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

Optimal. Leaf size=29 \[ \log \left (e^x-\frac {4+e^4-x-x^2-e (6+x)}{x}\right ) \]

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Rubi [F]  time = 1.32, 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 {4-6 e+e^4+x^2+e^x x^2}{-4 x-e^4 x+x^2+e^x x^2+x^3+e \left (6 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x - (4 - 6*E + E^4)*Defer[Int][(4*(1 + (E*(-6 + E^3))/4) - E^x*x - (1 + E)*x - x^2)^(-1), x] - (4 - 6*E + E^4)
*Defer[Int][1/(x*(4*(1 + (E*(-6 + E^3))/4) - E^x*x - (1 + E)*x - x^2)), x] + E*Defer[Int][x/(4*(1 + (E*(-6 + E
^3))/4) - E^x*x - (1 + E)*x - x^2), x] + Defer[Int][x^2/(4*(1 + (E*(-6 + E^3))/4) - E^x*x - (1 + E)*x - x^2),
x]

Rubi steps

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

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Mathematica [A]  time = 0.44, size = 32, normalized size = 1.10 \begin {gather*} -\log (x)+\log \left (4-6 e+e^4-x-e x-e^x x-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[x] + Log[4 - 6*E + E^4 - x - E*x - E^x*x - x^2]

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fricas [A]  time = 0.66, size = 25, normalized size = 0.86 \begin {gather*} \log \left (\frac {x^{2} + {\left (x + 6\right )} e + x e^{x} + x - e^{4} - 4}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x^2+exp(4)-6*exp(1)+x^2+4)/(exp(x)*x^2-x*exp(4)+(x^2+6*x)*exp(1)+x^3+x^2-4*x),x, algorithm="
fricas")

[Out]

log((x^2 + (x + 6)*e + x*e^x + x - e^4 - 4)/x)

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giac [A]  time = 0.32, size = 28, normalized size = 0.97 \begin {gather*} \log \left (x^{2} + x e + x e^{x} + x - e^{4} + 6 \, e - 4\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x^2+exp(4)-6*exp(1)+x^2+4)/(exp(x)*x^2-x*exp(4)+(x^2+6*x)*exp(1)+x^3+x^2-4*x),x, algorithm="
giac")

[Out]

log(x^2 + x*e + x*e^x + x - e^4 + 6*e - 4) - log(x)

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maple [A]  time = 0.05, size = 31, normalized size = 1.07

method	result	size
risch	\(\ln \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{4}-x \,{\mathrm e}-x^{2}-6 \,{\mathrm e}-x +4}{x}\right )\)	\(31\)
norman	\(-\ln \relax (x )+\ln \left (-x \,{\mathrm e}-x^{2}-{\mathrm e}^{x} x +{\mathrm e}^{4}-6 \,{\mathrm e}-x +4\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*x^2+exp(4)-6*exp(1)+x^2+4)/(exp(x)*x^2-x*exp(4)+(x^2+6*x)*exp(1)+x^3+x^2-4*x),x,method=_RETURNVERB
OSE)

[Out]

ln(exp(x)-(exp(4)-x*exp(1)-x^2-6*exp(1)-x+4)/x)

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maxima [A]  time = 0.64, size = 28, normalized size = 0.97 \begin {gather*} \log \left (\frac {x^{2} + x {\left (e + 1\right )} + x e^{x} - e^{4} + 6 \, e - 4}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x^2+exp(4)-6*exp(1)+x^2+4)/(exp(x)*x^2-x*exp(4)+(x^2+6*x)*exp(1)+x^3+x^2-4*x),x, algorithm="
maxima")

[Out]

log((x^2 + x*(e + 1) + x*e^x - e^4 + 6*e - 4)/x)

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mupad [B]  time = 0.31, size = 28, normalized size = 0.97 \begin {gather*} \ln \left (x+6\,\mathrm {e}-{\mathrm {e}}^4+x\,\mathrm {e}+x\,{\mathrm {e}}^x+x^2-4\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4) - 6*exp(1) + x^2*exp(x) + x^2 + 4)/(x^2*exp(x) - 4*x - x*exp(4) + exp(1)*(6*x + x^2) + x^2 + x^3),
x)

[Out]

log(x + 6*exp(1) - exp(4) + x*exp(1) + x*exp(x) + x^2 - 4) - log(x)

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sympy [A]  time = 0.24, size = 26, normalized size = 0.90 \begin {gather*} \log {\left (e^{x} + \frac {x^{2} + x + e x - e^{4} - 4 + 6 e}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x**2+exp(4)-6*exp(1)+x**2+4)/(exp(x)*x**2-x*exp(4)+(x**2+6*x)*exp(1)+x**3+x**2-4*x),x)

[Out]

log(exp(x) + (x**2 + x + E*x - exp(4) - 4 + 6*E)/x)

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