3.38.94 \(\int \frac {-4-2 x+2 x^3+(-2-2 x^2) \log (5)+e^x (-4+2 x+2 x^2+(-2-2 x) \log (5))}{-x-x^2+(1+x) \log (5)+e^x (-x+\log (5))} \, dx\)

Optimal. Leaf size=36 \[ -2 x-x^2+4 \left (x+\log \left (\frac {x-\log (5)}{5 \left (-1-e^x-x\right )}\right )\right ) \]

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Rubi [F]  time = 0.68, 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 {-4-2 x+2 x^3+\left (-2-2 x^2\right ) \log (5)+e^x \left (-4+2 x+2 x^2+(-2-2 x) \log (5)\right )}{-x-x^2+(1+x) \log (5)+e^x (-x+\log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 - 2*x + 2*x^3 + (-2 - 2*x^2)*Log[5] + E^x*(-4 + 2*x + 2*x^2 + (-2 - 2*x)*Log[5]))/(-x - x^2 + (1 + x)*
Log[5] + E^x*(-x + Log[5])),x]

[Out]

-2*x - x^2 + (4 - 2*Log[5]^2 + Log[5]*Log[25])*Log[x - Log[5]] + 4*Defer[Int][x/(1 + E^x + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4+2 x-2 x^3-\left (-2-2 x^2\right ) \log (5)-e^x \left (-4+2 x+2 x^2+(-2-2 x) \log (5)\right )}{\left (1+e^x+x\right ) (x-\log (5))} \, dx\\ &=\int \left (\frac {4 x}{1+e^x+x}+\frac {4-2 x^2-x (2-\log (25))+\log (25)}{x-\log (5)}\right ) \, dx\\ &=4 \int \frac {x}{1+e^x+x} \, dx+\int \frac {4-2 x^2-x (2-\log (25))+\log (25)}{x-\log (5)} \, dx\\ &=4 \int \frac {x}{1+e^x+x} \, dx+\int \left (-2-2 x+\frac {4-2 \log ^2(5)+\log (5) \log (25)}{x-\log (5)}\right ) \, dx\\ &=-2 x-x^2+\left (4-2 \log ^2(5)+\log (5) \log (25)\right ) \log (x-\log (5))+4 \int \frac {x}{1+e^x+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 29, normalized size = 0.81 \begin {gather*} 2 \left (x-\frac {x^2}{2}-2 \log \left (1+e^x+x\right )+2 \log (x-\log (5))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 2*x + 2*x^3 + (-2 - 2*x^2)*Log[5] + E^x*(-4 + 2*x + 2*x^2 + (-2 - 2*x)*Log[5]))/(-x - x^2 + (1
+ x)*Log[5] + E^x*(-x + Log[5])),x]

[Out]

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

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fricas [A]  time = 0.60, size = 26, normalized size = 0.72 \begin {gather*} -x^{2} + 2 \, x - 4 \, \log \left (x + e^{x} + 1\right ) + 4 \, \log \left (x - \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-2)*log(5)+2*x^2+2*x-4)*exp(x)+(-2*x^2-2)*log(5)+2*x^3-2*x-4)/((log(5)-x)*exp(x)+(x+1)*log(5)
-x^2-x),x, algorithm="fricas")

[Out]

-x^2 + 2*x - 4*log(x + e^x + 1) + 4*log(x - log(5))

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giac [A]  time = 0.19, size = 26, normalized size = 0.72 \begin {gather*} -x^{2} + 2 \, x - 4 \, \log \left (x + e^{x} + 1\right ) + 4 \, \log \left (x - \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-2)*log(5)+2*x^2+2*x-4)*exp(x)+(-2*x^2-2)*log(5)+2*x^3-2*x-4)/((log(5)-x)*exp(x)+(x+1)*log(5)
-x^2-x),x, algorithm="giac")

[Out]

-x^2 + 2*x - 4*log(x + e^x + 1) + 4*log(x - log(5))

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maple [A]  time = 0.09, size = 27, normalized size = 0.75

method	result	size
norman	\(-x^{2}+2 x +4 \ln \left (\ln \relax (5)-x \right )-4 \ln \left ({\mathrm e}^{x}+1+x \right )\)	\(27\)
risch	\(-x^{2}+2 x +4 \ln \left (-\ln \relax (5)+x \right )-4 \ln \left ({\mathrm e}^{x}+1+x \right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x-2)*ln(5)+2*x^2+2*x-4)*exp(x)+(-2*x^2-2)*ln(5)+2*x^3-2*x-4)/((ln(5)-x)*exp(x)+(x+1)*ln(5)-x^2-x),x,
method=_RETURNVERBOSE)

[Out]

-x^2+2*x+4*ln(ln(5)-x)-4*ln(exp(x)+1+x)

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maxima [A]  time = 0.61, size = 26, normalized size = 0.72 \begin {gather*} -x^{2} + 2 \, x - 4 \, \log \left (x + e^{x} + 1\right ) + 4 \, \log \left (x - \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-2)*log(5)+2*x^2+2*x-4)*exp(x)+(-2*x^2-2)*log(5)+2*x^3-2*x-4)/((log(5)-x)*exp(x)+(x+1)*log(5)
-x^2-x),x, algorithm="maxima")

[Out]

-x^2 + 2*x - 4*log(x + e^x + 1) + 4*log(x - log(5))

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mupad [B]  time = 0.16, size = 26, normalized size = 0.72 \begin {gather*} 2\,x+4\,\ln \left (x-\ln \relax (5)\right )-4\,\ln \left (x+{\mathrm {e}}^x+1\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(5)*(2*x^2 + 2) - exp(x)*(2*x - log(5)*(2*x + 2) + 2*x^2 - 4) - 2*x^3 + 4)/(x - log(5)*(x + 1) +
x^2 + exp(x)*(x - log(5))),x)

[Out]

2*x + 4*log(x - log(5)) - 4*log(x + exp(x) + 1) - x^2

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sympy [A]  time = 0.18, size = 24, normalized size = 0.67 \begin {gather*} - x^{2} + 2 x + 4 \log {\left (x - \log {\relax (5 )} \right )} - 4 \log {\left (x + e^{x} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-2)*ln(5)+2*x**2+2*x-4)*exp(x)+(-2*x**2-2)*ln(5)+2*x**3-2*x-4)/((ln(5)-x)*exp(x)+(x+1)*ln(5)-
x**2-x),x)

[Out]

-x**2 + 2*x + 4*log(x - log(5)) - 4*log(x + exp(x) + 1)

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