3.52.18 \(\int \frac {-25-5 x+10 x^2+2 x^3+(-25-10 x+5 x^2+2 x^3+(-25-10 x) \log (x)) \log (\frac {1}{5} (5-x^2+5 \log (x)))}{5-x^2+5 \log (x)} \, dx\)

Optimal. Leaf size=19 \[ (-5-x) x \log \left (1-\frac {x^2}{5}+\log (x)\right ) \]

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Rubi [F]  time = 0.41, 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 {-25-5 x+10 x^2+2 x^3+\left (-25-10 x+5 x^2+2 x^3+(-25-10 x) \log (x)\right ) \log \left (\frac {1}{5} \left (5-x^2+5 \log (x)\right )\right )}{5-x^2+5 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25 - 5*x + 10*x^2 + 2*x^3 + (-25 - 10*x + 5*x^2 + 2*x^3 + (-25 - 10*x)*Log[x])*Log[(5 - x^2 + 5*Log[x])/
5])/(5 - x^2 + 5*Log[x]),x]

[Out]

25*Defer[Int][(-5 + x^2 - 5*Log[x])^(-1), x] + 5*Defer[Int][x/(-5 + x^2 - 5*Log[x]), x] - 10*Defer[Int][x^2/(-
5 + x^2 - 5*Log[x]), x] - 2*Defer[Int][x^3/(-5 + x^2 - 5*Log[x]), x] - 5*Defer[Int][Log[1 - x^2/5 + Log[x]], x
] - 2*Defer[Int][x*Log[1 - x^2/5 + Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {25+5 x-10 x^2-2 x^3}{-5+x^2-5 \log (x)}-(5+2 x) \log \left (1-\frac {x^2}{5}+\log (x)\right )\right ) \, dx\\ &=\int \frac {25+5 x-10 x^2-2 x^3}{-5+x^2-5 \log (x)} \, dx-\int (5+2 x) \log \left (1-\frac {x^2}{5}+\log (x)\right ) \, dx\\ &=\int \left (\frac {25}{-5+x^2-5 \log (x)}+\frac {5 x}{-5+x^2-5 \log (x)}-\frac {10 x^2}{-5+x^2-5 \log (x)}-\frac {2 x^3}{-5+x^2-5 \log (x)}\right ) \, dx-\int \left (5 \log \left (1-\frac {x^2}{5}+\log (x)\right )+2 x \log \left (1-\frac {x^2}{5}+\log (x)\right )\right ) \, dx\\ &=-\left (2 \int \frac {x^3}{-5+x^2-5 \log (x)} \, dx\right )-2 \int x \log \left (1-\frac {x^2}{5}+\log (x)\right ) \, dx+5 \int \frac {x}{-5+x^2-5 \log (x)} \, dx-5 \int \log \left (1-\frac {x^2}{5}+\log (x)\right ) \, dx-10 \int \frac {x^2}{-5+x^2-5 \log (x)} \, dx+25 \int \frac {1}{-5+x^2-5 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 18, normalized size = 0.95 \begin {gather*} -x (5+x) \log \left (1-\frac {x^2}{5}+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 5*x + 10*x^2 + 2*x^3 + (-25 - 10*x + 5*x^2 + 2*x^3 + (-25 - 10*x)*Log[x])*Log[(5 - x^2 + 5*Lo
g[x])/5])/(5 - x^2 + 5*Log[x]),x]

[Out]

-(x*(5 + x)*Log[1 - x^2/5 + Log[x]])

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fricas [A]  time = 0.56, size = 19, normalized size = 1.00 \begin {gather*} -{\left (x^{2} + 5 \, x\right )} \log \left (-\frac {1}{5} \, x^{2} + \log \relax (x) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-25)*log(x)+2*x^3+5*x^2-10*x-25)*log(log(x)-1/5*x^2+1)+2*x^3+10*x^2-5*x-25)/(5*log(x)-x^2+5)
,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.17, size = 33, normalized size = 1.74 \begin {gather*} x^{2} \log \relax (5) + 5 \, x \log \relax (5) - {\left (x^{2} + 5 \, x\right )} \log \left (-x^{2} + 5 \, \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-25)*log(x)+2*x^3+5*x^2-10*x-25)*log(log(x)-1/5*x^2+1)+2*x^3+10*x^2-5*x-25)/(5*log(x)-x^2+5)
,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 21, normalized size = 1.11

method	result	size
risch	\(\left (-x^{2}-5 x \right ) \ln \left (\ln \relax (x )-\frac {x^{2}}{5}+1\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x-25)*ln(x)+2*x^3+5*x^2-10*x-25)*ln(ln(x)-1/5*x^2+1)+2*x^3+10*x^2-5*x-25)/(5*ln(x)-x^2+5),x,method=
_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.46, size = 33, normalized size = 1.74 \begin {gather*} x^{2} \log \relax (5) + 5 \, x \log \relax (5) - {\left (x^{2} + 5 \, x\right )} \log \left (-x^{2} + 5 \, \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-25)*log(x)+2*x^3+5*x^2-10*x-25)*log(log(x)-1/5*x^2+1)+2*x^3+10*x^2-5*x-25)/(5*log(x)-x^2+5)
,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.56, size = 16, normalized size = 0.84 \begin {gather*} -x\,\ln \left (\ln \relax (x)-\frac {x^2}{5}+1\right )\,\left (x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + log(log(x) - x^2/5 + 1)*(10*x + log(x)*(10*x + 25) - 5*x^2 - 2*x^3 + 25) - 10*x^2 - 2*x^3 + 25)/(5
*log(x) - x^2 + 5),x)

[Out]

-x*log(log(x) - x^2/5 + 1)*(x + 5)

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sympy [A]  time = 0.40, size = 19, normalized size = 1.00 \begin {gather*} \left (- x^{2} - 5 x\right ) \log {\left (- \frac {x^{2}}{5} + \log {\relax (x )} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-25)*ln(x)+2*x**3+5*x**2-10*x-25)*ln(ln(x)-1/5*x**2+1)+2*x**3+10*x**2-5*x-25)/(5*ln(x)-x**2+
5),x)

[Out]

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

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