3.77.9 \(\int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx\)

Optimal. Leaf size=14 \[ \log \left (-5+e^x-\frac {1}{e x}\right ) \]

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Rubi [F]  time = 0.46, 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+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x + Defer[Int][(-1 - 5*E*x + E^(1 + x)*x)^(-1), x] + Defer[Int][1/(x*(-1 - 5*E*x + E^(1 + x)*x)), x] + 5*E*Def
er[Int][x/(-1 - 5*E*x + E^(1 + x)*x), x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 20, normalized size = 1.43 \begin {gather*} -\log (x)+\log \left (1+5 e x-e^{1+x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(1 + x)*x^2)/(-x - 5*E*x^2 + E^(1 + x)*x^2),x]

[Out]

-Log[x] + Log[1 + 5*E*x - E^(1 + x)*x]

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fricas [A]  time = 1.57, size = 20, normalized size = 1.43 \begin {gather*} \log \left (-\frac {5 \, x e - x e^{\left (x + 1\right )} + 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x, algorithm="fricas")

[Out]

log(-(5*x*e - x*e^(x + 1) + 1)/x)

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giac [A]  time = 0.13, size = 19, normalized size = 1.36 \begin {gather*} \log \left (-5 \, x e + x e^{\left (x + 1\right )} - 1\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x, algorithm="giac")

[Out]

log(-5*x*e + x*e^(x + 1) - 1) - log(x)

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maple [A]  time = 0.06, size = 19, normalized size = 1.36

method	result	size
risch	\(\ln \left ({\mathrm e}^{x}-\frac {\left (5 x \,{\mathrm e}+1\right ) {\mathrm e}^{-1}}{x}\right )\)	\(19\)
norman	\(-\ln \relax (x )+\ln \left (x \,{\mathrm e} \,{\mathrm e}^{x}-5 x \,{\mathrm e}-1\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x,method=_RETURNVERBOSE)

[Out]

ln(exp(x)-(5*x*exp(1)+1)/x*exp(-1))

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maxima [A]  time = 0.39, size = 22, normalized size = 1.57 \begin {gather*} \log \left (-\frac {{\left (5 \, x e - x e^{\left (x + 1\right )} + 1\right )} e^{\left (-1\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x, algorithm="maxima")

[Out]

log(-(5*x*e - x*e^(x + 1) + 1)*e^(-1)/x)

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mupad [B]  time = 0.20, size = 19, normalized size = 1.36 \begin {gather*} \ln \left (x\,\mathrm {e}\,{\mathrm {e}}^x-5\,x\,\mathrm {e}-1\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*exp(1)*exp(x) + 1)/(x + 5*x^2*exp(1) - x^2*exp(1)*exp(x)),x)

[Out]

log(x*exp(1)*exp(x) - 5*x*exp(1) - 1) - log(x)

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sympy [A]  time = 0.19, size = 19, normalized size = 1.36 \begin {gather*} \log {\left (e^{x} + \frac {- 5 e x - 1}{e x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*exp(1)*exp(x)+1)/(x**2*exp(1)*exp(x)-5*x**2*exp(1)-x),x)

[Out]

log(exp(x) + (-5*E*x - 1)*exp(-1)/x)

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