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

Optimal. Leaf size=29 \[ \log \left (\frac {1}{\left (3-e^x-x+x^2\right ) \left (x-\log \left (1+e^x\right )\right )}\right ) \]

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Rubi [F]  time = 1.24, 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 x+e^{2 x} x-3 x^2+e^x \left (1+2 x-2 x^2\right )+\left (-1-e^{2 x}+2 x+e^x (-2+2 x)\right ) \log \left (1+e^x\right )}{3 x-e^{2 x} x-x^2+x^3+e^x \left (2 x-x^2+x^3\right )+\left (-3+e^{2 x}+x-x^2+e^x \left (-2+x-x^2\right )\right ) \log \left (1+e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 30, normalized size = 1.03 \begin {gather*} -\log \left (3-e^x-x+x^2\right )-\log \left (x-\log \left (1+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[3 - E^x - x + x^2] - Log[x - Log[1 + E^x]]

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fricas [A]  time = 0.59, size = 26, normalized size = 0.90 \begin {gather*} -\log \left (-x^{2} + x + e^{x} - 3\right ) - \log \left (-x + \log \left (e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-x^2 + x + e^x - 3) - log(-x + log(e^x + 1))

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giac [A]  time = 0.25, size = 26, normalized size = 0.90 \begin {gather*} -\log \left (-x^{2} + x + e^{x} - 3\right ) - \log \left (-x + \log \left (e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-x^2 + x + e^x - 3) - log(-x + log(e^x + 1))

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maple [A]  time = 0.11, size = 27, normalized size = 0.93

method	result	size
risch	\(-\ln \left (-x^{2}+x +{\mathrm e}^{x}-3\right )-\ln \left (\ln \left ({\mathrm e}^{x}+1\right )-x \right )\)	\(27\)
norman	\(-\ln \left (x -\ln \left ({\mathrm e}^{x}+1\right )\right )-\ln \left (x^{2}-{\mathrm e}^{x}-x +3\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(-x^2+x+exp(x)-3)-ln(ln(exp(x)+1)-x)

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maxima [A]  time = 0.47, size = 26, normalized size = 0.90 \begin {gather*} -\log \left (-x^{2} + x + e^{x} - 3\right ) - \log \left (-x + \log \left (e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-x^2 + x + e^x - 3) - log(-x + log(e^x + 1))

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mupad [B]  time = 0.47, size = 26, normalized size = 0.90 \begin {gather*} -\ln \left (\ln \left ({\mathrm {e}}^x+1\right )-x\right )-\ln \left (x+{\mathrm {e}}^x-x^2-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- log(log(exp(x) + 1) - x) - log(x + exp(x) - x^2 - 3)

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sympy [A]  time = 0.26, size = 22, normalized size = 0.76 \begin {gather*} - \log {\left (- x + \log {\left (e^{x} + 1 \right )} \right )} - \log {\left (- x^{2} + x + e^{x} - 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-x + log(exp(x) + 1)) - log(-x**2 + x + exp(x) - 3)

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