{2\,x}+{\left (x^2-2\right )}^2+2\,x^2\,{\mathrm {e}}^x\,\left (x^2-2\right )\right )}-\frac {x^{16}-2\,x^{15}+x^{14}-24\,x^{11}+8\,x^{10}+16\,x^9+144\,x^6+192\,x^5+64\,x^4}{2\,x^4\,\left (-x^4+2\,x^2+4\,x\right )\,\left (x^6\,{\mathrm {e}}^{3\,x}+{\left (x^2-2\right )}^3+3\,x^4\,{\mathrm {e}}^{2\,x}\,\left (x^2-2\right )+3\,x^2\,{\mathrm {e}}^x\,{\left (x^2-2\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(x*exp(x) + x^2 + 2))/(2*(x^2*exp(x) + x^2 - 2)) + (x*log((4*x*exp(x) + 4*x^2 + 8)/x)*(x*exp(x) + x^2 + 2)
*(4*x^2*exp(x) + 2*x^3*exp(x) - 2*x^4*exp(x) + x^5*exp(x) + 4*x*exp(x) + 8*x^2 - x^4 + 4))/(2*(x^2*exp(x) + x^
2 - 2)^3))/log((4*x*exp(x) + 4*x^2 + 8)/x) - exp(x + 2) - x + (x - (x*log((4*x*exp(x) + 4*x^2 + 8)/x)*(x*exp(x
) + x^2 + 2))/(2*(x^2*exp(x) + x^2 - 2)))/log((4*x*exp(x) + 4*x^2 + 8)/x)^2 - (24*x + 16*x^2 - 8*x^3 - 4*x^4 +
2*x^6 - x^7 + 16)/(2*(x^2*exp(x) + x^2 - 2)*(4*x + 2*x^2 - x^4)) - (64*x^2 + 144*x^3 + 88*x^4 - 28*x^6 + 20*x
^7 - 2*x^8 - 4*x^9 - 5*x^10 + 4*x^11 - x^12)/(2*x^2*(4*x + 2*x^2 - x^4)*(x^4*exp(2*x) + (x^2 - 2)^2 + 2*x^2*ex
p(x)*(x^2 - 2))) - (64*x^4 + 192*x^5 + 144*x^6 + 16*x^9 + 8*x^10 - 24*x^11 + x^14 - 2*x^15 + x^16)/(2*x^4*(4*x
+ 2*x^2 - x^4)*(x^6*exp(3*x) + (x^2 - 2)^3 + 3*x^4*exp(2*x)*(x^2 - 2) + 3*x^2*exp(x)*(x^2 - 2)^2))

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sympy [A]  time = 0.32, size = 27, normalized size = 0.93 \begin {gather*} - x + \frac {x}{\log {\left (\frac {4 x^{2} + 4 x e^{x} + 8}{x} \right )}^{2}} - e^{2} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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