3.5.88 \(\int \frac {-2 x+e^{\frac {1}{2} (2+e^x)} (4 x-e^x x^2)+4 x \log (9 x)}{e^{2+e^x}+2 e^{\frac {1}{2} (2+e^x)} \log (9 x)+\log ^2(9 x)} \, dx\)

Optimal. Leaf size=25 \[ 4+\frac {2 x^2}{e^{1+\frac {e^x}{2}}+\log (9 x)} \]

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Rubi [F]  time = 1.40, 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 x+e^{\frac {1}{2} \left (2+e^x\right )} \left (4 x-e^x x^2\right )+4 x \log (9 x)}{e^{2+e^x}+2 e^{\frac {1}{2} \left (2+e^x\right )} \log (9 x)+\log ^2(9 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-2*Defer[Int][x/(E^(1 + E^x/2) + Log[9*x])^2, x] - Defer[Int][(E^((2 + E^x + 2*x)/2)*x^2)/(E^(1 + E^x/2) + Log
[9*x])^2, x] + 4*Defer[Int][x/(E^(1 + E^x/2) + Log[9*x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x+e^{\frac {1}{2} \left (2+e^x\right )} \left (4 x-e^x x^2\right )+4 x \log (9 x)}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ &=\int \left (-\frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2}+\frac {2 x \left (-1+2 e^{1+\frac {e^x}{2}}+2 \log (9 x)\right )}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2}\right ) \, dx\\ &=2 \int \frac {x \left (-1+2 e^{1+\frac {e^x}{2}}+2 \log (9 x)\right )}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx-\int \frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ &=2 \int \left (-\frac {x}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2}+\frac {2 x}{e^{1+\frac {e^x}{2}}+\log (9 x)}\right ) \, dx-\int \frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ &=-\left (2 \int \frac {x}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\right )+4 \int \frac {x}{e^{1+\frac {e^x}{2}}+\log (9 x)} \, dx-\int \frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 x^2}{e^{1+\frac {e^x}{2}}+\log (9 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^2)/(E^(1 + E^x/2) + Log[9*x])

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fricas [A]  time = 0.89, size = 19, normalized size = 0.76 \begin {gather*} \frac {2 \, x^{2}}{e^{\left (\frac {1}{2} \, e^{x} + 1\right )} + \log \left (9 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+4*x)*exp(1/2*exp(x)+1)+4*x*log(9*x)-2*x)/(exp(1/2*exp(x)+1)^2+2*log(9*x)*exp(1/2*exp(x
)+1)+log(9*x)^2),x, algorithm="fricas")

[Out]

2*x^2/(e^(1/2*e^x + 1) + log(9*x))

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giac [B]  time = 0.62, size = 139, normalized size = 5.56 \begin {gather*} \frac {2 \, {\left (x^{3} e^{\left (2 \, x\right )} \log \left (9 \, x\right )^{2} + x^{3} e^{\left (2 \, x + \frac {1}{2} \, e^{x} + 1\right )} \log \left (9 \, x\right ) - 2 \, x^{2} e^{x} \log \left (9 \, x\right ) - 2 \, x^{2} e^{\left (x + \frac {1}{2} \, e^{x} + 1\right )}\right )}}{x e^{\left (2 \, x\right )} \log \left (9 \, x\right )^{3} + 2 \, x e^{\left (2 \, x + \frac {1}{2} \, e^{x} + 1\right )} \log \left (9 \, x\right )^{2} + x e^{\left (2 \, x + e^{x} + 2\right )} \log \left (9 \, x\right ) - 2 \, e^{x} \log \left (9 \, x\right )^{2} - 4 \, e^{\left (x + \frac {1}{2} \, e^{x} + 1\right )} \log \left (9 \, x\right ) - 2 \, e^{\left (x + e^{x} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+4*x)*exp(1/2*exp(x)+1)+4*x*log(9*x)-2*x)/(exp(1/2*exp(x)+1)^2+2*log(9*x)*exp(1/2*exp(x
)+1)+log(9*x)^2),x, algorithm="giac")

[Out]

2*(x^3*e^(2*x)*log(9*x)^2 + x^3*e^(2*x + 1/2*e^x + 1)*log(9*x) - 2*x^2*e^x*log(9*x) - 2*x^2*e^(x + 1/2*e^x + 1
))/(x*e^(2*x)*log(9*x)^3 + 2*x*e^(2*x + 1/2*e^x + 1)*log(9*x)^2 + x*e^(2*x + e^x + 2)*log(9*x) - 2*e^x*log(9*x
)^2 - 4*e^(x + 1/2*e^x + 1)*log(9*x) - 2*e^(x + e^x + 2))

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maple [A]  time = 0.10, size = 20, normalized size = 0.80

method	result	size
risch	\(\frac {2 x^{2}}{{\mathrm e}^{\frac {{\mathrm e}^{x}}{2}+1}+\ln \left (9 x \right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x^2+4*x)*exp(1/2*exp(x)+1)+4*x*ln(9*x)-2*x)/(exp(1/2*exp(x)+1)^2+2*ln(9*x)*exp(1/2*exp(x)+1)+ln(
9*x)^2),x,method=_RETURNVERBOSE)

[Out]

2*x^2/(exp(1/2*exp(x)+1)+ln(9*x))

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maxima [A]  time = 0.95, size = 21, normalized size = 0.84 \begin {gather*} \frac {2 \, x^{2}}{e^{\left (\frac {1}{2} \, e^{x} + 1\right )} + 2 \, \log \relax (3) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+4*x)*exp(1/2*exp(x)+1)+4*x*log(9*x)-2*x)/(exp(1/2*exp(x)+1)^2+2*log(9*x)*exp(1/2*exp(x
)+1)+log(9*x)^2),x, algorithm="maxima")

[Out]

2*x^2/(e^(1/2*e^x + 1) + 2*log(3) + log(x))

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mupad [B]  time = 0.60, size = 19, normalized size = 0.76 \begin {gather*} \frac {2\,x^2}{{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{2}+1}+\ln \left (9\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*log(9*x) - 2*x + exp(exp(x)/2 + 1)*(4*x - x^2*exp(x)))/(exp(exp(x) + 2) + log(9*x)^2 + 2*log(9*x)*exp
(exp(x)/2 + 1)),x)

[Out]

(2*x^2)/(exp(exp(x)/2 + 1) + log(9*x))

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sympy [A]  time = 0.31, size = 17, normalized size = 0.68 \begin {gather*} \frac {2 x^{2}}{e^{\frac {e^{x}}{2} + 1} + \log {\left (9 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x**2+4*x)*exp(1/2*exp(x)+1)+4*x*ln(9*x)-2*x)/(exp(1/2*exp(x)+1)**2+2*ln(9*x)*exp(1/2*exp(x
)+1)+ln(9*x)**2),x)

[Out]

2*x**2/(exp(exp(x)/2 + 1) + log(9*x))

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