3.87.66 \(\int \frac {-11 x^2-4 x^3+e^x (4 x^2+4 x^3)+(-12 x-3 x^2+e^x (4 x+4 x^2)) \log (x)+(-3+e^x (1+x)) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ x \left (-3+e^x-\frac {x^2}{2 x+\log (x)}\right ) \]

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Rubi [F]  time = 1.01, 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 {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-11*x^2 - 4*x^3 + E^x*(4*x^2 + 4*x^3) + (-12*x - 3*x^2 + E^x*(4*x + 4*x^2))*Log[x] + (-3 + E^x*(1 + x))*L
og[x]^2)/(4*x^2 + 4*x*Log[x] + Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2 \left (-11-4 x+4 e^x (1+x)\right )+x \left (4 e^x (1+x)-3 (4+x)\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{(2 x+\log (x))^2} \, dx\\ &=\int \left (e^x (1+x)-\frac {11 x^2}{(2 x+\log (x))^2}-\frac {4 x^3}{(2 x+\log (x))^2}-\frac {3 x (4+x) \log (x)}{(2 x+\log (x))^2}-\frac {3 \log ^2(x)}{(2 x+\log (x))^2}\right ) \, dx\\ &=-\left (3 \int \frac {x (4+x) \log (x)}{(2 x+\log (x))^2} \, dx\right )-3 \int \frac {\log ^2(x)}{(2 x+\log (x))^2} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+\int e^x (1+x) \, dx\\ &=e^x (1+x)-3 \int \left (1+\frac {4 x^2}{(2 x+\log (x))^2}-\frac {4 x}{2 x+\log (x)}\right ) \, dx-3 \int \left (-\frac {2 x^2 (4+x)}{(2 x+\log (x))^2}+\frac {x (4+x)}{2 x+\log (x)}\right ) \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-\int e^x \, dx\\ &=-e^x-3 x+e^x (1+x)-3 \int \frac {x (4+x)}{2 x+\log (x)} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \frac {x^2 (4+x)}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+12 \int \frac {x}{2 x+\log (x)} \, dx\\ &=-e^x-3 x+e^x (1+x)-3 \int \left (\frac {4 x}{2 x+\log (x)}+\frac {x^2}{2 x+\log (x)}\right ) \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \left (\frac {4 x^2}{(2 x+\log (x))^2}+\frac {x^3}{(2 x+\log (x))^2}\right ) \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+12 \int \frac {x}{2 x+\log (x)} \, dx\\ &=-e^x-3 x+e^x (1+x)-3 \int \frac {x^2}{2 x+\log (x)} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+24 \int \frac {x^2}{(2 x+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 22, normalized size = 1.10 \begin {gather*} -3 x+e^x x-\frac {x^3}{2 x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11*x^2 - 4*x^3 + E^x*(4*x^2 + 4*x^3) + (-12*x - 3*x^2 + E^x*(4*x + 4*x^2))*Log[x] + (-3 + E^x*(1 +
 x))*Log[x]^2)/(4*x^2 + 4*x*Log[x] + Log[x]^2),x]

[Out]

-3*x + E^x*x - x^3/(2*x + Log[x])

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fricas [A]  time = 0.68, size = 38, normalized size = 1.90 \begin {gather*} -\frac {x^{3} - 2 \, x^{2} e^{x} + 6 \, x^{2} - {\left (x e^{x} - 3 \, x\right )} \log \relax (x)}{2 \, x + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-3)*log(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*log(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)
/(log(x)^2+4*x*log(x)+4*x^2),x, algorithm="fricas")

[Out]

-(x^3 - 2*x^2*e^x + 6*x^2 - (x*e^x - 3*x)*log(x))/(2*x + log(x))

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giac [A]  time = 0.22, size = 38, normalized size = 1.90 \begin {gather*} -\frac {x^{3} - 2 \, x^{2} e^{x} - x e^{x} \log \relax (x) + 6 \, x^{2} + 3 \, x \log \relax (x)}{2 \, x + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-3)*log(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*log(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)
/(log(x)^2+4*x*log(x)+4*x^2),x, algorithm="giac")

[Out]

-(x^3 - 2*x^2*e^x - x*e^x*log(x) + 6*x^2 + 3*x*log(x))/(2*x + log(x))

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maple [A]  time = 0.05, size = 22, normalized size = 1.10




method result size



risch \({\mathrm e}^{x} x -3 x -\frac {x^{3}}{2 x +\ln \relax (x )}\) \(22\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(x)-3)*ln(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*ln(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)/(ln(x)^
2+4*x*ln(x)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 38, normalized size = 1.90 \begin {gather*} -\frac {x^{3} + 6 \, x^{2} - {\left (2 \, x^{2} + x \log \relax (x)\right )} e^{x} + 3 \, x \log \relax (x)}{2 \, x + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-3)*log(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*log(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)
/(log(x)^2+4*x*log(x)+4*x^2),x, algorithm="maxima")

[Out]

-(x^3 + 6*x^2 - (2*x^2 + x*log(x))*e^x + 3*x*log(x))/(2*x + log(x))

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mupad [B]  time = 5.89, size = 65, normalized size = 3.25 \begin {gather*} \frac {3}{16\,\left (x+\frac {1}{2}\right )}-\frac {9\,x}{4}+x\,{\mathrm {e}}^x+\frac {\frac {3\,x^3\,\ln \relax (x)}{2\,x+1}-\frac {x\,\left (x^2-4\,x^3\right )}{2\,x+1}}{2\,x+\ln \relax (x)}-\frac {3\,x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(12*x - exp(x)*(4*x + 4*x^2) + 3*x^2) - log(x)^2*(exp(x)*(x + 1) - 3) - exp(x)*(4*x^2 + 4*x^3) +
11*x^2 + 4*x^3)/(log(x)^2 + 4*x*log(x) + 4*x^2),x)

[Out]

3/(16*(x + 1/2)) - (9*x)/4 + x*exp(x) + ((3*x^3*log(x))/(2*x + 1) - (x*(x^2 - 4*x^3))/(2*x + 1))/(2*x + log(x)
) - (3*x^2)/2

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sympy [A]  time = 0.32, size = 17, normalized size = 0.85 \begin {gather*} - \frac {x^{3}}{2 x + \log {\relax (x )}} + x e^{x} - 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-3)*ln(x)**2+((4*x**2+4*x)*exp(x)-3*x**2-12*x)*ln(x)+(4*x**3+4*x**2)*exp(x)-4*x**3-11*
x**2)/(ln(x)**2+4*x*ln(x)+4*x**2),x)

[Out]

-x**3/(2*x + log(x)) + x*exp(x) - 3*x

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