3.89.2 \(\int \frac {2 e^{3+3 x} x \log (4)+e^{2 x} (-6 x-6 x^2) \log (4)}{-27+27 e^{3+x}-9 e^{6+2 x}+e^{9+3 x}} \, dx\)

Optimal. Leaf size=21 \[ \frac {x^2 \log (4)}{\left (-e^3+3 e^{-x}\right )^2} \]

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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 {2 e^{3+3 x} x \log (4)+e^{2 x} \left (-6 x-6 x^2\right ) \log (4)}{-27+27 e^{3+x}-9 e^{6+2 x}+e^{9+3 x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*E^(3 + 3*x)*x*Log[4] + E^(2*x)*(-6*x - 6*x^2)*Log[4])/(-27 + 27*E^(3 + x) - 9*E^(6 + 2*x) + E^(9 + 3*x)
),x]

[Out]

2*Log[4]*Defer[Int][(E^(2*x)*x)/(-3 + E^(3 + x))^2, x] - 6*Log[4]*Defer[Int][(E^(2*x)*x^2)/(-3 + E^(3 + x))^3,
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{2 x} x \left (-e^{3+x}+3 (1+x)\right ) \log (4)}{\left (3-e^{3+x}\right )^3} \, dx\\ &=(2 \log (4)) \int \frac {e^{2 x} x \left (-e^{3+x}+3 (1+x)\right )}{\left (3-e^{3+x}\right )^3} \, dx\\ &=(2 \log (4)) \int \left (\frac {e^{2 x} x}{\left (-3+e^{3+x}\right )^2}-\frac {3 e^{2 x} x^2}{\left (-3+e^{3+x}\right )^3}\right ) \, dx\\ &=(2 \log (4)) \int \frac {e^{2 x} x}{\left (-3+e^{3+x}\right )^2} \, dx-(6 \log (4)) \int \frac {e^{2 x} x^2}{\left (-3+e^{3+x}\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 20, normalized size = 0.95 \begin {gather*} \frac {e^{2 x} x^2 \log (4)}{\left (-3+e^{3+x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(3 + 3*x)*x*Log[4] + E^(2*x)*(-6*x - 6*x^2)*Log[4])/(-27 + 27*E^(3 + x) - 9*E^(6 + 2*x) + E^(9
+ 3*x)),x]

[Out]

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

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fricas [A]  time = 0.64, size = 32, normalized size = 1.52 \begin {gather*} \frac {2 \, x^{2} e^{\left (2 \, x + 6\right )} \log \relax (2)}{9 \, e^{6} + e^{\left (2 \, x + 12\right )} - 6 \, e^{\left (x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(3)*log(2)*exp(x)^3+2*(-6*x^2-6*x)*log(2)*exp(x)^2)/(exp(3)^3*exp(x)^3-9*exp(3)^2*exp(x)^2+2
7*exp(x)*exp(3)-27),x, algorithm="fricas")

[Out]

2*x^2*e^(2*x + 6)*log(2)/(9*e^6 + e^(2*x + 12) - 6*e^(x + 9))

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giac [A]  time = 0.14, size = 27, normalized size = 1.29 \begin {gather*} \frac {2 \, x^{2} e^{\left (2 \, x\right )} \log \relax (2)}{e^{\left (2 \, x + 6\right )} - 6 \, e^{\left (x + 3\right )} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(3)*log(2)*exp(x)^3+2*(-6*x^2-6*x)*log(2)*exp(x)^2)/(exp(3)^3*exp(x)^3-9*exp(3)^2*exp(x)^2+2
7*exp(x)*exp(3)-27),x, algorithm="giac")

[Out]

2*x^2*e^(2*x)*log(2)/(e^(2*x + 6) - 6*e^(x + 3) + 9)

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maple [A]  time = 0.18, size = 21, normalized size = 1.00

method	result	size
norman	\(\frac {2 x^{2} \ln \relax (2) {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} {\mathrm e}^{3}-3\right )^{2}}\)	\(21\)
risch	\(2 \,{\mathrm e}^{-6} x^{2} \ln \relax (2)+\frac {6 \ln \relax (2) x^{2} \left (2 \,{\mathrm e}^{3+x}-3\right ) {\mathrm e}^{-6}}{\left ({\mathrm e}^{3+x}-3\right )^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*exp(3)*ln(2)*exp(x)^3+2*(-6*x^2-6*x)*ln(2)*exp(x)^2)/(exp(3)^3*exp(x)^3-9*exp(3)^2*exp(x)^2+27*exp(x)
*exp(3)-27),x,method=_RETURNVERBOSE)

[Out]

2*x^2*ln(2)*exp(x)^2/(exp(x)*exp(3)-3)^2

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maxima [B]  time = 0.49, size = 127, normalized size = 6.05 \begin {gather*} 2 \, e^{\left (-6\right )} \log \relax (2) \log \left ({\left (e^{\left (x + 3\right )} - 3\right )} e^{\left (-3\right )}\right ) - 2 \, {\left (e^{\left (-6\right )} \log \left ({\left (e^{\left (x + 3\right )} - 3\right )} e^{\left (-3\right )}\right ) - \frac {x e^{\left (2 \, x + 6\right )} + 3 \, e^{\left (x + 3\right )} - 9}{9 \, e^{6} + e^{\left (2 \, x + 12\right )} - 6 \, e^{\left (x + 9\right )}}\right )} \log \relax (2) + \frac {2 \, {\left ({\left (x^{2} e^{6} \log \relax (2) - x e^{6} \log \relax (2)\right )} e^{\left (2 \, x\right )} - 3 \, e^{\left (x + 3\right )} \log \relax (2) + 9 \, \log \relax (2)\right )}}{9 \, e^{6} + e^{\left (2 \, x + 12\right )} - 6 \, e^{\left (x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(3)*log(2)*exp(x)^3+2*(-6*x^2-6*x)*log(2)*exp(x)^2)/(exp(3)^3*exp(x)^3-9*exp(3)^2*exp(x)^2+2
7*exp(x)*exp(3)-27),x, algorithm="maxima")

[Out]

2*e^(-6)*log(2)*log((e^(x + 3) - 3)*e^(-3)) - 2*(e^(-6)*log((e^(x + 3) - 3)*e^(-3)) - (x*e^(2*x + 6) + 3*e^(x
+ 3) - 9)/(9*e^6 + e^(2*x + 12) - 6*e^(x + 9)))*log(2) + 2*((x^2*e^6*log(2) - x*e^6*log(2))*e^(2*x) - 3*e^(x +
3)*log(2) + 9*log(2))/(9*e^6 + e^(2*x + 12) - 6*e^(x + 9))

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mupad [B]  time = 0.23, size = 33, normalized size = 1.57 \begin {gather*} \frac {2\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^6\,\ln \relax (2)}{9\,{\mathrm {e}}^6+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{12}-6\,{\mathrm {e}}^9\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2*x)*log(2)*(6*x + 6*x^2) - 4*x*exp(3*x)*exp(3)*log(2))/(9*exp(2*x)*exp(6) - exp(3*x)*exp(9) - 27*e
xp(3)*exp(x) + 27),x)

[Out]

(2*x^2*exp(2*x)*exp(6)*log(2))/(9*exp(6) + exp(2*x)*exp(12) - 6*exp(9)*exp(x))

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sympy [B]  time = 0.15, size = 56, normalized size = 2.67 \begin {gather*} \frac {2 x^{2} \log {\relax (2 )}}{e^{6}} + \frac {12 x^{2} e^{3} e^{x} \log {\relax (2 )} - 18 x^{2} \log {\relax (2 )}}{e^{12} e^{2 x} - 6 e^{9} e^{x} + 9 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(3)*ln(2)*exp(x)**3+2*(-6*x**2-6*x)*ln(2)*exp(x)**2)/(exp(3)**3*exp(x)**3-9*exp(3)**2*exp(x)
**2+27*exp(x)*exp(3)-27),x)

[Out]

2*x**2*exp(-6)*log(2) + (12*x**2*exp(3)*exp(x)*log(2) - 18*x**2*log(2))/(exp(12)*exp(2*x) - 6*exp(9)*exp(x) +
9*exp(6))

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