3.83.79 \(\int \frac {-1-3 x-x^2+e^{4 x} x (-x-3 x^2-3 x^3)}{-x-x^2+e^{4 x} x (x^2+x^3)} \, dx\)

Optimal. Leaf size=27 \[ x-\log \left (\frac {-\frac {1}{x}+e^{4 x} x}{2 (1+x)}\right ) \]

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Rubi [F]  time = 1.23, 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 {-1-3 x-x^2+e^{4 x} x \left (-x-3 x^2-3 x^3\right )}{-x-x^2+e^{4 x} x \left (x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 23, normalized size = 0.85 \begin {gather*} x+\log (x)+\log (1+x)-\log \left (1-e^{4 x} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + Log[x] + Log[1 + x] - Log[1 - E^(4*x)*x^2]

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fricas [A]  time = 0.77, size = 24, normalized size = 0.89 \begin {gather*} x + \log \left (x + 1\right ) - \log \left (\frac {x e^{\left (4 \, x + \log \relax (x)\right )} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.87, size = 21, normalized size = 0.78 \begin {gather*} x - \log \left (x^{2} e^{\left (4 \, x\right )} - 1\right ) + \log \left (x + 1\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 22, normalized size = 0.81

method	result	size
risch	\(x +\ln \left (x +1\right )-\ln \left (x \,{\mathrm e}^{4 x}-\frac {1}{x}\right )\)	\(22\)
norman	\(x +\ln \relax (x )-\ln \left (x \,{\mathrm e}^{4 x +\ln \relax (x )}-1\right )+\ln \left (x +1\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+ln(x+1)-ln(x*exp(4*x)-1/x)

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maxima [A]  time = 0.41, size = 40, normalized size = 1.48 \begin {gather*} x + \log \left (x + 1\right ) - \log \relax (x) - \log \left (\frac {x e^{\left (2 \, x\right )} + 1}{x}\right ) - \log \left (\frac {x e^{\left (2 \, x\right )} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.37, size = 23, normalized size = 0.85 \begin {gather*} x+\ln \left (x+1\right )-\ln \left (\frac {x^2\,{\mathrm {e}}^{4\,x}-1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.21, size = 20, normalized size = 0.74 \begin {gather*} x - \log {\relax (x )} + \log {\left (x + 1 \right )} - \log {\left (e^{4 x} - \frac {1}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(x) + log(x + 1) - log(exp(4*x) - 1/x**2)

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