3.20.50 \(\int \frac {-5 x+(10 x+4 x^2) \log (\frac {x}{5+2 x})+(25-4 x^2) \log ^2(\frac {x}{5+2 x})}{(5+2 x) \log ^2(\frac {x}{5+2 x})} \, dx\)

Optimal. Leaf size=28 \[ -4-(2-x)^2+x+\frac {x^2}{\log \left (\frac {x}{5+2 x}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 0.20, 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 {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*x + (10*x + 4*x^2)*Log[x/(5 + 2*x)] + (25 - 4*x^2)*Log[x/(5 + 2*x)]^2)/((5 + 2*x)*Log[x/(5 + 2*x)]^2),
x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 1.57, size = 25, normalized size = 0.89 \begin {gather*} 5 x-x^2+\frac {x^2}{\log \left (\frac {x}{5+2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x + (10*x + 4*x^2)*Log[x/(5 + 2*x)] + (25 - 4*x^2)*Log[x/(5 + 2*x)]^2)/((5 + 2*x)*Log[x/(5 + 2*x
)]^2),x]

[Out]

5*x - x^2 + x^2/Log[x/(5 + 2*x)]

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 36, normalized size = 1.29 \begin {gather*} \frac {x^{2} - {\left (x^{2} - 5 \, x\right )} \log \left (\frac {x}{2 \, x + 5}\right )}{\log \left (\frac {x}{2 \, x + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+25)*log(x/(5+2*x))^2+(4*x^2+10*x)*log(x/(5+2*x))-5*x)/(5+2*x)/log(x/(5+2*x))^2,x, algorithm
="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.27, size = 110, normalized size = 3.93 \begin {gather*} \frac {25 \, {\left (\frac {8 \, x}{2 \, x + 5} - 3\right )}}{4 \, {\left (\frac {4 \, x}{2 \, x + 5} - \frac {4 \, x^{2}}{{\left (2 \, x + 5\right )}^{2}} - 1\right )}} - \frac {25 \, x^{2}}{{\left (2 \, x + 5\right )}^{2} {\left (\frac {4 \, x \log \left (\frac {x}{2 \, x + 5}\right )}{2 \, x + 5} - \frac {4 \, x^{2} \log \left (\frac {x}{2 \, x + 5}\right )}{{\left (2 \, x + 5\right )}^{2}} - \log \left (\frac {x}{2 \, x + 5}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+25)*log(x/(5+2*x))^2+(4*x^2+10*x)*log(x/(5+2*x))-5*x)/(5+2*x)/log(x/(5+2*x))^2,x, algorithm
="giac")

[Out]

25/4*(8*x/(2*x + 5) - 3)/(4*x/(2*x + 5) - 4*x^2/(2*x + 5)^2 - 1) - 25*x^2/((2*x + 5)^2*(4*x*log(x/(2*x + 5))/(
2*x + 5) - 4*x^2*log(x/(2*x + 5))/(2*x + 5)^2 - log(x/(2*x + 5))))

________________________________________________________________________________________

maple [A]  time = 0.33, size = 26, normalized size = 0.93

method	result	size
risch	\(-x^{2}+5 x +\frac {x^{2}}{\ln \left (\frac {x}{5+2 x}\right )}\)	\(26\)
norman	\(\frac {x^{2}+5 \ln \left (\frac {x}{5+2 x}\right ) x -\ln \left (\frac {x}{5+2 x}\right ) x^{2}}{\ln \left (\frac {x}{5+2 x}\right )}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2+25)*ln(x/(5+2*x))^2+(4*x^2+10*x)*ln(x/(5+2*x))-5*x)/(5+2*x)/ln(x/(5+2*x))^2,x,method=_RETURNVERBO
SE)

[Out]

-x^2+5*x+x^2/ln(x/(5+2*x))

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 44, normalized size = 1.57 \begin {gather*} -\frac {x^{2} + {\left (x^{2} - 5 \, x\right )} \log \left (2 \, x + 5\right ) - {\left (x^{2} - 5 \, x\right )} \log \relax (x)}{\log \left (2 \, x + 5\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+25)*log(x/(5+2*x))^2+(4*x^2+10*x)*log(x/(5+2*x))-5*x)/(5+2*x)/log(x/(5+2*x))^2,x, algorithm
="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.32, size = 25, normalized size = 0.89 \begin {gather*} 5\,x+\frac {x^2}{\ln \left (\frac {x}{2\,x+5}\right )}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + log(x/(2*x + 5))^2*(4*x^2 - 25) - log(x/(2*x + 5))*(10*x + 4*x^2))/(log(x/(2*x + 5))^2*(2*x + 5)),
x)

[Out]

5*x + x^2/log(x/(2*x + 5)) - x^2

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 17, normalized size = 0.61 \begin {gather*} - x^{2} + \frac {x^{2}}{\log {\left (\frac {x}{2 x + 5} \right )}} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2+25)*ln(x/(5+2*x))**2+(4*x**2+10*x)*ln(x/(5+2*x))-5*x)/(5+2*x)/ln(x/(5+2*x))**2,x)

[Out]

-x**2 + x**2/log(x/(2*x + 5)) + 5*x

________________________________________________________________________________________