3.35.99 \(\int \frac {9 e^x-9 x+(e^x (-9-18 x)+27 x) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx\)

Optimal. Leaf size=20 \[ \frac {9 \log (2 x)}{2 x \left (-e^x+x\right )^2} \]

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Rubi [F]  time = 1.35, 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 {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(9*E^x - 9*x + (E^x*(-9 - 18*x) + 27*x)*Log[2*x])/(2*E^(3*x)*x^2 - 6*E^(2*x)*x^3 + 6*E^x*x^4 - 2*x^5),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 20, normalized size = 1.00 \begin {gather*} \frac {9 \log (2 x)}{2 \left (e^x-x\right )^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E^x - 9*x + (E^x*(-9 - 18*x) + 27*x)*Log[2*x])/(2*E^(3*x)*x^2 - 6*E^(2*x)*x^3 + 6*E^x*x^4 - 2*x^5
),x]

[Out]

(9*Log[2*x])/(2*(E^x - x)^2*x)

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fricas [A]  time = 0.99, size = 25, normalized size = 1.25 \begin {gather*} \frac {9 \, \log \left (2 \, x\right )}{2 \, {\left (x^{3} - 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-9)*exp(x)+27*x)*log(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x
, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 26, normalized size = 1.30 \begin {gather*} \frac {9 \, {\left (\log \relax (2) + \log \relax (x)\right )}}{2 \, {\left (x^{3} - 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-9)*exp(x)+27*x)*log(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x
, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 18, normalized size = 0.90

method	result	size
risch	\(\frac {9 \ln \left (2 x \right )}{2 x \left (x -{\mathrm e}^{x}\right )^{2}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-18*x-9)*exp(x)+27*x)*ln(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x,method
=_RETURNVERBOSE)

[Out]

9/2*ln(2*x)/x/(x-exp(x))^2

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maxima [A]  time = 0.70, size = 26, normalized size = 1.30 \begin {gather*} \frac {9 \, {\left (\log \relax (2) + \log \relax (x)\right )}}{2 \, {\left (x^{3} - 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-9)*exp(x)+27*x)*log(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x
, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.34, size = 28, normalized size = 1.40 \begin {gather*} \frac {9\,\ln \left (2\,x\right )}{2\,x\,{\mathrm {e}}^{2\,x}-4\,x^2\,{\mathrm {e}}^x+2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*exp(x) - 9*x + log(2*x)*(27*x - exp(x)*(18*x + 9)))/(6*x^4*exp(x) + 2*x^2*exp(3*x) - 6*x^3*exp(2*x) - 2
*x^5),x)

[Out]

(9*log(2*x))/(2*x*exp(2*x) - 4*x^2*exp(x) + 2*x^3)

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sympy [A]  time = 0.27, size = 27, normalized size = 1.35 \begin {gather*} \frac {9 \log {\left (2 x \right )}}{2 x^{3} - 4 x^{2} e^{x} + 2 x e^{2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-9)*exp(x)+27*x)*ln(2*x)+9*exp(x)-9*x)/(2*x**2*exp(x)**3-6*exp(x)**2*x**3+6*exp(x)*x**4-2*x*
*5),x)

[Out]

9*log(2*x)/(2*x**3 - 4*x**2*exp(x) + 2*x*exp(2*x))

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