3.9.85 \(\int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x (-20+30 x+10 x^2)+(20-10 e^x-30 x-10 x^2) \log (x)+5 \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {18 x}{5 \left (-2+e^x+x (3+x)-\log (x)\right )} \]

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Rubi [F]  time = 1.27, 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 {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-18 + E^x*(18 - 18*x) - 18*x^2 - 18*Log[x])/(20 + 5*E^(2*x) - 60*x + 25*x^2 + 30*x^3 + 5*x^4 + E^x*(-20 +
30*x + 10*x^2) + (20 - 10*E^x - 30*x - 10*x^2)*Log[x] + 5*Log[x]^2),x]

[Out]

(18*Defer[Int][(-2 + E^x + 3*x + x^2 - Log[x])^(-2), x])/5 - 18*Defer[Int][x/(-2 + E^x + 3*x + x^2 - Log[x])^2
, x] + (18*Defer[Int][x^2/(-2 + E^x + 3*x + x^2 - Log[x])^2, x])/5 + (18*Defer[Int][x^3/(-2 + E^x + 3*x + x^2
- Log[x])^2, x])/5 + (18*Defer[Int][(-2 + E^x + 3*x + x^2 - Log[x])^(-1), x])/5 - (18*Defer[Int][x/(-2 + E^x +
3*x + x^2 - Log[x]), x])/5 - (18*Defer[Int][(x*Log[x])/(-2 + E^x + 3*x + x^2 - Log[x])^2, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {18 \left (-1-e^x (-1+x)-x^2-\log (x)\right )}{5 \left (2-e^x-3 x-x^2+\log (x)\right )^2} \, dx\\ &=\frac {18}{5} \int \frac {-1-e^x (-1+x)-x^2-\log (x)}{\left (2-e^x-3 x-x^2+\log (x)\right )^2} \, dx\\ &=\frac {18}{5} \int \left (-\frac {-1+x}{-2+e^x+3 x+x^2-\log (x)}+\frac {1-5 x+x^2+x^3-x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}\right ) \, dx\\ &=-\left (\frac {18}{5} \int \frac {-1+x}{-2+e^x+3 x+x^2-\log (x)} \, dx\right )+\frac {18}{5} \int \frac {1-5 x+x^2+x^3-x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx\\ &=-\left (\frac {18}{5} \int \left (-\frac {1}{-2+e^x+3 x+x^2-\log (x)}+\frac {x}{-2+e^x+3 x+x^2-\log (x)}\right ) \, dx\right )+\frac {18}{5} \int \left (\frac {1}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}-\frac {5 x}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}+\frac {x^2}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}+\frac {x^3}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}-\frac {x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}\right ) \, dx\\ &=\frac {18}{5} \int \frac {1}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx+\frac {18}{5} \int \frac {x^2}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx+\frac {18}{5} \int \frac {x^3}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx+\frac {18}{5} \int \frac {1}{-2+e^x+3 x+x^2-\log (x)} \, dx-\frac {18}{5} \int \frac {x}{-2+e^x+3 x+x^2-\log (x)} \, dx-\frac {18}{5} \int \frac {x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx-18 \int \frac {x}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 24, normalized size = 1.14 \begin {gather*} -\frac {18 x}{5 \left (2-e^x-3 x-x^2+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18 + E^x*(18 - 18*x) - 18*x^2 - 18*Log[x])/(20 + 5*E^(2*x) - 60*x + 25*x^2 + 30*x^3 + 5*x^4 + E^x*
(-20 + 30*x + 10*x^2) + (20 - 10*E^x - 30*x - 10*x^2)*Log[x] + 5*Log[x]^2),x]

[Out]

(-18*x)/(5*(2 - E^x - 3*x - x^2 + Log[x]))

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fricas [A]  time = 0.93, size = 19, normalized size = 0.90 \begin {gather*} \frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \relax (x) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^2-30*x+20)*log(x)+5*exp(x)^2+(
10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x, algorithm="fricas")

[Out]

18/5*x/(x^2 + 3*x + e^x - log(x) - 2)

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giac [A]  time = 0.41, size = 19, normalized size = 0.90 \begin {gather*} \frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \relax (x) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^2-30*x+20)*log(x)+5*exp(x)^2+(
10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x, algorithm="giac")

[Out]

18/5*x/(x^2 + 3*x + e^x - log(x) - 2)

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maple [A]  time = 0.03, size = 20, normalized size = 0.95

method	result	size
risch	\(\frac {18 x}{5 \left (x^{2}+3 x +{\mathrm e}^{x}-\ln \relax (x )-2\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-18*ln(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*ln(x)^2+(-10*exp(x)-10*x^2-30*x+20)*ln(x)+5*exp(x)^2+(10*x^2+30
*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x,method=_RETURNVERBOSE)

[Out]

18/5*x/(x^2+3*x+exp(x)-ln(x)-2)

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maxima [A]  time = 0.54, size = 19, normalized size = 0.90 \begin {gather*} \frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \relax (x) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^2-30*x+20)*log(x)+5*exp(x)^2+(
10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x, algorithm="maxima")

[Out]

18/5*x/(x^2 + 3*x + e^x - log(x) - 2)

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mupad [B]  time = 1.11, size = 23, normalized size = 1.10 \begin {gather*} \frac {18\,x}{5\,\left (3\,x+{\mathrm {e}}^x-\ln \relax (x)+x^2-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(18*log(x) + exp(x)*(18*x - 18) + 18*x^2 + 18)/(5*exp(2*x) - 60*x + 5*log(x)^2 + exp(x)*(30*x + 10*x^2 -
20) + 25*x^2 + 30*x^3 + 5*x^4 - log(x)*(30*x + 10*exp(x) + 10*x^2 - 20) + 20),x)

[Out]

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

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sympy [A]  time = 0.31, size = 22, normalized size = 1.05 \begin {gather*} \frac {18 x}{5 x^{2} + 15 x + 5 e^{x} - 5 \log {\relax (x )} - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*ln(x)+(-18*x+18)*exp(x)-18*x**2-18)/(5*ln(x)**2+(-10*exp(x)-10*x**2-30*x+20)*ln(x)+5*exp(x)**2+
(10*x**2+30*x-20)*exp(x)+5*x**4+30*x**3+25*x**2-60*x+20),x)

[Out]

18*x/(5*x**2 + 15*x + 5*exp(x) - 5*log(x) - 10)

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