3.16.63 \(\int \frac {-90+e^x (-40-20 x) x}{243 x^2+108 e^x x^3+12 e^{2 x} x^4} \, dx\)

Optimal. Leaf size=18 \[ \frac {5}{3 x \left (\frac {9}{2}+e^x x\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6688, 12, 6687} \begin {gather*} \frac {10}{3 x \left (2 e^x x+9\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-90 + E^x*(-40 - 20*x)*x)/(243*x^2 + 108*E^x*x^3 + 12*E^(2*x)*x^4),x]

[Out]

10/(3*x*(9 + 2*E^x*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 17, normalized size = 0.94 \begin {gather*} \frac {10}{3 x \left (9+2 e^x x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-90 + E^x*(-40 - 20*x)*x)/(243*x^2 + 108*E^x*x^3 + 12*E^(2*x)*x^4),x]

[Out]

10/(3*x*(9 + 2*E^x*x))

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fricas [A]  time = 0.63, size = 16, normalized size = 0.89 \begin {gather*} \frac {10}{3 \, {\left (2 \, x e^{\left (x + \log \relax (x)\right )} + 9 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-40)*exp(x+log(x))-90)/(12*x^2*exp(x+log(x))^2+108*x^2*exp(x+log(x))+243*x^2),x, algorithm="f
ricas")

[Out]

10/3/(2*x*e^(x + log(x)) + 9*x)

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giac [A]  time = 0.66, size = 15, normalized size = 0.83 \begin {gather*} \frac {10}{3 \, {\left (2 \, x^{2} e^{x} + 9 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-40)*exp(x+log(x))-90)/(12*x^2*exp(x+log(x))^2+108*x^2*exp(x+log(x))+243*x^2),x, algorithm="g
iac")

[Out]

10/3/(2*x^2*e^x + 9*x)

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maple [A]  time = 0.04, size = 15, normalized size = 0.83

method	result	size
risch	\(\frac {10}{3 x \left (2 \,{\mathrm e}^{x} x +9\right )}\)	\(15\)
norman	\(\frac {10}{3 x \left (2 \,{\mathrm e}^{x +\ln \relax (x )}+9\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x-40)*exp(x+ln(x))-90)/(12*x^2*exp(x+ln(x))^2+108*x^2*exp(x+ln(x))+243*x^2),x,method=_RETURNVERBOSE)

[Out]

10/3/x/(2*exp(x)*x+9)

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maxima [A]  time = 0.67, size = 15, normalized size = 0.83 \begin {gather*} \frac {10}{3 \, {\left (2 \, x^{2} e^{x} + 9 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-40)*exp(x+log(x))-90)/(12*x^2*exp(x+log(x))^2+108*x^2*exp(x+log(x))+243*x^2),x, algorithm="m
axima")

[Out]

10/3/(2*x^2*e^x + 9*x)

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mupad [B]  time = 1.11, size = 14, normalized size = 0.78 \begin {gather*} \frac {10}{3\,x\,\left (2\,x\,{\mathrm {e}}^x+9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + log(x))*(20*x + 40) + 90)/(12*x^2*exp(2*x + 2*log(x)) + 108*x^2*exp(x + log(x)) + 243*x^2),x)

[Out]

10/(3*x*(2*x*exp(x) + 9))

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sympy [A]  time = 0.11, size = 12, normalized size = 0.67 \begin {gather*} \frac {10}{6 x^{2} e^{x} + 27 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-40)*exp(x+ln(x))-90)/(12*x**2*exp(x+ln(x))**2+108*x**2*exp(x+ln(x))+243*x**2),x)

[Out]

10/(6*x**2*exp(x) + 27*x)

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