Integrand size = 47, antiderivative size = 142 \[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=-\frac {1}{4} \text {arctanh}\left (\frac {(1-3 y) \sqrt {1-5 y-5 y^2}}{(1-5 y) \sqrt {1-y-y^2}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {(4+3 y) \sqrt {1-5 y-5 y^2}}{(6+5 y) \sqrt {1-y-y^2}}\right )+\frac {9}{4} \text {arctanh}\left (\frac {(11+7 y) \sqrt {1-5 y-5 y^2}}{3 (7+5 y) \sqrt {1-y-y^2}}\right ) \]
-1/4*arctanh((1-3*y)*(-5*y^2-5*y+1)^(1/2)/(1-5*y)/(-y^2-y+1)^(1/2))-1/2*ar ctanh((4+3*y)*(-5*y^2-5*y+1)^(1/2)/(6+5*y)/(-y^2-y+1)^(1/2))+9/4*arctanh(1 /3*(11+7*y)*(-5*y^2-5*y+1)^(1/2)/(7+5*y)/(-y^2-y+1)^(1/2))
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 3.38 (sec) , antiderivative size = 630, normalized size of antiderivative = 4.44 \[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=\frac {\left (-1-\frac {2}{\sqrt {5}}\right ) \left (1+\sqrt {5}+2 y\right )^2 \sqrt {\frac {5+3 \sqrt {5}+10 y}{5+5 \sqrt {5}+10 y}} \left (20 \left (-4 \sqrt {\frac {-5+3 \sqrt {5}-10 y}{1+\sqrt {5}+2 y}} \sqrt {\frac {-1+\sqrt {5}-2 y}{1+\sqrt {5}+2 y}}+\sqrt {5} \sqrt {\frac {-5+3 \sqrt {5}-10 y}{1+\sqrt {5}+2 y}} \sqrt {\frac {-1+\sqrt {5}-2 y}{1+\sqrt {5}+2 y}}+5 \sqrt {-\frac {-5+\sqrt {5}+2 \sqrt {5} y}{1+\sqrt {5}+2 y}} \sqrt {-\frac {-3+\sqrt {5}+2 \sqrt {5} y}{1+\sqrt {5}+2 y}}-2 \sqrt {5} \sqrt {-\frac {-5+\sqrt {5}+2 \sqrt {5} y}{1+\sqrt {5}+2 y}} \sqrt {-\frac {-3+\sqrt {5}+2 \sqrt {5} y}{1+\sqrt {5}+2 y}}\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {2 \sqrt {\frac {5+3 \sqrt {5}+10 y}{1+\sqrt {5}+2 y}}}{\sqrt {15}}\right ),\frac {15}{16}\right )+\sqrt {\frac {-5+3 \sqrt {5}-10 y}{1+\sqrt {5}+2 y}} \sqrt {\frac {-1+\sqrt {5}-2 y}{1+\sqrt {5}+2 y}} \left (9 \sqrt {5} \operatorname {EllipticPi}\left (\frac {5}{8}-\frac {\sqrt {5}}{8},\arcsin \left (\frac {2 \sqrt {\frac {5+3 \sqrt {5}+10 y}{1+\sqrt {5}+2 y}}}{\sqrt {15}}\right ),\frac {15}{16}\right )+\left (-20+9 \sqrt {5}\right ) \operatorname {EllipticPi}\left (-\frac {3}{8} \left (-5+\sqrt {5}\right ),\arcsin \left (\frac {2 \sqrt {\frac {5+3 \sqrt {5}+10 y}{1+\sqrt {5}+2 y}}}{\sqrt {15}}\right ),\frac {15}{16}\right )+2 \sqrt {5} \operatorname {EllipticPi}\left (\frac {3}{8} \left (5+\sqrt {5}\right ),\arcsin \left (\frac {2 \sqrt {\frac {5+3 \sqrt {5}+10 y}{1+\sqrt {5}+2 y}}}{\sqrt {15}}\right ),\frac {15}{16}\right )\right )\right )}{16 \sqrt {1-5 y-5 y^2} \sqrt {1-y-y^2}} \]
((-1 - 2/Sqrt[5])*(1 + Sqrt[5] + 2*y)^2*Sqrt[(5 + 3*Sqrt[5] + 10*y)/(5 + 5 *Sqrt[5] + 10*y)]*(20*(-4*Sqrt[(-5 + 3*Sqrt[5] - 10*y)/(1 + Sqrt[5] + 2*y) ]*Sqrt[(-1 + Sqrt[5] - 2*y)/(1 + Sqrt[5] + 2*y)] + Sqrt[5]*Sqrt[(-5 + 3*Sq rt[5] - 10*y)/(1 + Sqrt[5] + 2*y)]*Sqrt[(-1 + Sqrt[5] - 2*y)/(1 + Sqrt[5] + 2*y)] + 5*Sqrt[-((-5 + Sqrt[5] + 2*Sqrt[5]*y)/(1 + Sqrt[5] + 2*y))]*Sqrt [-((-3 + Sqrt[5] + 2*Sqrt[5]*y)/(1 + Sqrt[5] + 2*y))] - 2*Sqrt[5]*Sqrt[-(( -5 + Sqrt[5] + 2*Sqrt[5]*y)/(1 + Sqrt[5] + 2*y))]*Sqrt[-((-3 + Sqrt[5] + 2 *Sqrt[5]*y)/(1 + Sqrt[5] + 2*y))])*EllipticF[ArcSin[(2*Sqrt[(5 + 3*Sqrt[5] + 10*y)/(1 + Sqrt[5] + 2*y)])/Sqrt[15]], 15/16] + Sqrt[(-5 + 3*Sqrt[5] - 10*y)/(1 + Sqrt[5] + 2*y)]*Sqrt[(-1 + Sqrt[5] - 2*y)/(1 + Sqrt[5] + 2*y)]* (9*Sqrt[5]*EllipticPi[5/8 - Sqrt[5]/8, ArcSin[(2*Sqrt[(5 + 3*Sqrt[5] + 10* y)/(1 + Sqrt[5] + 2*y)])/Sqrt[15]], 15/16] + (-20 + 9*Sqrt[5])*EllipticPi[ (-3*(-5 + Sqrt[5]))/8, ArcSin[(2*Sqrt[(5 + 3*Sqrt[5] + 10*y)/(1 + Sqrt[5] + 2*y)])/Sqrt[15]], 15/16] + 2*Sqrt[5]*EllipticPi[(3*(5 + Sqrt[5]))/8, Arc Sin[(2*Sqrt[(5 + 3*Sqrt[5] + 10*y)/(1 + Sqrt[5] + 2*y)])/Sqrt[15]], 15/16] )))/(16*Sqrt[1 - 5*y - 5*y^2]*Sqrt[1 - y - y^2])
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.
| \(\displaystyle \int \frac {(2 y+1) \sqrt {-5 y^2-5 y+1}}{y (y+1) (y+2) \sqrt {-y^2-y+1}} \, dy\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {\sqrt {-5 y^2-5 y+1}}{2 y \sqrt {-y^2-y+1}}+\frac {\sqrt {-5 y^2-5 y+1}}{(y+1) \sqrt {-y^2-y+1}}-\frac {3 \sqrt {-5 y^2-5 y+1}}{2 (y+2) \sqrt {-y^2-y+1}}\right )dy\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {-5 y^2-5 y+1}}{y \sqrt {-y^2-y+1}}dy+\int \frac {\sqrt {-5 y^2-5 y+1}}{(y+1) \sqrt {-y^2-y+1}}dy-\frac {3}{2} \int \frac {\sqrt {-5 y^2-5 y+1}}{(y+2) \sqrt {-y^2-y+1}}dy\) |
3.3.79.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 3.29 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.48
| method | result | size |
| default | \(-\frac {300 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}\, \sqrt {-\frac {10 y +5+3 \sqrt {5}}{-10 y -5+3 \sqrt {5}}}\, \left (-10 y -5+3 \sqrt {5}\right )^{2} \sqrt {\frac {-2 y +\sqrt {5}-1}{-10 y -5+3 \sqrt {5}}}\, \sqrt {5}\, \sqrt {\frac {2 y +1+\sqrt {5}}{-10 y -5+3 \sqrt {5}}}\, \left (\Pi \left (2 \sqrt {-\frac {10 y +5+3 \sqrt {5}}{-10 y -5+3 \sqrt {5}}}, -\frac {3 \sqrt {5}-5}{4 \left (5+3 \sqrt {5}\right )}, \frac {1}{4}\right )+2 \Pi \left (2 \sqrt {-\frac {10 y +5+3 \sqrt {5}}{-10 y -5+3 \sqrt {5}}}, -\frac {5+3 \sqrt {5}}{4 \left (3 \sqrt {5}-5\right )}, \frac {1}{4}\right )-3 \Pi \left (2 \sqrt {-\frac {10 y +5+3 \sqrt {5}}{-10 y -5+3 \sqrt {5}}}, -\frac {5+\sqrt {5}}{4 \left (\sqrt {5}-5\right )}, \frac {1}{4}\right )\right )}{\sqrt {5 y^{4}+10 y^{3}-y^{2}-6 y +1}\, \sqrt {\left (10 y +5+3 \sqrt {5}\right ) \left (-10 y -5+3 \sqrt {5}\right ) \left (-2 y +\sqrt {5}-1\right ) \left (2 y +1+\sqrt {5}\right )}\, \left (3 \sqrt {5}-5\right ) \left (5+3 \sqrt {5}\right ) \left (5+\sqrt {5}\right ) \left (\sqrt {5}-5\right )}\) | \(352\) |
| elliptic | \(\text {Expression too large to display}\) | \(7002\) |
-300*(-5*y^2-5*y+1)^(1/2)*(-y^2-y+1)^(1/2)*(-(10*y+5+3*5^(1/2))/(-10*y-5+3 *5^(1/2)))^(1/2)*(-10*y-5+3*5^(1/2))^2*((-2*y+5^(1/2)-1)/(-10*y-5+3*5^(1/2 )))^(1/2)*5^(1/2)*((2*y+1+5^(1/2))/(-10*y-5+3*5^(1/2)))^(1/2)*(EllipticPi( 2*(-(10*y+5+3*5^(1/2))/(-10*y-5+3*5^(1/2)))^(1/2),-1/4*(3*5^(1/2)-5)/(5+3* 5^(1/2)),1/4)+2*EllipticPi(2*(-(10*y+5+3*5^(1/2))/(-10*y-5+3*5^(1/2)))^(1/ 2),-1/4*(5+3*5^(1/2))/(3*5^(1/2)-5),1/4)-3*EllipticPi(2*(-(10*y+5+3*5^(1/2 ))/(-10*y-5+3*5^(1/2)))^(1/2),-1/4*(5+5^(1/2))/(5^(1/2)-5),1/4))/(5*y^4+10 *y^3-y^2-6*y+1)^(1/2)/((10*y+5+3*5^(1/2))*(-10*y-5+3*5^(1/2))*(-2*y+5^(1/2 )-1)*(2*y+1+5^(1/2)))^(1/2)/(3*5^(1/2)-5)/(5+3*5^(1/2))/(5+5^(1/2))/(5^(1/ 2)-5)
Time = 0.32 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.57 \[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=\frac {9}{8} \, \log \left (-\frac {235 \, y^{4} + 935 \, y^{3} - 3 \, {\left (35 \, y^{2} + 104 \, y + 77\right )} \sqrt {-y^{2} - y + 1} \sqrt {-5 \, y^{2} - 5 \, y + 1} + 1086 \, y^{2} + 131 \, y - 281}{y^{4} + 8 \, y^{3} + 24 \, y^{2} + 32 \, y + 16}\right ) + \frac {1}{4} \, \log \left (\frac {35 \, y^{4} + 125 \, y^{3} + {\left (15 \, y^{2} + 38 \, y + 24\right )} \sqrt {-y^{2} - y + 1} \sqrt {-5 \, y^{2} - 5 \, y + 1} + 131 \, y^{2} + 16 \, y - 26}{y^{4} + 4 \, y^{3} + 6 \, y^{2} + 4 \, y + 1}\right ) + \frac {1}{8} \, \log \left (\frac {35 \, y^{4} + 15 \, y^{3} + {\left (15 \, y^{2} - 8 \, y + 1\right )} \sqrt {-y^{2} - y + 1} \sqrt {-5 \, y^{2} - 5 \, y + 1} - 34 \, y^{2} + 11 \, y - 1}{y^{4}}\right ) \]
9/8*log(-(235*y^4 + 935*y^3 - 3*(35*y^2 + 104*y + 77)*sqrt(-y^2 - y + 1)*s qrt(-5*y^2 - 5*y + 1) + 1086*y^2 + 131*y - 281)/(y^4 + 8*y^3 + 24*y^2 + 32 *y + 16)) + 1/4*log((35*y^4 + 125*y^3 + (15*y^2 + 38*y + 24)*sqrt(-y^2 - y + 1)*sqrt(-5*y^2 - 5*y + 1) + 131*y^2 + 16*y - 26)/(y^4 + 4*y^3 + 6*y^2 + 4*y + 1)) + 1/8*log((35*y^4 + 15*y^3 + (15*y^2 - 8*y + 1)*sqrt(-y^2 - y + 1)*sqrt(-5*y^2 - 5*y + 1) - 34*y^2 + 11*y - 1)/y^4)
\[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=\int \frac {\left (2 y + 1\right ) \sqrt {- 5 y^{2} - 5 y + 1}}{y \left (y + 1\right ) \left (y + 2\right ) \sqrt {- y^{2} - y + 1}}\, dy \]
\[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=\int { \frac {\sqrt {-5 \, y^{2} - 5 \, y + 1} {\left (2 \, y + 1\right )}}{\sqrt {-y^{2} - y + 1} {\left (y + 2\right )} {\left (y + 1\right )} y} \,d y } \]
\[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=\int { \frac {\sqrt {-5 \, y^{2} - 5 \, y + 1} {\left (2 \, y + 1\right )}}{\sqrt {-y^{2} - y + 1} {\left (y + 2\right )} {\left (y + 1\right )} y} \,d y } \]
Timed out. \[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=\int \frac {\left (2\,y+1\right )\,\sqrt {-5\,y^2-5\,y+1}}{y\,\left (y+1\right )\,\left (y+2\right )\,\sqrt {-y^2-y+1}} \,d y \]
Time = 0.06 (sec) , antiderivative size = 706, normalized size of antiderivative = 4.97 \[ \int \frac {(1+2 y) \sqrt {1-5 y-5 y^2}}{y (1+y) (2+y) \sqrt {1-y-y^2}} \, dy=-2 \,\mathrm {log}\left (y^{2}-2 y +1\right )-\frac {\mathrm {log}\left (y^{2}+4 y +4\right )}{2}+\mathrm {log}\left (-3 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}\, i +4 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}-15 \sqrt {-y^{2}-y +1}\, i y +3 \sqrt {-y^{2}-y +1}\, i -5 \sqrt {-y^{2}-y +1}\, y -4 \sqrt {-y^{2}-y +1}-\sqrt {-5 y^{2}-5 y +1}\, i y -3 \sqrt {-5 y^{2}-5 y +1}\, i -7 \sqrt {-5 y^{2}-5 y +1}\, y +4 \sqrt {-5 y^{2}-5 y +1}-5 i \,y^{2}+11 i y +3 i -10 y^{2}+2 y -4\right )+\frac {\mathrm {log}\left (-2 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}\, i -5 \sqrt {-y^{2}-y +1}\, y -4 \sqrt {-y^{2}-y +1}+3 \sqrt {-5 y^{2}-5 y +1}\, y +4 \sqrt {-5 y^{2}-5 y +1}-5 i \,y^{2}-6 i y +2 i \right )}{2}+\mathrm {log}\left (\sqrt {-y^{2}-y +1}-\sqrt {-5 y^{2}-5 y +1}-4 i y -4 i \right )+\frac {\mathrm {log}\left (\sqrt {-y^{2}-y +1}\, i -\sqrt {-5 y^{2}-5 y +1}\, i +2 y \right )}{2}+\mathrm {log}\left (6 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}\, i +8 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}-20 \sqrt {-y^{2}-y +1}\, i y -4 \sqrt {-y^{2}-y +1}\, i +15 \sqrt {-y^{2}-y +1}\, y +3 \sqrt {-y^{2}-y +1}-12 \sqrt {-5 y^{2}-5 y +1}\, i y +4 \sqrt {-5 y^{2}-5 y +1}\, i +9 \sqrt {-5 y^{2}-5 y +1}\, y -3 \sqrt {-5 y^{2}-5 y +1}-10 i \,y^{2}-22 i y +14 i -30 y^{2}+4 y +2\right )+\mathrm {log}\left (6 \sqrt {-5 y^{2}-5 y +1}\, \sqrt {-y^{2}-y +1}-15 \sqrt {-y^{2}-y +1}\, i y -3 \sqrt {-y^{2}-y +1}\, i -9 \sqrt {-5 y^{2}-5 y +1}\, i y +3 \sqrt {-5 y^{2}-5 y +1}\, i -10 y^{2}-22 y +14\right )+\frac {9 \,\mathrm {log}\left (3 \sqrt {-y^{2}-y +1}\, i +\sqrt {-5 y^{2}-5 y +1}\, i -2 y -4\right )}{2}-\mathrm {log}\left (y -1\right )-\frac {9 \,\mathrm {log}\left (y +2\right )}{2}-\mathrm {log}\left (y +1\right )-\frac {3 \,\mathrm {log}\left (y \right )}{2} \]
( - 4*log(y**2 - 2*y + 1) - log(y**2 + 4*y + 4) + 2*log( - 3*sqrt( - 5*y** 2 - 5*y + 1)*sqrt( - y**2 - y + 1)*i + 4*sqrt( - 5*y**2 - 5*y + 1)*sqrt( - y**2 - y + 1) - 15*sqrt( - y**2 - y + 1)*i*y + 3*sqrt( - y**2 - y + 1)*i - 5*sqrt( - y**2 - y + 1)*y - 4*sqrt( - y**2 - y + 1) - sqrt( - 5*y**2 - 5 *y + 1)*i*y - 3*sqrt( - 5*y**2 - 5*y + 1)*i - 7*sqrt( - 5*y**2 - 5*y + 1)* y + 4*sqrt( - 5*y**2 - 5*y + 1) - 5*i*y**2 + 11*i*y + 3*i - 10*y**2 + 2*y - 4) + log( - 2*sqrt( - 5*y**2 - 5*y + 1)*sqrt( - y**2 - y + 1)*i - 5*sqrt ( - y**2 - y + 1)*y - 4*sqrt( - y**2 - y + 1) + 3*sqrt( - 5*y**2 - 5*y + 1 )*y + 4*sqrt( - 5*y**2 - 5*y + 1) - 5*i*y**2 - 6*i*y + 2*i) + 2*log(sqrt( - y**2 - y + 1) - sqrt( - 5*y**2 - 5*y + 1) - 4*i*y - 4*i) + log(sqrt( - y **2 - y + 1)*i - sqrt( - 5*y**2 - 5*y + 1)*i + 2*y) + 2*log(6*sqrt( - 5*y* *2 - 5*y + 1)*sqrt( - y**2 - y + 1)*i + 8*sqrt( - 5*y**2 - 5*y + 1)*sqrt( - y**2 - y + 1) - 20*sqrt( - y**2 - y + 1)*i*y - 4*sqrt( - y**2 - y + 1)*i + 15*sqrt( - y**2 - y + 1)*y + 3*sqrt( - y**2 - y + 1) - 12*sqrt( - 5*y** 2 - 5*y + 1)*i*y + 4*sqrt( - 5*y**2 - 5*y + 1)*i + 9*sqrt( - 5*y**2 - 5*y + 1)*y - 3*sqrt( - 5*y**2 - 5*y + 1) - 10*i*y**2 - 22*i*y + 14*i - 30*y**2 + 4*y + 2) + 2*log(6*sqrt( - 5*y**2 - 5*y + 1)*sqrt( - y**2 - y + 1) - 15 *sqrt( - y**2 - y + 1)*i*y - 3*sqrt( - y**2 - y + 1)*i - 9*sqrt( - 5*y**2 - 5*y + 1)*i*y + 3*sqrt( - 5*y**2 - 5*y + 1)*i - 10*y**2 - 22*y + 14) + 9* log(3*sqrt( - y**2 - y + 1)*i + sqrt( - 5*y**2 - 5*y + 1)*i - 2*y - 4) ...