Integrand size = 32, antiderivative size = 105 \[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=-2 \sqrt {\frac {10}{-2+\sqrt {35}}} \arctan \left (\frac {\sqrt {2+\sqrt {35}}-\sqrt {10+20 x}}{\sqrt {-2+\sqrt {35}}}\right )+2 \sqrt {\frac {10}{-2+\sqrt {35}}} \arctan \left (\frac {\sqrt {2+\sqrt {35}}+\sqrt {10+20 x}}{\sqrt {-2+\sqrt {35}}}\right ) \] Output:
-2*10^(1/2)/(-2+35^(1/2))^(1/2)*arctan(((2+35^(1/2))^(1/2)-(10+20*x)^(1/2) )/(-2+35^(1/2))^(1/2))+2*10^(1/2)/(-2+35^(1/2))^(1/2)*arctan(((2+35^(1/2)) ^(1/2)+(10+20*x)^(1/2))/(-2+35^(1/2))^(1/2))
Time = 0.51 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.61 \[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=2 \sqrt {\frac {10}{31} \left (2+\sqrt {35}\right )} \arctan \left (\frac {\sqrt {\frac {1}{62} \left (2+\sqrt {35}\right )} \left (\sqrt {5}-\sqrt {7}+2 \sqrt {5} x\right )}{\sqrt {1+2 x}}\right ) \] Input:
Integrate[(5 + Sqrt[35] + 10*x)/(Sqrt[1 + 2*x]*(2 + 3*x + 5*x^2)),x]
Output:
2*Sqrt[(10*(2 + Sqrt[35]))/31]*ArcTan[(Sqrt[(2 + Sqrt[35])/62]*(Sqrt[5] - Sqrt[7] + 2*Sqrt[5]*x))/Sqrt[1 + 2*x]]
Time = 0.55 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1197, 27, 1475, 1083, 217}
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.
| \(\displaystyle \int \frac {10 x+\sqrt {35}+5}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )} \, dx\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle 2 \int \frac {2 \left (5 (2 x+1)+\sqrt {35}\right )}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \int \frac {5 (2 x+1)+\sqrt {35}}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\) |
| \(\Big \downarrow \) 1475 |
| \(\displaystyle 4 \left (\frac {1}{2} \int \frac {1}{2 x-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {\frac {7}{5}}+1}d\sqrt {2 x+1}+\frac {1}{2} \int \frac {1}{2 x+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {\frac {7}{5}}+1}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle 4 \left (-\int \frac {1}{-2 x+\frac {2}{5} \left (2-\sqrt {35}\right )-1}d\left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )-\int \frac {1}{-2 x+\frac {2}{5} \left (2-\sqrt {35}\right )-1}d\left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 4 \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )+\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )\right )\) |
Input:
Int[(5 + Sqrt[35] + 10*x)/(Sqrt[1 + 2*x]*(2 + 3*x + 5*x^2)),x]
Output:
4*(Sqrt[5/(2*(-2 + Sqrt[35]))]*ArcTan[Sqrt[5/(2*(-2 + Sqrt[35]))]*(-Sqrt[( 2*(2 + Sqrt[35]))/5] + 2*Sqrt[1 + 2*x])] + Sqrt[5/(2*(-2 + Sqrt[35]))]*Arc Tan[Sqrt[5/(2*(-2 + Sqrt[35]))]*(Sqrt[(2*(2 + Sqrt[35]))/5] + 2*Sqrt[1 + 2 *x])])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^ 2, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Time = 4.44 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90
| method | result | size |
| pseudoelliptic | \(-\frac {20 \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )}{\sqrt {-20+10 \sqrt {35}}}\) | \(95\) |
| derivativedivides | \(\frac {20 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {20 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(111\) |
| default | \(\frac {20 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {20 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(111\) |
| trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right ) \ln \left (-\frac {27325 \operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right ) x^{2}+11445 \operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right ) x -1694150 \sqrt {1+2 x}\, x +1915 \sqrt {35}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right ) x^{2}-8249 \sqrt {35}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right ) x -118730 \sqrt {1+2 x}\, \sqrt {35}\, x -3933 \operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right ) \sqrt {35}+110050 \sqrt {1+2 x}\, \sqrt {35}+8455 \operatorname {RootOf}\left (\textit {\_Z}^{2}+620+310 \sqrt {35}\right )-431520 \sqrt {1+2 x}}{5 x^{2}+3 x +2}\right )}{31}\) | \(169\) |
Input:
int((5+35^(1/2)+10*x)/(1+2*x)^(1/2)/(5*x^2+3*x+2),x,method=_RETURNVERBOSE)
Output:
-20*(arctan((5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-10*(1+2*x)^(1/2))/(10*5^( 1/2)*7^(1/2)-20)^(1/2))-arctan((5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(1+ 2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2)))/(-20+10*35^(1/2))^(1/2)
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.38 \[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=-2 \, \sqrt {\frac {10}{31} \, \sqrt {35} + \frac {20}{31}} \arctan \left (-\frac {{\left (10 \, x - \sqrt {35} + 5\right )} \sqrt {\frac {10}{31} \, \sqrt {35} + \frac {20}{31}}}{10 \, \sqrt {2 \, x + 1}}\right ) \] Input:
integrate((5+35^(1/2)+10*x)/(1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="fric as")
Output:
-2*sqrt(10/31*sqrt(35) + 20/31)*arctan(-1/10*(10*x - sqrt(35) + 5)*sqrt(10 /31*sqrt(35) + 20/31)/sqrt(2*x + 1))
\[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\int \frac {10 x + 5 + \sqrt {35}}{\sqrt {2 x + 1} \cdot \left (5 x^{2} + 3 x + 2\right )}\, dx \] Input:
integrate((5+35**(1/2)+10*x)/(1+2*x)**(1/2)/(5*x**2+3*x+2),x)
Output:
Integral((10*x + 5 + sqrt(35))/(sqrt(2*x + 1)*(5*x**2 + 3*x + 2)), x)
\[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\int { \frac {10 \, x + \sqrt {35} + 5}{{\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {2 \, x + 1}} \,d x } \] Input:
integrate((5+35^(1/2)+10*x)/(1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="maxi ma")
Output:
integrate((10*x + sqrt(35) + 5)/((5*x^2 + 3*x + 2)*sqrt(2*x + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (76) = 152\).
Time = 0.58 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.98 \[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {1}{7443100} \, \sqrt {31} {\left (210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} - \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )} + 980 \, \sqrt {35} \sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 1960 \, \sqrt {35} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{7443100} \, \sqrt {31} {\left (210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} - \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )} + 980 \, \sqrt {35} \sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 1960 \, \sqrt {35} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (-\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} - \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) \] Input:
integrate((5+35^(1/2)+10*x)/(1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="giac ")
Output:
1/7443100*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*s qrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7 /5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 980*sqrt(35)*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sq rt(35) + 2450) + 1960*sqrt(35)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arct an(5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) + sqrt(2*x + 1)) /sqrt(-1/35*sqrt(35) + 1/2)) + 1/7443100*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4 )*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-14 0*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420 *(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 980*sqrt(35)*sq rt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 1960*sqrt(35)*(7/5)^(1/4)* sqrt(140*sqrt(35) + 2450))*arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35* sqrt(35) + 1/2) - sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2))
Time = 12.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.36 \[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=2\,\sqrt {\frac {10\,\sqrt {35}}{31}+\frac {20}{31}}\,\left (\mathrm {atan}\left (\frac {\sqrt {434}\,\left (39\,\sqrt {35}+140\right )\,\sqrt {2\,x+1}\,{\left (\sqrt {35}-2\right )}^2\,\sqrt {\sqrt {35}+2}}{417074}\right )+\mathrm {atan}\left (\frac {31\,\sqrt {2\,x+1}\,\left (\frac {\sqrt {\frac {10\,\sqrt {35}}{31}+\frac {20}{31}}\,\left (10000\,\sqrt {35}+20000\right )}{39\,\sqrt {35}+140}-\frac {\sqrt {434}\,\left (\frac {390000\,\sqrt {35}}{31}+\frac {1400000}{31}\right )\,{\left (\sqrt {35}-2\right )}^2\,\sqrt {\sqrt {35}+2}}{417074}\right )}{10000}+\frac {\sqrt {434}\,\left (\frac {200000\,\sqrt {35}}{31}+\frac {1950000}{31}\right )\,{\left (2\,x+1\right )}^{3/2}\,{\left (\sqrt {35}-2\right )}^2\,\sqrt {\sqrt {35}+2}}{134540000}\right )\right ) \] Input:
int((10*x + 35^(1/2) + 5)/((2*x + 1)^(1/2)*(3*x + 5*x^2 + 2)),x)
Output:
2*((10*35^(1/2))/31 + 20/31)^(1/2)*(atan((434^(1/2)*(39*35^(1/2) + 140)*(2 *x + 1)^(1/2)*(35^(1/2) - 2)^2*(35^(1/2) + 2)^(1/2))/417074) + atan((31*(2 *x + 1)^(1/2)*((((10*35^(1/2))/31 + 20/31)^(1/2)*(10000*35^(1/2) + 20000)) /(39*35^(1/2) + 140) - (434^(1/2)*((390000*35^(1/2))/31 + 1400000/31)*(35^ (1/2) - 2)^2*(35^(1/2) + 2)^(1/2))/417074))/10000 + (434^(1/2)*((200000*35 ^(1/2))/31 + 1950000/31)*(2*x + 1)^(3/2)*(35^(1/2) - 2)^2*(35^(1/2) + 2)^( 1/2))/134540000))
Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.47 \[ \int \frac {5+\sqrt {35}+10 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {2 \sqrt {\sqrt {35}-2}\, \sqrt {2}\, \left (-5 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )-2 \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )+5 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )+2 \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )\right )}{31} \] Input:
int((5+35^(1/2)+10*x)/(1+2*x)^(1/2)/(5*x^2+3*x+2),x)
Output:
(2*sqrt(sqrt(35) - 2)*sqrt(2)*( - 5*sqrt(7)*atan((sqrt(sqrt(35) + 2)*sqrt( 2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) - 2*sqrt(5)*at an((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 5*sqrt(7)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 2*sqrt(5)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))))/31