3.44.75 \(\int \frac {e^{\frac {6}{-9-5 x+3 \log (x)}} (-36-39 x-60 x^2-25 x^3+(54 x+30 x^2) \log (x)-9 x \log ^2(x))}{324 x+684 x^2+541 x^3+190 x^4+25 x^5+(-216 x-336 x^2-174 x^3-30 x^4) \log (x)+(36 x+36 x^2+9 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ 4-e+\frac {e^{\frac {2}{-4-2 x+\frac {3+x}{3}+\log (x)}}}{2+x} \]

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Rubi [F]  time = 3.70, 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 {e^{\frac {6}{-9-5 x+3 \log (x)}} \left (-36-39 x-60 x^2-25 x^3+\left (54 x+30 x^2\right ) \log (x)-9 x \log ^2(x)\right )}{324 x+684 x^2+541 x^3+190 x^4+25 x^5+\left (-216 x-336 x^2-174 x^3-30 x^4\right ) \log (x)+\left (36 x+36 x^2+9 x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(6/(-9 - 5*x + 3*Log[x]))*(-36 - 39*x - 60*x^2 - 25*x^3 + (54*x + 30*x^2)*Log[x] - 9*x*Log[x]^2))/(324*
x + 684*x^2 + 541*x^3 + 190*x^4 + 25*x^5 + (-216*x - 336*x^2 - 174*x^3 - 30*x^4)*Log[x] + (36*x + 36*x^2 + 9*x
^3)*Log[x]^2),x]

[Out]

-Defer[Int][1/(E^(6/(9 + 5*x - 3*Log[x]))*(2 + x)^2), x] - 9*Defer[Int][1/(E^(6/(9 + 5*x - 3*Log[x]))*x*(9 + 5
*x - 3*Log[x])^2), x] + 39*Defer[Int][1/(E^(6/(9 + 5*x - 3*Log[x]))*(2 + x)*(9 + 5*x - 3*Log[x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}} \left (-36-39 x-60 x^2-25 x^3+6 x (9+5 x) \log (x)-9 x \log ^2(x)\right )}{x (2+x)^2 (9+5 x-3 \log (x))^2} \, dx\\ &=\int \left (-\frac {e^{-\frac {6}{9+5 x-3 \log (x)}}}{(2+x)^2}+\frac {6 e^{-\frac {6}{9+5 x-3 \log (x)}} (-3+5 x)}{x (2+x) (9+5 x-3 \log (x))^2}\right ) \, dx\\ &=6 \int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}} (-3+5 x)}{x (2+x) (9+5 x-3 \log (x))^2} \, dx-\int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}}}{(2+x)^2} \, dx\\ &=6 \int \left (-\frac {3 e^{-\frac {6}{9+5 x-3 \log (x)}}}{2 x (9+5 x-3 \log (x))^2}+\frac {13 e^{-\frac {6}{9+5 x-3 \log (x)}}}{2 (2+x) (9+5 x-3 \log (x))^2}\right ) \, dx-\int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}}}{(2+x)^2} \, dx\\ &=-\left (9 \int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}}}{x (9+5 x-3 \log (x))^2} \, dx\right )+39 \int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}}}{(2+x) (9+5 x-3 \log (x))^2} \, dx-\int \frac {e^{-\frac {6}{9+5 x-3 \log (x)}}}{(2+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.77, size = 21, normalized size = 0.68 \begin {gather*} \frac {e^{\frac {6}{-9-5 x+3 \log (x)}}}{2+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(6/(-9 - 5*x + 3*Log[x]))*(-36 - 39*x - 60*x^2 - 25*x^3 + (54*x + 30*x^2)*Log[x] - 9*x*Log[x]^2))
/(324*x + 684*x^2 + 541*x^3 + 190*x^4 + 25*x^5 + (-216*x - 336*x^2 - 174*x^3 - 30*x^4)*Log[x] + (36*x + 36*x^2
+ 9*x^3)*Log[x]^2),x]

[Out]

E^(6/(-9 - 5*x + 3*Log[x]))/(2 + x)

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fricas [A]  time = 0.77, size = 20, normalized size = 0.65 \begin {gather*} \frac {e^{\left (-\frac {6}{5 \, x - 3 \, \log \relax (x) + 9}\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x*log(x)^2+(30*x^2+54*x)*log(x)-25*x^3-60*x^2-39*x-36)*exp(6/(3*log(x)-5*x-9))/((9*x^3+36*x^2+36
*x)*log(x)^2+(-30*x^4-174*x^3-336*x^2-216*x)*log(x)+25*x^5+190*x^4+541*x^3+684*x^2+324*x),x, algorithm="fricas
")

[Out]

e^(-6/(5*x - 3*log(x) + 9))/(x + 2)

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giac [A]  time = 0.36, size = 30, normalized size = 0.97 \begin {gather*} \frac {e^{\left (\frac {2 \, {\left (5 \, x - 3 \, \log \relax (x)\right )}}{3 \, {\left (5 \, x - 3 \, \log \relax (x) + 9\right )}} - \frac {2}{3}\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x*log(x)^2+(30*x^2+54*x)*log(x)-25*x^3-60*x^2-39*x-36)*exp(6/(3*log(x)-5*x-9))/((9*x^3+36*x^2+36
*x)*log(x)^2+(-30*x^4-174*x^3-336*x^2-216*x)*log(x)+25*x^5+190*x^4+541*x^3+684*x^2+324*x),x, algorithm="giac")

[Out]

e^(2/3*(5*x - 3*log(x))/(5*x - 3*log(x) + 9) - 2/3)/(x + 2)

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maple [A]  time = 0.03, size = 21, normalized size = 0.68

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {6}{3 \ln \relax (x )-5 x -9}}}{2+x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*x*ln(x)^2+(30*x^2+54*x)*ln(x)-25*x^3-60*x^2-39*x-36)*exp(6/(3*ln(x)-5*x-9))/((9*x^3+36*x^2+36*x)*ln(x)
^2+(-30*x^4-174*x^3-336*x^2-216*x)*ln(x)+25*x^5+190*x^4+541*x^3+684*x^2+324*x),x,method=_RETURNVERBOSE)

[Out]

1/(2+x)*exp(6/(3*ln(x)-5*x-9))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {25 \, x^{3} e^{\left (-\frac {6}{5 \, x - 3 \, \log \relax (x) + 9}\right )}}{6 \, {\left (5 \, x^{3} + 17 \, x^{2} + 8 \, x - 12\right )}} - \frac {10 \, x^{2} e^{\left (-\frac {6}{5 \, x - 3 \, \log \relax (x) + 9}\right )}}{5 \, x^{3} + 17 \, x^{2} + 8 \, x - 12} - \frac {13 \, x e^{\left (-\frac {6}{5 \, x - 3 \, \log \relax (x) + 9}\right )}}{2 \, {\left (5 \, x^{3} + 17 \, x^{2} + 8 \, x - 12\right )}} - \frac {6 \, e^{\left (-\frac {6}{5 \, x - 3 \, \log \relax (x) + 9}\right )}}{5 \, x^{3} + 17 \, x^{2} + 8 \, x - 12} - \int -\frac {3 \, {\left (2 \, {\left (5 \, x + 9\right )} \log \relax (x) - 3 \, \log \relax (x)^{2}\right )} e^{\left (-\frac {6}{5 \, x - 3 \, \log \relax (x) + 9}\right )}}{25 \, x^{4} + 190 \, x^{3} + 9 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (x)^{2} + 541 \, x^{2} - 6 \, {\left (5 \, x^{3} + 29 \, x^{2} + 56 \, x + 36\right )} \log \relax (x) + 684 \, x + 324}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x*log(x)^2+(30*x^2+54*x)*log(x)-25*x^3-60*x^2-39*x-36)*exp(6/(3*log(x)-5*x-9))/((9*x^3+36*x^2+36
*x)*log(x)^2+(-30*x^4-174*x^3-336*x^2-216*x)*log(x)+25*x^5+190*x^4+541*x^3+684*x^2+324*x),x, algorithm="maxima
")

[Out]

-25/6*x^3*e^(-6/(5*x - 3*log(x) + 9))/(5*x^3 + 17*x^2 + 8*x - 12) - 10*x^2*e^(-6/(5*x - 3*log(x) + 9))/(5*x^3
+ 17*x^2 + 8*x - 12) - 13/2*x*e^(-6/(5*x - 3*log(x) + 9))/(5*x^3 + 17*x^2 + 8*x - 12) - 6*e^(-6/(5*x - 3*log(x
) + 9))/(5*x^3 + 17*x^2 + 8*x - 12) - integrate(-3*(2*(5*x + 9)*log(x) - 3*log(x)^2)*e^(-6/(5*x - 3*log(x) + 9
))/(25*x^4 + 190*x^3 + 9*(x^2 + 4*x + 4)*log(x)^2 + 541*x^2 - 6*(5*x^3 + 29*x^2 + 56*x + 36)*log(x) + 684*x +
324), x)

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mupad [B]  time = 3.66, size = 20, normalized size = 0.65 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {6}{5\,x-3\,\ln \relax (x)+9}}}{x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-6/(5*x - 3*log(x) + 9))*(39*x + 9*x*log(x)^2 - log(x)*(54*x + 30*x^2) + 60*x^2 + 25*x^3 + 36))/(324
*x - log(x)*(216*x + 336*x^2 + 174*x^3 + 30*x^4) + log(x)^2*(36*x + 36*x^2 + 9*x^3) + 684*x^2 + 541*x^3 + 190*
x^4 + 25*x^5),x)

[Out]

exp(-6/(5*x - 3*log(x) + 9))/(x + 2)

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sympy [A]  time = 0.49, size = 15, normalized size = 0.48 \begin {gather*} \frac {e^{\frac {6}{- 5 x + 3 \log {\relax (x )} - 9}}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x*ln(x)**2+(30*x**2+54*x)*ln(x)-25*x**3-60*x**2-39*x-36)*exp(6/(3*ln(x)-5*x-9))/((9*x**3+36*x**2
+36*x)*ln(x)**2+(-30*x**4-174*x**3-336*x**2-216*x)*ln(x)+25*x**5+190*x**4+541*x**3+684*x**2+324*x),x)

[Out]

exp(6/(-5*x + 3*log(x) - 9))/(x + 2)

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