3.27.24 \(\int \frac {1-x^2+\log (x)}{20+20 x+25 x^2+10 x^3+5 x^4+(20 x+10 x^2+10 x^3) \log (3)+5 x^2 \log ^2(3)+(20 x+10 x^2+10 x^3+10 x^2 \log (3)) \log (25)+5 x^2 \log ^2(25)+(20+10 x+10 x^2+10 x \log (3)+10 x \log (25)) \log (x)+5 \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {x}{5 \left (2+x+x^2+x (\log (3)+\log (25))+\log (x)\right )} \]

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Rubi [F]  time = 0.52, 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 {1-x^2+\log (x)}{20+20 x+25 x^2+10 x^3+5 x^4+\left (20 x+10 x^2+10 x^3\right ) \log (3)+5 x^2 \log ^2(3)+\left (20 x+10 x^2+10 x^3+10 x^2 \log (3)\right ) \log (25)+5 x^2 \log ^2(25)+\left (20+10 x+10 x^2+10 x \log (3)+10 x \log (25)\right ) \log (x)+5 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 - x^2 + Log[x])/(20 + 20*x + 25*x^2 + 10*x^3 + 5*x^4 + (20*x + 10*x^2 + 10*x^3)*Log[3] + 5*x^2*Log[3]^2
+ (20*x + 10*x^2 + 10*x^3 + 10*x^2*Log[3])*Log[25] + 5*x^2*Log[25]^2 + (20 + 10*x + 10*x^2 + 10*x*Log[3] + 10
*x*Log[25])*Log[x] + 5*Log[x]^2),x]

[Out]

-1/5*Defer[Int][(2 + x^2 + x*(1 + Log[75]) + Log[x])^(-2), x] - ((1 + Log[75])*Defer[Int][x/(2 + x^2 + x*(1 +
Log[75]) + Log[x])^2, x])/5 - (2*Defer[Int][x^2/(2 + x^2 + x*(1 + Log[75]) + Log[x])^2, x])/5 + Defer[Int][(2
+ x^2 + x*(1 + Log[75]) + Log[x])^(-1), x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-x^2+\log (x)}{20+20 x+10 x^3+5 x^4+\left (20 x+10 x^2+10 x^3\right ) \log (3)+x^2 \left (25+5 \log ^2(3)\right )+\left (20 x+10 x^2+10 x^3+10 x^2 \log (3)\right ) \log (25)+5 x^2 \log ^2(25)+\left (20+10 x+10 x^2+10 x \log (3)+10 x \log (25)\right ) \log (x)+5 \log ^2(x)} \, dx\\ &=\int \frac {1-x^2+\log (x)}{20+20 x+10 x^3+5 x^4+\left (20 x+10 x^2+10 x^3\right ) \log (3)+\left (20 x+10 x^2+10 x^3+10 x^2 \log (3)\right ) \log (25)+x^2 \left (25+5 \log ^2(3)+5 \log ^2(25)\right )+\left (20+10 x+10 x^2+10 x \log (3)+10 x \log (25)\right ) \log (x)+5 \log ^2(x)} \, dx\\ &=\int \frac {1-x^2+\log (x)}{5 \left (2+x^2+x (1+\log (75))+\log (x)\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {1-x^2+\log (x)}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {-1-2 x^2-x (1+\log (75))}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2}+\frac {1}{2+x^2+x (1+\log (75))+\log (x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-1-2 x^2-x (1+\log (75))}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {1}{2+x^2+x (1+\log (75))+\log (x)} \, dx\\ &=\frac {1}{5} \int \frac {1}{2+x^2+x (1+\log (75))+\log (x)} \, dx+\frac {1}{5} \int \left (-\frac {1}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2}-\frac {2 x^2}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2}+\frac {x (-1-\log (75))}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {1}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2} \, dx\right )+\frac {1}{5} \int \frac {1}{2+x^2+x (1+\log (75))+\log (x)} \, dx-\frac {2}{5} \int \frac {x^2}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2} \, dx+\frac {1}{5} (-1-\log (75)) \int \frac {x}{\left (2+x^2+x (1+\log (75))+\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 19, normalized size = 0.86 \begin {gather*} \frac {x}{5 \left (2+x+x^2+x \log (75)+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^2 + Log[x])/(20 + 20*x + 25*x^2 + 10*x^3 + 5*x^4 + (20*x + 10*x^2 + 10*x^3)*Log[3] + 5*x^2*Lo
g[3]^2 + (20*x + 10*x^2 + 10*x^3 + 10*x^2*Log[3])*Log[25] + 5*x^2*Log[25]^2 + (20 + 10*x + 10*x^2 + 10*x*Log[3
] + 10*x*Log[25])*Log[x] + 5*Log[x]^2),x]

[Out]

x/(5*(2 + x + x^2 + x*Log[75] + Log[x]))

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fricas [A]  time = 0.68, size = 22, normalized size = 1.00 \begin {gather*} \frac {x}{5 \, {\left (x^{2} + 2 \, x \log \relax (5) + x \log \relax (3) + x + \log \relax (x) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-x^2+1)/(5*log(x)^2+(20*x*log(5)+10*x*log(3)+10*x^2+10*x+20)*log(x)+20*x^2*log(5)^2+2*(10*x^2
*log(3)+10*x^3+10*x^2+20*x)*log(5)+5*x^2*log(3)^2+(10*x^3+10*x^2+20*x)*log(3)+5*x^4+10*x^3+25*x^2+20*x+20),x,
algorithm="fricas")

[Out]

1/5*x/(x^2 + 2*x*log(5) + x*log(3) + x + log(x) + 2)

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giac [A]  time = 0.50, size = 22, normalized size = 1.00 \begin {gather*} \frac {x}{5 \, {\left (x^{2} + 2 \, x \log \relax (5) + x \log \relax (3) + x + \log \relax (x) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-x^2+1)/(5*log(x)^2+(20*x*log(5)+10*x*log(3)+10*x^2+10*x+20)*log(x)+20*x^2*log(5)^2+2*(10*x^2
*log(3)+10*x^3+10*x^2+20*x)*log(5)+5*x^2*log(3)^2+(10*x^3+10*x^2+20*x)*log(3)+5*x^4+10*x^3+25*x^2+20*x+20),x,
algorithm="giac")

[Out]

1/5*x/(x^2 + 2*x*log(5) + x*log(3) + x + log(x) + 2)

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maple [A]  time = 0.19, size = 23, normalized size = 1.05

method	result	size
norman	\(\frac {x}{10 x \ln \relax (5)+5 x \ln \relax (3)+5 x^{2}+5 \ln \relax (x )+5 x +10}\)	\(23\)
risch	\(\frac {x}{10 x \ln \relax (5)+5 x \ln \relax (3)+5 x^{2}+5 \ln \relax (x )+5 x +10}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)-x^2+1)/(5*ln(x)^2+(20*x*ln(5)+10*x*ln(3)+10*x^2+10*x+20)*ln(x)+20*x^2*ln(5)^2+2*(10*x^2*ln(3)+10*x^
3+10*x^2+20*x)*ln(5)+5*x^2*ln(3)^2+(10*x^3+10*x^2+20*x)*ln(3)+5*x^4+10*x^3+25*x^2+20*x+20),x,method=_RETURNVER
BOSE)

[Out]

1/5*x/(2*x*ln(5)+x*ln(3)+x^2+ln(x)+x+2)

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maxima [A]  time = 1.08, size = 22, normalized size = 1.00 \begin {gather*} \frac {x}{5 \, {\left (x^{2} + x {\left (2 \, \log \relax (5) + \log \relax (3) + 1\right )} + \log \relax (x) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-x^2+1)/(5*log(x)^2+(20*x*log(5)+10*x*log(3)+10*x^2+10*x+20)*log(x)+20*x^2*log(5)^2+2*(10*x^2
*log(3)+10*x^3+10*x^2+20*x)*log(5)+5*x^2*log(3)^2+(10*x^3+10*x^2+20*x)*log(3)+5*x^4+10*x^3+25*x^2+20*x+20),x,
algorithm="maxima")

[Out]

1/5*x/(x^2 + x*(2*log(5) + log(3) + 1) + log(x) + 2)

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mupad [B]  time = 2.44, size = 22, normalized size = 1.00 \begin {gather*} \frac {x}{5\,\left (\ln \relax (x)+x\,\left (\ln \left (75\right )+1\right )+x^2+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) - x^2 + 1)/(20*x + 5*x^2*log(3)^2 + 20*x^2*log(5)^2 + 2*log(5)*(20*x + 10*x^2*log(3) + 10*x^2 + 10
*x^3) + log(x)*(10*x + 10*x*log(3) + 20*x*log(5) + 10*x^2 + 20) + 5*log(x)^2 + log(3)*(20*x + 10*x^2 + 10*x^3)
+ 25*x^2 + 10*x^3 + 5*x^4 + 20),x)

[Out]

x/(5*(log(x) + x*(log(75) + 1) + x^2 + 2))

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sympy [A]  time = 0.18, size = 29, normalized size = 1.32 \begin {gather*} \frac {x}{5 x^{2} + 5 x + 5 x \log {\relax (3 )} + 10 x \log {\relax (5 )} + 5 \log {\relax (x )} + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)-x**2+1)/(5*ln(x)**2+(20*x*ln(5)+10*x*ln(3)+10*x**2+10*x+20)*ln(x)+20*x**2*ln(5)**2+2*(10*x**2
*ln(3)+10*x**3+10*x**2+20*x)*ln(5)+5*x**2*ln(3)**2+(10*x**3+10*x**2+20*x)*ln(3)+5*x**4+10*x**3+25*x**2+20*x+20
),x)

[Out]

x/(5*x**2 + 5*x + 5*x*log(3) + 10*x*log(5) + 5*log(x) + 10)

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