3.65.9 \(\int \frac {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx\)

Optimal. Leaf size=13 \[ \log \left (5+e^{2 x}-\frac {3}{x}\right ) \]

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Rubi [F]  time = 0.44, 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 {3+2 e^{2 x} x^2}{-3 x+5 x^2+e^{2 x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

2*x + 6*Defer[Int][(-3 + 5*x + E^(2*x)*x)^(-1), x] + 3*Defer[Int][1/(x*(-3 + 5*x + E^(2*x)*x)), x] - 10*Defer[
Int][x/(-3 + 5*x + E^(2*x)*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2-\frac {-3-6 x+10 x^2}{x \left (-3+5 x+e^{2 x} x\right )}\right ) \, dx\\ &=2 x-\int \frac {-3-6 x+10 x^2}{x \left (-3+5 x+e^{2 x} x\right )} \, dx\\ &=2 x-\int \left (-\frac {6}{-3+5 x+e^{2 x} x}-\frac {3}{x \left (-3+5 x+e^{2 x} x\right )}+\frac {10 x}{-3+5 x+e^{2 x} x}\right ) \, dx\\ &=2 x+3 \int \frac {1}{x \left (-3+5 x+e^{2 x} x\right )} \, dx+6 \int \frac {1}{-3+5 x+e^{2 x} x} \, dx-10 \int \frac {x}{-3+5 x+e^{2 x} x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 19, normalized size = 1.46 \begin {gather*} -\log (x)+\log \left (3-5 x-e^{2 x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 2*E^(2*x)*x^2)/(-3*x + 5*x^2 + E^(2*x)*x^2),x]

[Out]

-Log[x] + Log[3 - 5*x - E^(2*x)*x]

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fricas [A]  time = 0.78, size = 16, normalized size = 1.23 \begin {gather*} \log \left (\frac {x e^{\left (2 \, x\right )} + 5 \, x - 3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x, algorithm="fricas")

[Out]

log((x*e^(2*x) + 5*x - 3)/x)

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giac [A]  time = 0.12, size = 17, normalized size = 1.31 \begin {gather*} \log \left (x e^{\left (2 \, x\right )} + 5 \, x - 3\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x, algorithm="giac")

[Out]

log(x*e^(2*x) + 5*x - 3) - log(x)

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maple [A]  time = 0.04, size = 16, normalized size = 1.23

method	result	size
risch	\(\ln \left ({\mathrm e}^{2 x}+\frac {5 x -3}{x}\right )\)	\(16\)
norman	\(-\ln \relax (x )+\ln \left (x \,{\mathrm e}^{2 x}+5 x -3\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.38, size = 16, normalized size = 1.23 \begin {gather*} \log \left (\frac {x e^{\left (2 \, x\right )} + 5 \, x - 3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)^2*x^2+3)/(exp(x)^2*x^2+5*x^2-3*x),x, algorithm="maxima")

[Out]

log((x*e^(2*x) + 5*x - 3)/x)

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mupad [B]  time = 4.23, size = 17, normalized size = 1.31 \begin {gather*} \ln \left (5\,x+x\,{\mathrm {e}}^{2\,x}-3\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*exp(2*x) + 3)/(x^2*exp(2*x) - 3*x + 5*x^2),x)

[Out]

log(5*x + x*exp(2*x) - 3) - log(x)

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sympy [A]  time = 0.16, size = 12, normalized size = 0.92 \begin {gather*} \log {\left (e^{2 x} + \frac {5 x - 3}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)**2*x**2+3)/(exp(x)**2*x**2+5*x**2-3*x),x)

[Out]

log(exp(2*x) + (5*x - 3)/x)

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