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

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

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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 = {} $$\begin{array}{c}\int \frac{2e^{3+3x}x\log (4)+e^{2x}(-6x-6x^2)\log (4)}{-27+27e^{3+x}-9e^{6+2x}+e^{9+3x}}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2e^{2x}x(-e^{3+x}+3(1+x))\log (4)}{{(3-e^{3+x})}^3}dx\\ & =(2\log (4))\int \frac{e^{2x}x(-e^{3+x}+3(1+x))}{{(3-e^{3+x})}^3}dx\\ & =(2\log (4))\int (\frac{e^{2x}x}{{(-3+e^{3+x})}^2}-\frac{3e^{2x}{x}^2}{{(-3+e^{3+x})}^3})dx\\ & =(2\log (4))\int \frac{e^{2x}x}{{(-3+e^{3+x})}^2}dx-(6\log (4))\int \frac{e^{2x}{x}^2}{{(-3+e^{3+x})}^3}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{2x^2{e}^{(2x+6)}\log (2)}{9e^6+e^{(2x+12)}-6e^{(x+9)}}\end{array}$$

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 $$\begin{array}{c}\frac{2x^2{e}^{\left(2x\right)}\log (2)}{e^{(2x+6)}-6e^{(x+3)}+9}\end{array}$$

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	$\frac{2x^2\ln (2){\mathrm{e}}^{2x}}{{({\mathrm{e}}^x{\mathrm{e}}^3-3)}^2}$	$21$
risch	$2{\mathrm{e}}^{-6}{x}^2\ln (2)+\frac{6\ln (2)x^2(2{\mathrm{e}}^{3+x}-3){\mathrm{e}}^{-6}}{{({\mathrm{e}}^{3+x}-3)}^2}$	$36$

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 $$\begin{array}{c}2e^{(-6)}\log (2)\log \left((e^{(x+3)}-3)e^{(-3)}\right)-2(e^{(-6)}\log \left((e^{(x+3)}-3)e^{(-3)}\right)-\frac{x{e}^{(2x+6)}+3e^{(x+3)}-9}{9e^6+e^{(2x+12)}-6e^{(x+9)}})\log (2)+\frac{2((x^2{e}^6\log (2)-x{e}^6\log (2))e^{\left(2x\right)}-3e^{(x+3)}\log (2)+9\log (2))}{9e^6+e^{(2x+12)}-6e^{(x+9)}}\end{array}$$

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 $$\begin{array}{c}\frac{2x^2{\mathrm{e}}^{2x}{\mathrm{e}}^6\ln (2)}{9{\mathrm{e}}^6+{\mathrm{e}}^{2x}{\mathrm{e}}^{12}-6{\mathrm{e}}^9{\mathrm{e}}^x}\end{array}$$

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 $$\begin{array}{c}\frac{2x^2\log (2)}{e^6}+\frac{12x^2{e}^3{e}^x\log (2)-18x^2\log (2)}{e^{12}{e}^{2x}-6e^9{e}^x+9e^6}\end{array}$$

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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