3.41.58 \(\int \frac {-10-2 x+e^{-2-6 x} (5+x)^2 (10+30 x+6 x^2)}{-5 x-x^2+e^{-2-6 x} (5+x)^2 (5 x+x^2)} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {x^2}{-1+e^{-8+6 (1-x)} (5+x)^2}\right ) \]

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Rubi [F]  time = 0.75, 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 {-10-2 x+e^{-2-6 x} (5+x)^2 \left (10+30 x+6 x^2\right )}{-5 x-x^2+e^{-2-6 x} (5+x)^2 \left (5 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 - 2*x + E^(-2 - 6*x)*(5 + x)^2*(10 + 30*x + 6*x^2))/(-5*x - x^2 + E^(-2 - 6*x)*(5 + x)^2*(5*x + x^2))
,x]

[Out]

2*Log[x] - 14*Defer[Int][(-5 + E^(1 + 3*x) - x)^(-1), x] + 3*Defer[Int][x/(5 - E^(1 + 3*x) + x), x] + 14*Defer
[Int][(5 + E^(1 + 3*x) + x)^(-1), x] + 3*Defer[Int][x/(5 + E^(1 + 3*x) + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-25+e^{2+6 x}-80 x-30 x^2-3 x^3\right )}{x \left (e^{2+6 x}-(5+x)^2\right )} \, dx\\ &=2 \int \frac {-25+e^{2+6 x}-80 x-30 x^2-3 x^3}{x \left (e^{2+6 x}-(5+x)^2\right )} \, dx\\ &=2 \int \left (\frac {1}{x}+\frac {14+3 x}{2 \left (5-e^{1+3 x}+x\right )}+\frac {14+3 x}{2 \left (5+e^{1+3 x}+x\right )}\right ) \, dx\\ &=2 \log (x)+\int \frac {14+3 x}{5-e^{1+3 x}+x} \, dx+\int \frac {14+3 x}{5+e^{1+3 x}+x} \, dx\\ &=2 \log (x)+\int \left (-\frac {14}{-5+e^{1+3 x}-x}+\frac {3 x}{5-e^{1+3 x}+x}\right ) \, dx+\int \left (\frac {14}{5+e^{1+3 x}+x}+\frac {3 x}{5+e^{1+3 x}+x}\right ) \, dx\\ &=2 \log (x)+3 \int \frac {x}{5-e^{1+3 x}+x} \, dx+3 \int \frac {x}{5+e^{1+3 x}+x} \, dx-14 \int \frac {1}{-5+e^{1+3 x}-x} \, dx+14 \int \frac {1}{5+e^{1+3 x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 28, normalized size = 1.08 \begin {gather*} 6 x+2 \log (x)-\log \left (25-e^{2+6 x}+10 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 2*x + E^(-2 - 6*x)*(5 + x)^2*(10 + 30*x + 6*x^2))/(-5*x - x^2 + E^(-2 - 6*x)*(5 + x)^2*(5*x +
x^2)),x]

[Out]

6*x + 2*Log[x] - Log[25 - E^(2 + 6*x) + 10*x + x^2]

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fricas [A]  time = 0.93, size = 22, normalized size = 0.85 \begin {gather*} 2 \, \log \relax (x) - \log \left (e^{\left (-6 \, x + 2 \, \log \left (x + 5\right ) - 2\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+30*x+10)*exp(log(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(log(5+x)-3*x-1)^2-x^2-5*x),x, algorith
m="fricas")

[Out]

2*log(x) - log(e^(-6*x + 2*log(x + 5) - 2) - 1)

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giac [A]  time = 0.31, size = 27, normalized size = 1.04 \begin {gather*} 6 \, x - \log \left (-x^{2} - 10 \, x + e^{\left (6 \, x + 2\right )} - 25\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+30*x+10)*exp(log(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(log(5+x)-3*x-1)^2-x^2-5*x),x, algorith
m="giac")

[Out]

6*x - log(-x^2 - 10*x + e^(6*x + 2) - 25) + 2*log(x)

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maple [A]  time = 0.11, size = 24, normalized size = 0.92

method	result	size
risch	\(2 \ln \relax (x )-2-\ln \left (\left (5+x \right )^{2} {\mathrm e}^{-6 x -2}-1\right )\)	\(24\)
norman	\(2 \ln \relax (x )-\ln \left ({\mathrm e}^{\ln \left (5+x \right )-3 x -1}-1\right )-\ln \left ({\mathrm e}^{\ln \left (5+x \right )-3 x -1}+1\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2+30*x+10)*exp(ln(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(ln(5+x)-3*x-1)^2-x^2-5*x),x,method=_RETURNVER
BOSE)

[Out]

2*ln(x)-2-ln((5+x)^2*exp(-6*x-2)-1)

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maxima [A]  time = 0.45, size = 41, normalized size = 1.58 \begin {gather*} 6 \, x - \log \left ({\left (x + e^{\left (3 \, x + 1\right )} + 5\right )} e^{\left (-1\right )}\right ) - \log \left (-{\left (x - e^{\left (3 \, x + 1\right )} + 5\right )} e^{\left (-1\right )}\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+30*x+10)*exp(log(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(log(5+x)-3*x-1)^2-x^2-5*x),x, algorith
m="maxima")

[Out]

6*x - log((x + e^(3*x + 1) + 5)*e^(-1)) - log(-(x - e^(3*x + 1) + 5)*e^(-1)) + 2*log(x)

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mupad [B]  time = 0.24, size = 37, normalized size = 1.42 \begin {gather*} 2\,\ln \relax (x)-\ln \left (25\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{-2}+10\,x\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{-2}+x^2\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{-2}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - exp(2*log(x + 5) - 6*x - 2)*(30*x + 6*x^2 + 10) + 10)/(5*x - exp(2*log(x + 5) - 6*x - 2)*(5*x + x^2
) + x^2),x)

[Out]

2*log(x) - log(25*exp(-6*x)*exp(-2) + 10*x*exp(-6*x)*exp(-2) + x^2*exp(-6*x)*exp(-2) - 1)

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sympy [A]  time = 0.36, size = 31, normalized size = 1.19 \begin {gather*} 2 \log {\relax (x )} - 2 \log {\left (x + 5 \right )} - \log {\left (e^{- 6 x - 2} - \frac {1}{x^{2} + 10 x + 25} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2+30*x+10)*exp(ln(5+x)-3*x-1)**2-2*x-10)/((x**2+5*x)*exp(ln(5+x)-3*x-1)**2-x**2-5*x),x)

[Out]

2*log(x) - 2*log(x + 5) - log(exp(-6*x - 2) - 1/(x**2 + 10*x + 25))

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