3.51.27 \(\int \frac {(-3 x-2 x^2) \log (x^2)+(4+6 x+2 x^2) \log (2+3 x+x^2)+(-2 x-3 x^2-x^3) \log ^2(2+3 x+x^2)+e^{e^x} (3 x+2 x^2+e^x (-2 x-3 x^2-x^3) \log (2+3 x+x^2))}{(2 x+3 x^2+x^3) \log ^2(2+3 x+x^2)} \, dx\)

Optimal. Leaf size=32 \[ -x+\frac {-e^{e^x}+\log \left (x^2\right )}{\log \left (\frac {(2+x) \left (x+x^2\right )}{x}\right )} \]

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Rubi [F]  time = 2.87, 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 {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{x \left (2+3 x+x^2\right ) \log ^2\left (2+3 x+x^2\right )} \, dx\\ &=\int \left (-1+\frac {(3+2 x) \left (e^{e^x}-\log \left (x^2\right )\right )}{\left (2+3 x+x^2\right ) \log ^2\left (2+3 x+x^2\right )}+\frac {2-e^{e^x+x} x}{x \log \left (2+3 x+x^2\right )}\right ) \, dx\\ &=-x+\int \frac {(3+2 x) \left (e^{e^x}-\log \left (x^2\right )\right )}{\left (2+3 x+x^2\right ) \log ^2\left (2+3 x+x^2\right )} \, dx+\int \frac {2-e^{e^x+x} x}{x \log \left (2+3 x+x^2\right )} \, dx\\ &=-x+\int \left (\frac {e^{e^x} (3+2 x)}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )}-\frac {(3+2 x) \log \left (x^2\right )}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )}\right ) \, dx+\int \left (-\frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )}+\frac {2}{x \log \left (2+3 x+x^2\right )}\right ) \, dx\\ &=-x+2 \int \frac {1}{x \log \left (2+3 x+x^2\right )} \, dx+\int \frac {e^{e^x} (3+2 x)}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {(3+2 x) \log \left (x^2\right )}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )} \, dx\\ &=-x+2 \int \frac {1}{x \log \left (2+3 x+x^2\right )} \, dx+\int \left (\frac {e^{e^x}}{(1+x) \log ^2\left (2+3 x+x^2\right )}+\frac {e^{e^x}}{(2+x) \log ^2\left (2+3 x+x^2\right )}\right ) \, dx-\int \left (\frac {\log \left (x^2\right )}{(1+x) \log ^2\left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{(2+x) \log ^2\left (2+3 x+x^2\right )}\right ) \, dx-\int \frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )} \, dx\\ &=-x+2 \int \frac {1}{x \log \left (2+3 x+x^2\right )} \, dx+\int \frac {e^{e^x}}{(1+x) \log ^2\left (2+3 x+x^2\right )} \, dx+\int \frac {e^{e^x}}{(2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {\log \left (x^2\right )}{(1+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {\log \left (x^2\right )}{(2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 38, normalized size = 1.19 \begin {gather*} -x-\frac {e^{e^x}}{\log \left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{\log \left (2+3 x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 34, normalized size = 1.06 \begin {gather*} -\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2
*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 34, normalized size = 1.06 \begin {gather*} -\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2
*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.15, size = 88, normalized size = 2.75

method	result	size
risch	\(-x +\frac {4 \ln \relax (x )-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2 \ln \left (x^{2}+3 x +2\right )}-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (x^{2}+3 x +2\right )}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3-3*x^2-2*x)*exp(x)*ln(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*ln(x^2+3*x+2)^2+(2*x^2+6*x
+4)*ln(x^2+3*x+2)+(-2*x^2-3*x)*ln(x^2))/(x^3+3*x^2+2*x)/ln(x^2+3*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

-x+1/2*(4*ln(x)-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2)^3)/ln(x^2+3*x+2)-
1/ln(x^2+3*x+2)*exp(exp(x))

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maxima [A]  time = 0.40, size = 33, normalized size = 1.03 \begin {gather*} -\frac {x \log \left (x + 2\right ) + x \log \left (x + 1\right ) + e^{\left (e^{x}\right )} - 2 \, \log \relax (x)}{\log \left (x + 2\right ) + \log \left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2
*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x+2\,x^2-{\mathrm {e}}^x\,\ln \left (x^2+3\,x+2\right )\,\left (x^3+3\,x^2+2\,x\right )\right )-\ln \left (x^2\right )\,\left (2\,x^2+3\,x\right )+\ln \left (x^2+3\,x+2\right )\,\left (2\,x^2+6\,x+4\right )-{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )}{{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x^3)) - log(x^2)*(3*x + 2*x^2) + log(
3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^
2 + x^3)),x)

[Out]

int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x^3)) - log(x^2)*(3*x + 2*x^2) + log(
3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^
2 + x^3)), x)

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sympy [A]  time = 0.73, size = 31, normalized size = 0.97 \begin {gather*} - x - \frac {e^{e^{x}}}{\log {\left (x^{2} + 3 x + 2 \right )}} + \frac {\log {\left (x^{2} \right )}}{\log {\left (x^{2} + 3 x + 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3-3*x**2-2*x)*exp(x)*ln(x**2+3*x+2)+2*x**2+3*x)*exp(exp(x))+(-x**3-3*x**2-2*x)*ln(x**2+3*x+2)
**2+(2*x**2+6*x+4)*ln(x**2+3*x+2)+(-2*x**2-3*x)*ln(x**2))/(x**3+3*x**2+2*x)/ln(x**2+3*x+2)**2,x)

[Out]

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

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