Integrand size = 35, antiderivative size = 84 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{2 x}+\frac {5}{12} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {3 \log (x)}{4}+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{24} \log \left (2-x+2 x^2\right ) \] Output:
-5/2/x+5/36*15^(1/2)*arctan(1/15*(1-4*x)*15^(1/2))-10/9*arctan(1/3*(1+2*x) *3^(1/2))*3^(1/2)-3/4*ln(x)+1/3*ln(x^2+x+1)+1/24*ln(2*x^2-x+2)
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {180+80 \sqrt {3} x \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+10 \sqrt {15} x \arctan \left (\frac {-1+4 x}{\sqrt {15}}\right )+54 x \log (x)-24 x \log \left (1+x+x^2\right )-3 x \log \left (2-x+2 x^2\right )}{72 x} \] Input:
Integrate[(5 + x + 3*x^2 + 2*x^3)/(x^2*(2 + x + 3*x^2 + x^3 + 2*x^4)),x]
Output:
-1/72*(180 + 80*Sqrt[3]*x*ArcTan[(1 + 2*x)/Sqrt[3]] + 10*Sqrt[15]*x*ArcTan [(-1 + 4*x)/Sqrt[15]] + 54*x*Log[x] - 24*x*Log[1 + x + x^2] - 3*x*Log[2 - x + 2*x^2])/x
Time = 0.50 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2462, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {2 x^3+3 x^2+x+5}{x^2 \left (2 x^4+x^3+3 x^2+x+2\right )} \, dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {2 (x-2)}{3 \left (x^2+x+1\right )}+\frac {2 x-13}{12 \left (2 x^2-x+2\right )}+\frac {5}{2 x^2}-\frac {3}{4 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{12} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} \log \left (x^2+x+1\right )+\frac {1}{24} \log \left (2 x^2-x+2\right )-\frac {5}{2 x}-\frac {3 \log (x)}{4}\) |
Input:
Int[(5 + x + 3*x^2 + 2*x^3)/(x^2*(2 + x + 3*x^2 + x^3 + 2*x^4)),x]
Output:
-5/(2*x) + (5*Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/12 - (10*ArcTan[(1 + 2 *x)/Sqrt[3]])/(3*Sqrt[3]) - (3*Log[x])/4 + Log[1 + x + x^2]/3 + Log[2 - x + 2*x^2]/24
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77
method | result | size |
default | \(\frac {\ln \left (x^{2}+x +1\right )}{3}-\frac {10 \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}+\frac {\ln \left (2 x^{2}-x +2\right )}{24}-\frac {5 \sqrt {15}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {15}}{15}\right )}{36}-\frac {5}{2 x}-\frac {3 \ln \left (x \right )}{4}\) | \(65\) |
risch | \(-\frac {5}{2 x}-\frac {3 \ln \left (x \right )}{4}+\frac {\ln \left (25 x^{2}+25 x +25\right )}{3}-\frac {10 \sqrt {3}\, \arctan \left (\frac {2 \left (5 x +\frac {5}{2}\right ) \sqrt {3}}{15}\right )}{9}+\frac {\ln \left (100 x^{2}-50 x +100\right )}{24}-\frac {5 \sqrt {15}\, \arctan \left (\frac {2 \left (10 x -\frac {5}{2}\right ) \sqrt {15}}{75}\right )}{36}\) | \(69\) |
Input:
int((2*x^3+3*x^2+x+5)/x^2/(2*x^4+x^3+3*x^2+x+2),x,method=_RETURNVERBOSE)
Output:
1/3*ln(x^2+x+1)-10/9*arctan(1/3*(2*x+1)*3^(1/2))*3^(1/2)+1/24*ln(2*x^2-x+2 )-5/36*15^(1/2)*arctan(1/15*(4*x-1)*15^(1/2))-5/2/x-3/4*ln(x)
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {30 \, \sqrt {\frac {5}{3}} x \arctan \left (\frac {1}{5} \, \sqrt {\frac {5}{3}} {\left (4 \, x - 1\right )}\right ) + 80 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 3 \, x \log \left (2 \, x^{2} - x + 2\right ) - 24 \, x \log \left (x^{2} + x + 1\right ) + 54 \, x \log \left (x\right ) + 180}{72 \, x} \] Input:
integrate((2*x^3+3*x^2+x+5)/x^2/(2*x^4+x^3+3*x^2+x+2),x, algorithm="fricas ")
Output:
-1/72*(30*sqrt(5/3)*x*arctan(1/5*sqrt(5/3)*(4*x - 1)) + 80*sqrt(3)*x*arcta n(1/3*sqrt(3)*(2*x + 1)) - 3*x*log(2*x^2 - x + 2) - 24*x*log(x^2 + x + 1) + 54*x*log(x) + 180)/x
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=- \frac {3 \log {\left (x \right )}}{4} + \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{24} + \frac {\log {\left (x^{2} + x + 1 \right )}}{3} - \frac {5 \sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{36} - \frac {10 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} - \frac {5}{2 x} \] Input:
integrate((2*x**3+3*x**2+x+5)/x**2/(2*x**4+x**3+3*x**2+x+2),x)
Output:
-3*log(x)/4 + log(x**2 - x/2 + 1)/24 + log(x**2 + x + 1)/3 - 5*sqrt(15)*at an(4*sqrt(15)*x/15 - sqrt(15)/15)/36 - 10*sqrt(3)*atan(2*sqrt(3)*x/3 + sqr t(3)/3)/9 - 5/(2*x)
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {5}{2 \, x} + \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {3}{4} \, \log \left (x\right ) \] Input:
integrate((2*x^3+3*x^2+x+5)/x^2/(2*x^4+x^3+3*x^2+x+2),x, algorithm="maxima ")
Output:
-5/36*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) - 10/9*sqrt(3)*arctan(1/3*s qrt(3)*(2*x + 1)) - 5/2/x + 1/24*log(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1) - 3/4*log(x)
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {5}{2 \, x} + \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {3}{4} \, \log \left ({\left | x \right |}\right ) \] Input:
integrate((2*x^3+3*x^2+x+5)/x^2/(2*x^4+x^3+3*x^2+x+2),x, algorithm="giac")
Output:
-5/36*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) - 10/9*sqrt(3)*arctan(1/3*s qrt(3)*(2*x + 1)) - 5/2/x + 1/24*log(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1) - 3/4*log(abs(x))
Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.05 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {3\,\ln \left (x\right )}{4}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\frac {5}{2\,x} \] Input:
int((x + 3*x^2 + 2*x^3 + 5)/(x^2*(x + 3*x^2 + x^3 + 2*x^4 + 2)),x)
Output:
log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*5i)/9 + 1/3) - (3*log(x))/4 - log( x + (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*5i)/9 - 1/3) + log(x - (15^(1/2)*1i)/4 - 1/4)*((15^(1/2)*5i)/72 + 1/24) - log(x + (15^(1/2)*1i)/4 - 1/4)*((15^(1 /2)*5i)/72 - 1/24) - 5/(2*x)
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {-10 \sqrt {15}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {15}}\right ) x -80 \sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) x +24 \,\mathrm {log}\left (x^{2}+x +1\right ) x +3 \,\mathrm {log}\left (2 x^{2}-x +2\right ) x -54 \,\mathrm {log}\left (x \right ) x -180}{72 x} \] Input:
int((2*x^3+3*x^2+x+5)/x^2/(2*x^4+x^3+3*x^2+x+2),x)
Output:
( - 10*sqrt(15)*atan((4*x - 1)/sqrt(15))*x - 80*sqrt(3)*atan((2*x + 1)/sqr t(3))*x + 24*log(x**2 + x + 1)*x + 3*log(2*x**2 - x + 2)*x - 54*log(x)*x - 180)/(72*x)