3.63.39 \(\int \frac {9-10 e^x+e^{2 x}+(16-8 e^x x) \log (x)+(20-4 e^x) \log (\frac {3 x}{2})+4 \log ^2(\frac {3 x}{2})}{81 x-18 e^x x+e^{2 x} x+(36 x-4 e^x x) \log (\frac {3 x}{2})+4 x \log ^2(\frac {3 x}{2})} \, dx\)

Optimal. Leaf size=26 \[ \frac {\log (x)}{1+\frac {8}{1-e^x+2 \log \left (\frac {3 x}{2}\right )}} \]

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Rubi [F]  time = 4.07, 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 {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(9 - 10*E^x + E^(2*x) + (16 - 8*E^x*x)*Log[x] + (20 - 4*E^x)*Log[(3*x)/2] + 4*Log[(3*x)/2]^2)/(81*x - 18*E
^x*x + E^(2*x)*x + (36*x - 4*E^x*x)*Log[(3*x)/2] + 4*x*Log[(3*x)/2]^2),x]

[Out]

Log[x] - 8*Defer[Int][1/(x*(-E^x + 9*(1 + (2*Log[3/2])/9) + 2*Log[x])), x] + 16*Defer[Int][Log[x]/(x*(-9 + E^x
- 2*Log[(3*x)/2])^2), x] - 144*Defer[Subst][Defer[Int][Log[2*x]/(-9 + E^(2*x) - 2*Log[3*x])^2, x], x, x/2] -
16*Defer[Subst][Defer[Int][Log[2*x]/(-9 + E^(2*x) - 2*Log[3*x]), x], x, x/2] - 32*Defer[Subst][Defer[Int][(Log
[2*x]*Log[3*x])/(-9 + E^(2*x) - 2*Log[3*x])^2, x], x, x/2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9-10 e^x+e^{2 x}-8 \left (-2+e^x x\right ) \log (x)-4 \left (-5+e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{x \left (9-e^x+2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx\\ &=\int \left (\frac {1}{x}-\frac {8 (-1+x \log (x))}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )}-\frac {8 \log (x) \left (-2+9 x+2 x \log \left (\frac {3 x}{2}\right )\right )}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}\right ) \, dx\\ &=\log (x)-8 \int \frac {-1+x \log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )} \, dx-8 \int \frac {\log (x) \left (-2+9 x+2 x \log \left (\frac {3 x}{2}\right )\right )}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx\\ &=\log (x)-8 \int \left (\frac {1}{x \left (-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )+2 \log (x)\right )}+\frac {\log (x)}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )}\right ) \, dx-8 \int \left (\frac {9 \log (x)}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}-\frac {2 \log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}+\frac {2 \log (x) \log \left (\frac {3 x}{2}\right )}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}\right ) \, dx\\ &=\log (x)-8 \int \frac {1}{x \left (-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )+2 \log (x)\right )} \, dx-8 \int \frac {\log (x)}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )} \, dx+16 \int \frac {\log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx-16 \int \frac {\log (x) \log \left (\frac {3 x}{2}\right )}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx-72 \int \frac {\log (x)}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx\\ &=\log (x)-8 \int \frac {1}{x \left (-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )+2 \log (x)\right )} \, dx+16 \int \frac {\log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx-16 \operatorname {Subst}\left (\int \frac {\log (2 x)}{-9+e^{2 x}-2 \log (3 x)} \, dx,x,\frac {x}{2}\right )-32 \operatorname {Subst}\left (\int \frac {\log (2 x) \log (3 x)}{\left (-9+e^{2 x}-2 \log (3 x)\right )^2} \, dx,x,\frac {x}{2}\right )-144 \operatorname {Subst}\left (\int \frac {\log (2 x)}{\left (-9+e^{2 x}-2 \log (3 x)\right )^2} \, dx,x,\frac {x}{2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 31, normalized size = 1.19 \begin {gather*} \log (x)+\frac {4 \left (-9+e^x-\log \left (\frac {9}{4}\right )\right )}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 - 10*E^x + E^(2*x) + (16 - 8*E^x*x)*Log[x] + (20 - 4*E^x)*Log[(3*x)/2] + 4*Log[(3*x)/2]^2)/(81*x
- 18*E^x*x + E^(2*x)*x + (36*x - 4*E^x*x)*Log[(3*x)/2] + 4*x*Log[(3*x)/2]^2),x]

[Out]

Log[x] + (4*(-9 + E^x - Log[9/4]))/(-9 + E^x - 2*Log[(3*x)/2])

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fricas [A]  time = 0.93, size = 42, normalized size = 1.62 \begin {gather*} \frac {{\left (e^{x} - 2 \, \log \left (\frac {3}{2}\right ) - 9\right )} \log \relax (x) - 2 \, \log \relax (x)^{2} + 4 \, e^{x} - 8 \, \log \left (\frac {3}{2}\right ) - 36}{e^{x} - 2 \, \log \left (\frac {3}{2}\right ) - 2 \, \log \relax (x) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(3/2*x)^2+(-4*exp(x)+20)*log(3/2*x)+(-8*exp(x)*x+16)*log(x)+exp(x)^2-10*exp(x)+9)/(4*x*log(3/2
*x)^2+(-4*exp(x)*x+36*x)*log(3/2*x)+x*exp(x)^2-18*exp(x)*x+81*x),x, algorithm="fricas")

[Out]

((e^x - 2*log(3/2) - 9)*log(x) - 2*log(x)^2 + 4*e^x - 8*log(3/2) - 36)/(e^x - 2*log(3/2) - 2*log(x) - 9)

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giac [B]  time = 0.21, size = 60, normalized size = 2.31 \begin {gather*} \frac {e^{x} \log \relax (x) - 2 \, \log \relax (3) \log \relax (x) + 2 \, \log \relax (2) \log \relax (x) - 2 \, \log \relax (x)^{2} + 4 \, e^{x} - 8 \, \log \relax (3) + 8 \, \log \relax (2) - 9 \, \log \relax (x) - 36}{e^{x} - 2 \, \log \relax (3) + 2 \, \log \relax (2) - 2 \, \log \relax (x) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(3/2*x)^2+(-4*exp(x)+20)*log(3/2*x)+(-8*exp(x)*x+16)*log(x)+exp(x)^2-10*exp(x)+9)/(4*x*log(3/2
*x)^2+(-4*exp(x)*x+36*x)*log(3/2*x)+x*exp(x)^2-18*exp(x)*x+81*x),x, algorithm="giac")

[Out]

(e^x*log(x) - 2*log(3)*log(x) + 2*log(2)*log(x) - 2*log(x)^2 + 4*e^x - 8*log(3) + 8*log(2) - 9*log(x) - 36)/(e
^x - 2*log(3) + 2*log(2) - 2*log(x) - 9)

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maple [A]  time = 0.13, size = 36, normalized size = 1.38

method	result	size
risch	\(\ln \relax (x )+\frac {-36+8 \ln \relax (2)-8 \ln \relax (3)+4 \,{\mathrm e}^{x}}{-9+2 \ln \relax (2)-2 \ln \relax (3)+{\mathrm e}^{x}-2 \ln \relax (x )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(3/2*x)^2+(-4*exp(x)+20)*ln(3/2*x)+(-8*exp(x)*x+16)*ln(x)+exp(x)^2-10*exp(x)+9)/(4*x*ln(3/2*x)^2+(-4*
exp(x)*x+36*x)*ln(3/2*x)+x*exp(x)^2-18*exp(x)*x+81*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 25, normalized size = 0.96 \begin {gather*} \frac {8 \, \log \relax (x)}{e^{x} - 2 \, \log \relax (3) + 2 \, \log \relax (2) - 2 \, \log \relax (x) - 9} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(3/2*x)^2+(-4*exp(x)+20)*log(3/2*x)+(-8*exp(x)*x+16)*log(x)+exp(x)^2-10*exp(x)+9)/(4*x*log(3/2
*x)^2+(-4*exp(x)*x+36*x)*log(3/2*x)+x*exp(x)^2-18*exp(x)*x+81*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.50, size = 29, normalized size = 1.12 \begin {gather*} \frac {\ln \relax (x)\,\left (2\,\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^x+1\right )}{2\,\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^x+9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x) - 10*exp(x) - log((3*x)/2)*(4*exp(x) - 20) - log(x)*(8*x*exp(x) - 16) + 4*log((3*x)/2)^2 + 9)/(8
1*x + x*exp(2*x) + 4*x*log((3*x)/2)^2 + log((3*x)/2)*(36*x - 4*x*exp(x)) - 18*x*exp(x)),x)

[Out]

(log(x)*(2*log((3*x)/2) - exp(x) + 1))/(2*log((3*x)/2) - exp(x) + 9)

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sympy [A]  time = 0.34, size = 27, normalized size = 1.04 \begin {gather*} \log {\relax (x )} + \frac {8 \log {\relax (x )}}{e^{x} - 2 \log {\relax (x )} - 9 - 2 \log {\relax (3 )} + 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(3/2*x)**2+(-4*exp(x)+20)*ln(3/2*x)+(-8*exp(x)*x+16)*ln(x)+exp(x)**2-10*exp(x)+9)/(4*x*ln(3/2*x
)**2+(-4*exp(x)*x+36*x)*ln(3/2*x)+x*exp(x)**2-18*exp(x)*x+81*x),x)

[Out]

log(x) + 8*log(x)/(exp(x) - 2*log(x) - 9 - 2*log(3) + 2*log(2))

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