3.17.57 \(\int \frac {(5 x+4 e^3 x) \log (x)+(-19+e^3 (8-4 x)-5 x) \log (\frac {1}{5} (38+10 x+e^3 (-16+8 x)))}{(19 x+5 x^2+e^3 (-8 x+4 x^2)) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {\log \left (2 \left (3-\frac {4}{5} \left (-1+e^3 (2-x)\right )+x\right )\right )}{\log (x)} \]

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Rubi [F]  time = 0.31, 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 {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((5*x + 4*E^3*x)*Log[x] + (-19 + E^3*(8 - 4*x) - 5*x)*Log[(38 + 10*x + E^3*(-16 + 8*x))/5])/((19*x + 5*x^2
+ E^3*(-8*x + 4*x^2))*Log[x]^2),x]

[Out]

(5 + 4*E^3)*Defer[Int][1/((19 - 8*E^3 + (5 + 4*E^3)*x)*Log[x]), x] - Defer[Int][Log[(2*(19 - 8*E^3))/5 + (2*(5
+ 4*E^3)*x)/5]/(x*Log[x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {\left (5+4 e^3\right ) \log (x)}{19+4 e^3 (-2+x)+5 x}-\frac {\log \left (\frac {2}{5} \left (19+4 e^3 (-2+x)+5 x\right )\right )}{x}}{\log ^2(x)} \, dx\\ &=\int \left (\frac {5+4 e^3}{\left (19-8 e^3+\left (5+4 e^3\right ) x\right ) \log (x)}-\frac {\log \left (\frac {2}{5} \left (19-8 e^3\right )+\frac {2}{5} \left (5+4 e^3\right ) x\right )}{x \log ^2(x)}\right ) \, dx\\ &=\left (5+4 e^3\right ) \int \frac {1}{\left (19-8 e^3+\left (5+4 e^3\right ) x\right ) \log (x)} \, dx-\int \frac {\log \left (\frac {2}{5} \left (19-8 e^3\right )+\frac {2}{5} \left (5+4 e^3\right ) x\right )}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 23, normalized size = 0.88 \begin {gather*} \frac {\log \left (\frac {2}{5} \left (19+4 e^3 (-2+x)+5 x\right )\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((5*x + 4*E^3*x)*Log[x] + (-19 + E^3*(8 - 4*x) - 5*x)*Log[(38 + 10*x + E^3*(-16 + 8*x))/5])/((19*x +
5*x^2 + E^3*(-8*x + 4*x^2))*Log[x]^2),x]

[Out]

Log[(2*(19 + 4*E^3*(-2 + x) + 5*x))/5]/Log[x]

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fricas [A]  time = 0.88, size = 18, normalized size = 0.69 \begin {gather*} \frac {\log \left (\frac {8}{5} \, {\left (x - 2\right )} e^{3} + 2 \, x + \frac {38}{5}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3)+5*x)*log(x))/((4*x^2-8*x)*ex
p(3)+5*x^2+19*x)/log(x)^2,x, algorithm="fricas")

[Out]

log(8/5*(x - 2)*e^3 + 2*x + 38/5)/log(x)

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giac [A]  time = 0.25, size = 26, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (5) - \log \left (8 \, x e^{3} + 10 \, x - 16 \, e^{3} + 38\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3)+5*x)*log(x))/((4*x^2-8*x)*ex
p(3)+5*x^2+19*x)/log(x)^2,x, algorithm="giac")

[Out]

-(log(5) - log(8*x*e^3 + 10*x - 16*e^3 + 38))/log(x)

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maple [A]  time = 0.33, size = 21, normalized size = 0.81

method	result	size
risch	\(\frac {\ln \left (\frac {\left (8 x -16\right ) {\mathrm e}^{3}}{5}+2 x +\frac {38}{5}\right )}{\ln \relax (x )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x+8)*exp(3)-5*x-19)*ln(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3)+5*x)*ln(x))/((4*x^2-8*x)*exp(3)+5*x
^2+19*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(x)*ln(1/5*(8*x-16)*exp(3)+2*x+38/5)

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maxima [A]  time = 0.53, size = 30, normalized size = 1.15 \begin {gather*} -\frac {\log \relax (5) - \log \relax (2) - \log \left (x {\left (4 \, e^{3} + 5\right )} - 8 \, e^{3} + 19\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3)+5*x)*log(x))/((4*x^2-8*x)*ex
p(3)+5*x^2+19*x)/log(x)^2,x, algorithm="maxima")

[Out]

-(log(5) - log(2) - log(x*(4*e^3 + 5) - 8*e^3 + 19))/log(x)

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mupad [B]  time = 1.59, size = 20, normalized size = 0.77 \begin {gather*} \frac {\ln \left (2\,x+\frac {{\mathrm {e}}^3\,\left (8\,x-16\right )}{5}+\frac {38}{5}\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x + (exp(3)*(8*x - 16))/5 + 38/5)*(5*x + exp(3)*(4*x - 8) + 19) - log(x)*(5*x + 4*x*exp(3)))/(log(
x)^2*(19*x - exp(3)*(8*x - 4*x^2) + 5*x^2)),x)

[Out]

log(2*x + (exp(3)*(8*x - 16))/5 + 38/5)/log(x)

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sympy [A]  time = 0.45, size = 22, normalized size = 0.85 \begin {gather*} \frac {\log {\left (2 x + \left (\frac {8 x}{5} - \frac {16}{5}\right ) e^{3} + \frac {38}{5} \right )}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*exp(3)-5*x-19)*ln(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3)+5*x)*ln(x))/((4*x**2-8*x)*exp
(3)+5*x**2+19*x)/ln(x)**2,x)

[Out]

log(2*x + (8*x/5 - 16/5)*exp(3) + 38/5)/log(x)

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