3.14.8 \(\int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+(-4 x-8 e^x x-4 e^{2 x} x) \log (x)+(x+2 e^x x+e^{2 x} x) \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^5 \left (2+\frac {x}{x+e^x x}\right )}{2-\log (x)} \]

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Rubi [F]  time = 1.64, 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 {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3*E^5 + 2*E^(5 + 2*x) + E^(5 + x)*(5 - 2*x) + E^(5 + x)*x*Log[x])/(4*x + 8*E^x*x + 4*E^(2*x)*x + (-4*x -
8*E^x*x - 4*E^(2*x)*x)*Log[x] + (x + 2*E^x*x + E^(2*x)*x)*Log[x]^2),x]

[Out]

(2*E^5)/(2 - Log[x]) - 2*E^5*Defer[Int][1/((1 + E^x)*(-2 + Log[x])^2), x] + E^5*Defer[Int][1/((1 + E^x)*x*(-2
+ Log[x])^2), x] - E^5*Defer[Int][1/((1 + E^x)^2*(-2 + Log[x])), x] + E^5*Defer[Int][Log[x]/((1 + E^x)*(-2 + L
og[x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 \left (3+2 e^{2 x}+e^x (5-2 x)+e^x x \log (x)\right )}{\left (1+e^x\right )^2 x (2-\log (x))^2} \, dx\\ &=e^5 \int \frac {3+2 e^{2 x}+e^x (5-2 x)+e^x x \log (x)}{\left (1+e^x\right )^2 x (2-\log (x))^2} \, dx\\ &=e^5 \int \left (\frac {2}{x (-2+\log (x))^2}-\frac {1}{\left (1+e^x\right )^2 (-2+\log (x))}+\frac {1-2 x+x \log (x)}{\left (1+e^x\right ) x (-2+\log (x))^2}\right ) \, dx\\ &=-\left (e^5 \int \frac {1}{\left (1+e^x\right )^2 (-2+\log (x))} \, dx\right )+e^5 \int \frac {1-2 x+x \log (x)}{\left (1+e^x\right ) x (-2+\log (x))^2} \, dx+\left (2 e^5\right ) \int \frac {1}{x (-2+\log (x))^2} \, dx\\ &=-\left (e^5 \int \frac {1}{\left (1+e^x\right )^2 (-2+\log (x))} \, dx\right )+e^5 \int \left (-\frac {2}{\left (1+e^x\right ) (-2+\log (x))^2}+\frac {1}{\left (1+e^x\right ) x (-2+\log (x))^2}+\frac {\log (x)}{\left (1+e^x\right ) (-2+\log (x))^2}\right ) \, dx+\left (2 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-2+\log (x)\right )\\ &=\frac {2 e^5}{2-\log (x)}+e^5 \int \frac {1}{\left (1+e^x\right ) x (-2+\log (x))^2} \, dx-e^5 \int \frac {1}{\left (1+e^x\right )^2 (-2+\log (x))} \, dx+e^5 \int \frac {\log (x)}{\left (1+e^x\right ) (-2+\log (x))^2} \, dx-\left (2 e^5\right ) \int \frac {1}{\left (1+e^x\right ) (-2+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 25, normalized size = 1.00 \begin {gather*} -\frac {e^5 \left (3+2 e^x\right )}{\left (1+e^x\right ) (-2+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^5 + 2*E^(5 + 2*x) + E^(5 + x)*(5 - 2*x) + E^(5 + x)*x*Log[x])/(4*x + 8*E^x*x + 4*E^(2*x)*x + (-
4*x - 8*E^x*x - 4*E^(2*x)*x)*Log[x] + (x + 2*E^x*x + E^(2*x)*x)*Log[x]^2),x]

[Out]

-((E^5*(3 + 2*E^x))/((1 + E^x)*(-2 + Log[x])))

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fricas [A]  time = 0.97, size = 36, normalized size = 1.44 \begin {gather*} -\frac {3 \, e^{10} + 2 \, e^{\left (x + 10\right )}}{{\left (e^{5} + e^{\left (x + 5\right )}\right )} \log \relax (x) - 2 \, e^{5} - 2 \, e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(-2*x+5)*exp(5)*exp(x)+3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x
)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm="fricas")

[Out]

-(3*e^10 + 2*e^(x + 10))/((e^5 + e^(x + 5))*log(x) - 2*e^5 - 2*e^(x + 5))

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giac [A]  time = 0.33, size = 28, normalized size = 1.12 \begin {gather*} -\frac {3 \, e^{5} + 2 \, e^{\left (x + 5\right )}}{e^{x} \log \relax (x) - 2 \, e^{x} + \log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(-2*x+5)*exp(5)*exp(x)+3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x
)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm="giac")

[Out]

-(3*e^5 + 2*e^(x + 5))/(e^x*log(x) - 2*e^x + log(x) - 2)

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maple [A]  time = 0.16, size = 23, normalized size = 0.92

method	result	size
risch	\(-\frac {\left (3+2 \,{\mathrm e}^{x}\right ) {\mathrm e}^{5}}{\left ({\mathrm e}^{x}+1\right ) \left (\ln \relax (x )-2\right )}\)	\(23\)
norman	\(\frac {-2 \,{\mathrm e}^{5} {\mathrm e}^{x}-3 \,{\mathrm e}^{5}}{\left (\ln \relax (x )-2\right ) \left ({\mathrm e}^{x}+1\right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(5)*exp(x)*ln(x)+2*exp(5)*exp(x)^2+(-2*x+5)*exp(5)*exp(x)+3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*ln(x)
^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*ln(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x,method=_RETURNVERBOSE)

[Out]

-(3+2*exp(x))*exp(5)/(exp(x)+1)/(ln(x)-2)

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maxima [A]  time = 0.46, size = 26, normalized size = 1.04 \begin {gather*} -\frac {3 \, e^{5} + 2 \, e^{\left (x + 5\right )}}{{\left (\log \relax (x) - 2\right )} e^{x} + \log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(-2*x+5)*exp(5)*exp(x)+3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x
)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm="maxima")

[Out]

-(3*e^5 + 2*e^(x + 5))/((log(x) - 2)*e^x + log(x) - 2)

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mupad [B]  time = 1.38, size = 100, normalized size = 4.00 \begin {gather*} \frac {\frac {{\mathrm {e}}^5}{2}+\frac {x\,{\mathrm {e}}^5}{2}}{{\mathrm {e}}^x+1}-\frac {\frac {{\mathrm {e}}^5\,\left (2\,{\mathrm {e}}^{2\,x}+5\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x+3\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}+\frac {x\,{\mathrm {e}}^{x+5}\,\ln \relax (x)}{{\left ({\mathrm {e}}^x+1\right )}^2}}{\ln \relax (x)-2}-\frac {\frac {{\mathrm {e}}^5\,\left (x+1\right )}{2}-{\mathrm {e}}^{x+5}\,\left (\frac {x}{2}-\frac {1}{2}\right )}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(5) + 2*exp(2*x)*exp(5) - exp(5)*exp(x)*(2*x - 5) + x*exp(5)*exp(x)*log(x))/(4*x + 4*x*exp(2*x) - lo
g(x)*(4*x + 4*x*exp(2*x) + 8*x*exp(x)) + log(x)^2*(x + x*exp(2*x) + 2*x*exp(x)) + 8*x*exp(x)),x)

[Out]

(exp(5)/2 + (x*exp(5))/2)/(exp(x) + 1) - ((exp(5)*(2*exp(2*x) + 5*exp(x) - 2*x*exp(x) + 3))/(exp(x) + 1)^2 + (
x*exp(x + 5)*log(x))/(exp(x) + 1)^2)/(log(x) - 2) - ((exp(5)*(x + 1))/2 - exp(x + 5)*(x/2 - 1/2))/(exp(2*x) +
2*exp(x) + 1)

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sympy [A]  time = 0.30, size = 27, normalized size = 1.08 \begin {gather*} - \frac {e^{5}}{\left (\log {\relax (x )} - 2\right ) e^{x} + \log {\relax (x )} - 2} - \frac {2 e^{5}}{\log {\relax (x )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(5)*exp(x)*ln(x)+2*exp(5)*exp(x)**2+(-2*x+5)*exp(5)*exp(x)+3*exp(5))/((x*exp(x)**2+2*exp(x)*x+
x)*ln(x)**2+(-4*x*exp(x)**2-8*exp(x)*x-4*x)*ln(x)+4*x*exp(x)**2+8*exp(x)*x+4*x),x)

[Out]

-exp(5)/((log(x) - 2)*exp(x) + log(x) - 2) - 2*exp(5)/(log(x) - 2)

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