3.70.38 \(\int \frac {-14+6 e^{2 x}+e^x (7+x)+(-2+e^x+e^{2 x}) \log (2+e^x)}{e^{3 x}+e^{2 x} (2-2 x)+2 x^2+e^x (-4 x+x^2)} \, dx\)

Optimal. Leaf size=18 \[ \frac {7+\log \left (2+e^x\right )}{-e^x+x} \]

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Rubi [F]  time = 3.16, 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 {-14+6 e^{2 x}+e^x (7+x)+\left (-2+e^x+e^{2 x}\right ) \log \left (2+e^x\right )}{e^{3 x}+e^{2 x} (2-2 x)+2 x^2+e^x \left (-4 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-14 + 6*E^(2*x) + E^x*(7 + x) + (-2 + E^x + E^(2*x))*Log[2 + E^x])/(E^(3*x) + E^(2*x)*(2 - 2*x) + 2*x^2 +
E^x*(-4*x + x^2)),x]

[Out]

-7*Defer[Int][(E^x - x)^(-2), x] - Log[2 + E^x]*Defer[Int][(E^x - x)^(-2), x] + 6*Defer[Int][(E^x - x)^(-1), x
] + Log[2 + E^x]*Defer[Int][(E^x - x)^(-1), x] + 7*Defer[Int][x/(E^x - x)^2, x] + Log[2 + E^x]*Defer[Int][x/(E
^x - x)^2, x] - 2*Defer[Int][1/((2 + E^x)*(2 + x)), x] + 2*Defer[Int][1/((E^x - x)*(2 + x)), x] + Defer[Int][(
E^x*Defer[Int][(E^x - x)^(-2), x])/(2 + E^x), x] - Defer[Int][(E^x*Defer[Int][(E^x - x)^(-1), x])/(2 + E^x), x
] - Defer[Int][(E^x*Defer[Int][x/(E^x - x)^2, x])/(2 + E^x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-14+6 e^{2 x}+e^x (7+x)+\left (-2+e^x+e^{2 x}\right ) \log \left (2+e^x\right )}{\left (2+e^x\right ) \left (e^x-x\right )^2} \, dx\\ &=\int \left (-\frac {2}{\left (2+e^x\right ) (2+x)}+\frac {(-1+x) \left (7+\log \left (2+e^x\right )\right )}{\left (e^x-x\right )^2}+\frac {14+6 x+2 \log \left (2+e^x\right )+x \log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )+\int \frac {(-1+x) \left (7+\log \left (2+e^x\right )\right )}{\left (e^x-x\right )^2} \, dx+\int \frac {14+6 x+2 \log \left (2+e^x\right )+x \log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)} \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )+\int \frac {14+6 x+(2+x) \log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)} \, dx+\int \left (-\frac {7+\log \left (2+e^x\right )}{\left (e^x-x\right )^2}+\frac {x \left (7+\log \left (2+e^x\right )\right )}{\left (e^x-x\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )-\int \frac {7+\log \left (2+e^x\right )}{\left (e^x-x\right )^2} \, dx+\int \frac {x \left (7+\log \left (2+e^x\right )\right )}{\left (e^x-x\right )^2} \, dx+\int \left (\frac {14}{\left (e^x-x\right ) (2+x)}+\frac {6 x}{\left (e^x-x\right ) (2+x)}+\frac {2 \log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)}+\frac {x \log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )+2 \int \frac {\log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)} \, dx+6 \int \frac {x}{\left (e^x-x\right ) (2+x)} \, dx+14 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx+\int \frac {x \log \left (2+e^x\right )}{\left (e^x-x\right ) (2+x)} \, dx-\int \left (\frac {7}{\left (e^x-x\right )^2}+\frac {\log \left (2+e^x\right )}{\left (e^x-x\right )^2}\right ) \, dx+\int \left (\frac {7 x}{\left (e^x-x\right )^2}+\frac {x \log \left (2+e^x\right )}{\left (e^x-x\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )-2 \int \frac {e^x \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx}{2+e^x} \, dx+6 \int \left (\frac {1}{e^x-x}-\frac {2}{\left (e^x-x\right ) (2+x)}\right ) \, dx-7 \int \frac {1}{\left (e^x-x\right )^2} \, dx+7 \int \frac {x}{\left (e^x-x\right )^2} \, dx+14 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx+\log \left (2+e^x\right ) \int \frac {1}{e^x-x} \, dx-\int \frac {\log \left (2+e^x\right )}{\left (e^x-x\right )^2} \, dx+\int \frac {x \log \left (2+e^x\right )}{\left (e^x-x\right )^2} \, dx-\int \frac {e^x \left (\int \frac {1}{e^x-x} \, dx-2 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx\right )}{2+e^x} \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )-2 \int \frac {e^x \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx}{2+e^x} \, dx+6 \int \frac {1}{e^x-x} \, dx-7 \int \frac {1}{\left (e^x-x\right )^2} \, dx+7 \int \frac {x}{\left (e^x-x\right )^2} \, dx-12 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx+14 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx-\log \left (2+e^x\right ) \int \frac {1}{\left (e^x-x\right )^2} \, dx+\log \left (2+e^x\right ) \int \frac {1}{e^x-x} \, dx+\log \left (2+e^x\right ) \int \frac {x}{\left (e^x-x\right )^2} \, dx+\int \frac {e^x \int \frac {1}{\left (e^x-x\right )^2} \, dx}{2+e^x} \, dx-\int \frac {e^x \int \frac {x}{\left (e^x-x\right )^2} \, dx}{2+e^x} \, dx-\int \left (\frac {e^x \int \frac {1}{e^x-x} \, dx}{2+e^x}-\frac {2 e^x \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx}{2+e^x}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (2+e^x\right ) (2+x)} \, dx\right )+6 \int \frac {1}{e^x-x} \, dx-7 \int \frac {1}{\left (e^x-x\right )^2} \, dx+7 \int \frac {x}{\left (e^x-x\right )^2} \, dx-12 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx+14 \int \frac {1}{\left (e^x-x\right ) (2+x)} \, dx-\log \left (2+e^x\right ) \int \frac {1}{\left (e^x-x\right )^2} \, dx+\log \left (2+e^x\right ) \int \frac {1}{e^x-x} \, dx+\log \left (2+e^x\right ) \int \frac {x}{\left (e^x-x\right )^2} \, dx+\int \frac {e^x \int \frac {1}{\left (e^x-x\right )^2} \, dx}{2+e^x} \, dx-\int \frac {e^x \int \frac {1}{e^x-x} \, dx}{2+e^x} \, dx-\int \frac {e^x \int \frac {x}{\left (e^x-x\right )^2} \, dx}{2+e^x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 19, normalized size = 1.06 \begin {gather*} -\frac {7+\log \left (2+e^x\right )}{e^x-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-14 + 6*E^(2*x) + E^x*(7 + x) + (-2 + E^x + E^(2*x))*Log[2 + E^x])/(E^(3*x) + E^(2*x)*(2 - 2*x) + 2
*x^2 + E^x*(-4*x + x^2)),x]

[Out]

-((7 + Log[2 + E^x])/(E^x - x))

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fricas [A]  time = 0.76, size = 16, normalized size = 0.89 \begin {gather*} \frac {\log \left (e^{x} + 2\right ) + 7}{x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)^2+exp(x)-2)*log(exp(x)+2)+6*exp(x)^2+(x+7)*exp(x)-14)/(exp(x)^3+(-2*x+2)*exp(x)^2+(x^2-4*x)
*exp(x)+2*x^2),x, algorithm="fricas")

[Out]

(log(e^x + 2) + 7)/(x - e^x)

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giac [A]  time = 0.24, size = 16, normalized size = 0.89 \begin {gather*} \frac {\log \left (e^{x} + 2\right ) + 7}{x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)^2+exp(x)-2)*log(exp(x)+2)+6*exp(x)^2+(x+7)*exp(x)-14)/(exp(x)^3+(-2*x+2)*exp(x)^2+(x^2-4*x)
*exp(x)+2*x^2),x, algorithm="giac")

[Out]

(log(e^x + 2) + 7)/(x - e^x)

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maple [A]  time = 0.03, size = 26, normalized size = 1.44

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{x}+2\right )}{x -{\mathrm e}^{x}}+\frac {7}{x -{\mathrm e}^{x}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)^2+exp(x)-2)*ln(exp(x)+2)+6*exp(x)^2+(x+7)*exp(x)-14)/(exp(x)^3+(-2*x+2)*exp(x)^2+(x^2-4*x)*exp(x)
+2*x^2),x,method=_RETURNVERBOSE)

[Out]

1/(x-exp(x))*ln(exp(x)+2)+7/(x-exp(x))

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maxima [A]  time = 0.80, size = 16, normalized size = 0.89 \begin {gather*} \frac {\log \left (e^{x} + 2\right ) + 7}{x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)^2+exp(x)-2)*log(exp(x)+2)+6*exp(x)^2+(x+7)*exp(x)-14)/(exp(x)^3+(-2*x+2)*exp(x)^2+(x^2-4*x)
*exp(x)+2*x^2),x, algorithm="maxima")

[Out]

(log(e^x + 2) + 7)/(x - e^x)

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mupad [B]  time = 0.36, size = 16, normalized size = 0.89 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^x+2\right )+7}{x-{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*exp(2*x) + exp(x)*(x + 7) + log(exp(x) + 2)*(exp(2*x) + exp(x) - 2) - 14)/(exp(3*x) - exp(x)*(4*x - x^2
) - exp(2*x)*(2*x - 2) + 2*x^2),x)

[Out]

(log(exp(x) + 2) + 7)/(x - exp(x))

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sympy [A]  time = 0.35, size = 17, normalized size = 0.94 \begin {gather*} \frac {\log {\left (e^{x} + 2 \right )}}{x - e^{x}} - \frac {7}{- x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)**2+exp(x)-2)*ln(exp(x)+2)+6*exp(x)**2+(x+7)*exp(x)-14)/(exp(x)**3+(-2*x+2)*exp(x)**2+(x**2-
4*x)*exp(x)+2*x**2),x)

[Out]

log(exp(x) + 2)/(x - exp(x)) - 7/(-x + exp(x))

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