3.42.39 \(\int \frac {-15+10 e^x x^2}{3 x-3 x^2+2 e^x x^2} \, dx\)

Optimal. Leaf size=23 \[ 5 \left (-e^{e^4}+\log \left (-1+\frac {2 e^x}{3}+\frac {1}{x}\right )\right ) \]

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Rubi [F]  time = 0.52, 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 {-15+10 e^x x^2}{3 x-3 x^2+2 e^x x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-15 + 10*E^x*x^2)/(3*x - 3*x^2 + 2*E^x*x^2),x]

[Out]

5*x - 15*Defer[Int][(3 - 3*x + 2*E^x*x)^(-1), x] - 15*Defer[Int][1/(x*(3 - 3*x + 2*E^x*x)), x] + 15*Defer[Int]
[x/(3 - 3*x + 2*E^x*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (-3+2 e^x x^2\right )}{3 x-3 x^2+2 e^x x^2} \, dx\\ &=5 \int \frac {-3+2 e^x x^2}{3 x-3 x^2+2 e^x x^2} \, dx\\ &=5 \int \left (1+\frac {3 \left (-1-x+x^2\right )}{x \left (3-3 x+2 e^x x\right )}\right ) \, dx\\ &=5 x+15 \int \frac {-1-x+x^2}{x \left (3-3 x+2 e^x x\right )} \, dx\\ &=5 x+15 \int \left (-\frac {1}{3-3 x+2 e^x x}-\frac {1}{x \left (3-3 x+2 e^x x\right )}+\frac {x}{3-3 x+2 e^x x}\right ) \, dx\\ &=5 x-15 \int \frac {1}{3-3 x+2 e^x x} \, dx-15 \int \frac {1}{x \left (3-3 x+2 e^x x\right )} \, dx+15 \int \frac {x}{3-3 x+2 e^x x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 19, normalized size = 0.83 \begin {gather*} 5 \left (-\log (x)+\log \left (3-3 x+2 e^x x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 + 10*E^x*x^2)/(3*x - 3*x^2 + 2*E^x*x^2),x]

[Out]

5*(-Log[x] + Log[3 - 3*x + 2*E^x*x])

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fricas [A]  time = 0.61, size = 17, normalized size = 0.74 \begin {gather*} 5 \, \log \left (\frac {2 \, x e^{x} - 3 \, x + 3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)*x^2-15)/(2*exp(x)*x^2-3*x^2+3*x),x, algorithm="fricas")

[Out]

5*log((2*x*e^x - 3*x + 3)/x)

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giac [A]  time = 0.14, size = 18, normalized size = 0.78 \begin {gather*} 5 \, \log \left (2 \, x e^{x} - 3 \, x + 3\right ) - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)*x^2-15)/(2*exp(x)*x^2-3*x^2+3*x),x, algorithm="giac")

[Out]

5*log(2*x*e^x - 3*x + 3) - 5*log(x)

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maple [A]  time = 0.02, size = 15, normalized size = 0.65

method	result	size
risch	\(5 \ln \left ({\mathrm e}^{x}-\frac {3 \left (x -1\right )}{2 x}\right )\)	\(15\)
norman	\(-5 \ln \relax (x )+5 \ln \left (2 \,{\mathrm e}^{x} x -3 x +3\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(x)*x^2-15)/(2*exp(x)*x^2-3*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 18, normalized size = 0.78 \begin {gather*} 5 \, \log \left (\frac {2 \, x e^{x} - 3 \, x + 3}{2 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)*x^2-15)/(2*exp(x)*x^2-3*x^2+3*x),x, algorithm="maxima")

[Out]

5*log(1/2*(2*x*e^x - 3*x + 3)/x)

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mupad [B]  time = 0.08, size = 18, normalized size = 0.78 \begin {gather*} 5\,\ln \left (2\,x\,{\mathrm {e}}^x-3\,x+3\right )-5\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x^2*exp(x) - 15)/(3*x + 2*x^2*exp(x) - 3*x^2),x)

[Out]

5*log(2*x*exp(x) - 3*x + 3) - 5*log(x)

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sympy [A]  time = 0.17, size = 14, normalized size = 0.61 \begin {gather*} 5 \log {\left (e^{x} + \frac {3 - 3 x}{2 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)*x**2-15)/(2*exp(x)*x**2-3*x**2+3*x),x)

[Out]

5*log(exp(x) + (3 - 3*x)/(2*x))

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