3.97.12 \(\int \frac {-10-20 x+(-5-10 x-10 x^2) \log (3)+e^x (-20-10 x+(-5-20 x-5 x^2+5 x^3) \log (3))}{x^2+2 x^3+x^4+e^{2 x} (1+2 x+x^2)+e^x (2 x+4 x^2+2 x^3)} \, dx\)

Optimal. Leaf size=27 \[ \frac {5 \left (2+\left (1+x-x^2\right ) \log (3)\right )}{(1+x) \left (e^x+x\right )} \]

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Rubi [F]  time = 1.49, 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 {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 - 20*x + (-5 - 10*x - 10*x^2)*Log[3] + E^x*(-20 - 10*x + (-5 - 20*x - 5*x^2 + 5*x^3)*Log[3]))/(x^2 +
2*x^3 + x^4 + E^(2*x)*(1 + 2*x + x^2) + E^x*(2*x + 4*x^2 + 2*x^3)),x]

[Out]

(5*Log[27])/(E^x + x) + 5*(2 - Log[27])*Defer[Int][(E^x + x)^(-2), x] + 5*Log[27]*Defer[Int][(E^x + x)^(-2), x
] - 5*Log[3]*Defer[Int][x^2/(E^x + x)^2, x] - 5*(4 - Log[9])*Defer[Int][1/((1 + x)*(E^x + x)^2), x] - 15*Log[3
]*Defer[Int][(E^x + x)^(-1), x] + 5*Log[27]*Defer[Int][(E^x + x)^(-1), x] + 5*Log[3]*Defer[Int][x/(E^x + x), x
] - 5*(2 - Log[3])*Defer[Int][1/((1 + x)^2*(E^x + x)), x] - 5*(2 - Log[3])*Defer[Int][1/((1 + x)*(E^x + x)), x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{(1+x)^2 \left (e^x+x\right )^2} \, dx\\ &=\int \left (-\frac {5 \left (2-2 x+\log (3)+x^3 \log (3)-x^2 \log (9)\right )}{(1+x) \left (e^x+x\right )^2}+\frac {5 \left (-4-\log (3)-x^2 \log (3)+x^3 \log (3)-2 x (1+\log (9))\right )}{(1+x)^2 \left (e^x+x\right )}\right ) \, dx\\ &=-\left (5 \int \frac {2-2 x+\log (3)+x^3 \log (3)-x^2 \log (9)}{(1+x) \left (e^x+x\right )^2} \, dx\right )+5 \int \frac {-4-\log (3)-x^2 \log (3)+x^3 \log (3)-2 x (1+\log (9))}{(1+x)^2 \left (e^x+x\right )} \, dx\\ &=5 \int \left (\frac {-2+\log (3)}{(1+x)^2 \left (e^x+x\right )}+\frac {-2+\log (3)}{(1+x) \left (e^x+x\right )}-\frac {3 \log (3)}{e^x+x}+\frac {x \log (3)}{e^x+x}\right ) \, dx-5 \int \left (\frac {x^2 \log (3)}{\left (e^x+x\right )^2}-\frac {-4+\log (9)}{(1+x) \left (e^x+x\right )^2}-\frac {2 \left (1-\frac {\log (27)}{2}\right )}{\left (e^x+x\right )^2}-\frac {x \log (27)}{\left (e^x+x\right )^2}\right ) \, dx\\ &=-\left ((5 (2-\log (3))) \int \frac {1}{(1+x)^2 \left (e^x+x\right )} \, dx\right )-(5 (2-\log (3))) \int \frac {1}{(1+x) \left (e^x+x\right )} \, dx-(5 \log (3)) \int \frac {x^2}{\left (e^x+x\right )^2} \, dx+(5 \log (3)) \int \frac {x}{e^x+x} \, dx-(15 \log (3)) \int \frac {1}{e^x+x} \, dx-(5 (4-\log (9))) \int \frac {1}{(1+x) \left (e^x+x\right )^2} \, dx+(5 (2-\log (27))) \int \frac {1}{\left (e^x+x\right )^2} \, dx+(5 \log (27)) \int \frac {x}{\left (e^x+x\right )^2} \, dx\\ &=\frac {5 \log (27)}{e^x+x}-(5 (2-\log (3))) \int \frac {1}{(1+x)^2 \left (e^x+x\right )} \, dx-(5 (2-\log (3))) \int \frac {1}{(1+x) \left (e^x+x\right )} \, dx-(5 \log (3)) \int \frac {x^2}{\left (e^x+x\right )^2} \, dx+(5 \log (3)) \int \frac {x}{e^x+x} \, dx-(15 \log (3)) \int \frac {1}{e^x+x} \, dx-(5 (4-\log (9))) \int \frac {1}{(1+x) \left (e^x+x\right )^2} \, dx+(5 (2-\log (27))) \int \frac {1}{\left (e^x+x\right )^2} \, dx+(5 \log (27)) \int \frac {1}{\left (e^x+x\right )^2} \, dx+(5 \log (27)) \int \frac {1}{e^x+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 49, normalized size = 1.81 \begin {gather*} -\frac {5 \left (2+\log (3)+x \log (3)-x^3 \log (3)+x^4 \log (3)-x^2 (2+\log (9))\right )}{(-1+x) (1+x)^2 \left (e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 20*x + (-5 - 10*x - 10*x^2)*Log[3] + E^x*(-20 - 10*x + (-5 - 20*x - 5*x^2 + 5*x^3)*Log[3]))/(
x^2 + 2*x^3 + x^4 + E^(2*x)*(1 + 2*x + x^2) + E^x*(2*x + 4*x^2 + 2*x^3)),x]

[Out]

(-5*(2 + Log[3] + x*Log[3] - x^3*Log[3] + x^4*Log[3] - x^2*(2 + Log[9])))/((-1 + x)*(1 + x)^2*(E^x + x))

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fricas [A]  time = 0.57, size = 28, normalized size = 1.04 \begin {gather*} -\frac {5 \, {\left ({\left (x^{2} - x - 1\right )} \log \relax (3) - 2\right )}}{x^{2} + {\left (x + 1\right )} e^{x} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*log(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2
+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 32, normalized size = 1.19 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \relax (3) - x \log \relax (3) - \log \relax (3) - 2\right )}}{x^{2} + x e^{x} + x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*log(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2
+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 31, normalized size = 1.15

method	result	size
risch	\(-\frac {5 \left (x^{2} \ln \relax (3)-x \ln \relax (3)-\ln \relax (3)-2\right )}{\left (x +1\right ) \left ({\mathrm e}^{x}+x \right )}\)	\(31\)
norman	\(\frac {5 \ln \relax (3) {\mathrm e}^{x}+10 x \ln \relax (3)+5 x \ln \relax (3) {\mathrm e}^{x}+10+5 \ln \relax (3)}{{\mathrm e}^{x} x +x^{2}+{\mathrm e}^{x}+x}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x^3-5*x^2-20*x-5)*ln(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*ln(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2+(2*x^3+
4*x^2+2*x)*exp(x)+x^4+2*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

-5*(x^2*ln(3)-x*ln(3)-ln(3)-2)/(x+1)/(exp(x)+x)

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maxima [A]  time = 0.50, size = 32, normalized size = 1.19 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \relax (3) - x \log \relax (3) - \log \relax (3) - 2\right )}}{x^{2} + {\left (x + 1\right )} e^{x} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*log(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2
+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.97, size = 27, normalized size = 1.00 \begin {gather*} \frac {-\ln \left (243\right )\,x^2+\ln \left (243\right )\,x+\ln \left (243\right )+10}{\left (x+{\mathrm {e}}^x\right )\,\left (x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + log(3)*(10*x + 10*x^2 + 5) + exp(x)*(10*x + log(3)*(20*x + 5*x^2 - 5*x^3 + 5) + 20) + 10)/(exp(2*
x)*(2*x + x^2 + 1) + x^2 + 2*x^3 + x^4 + exp(x)*(2*x + 4*x^2 + 2*x^3)),x)

[Out]

(log(243) + x*log(243) - x^2*log(243) + 10)/((x + exp(x))*(x + 1))

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sympy [A]  time = 0.16, size = 32, normalized size = 1.19 \begin {gather*} \frac {- 5 x^{2} \log {\relax (3 )} + 5 x \log {\relax (3 )} + 5 \log {\relax (3 )} + 10}{x^{2} + x + \left (x + 1\right ) e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x**3-5*x**2-20*x-5)*ln(3)-10*x-20)*exp(x)+(-10*x**2-10*x-5)*ln(3)-20*x-10)/((x**2+2*x+1)*exp(x)
**2+(2*x**3+4*x**2+2*x)*exp(x)+x**4+2*x**3+x**2),x)

[Out]

(-5*x**2*log(3) + 5*x*log(3) + 5*log(3) + 10)/(x**2 + x + (x + 1)*exp(x))

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