3.19.74 \(\int \frac {-10-8 x+2 x^2+(15+17 x-4 x^2) \log ^2(-\frac {x}{-5+x})+(-5-9 x+2 x^2) \log ^4(-\frac {x}{-5+x})+\log (x) (-5+x-10 \log (-\frac {x}{-5+x})+(5-x) \log ^2(-\frac {x}{-5+x}))}{-5+x+(10-2 x) \log ^2(-\frac {x}{-5+x})+(-5+x) \log ^4(-\frac {x}{-5+x})} \, dx\) [1874]

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

[Out]

x+x^2-x*ln(x)/(ln(x/(5-x))^2-1)

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Rubi [F]
time = 0.51, 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 {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 - 8*x + 2*x^2 + (15 + 17*x - 4*x^2)*Log[-(x/(-5 + x))]^2 + (-5 - 9*x + 2*x^2)*Log[-(x/(-5 + x))]^4 +
Log[x]*(-5 + x - 10*Log[-(x/(-5 + x))] + (5 - x)*Log[-(x/(-5 + x))]^2))/(-5 + x + (10 - 2*x)*Log[-(x/(-5 + x))
]^2 + (-5 + x)*Log[-(x/(-5 + x))]^4),x]

[Out]

x + x^2 + Defer[Int][(1 - Log[-(x/(-5 + x))])^(-1), x]/2 - (5*Defer[Int][Log[x]/((-5 + x)*(-1 + Log[-(x/(-5 +
x))])^2), x])/2 - Defer[Int][Log[x]/(-1 + Log[-(x/(-5 + x))]), x]/2 + (5*Defer[Int][Log[x]/((-5 + x)*(1 + Log[
-(x/(-5 + x))])^2), x])/2 + Defer[Int][(1 + Log[-(x/(-5 + x))])^(-1), x]/2 + Defer[Int][Log[x]/(1 + Log[-(x/(-
5 + x))]), x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10+8 x-2 x^2-\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )-\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )-\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{(5-x) \left (1-\log ^2\left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ &=\int \left (1+2 x-\frac {5 \log (x)}{2 (-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2}+\frac {-1-\log (x)}{2 \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )}+\frac {5 \log (x)}{2 (-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2}+\frac {1+\log (x)}{2 \left (1+\log \left (-\frac {x}{-5+x}\right )\right )}\right ) \, dx\\ &=x+x^2+\frac {1}{2} \int \frac {-1-\log (x)}{-1+\log \left (-\frac {x}{-5+x}\right )} \, dx+\frac {1}{2} \int \frac {1+\log (x)}{1+\log \left (-\frac {x}{-5+x}\right )} \, dx-\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ &=x+x^2+\frac {1}{2} \int \left (\frac {1}{1-\log \left (-\frac {x}{-5+x}\right )}-\frac {\log (x)}{-1+\log \left (-\frac {x}{-5+x}\right )}\right ) \, dx+\frac {1}{2} \int \left (\frac {1}{1+\log \left (-\frac {x}{-5+x}\right )}+\frac {\log (x)}{1+\log \left (-\frac {x}{-5+x}\right )}\right ) \, dx-\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ &=x+x^2+\frac {1}{2} \int \frac {1}{1-\log \left (-\frac {x}{-5+x}\right )} \, dx-\frac {1}{2} \int \frac {\log (x)}{-1+\log \left (-\frac {x}{-5+x}\right )} \, dx+\frac {1}{2} \int \frac {1}{1+\log \left (-\frac {x}{-5+x}\right )} \, dx+\frac {1}{2} \int \frac {\log (x)}{1+\log \left (-\frac {x}{-5+x}\right )} \, dx-\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.06, size = 25, normalized size = 0.96 \begin {gather*} x+x^2-\frac {x \log (x)}{-1+\log ^2\left (-\frac {x}{-5+x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 8*x + 2*x^2 + (15 + 17*x - 4*x^2)*Log[-(x/(-5 + x))]^2 + (-5 - 9*x + 2*x^2)*Log[-(x/(-5 + x))
]^4 + Log[x]*(-5 + x - 10*Log[-(x/(-5 + x))] + (5 - x)*Log[-(x/(-5 + x))]^2))/(-5 + x + (10 - 2*x)*Log[-(x/(-5
+ x))]^2 + (-5 + x)*Log[-(x/(-5 + x))]^4),x]

[Out]

x + x^2 - (x*Log[x])/(-1 + Log[-(x/(-5 + x))]^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 39.07, size = 740, normalized size = 28.46

method	result	size
risch	\(x^{2}+x -\frac {4 x \ln \left (x \right )}{-4-4 \pi ^{2}+4 \ln \left (x -5\right )^{2}+4 \ln \left (x \right )^{2}-4 i \ln \left (x \right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )+4 i \ln \left (x -5\right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )+4 i \ln \left (x \right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+4 i \ln \left (x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 i \ln \left (x -5\right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 i \ln \left (x -5\right ) \pi \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right )^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-8 \ln \left (x -5\right ) \ln \left (x \right )-4 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}+8 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+4 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{5}-\pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{6}-4 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-\pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}-2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{5}+4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}-\pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}-4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-2 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{5}-8 i \pi \ln \left (x -5\right )+8 i \pi \ln \left (x \right )+2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}+4 i \ln \left (x \right ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-8 i \ln \left (x \right ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+8 i \ln \left (x -5\right ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 i \ln \left (x -5\right ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}+4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )-\pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right )^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}}\)	\(740\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5-x)*ln(-x/(x-5))^2-10*ln(-x/(x-5))+x-5)*ln(x)+(2*x^2-9*x-5)*ln(-x/(x-5))^4+(-4*x^2+17*x+15)*ln(-x/(x-5
))^2+2*x^2-8*x-10)/((x-5)*ln(-x/(x-5))^4+(-2*x+10)*ln(-x/(x-5))^2+x-5),x,method=_RETURNVERBOSE)

[Out]

x^2+x-4*x*ln(x)/(-4-4*Pi^2+4*ln(x-5)^2+4*ln(x)^2-Pi^2*csgn(I/(x-5))^2*csgn(I*x/(x-5))^4-2*Pi^2*csgn(I/(x-5))*c
sgn(I*x/(x-5))^5+2*Pi^2*csgn(I/(x-5))*csgn(I*x)^2*csgn(I*x/(x-5))^3-4*Pi^2*csgn(I/(x-5))*csgn(I*x)*csgn(I*x/(x
-5))^3+4*Pi^2*csgn(I/(x-5))*csgn(I*x/(x-5))^4-Pi^2*csgn(I*x)^2*csgn(I*x/(x-5))^4+4*Pi^2*csgn(I*x)*csgn(I*x/(x-
5))^4-4*Pi^2*csgn(I/(x-5))*csgn(I*x/(x-5))^2-4*Pi^2*csgn(I*x)*csgn(I*x/(x-5))^2-2*Pi^2*csgn(I*x)*csgn(I*x/(x-5
))^5+4*I*ln(x)*Pi*csgn(I*x/(x-5))^3-8*I*ln(x)*Pi*csgn(I*x/(x-5))^2+8*I*ln(x-5)*Pi*csgn(I*x/(x-5))^2-4*I*ln(x-5
)*Pi*csgn(I*x/(x-5))^3-4*I*ln(x)*Pi*csgn(I*x)*csgn(I/(x-5))*csgn(I*x/(x-5))+4*I*ln(x-5)*Pi*csgn(I*x)*csgn(I/(x
-5))*csgn(I*x/(x-5))-8*ln(x-5)*ln(x)-4*Pi^2*csgn(I*x/(x-5))^4+8*Pi^2*csgn(I*x/(x-5))^2+4*Pi^2*csgn(I*x/(x-5))^
5-Pi^2*csgn(I*x/(x-5))^6-4*Pi^2*csgn(I*x/(x-5))^3-8*I*Pi*ln(x-5)+8*I*Pi*ln(x)+4*I*ln(x)*Pi*csgn(I*x)*csgn(I*x/
(x-5))^2+4*I*ln(x)*Pi*csgn(I/(x-5))*csgn(I*x/(x-5))^2-4*I*ln(x-5)*Pi*csgn(I*x)*csgn(I*x/(x-5))^2-4*I*ln(x-5)*P
i*csgn(I/(x-5))*csgn(I*x/(x-5))^2+4*Pi^2*csgn(I/(x-5))*csgn(I*x)*csgn(I*x/(x-5))-Pi^2*csgn(I/(x-5))^2*csgn(I*x
)^2*csgn(I*x/(x-5))^2+2*Pi^2*csgn(I/(x-5))^2*csgn(I*x)*csgn(I*x/(x-5))^3)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (25) = 50\).
time = 0.32, size = 80, normalized size = 3.08 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \left (x\right )^{2} - 2 \, {\left (x^{2} + x\right )} \log \left (x\right ) \log \left (-x + 5\right ) + {\left (x^{2} + x\right )} \log \left (-x + 5\right )^{2} - x^{2} - x \log \left (x\right ) - x}{\log \left (x\right )^{2} - 2 \, \log \left (x\right ) \log \left (-x + 5\right ) + \log \left (-x + 5\right )^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5-x)*log(-x/(-5+x))^2-10*log(-x/(-5+x))+x-5)*log(x)+(2*x^2-9*x-5)*log(-x/(-5+x))^4+(-4*x^2+17*x+1
5)*log(-x/(-5+x))^2+2*x^2-8*x-10)/((-5+x)*log(-x/(-5+x))^4+(-2*x+10)*log(-x/(-5+x))^2+x-5),x, algorithm="maxim
a")

[Out]

((x^2 + x)*log(x)^2 - 2*(x^2 + x)*log(x)*log(-x + 5) + (x^2 + x)*log(-x + 5)^2 - x^2 - x*log(x) - x)/(log(x)^2
- 2*log(x)*log(-x + 5) + log(-x + 5)^2 - 1)

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Fricas [A]
time = 0.37, size = 47, normalized size = 1.81 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \left (-\frac {x}{x - 5}\right )^{2} - x^{2} - x \log \left (x\right ) - x}{\log \left (-\frac {x}{x - 5}\right )^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5-x)*log(-x/(-5+x))^2-10*log(-x/(-5+x))+x-5)*log(x)+(2*x^2-9*x-5)*log(-x/(-5+x))^4+(-4*x^2+17*x+1
5)*log(-x/(-5+x))^2+2*x^2-8*x-10)/((-5+x)*log(-x/(-5+x))^4+(-2*x+10)*log(-x/(-5+x))^2+x-5),x, algorithm="frica
s")

[Out]

((x^2 + x)*log(-x/(x - 5))^2 - x^2 - x*log(x) - x)/(log(-x/(x - 5))^2 - 1)

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Sympy [A]
time = 0.07, size = 20, normalized size = 0.77 \begin {gather*} x^{2} + x - \frac {x \log {\left (x \right )}}{\log {\left (- \frac {x}{x - 5} \right )}^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5-x)*ln(-x/(-5+x))**2-10*ln(-x/(-5+x))+x-5)*ln(x)+(2*x**2-9*x-5)*ln(-x/(-5+x))**4+(-4*x**2+17*x+1
5)*ln(-x/(-5+x))**2+2*x**2-8*x-10)/((-5+x)*ln(-x/(-5+x))**4+(-2*x+10)*ln(-x/(-5+x))**2+x-5),x)

[Out]

x**2 + x - x*log(x)/(log(-x/(x - 5))**2 - 1)

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Giac [C] Result contains complex when optimal does not.
time = 0.49, size = 50, normalized size = 1.92 \begin {gather*} x^{2} + x + \frac {x \log \left (x\right )}{\pi ^{2} + 2 i \, \pi \log \left (x - 5\right ) - \log \left (x - 5\right )^{2} - 2 i \, \pi \log \left (x\right ) + 2 \, \log \left (x - 5\right ) \log \left (x\right ) - \log \left (x\right )^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5-x)*log(-x/(-5+x))^2-10*log(-x/(-5+x))+x-5)*log(x)+(2*x^2-9*x-5)*log(-x/(-5+x))^4+(-4*x^2+17*x+1
5)*log(-x/(-5+x))^2+2*x^2-8*x-10)/((-5+x)*log(-x/(-5+x))^4+(-2*x+10)*log(-x/(-5+x))^2+x-5),x, algorithm="giac"
)

[Out]

x^2 + x + x*log(x)/(pi^2 + 2*I*pi*log(x - 5) - log(x - 5)^2 - 2*I*pi*log(x) + 2*log(x - 5)*log(x) - log(x)^2 +
1)

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Mupad [B]
time = 1.45, size = 37, normalized size = 1.42 \begin {gather*} x+x^2+\frac {\ln \left (x\right )\,\left (5\,x-x^2\right )}{\left ({\ln \left (-\frac {x}{x-5}\right )}^2-1\right )\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + log(x)*(10*log(-x/(x - 5)) - x + log(-x/(x - 5))^2*(x - 5) + 5) + log(-x/(x - 5))^4*(9*x - 2*x^2 +
5) - log(-x/(x - 5))^2*(17*x - 4*x^2 + 15) - 2*x^2 + 10)/(x + log(-x/(x - 5))^4*(x - 5) - log(-x/(x - 5))^2*(
2*x - 10) - 5),x)

[Out]

x + x^2 + (log(x)*(5*x - x^2))/((log(-x/(x - 5))^2 - 1)*(x - 5))

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