3.34.60 \(\int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+(40 x-10 x^2) \log (x)+5 x \log ^2(x)} \, dx\)

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

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Rubi [A]  time = 0.28, antiderivative size = 40, normalized size of antiderivative = 1.54, number of steps used = 5, number of rules used = 3, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6, 6742, 6684} \begin {gather*} 2 \log (x)-\log \left (5 x^2-\left (39-e^2\right ) x+5 \log ^2(x)-10 x \log (x)+40 \log (x)+80\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120 - 29*x + E^2*x + (70 - 10*x)*Log[x] + 10*Log[x]^2)/(80*x - 39*x^2 + E^2*x^2 + 5*x^3 + (40*x - 10*x^2)
*Log[x] + 5*x*Log[x]^2),x]

[Out]

2*Log[x] - Log[80 - (39 - E^2)*x + 5*x^2 + 40*Log[x] - 10*x*Log[x] + 5*Log[x]^2]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {120+\left (-29+e^2\right ) x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx\\ &=\int \frac {120+\left (-29+e^2\right ) x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x+\left (-39+e^2\right ) x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx\\ &=\int \left (\frac {2}{x}+\frac {-40+49 \left (1-\frac {e^2}{49}\right ) x-10 x^2-10 \log (x)+10 x \log (x)}{x \left (80-39 \left (1-\frac {e^2}{39}\right ) x+5 x^2+40 \log (x)-10 x \log (x)+5 \log ^2(x)\right )}\right ) \, dx\\ &=2 \log (x)+\int \frac {-40+49 \left (1-\frac {e^2}{49}\right ) x-10 x^2-10 \log (x)+10 x \log (x)}{x \left (80-39 \left (1-\frac {e^2}{39}\right ) x+5 x^2+40 \log (x)-10 x \log (x)+5 \log ^2(x)\right )} \, dx\\ &=2 \log (x)-\log \left (80-\left (39-e^2\right ) x+5 x^2+40 \log (x)-10 x \log (x)+5 \log ^2(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 38, normalized size = 1.46 \begin {gather*} 2 \log (x)-\log \left (80-39 x+e^2 x+5 x^2+40 \log (x)-10 x \log (x)+5 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120 - 29*x + E^2*x + (70 - 10*x)*Log[x] + 10*Log[x]^2)/(80*x - 39*x^2 + E^2*x^2 + 5*x^3 + (40*x - 1
0*x^2)*Log[x] + 5*x*Log[x]^2),x]

[Out]

2*Log[x] - Log[80 - 39*x + E^2*x + 5*x^2 + 40*Log[x] - 10*x*Log[x] + 5*Log[x]^2]

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fricas [A]  time = 0.48, size = 35, normalized size = 1.35 \begin {gather*} -\log \left (5 \, x^{2} + x e^{2} - 10 \, {\left (x - 4\right )} \log \relax (x) + 5 \, \log \relax (x)^{2} - 39 \, x + 80\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+(-10*x^2+40*x)*log(x)+x^2*exp(2)+5*x
^3-39*x^2+80*x),x, algorithm="fricas")

[Out]

-log(5*x^2 + x*e^2 - 10*(x - 4)*log(x) + 5*log(x)^2 - 39*x + 80) + 2*log(x)

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giac [A]  time = 0.34, size = 38, normalized size = 1.46 \begin {gather*} -\log \left (-5 \, x^{2} - x e^{2} + 10 \, x \log \relax (x) - 5 \, \log \relax (x)^{2} + 39 \, x - 40 \, \log \relax (x) - 80\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+(-10*x^2+40*x)*log(x)+x^2*exp(2)+5*x
^3-39*x^2+80*x),x, algorithm="giac")

[Out]

-log(-5*x^2 - x*e^2 + 10*x*log(x) - 5*log(x)^2 + 39*x - 40*log(x) - 80) + 2*log(x)

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maple [A]  time = 0.05, size = 34, normalized size = 1.31

method	result	size
risch	\(2 \ln \relax (x )-\ln \left (\ln \relax (x )^{2}+\left (-2 x +8\right ) \ln \relax (x )+\frac {{\mathrm e}^{2} x}{5}+x^{2}-\frac {39 x}{5}+16\right )\)	\(34\)
norman	\(2 \ln \relax (x )-\ln \left ({\mathrm e}^{2} x +5 \ln \relax (x )^{2}-10 x \ln \relax (x )+5 x^{2}+40 \ln \relax (x )-39 x +80\right )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(x)^2+(-10*x+70)*ln(x)+exp(2)*x-29*x+120)/(5*x*ln(x)^2+(-10*x^2+40*x)*ln(x)+x^2*exp(2)+5*x^3-39*x^2+
80*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)-ln(ln(x)^2+(-2*x+8)*ln(x)+1/5*exp(2)*x+x^2-39/5*x+16)

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maxima [A]  time = 0.51, size = 31, normalized size = 1.19 \begin {gather*} -\log \left (x^{2} + \frac {1}{5} \, x {\left (e^{2} - 39\right )} - 2 \, {\left (x - 4\right )} \log \relax (x) + \log \relax (x)^{2} + 16\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+(-10*x^2+40*x)*log(x)+x^2*exp(2)+5*x
^3-39*x^2+80*x),x, algorithm="maxima")

[Out]

-log(x^2 + 1/5*x*(e^2 - 39) - 2*(x - 4)*log(x) + log(x)^2 + 16) + 2*log(x)

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mupad [B]  time = 2.12, size = 34, normalized size = 1.31 \begin {gather*} 2\,\ln \relax (x)-\ln \left (8\,\ln \relax (x)-\frac {39\,x}{5}+\frac {x\,{\mathrm {e}}^2}{5}+{\ln \relax (x)}^2-2\,x\,\ln \relax (x)+x^2+16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(2) - 29*x + 10*log(x)^2 - log(x)*(10*x - 70) + 120)/(80*x + 5*x*log(x)^2 + x^2*exp(2) + log(x)*(40*
x - 10*x^2) - 39*x^2 + 5*x^3),x)

[Out]

2*log(x) - log(8*log(x) - (39*x)/5 + (x*exp(2))/5 + log(x)^2 - 2*x*log(x) + x^2 + 16)

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sympy [A]  time = 0.23, size = 36, normalized size = 1.38 \begin {gather*} 2 \log {\relax (x )} - \log {\left (x^{2} - \frac {39 x}{5} + \frac {x e^{2}}{5} + \left (8 - 2 x\right ) \log {\relax (x )} + \log {\relax (x )}^{2} + 16 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*ln(x)**2+(-10*x+70)*ln(x)+exp(2)*x-29*x+120)/(5*x*ln(x)**2+(-10*x**2+40*x)*ln(x)+x**2*exp(2)+5*x
**3-39*x**2+80*x),x)

[Out]

2*log(x) - log(x**2 - 39*x/5 + x*exp(2)/5 + (8 - 2*x)*log(x) + log(x)**2 + 16)

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