Integrand size = 22, antiderivative size = 218 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=-\sqrt {\frac {2}{217} \left (2+\sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\sqrt {\frac {2}{217} \left (2+\sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {\log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{\sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{\sqrt {14 \left (2+\sqrt {35}\right )}} \]
-1/217*arctan((-10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2)) ^(1/2))*(868+434*35^(1/2))^(1/2)+1/217*arctan((10*(1+2*x)^(1/2)+(20+10*35^ (1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*(868+434*35^(1/2))^(1/2)-ln(5+10*x+ 35^(1/2)-(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))/(28+14*35^(1/2))^(1/2)+ln(5 +10*x+35^(1/2)+(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))/(28+14*35^(1/2))^(1/2 )
Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {2 \left (\sqrt {2+i \sqrt {31}} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )+\sqrt {2-i \sqrt {31}} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right )}{\sqrt {217}} \]
(2*(Sqrt[2 + I*Sqrt[31]]*ArcTan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] + Sqrt[2 - I*Sqrt[31]]*ArcTan[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x]])) /Sqrt[217]
Time = 0.46 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1149, 1407, 27, 1142, 25, 1083, 217, 1103}
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.
| \(\displaystyle \int \frac {1}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )} \, dx\) |
| \(\Big \downarrow \) 1149 |
| \(\displaystyle 4 \int \frac {1}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\) |
| \(\Big \downarrow \) 1407 |
| \(\displaystyle 4 \left (\frac {\int \frac {5 \left (\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-\sqrt {2 x+1}\right )}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\int \frac {5 \left (\sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {5 \int \frac {\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-\sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \int \frac {\sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle 4 \left (\frac {5 \left (\sqrt {\frac {1}{10} \left (2+\sqrt {35}\right )} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{10} \int -\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\sqrt {\frac {1}{10} \left (2+\sqrt {35}\right )} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 4 \left (\frac {5 \left (\sqrt {\frac {1}{10} \left (2+\sqrt {35}\right )} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\sqrt {\frac {1}{10} \left (2+\sqrt {35}\right )} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle 4 \left (\frac {5 \left (\frac {1}{10} \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{10} \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 4 \left (\frac {5 \left (\frac {1}{10} \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{5} \sqrt {\frac {2+\sqrt {35}}{\sqrt {35}-2}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{10} \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{5} \sqrt {\frac {2+\sqrt {35}}{\sqrt {35}-2}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle 4 \left (\frac {5 \left (\frac {1}{5} \sqrt {\frac {2+\sqrt {35}}{\sqrt {35}-2}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{5} \sqrt {\frac {2+\sqrt {35}}{\sqrt {35}-2}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
4*((5*((Sqrt[(2 + Sqrt[35])/(-2 + Sqrt[35])]*ArcTan[(-Sqrt[10*(2 + Sqrt[35 ])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/5 - Log[Sqrt[35] - Sqrt [10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/10))/(2*Sqrt[14*(2 + Sqrt [35])]) + (5*((Sqrt[(2 + Sqrt[35])/(-2 + Sqrt[35])]*ArcTan[(Sqrt[10*(2 + S qrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/5 + Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/10))/(2*Sqrt[14*(2 + Sqrt[35])]))
3.24.13.3.1 Defintions of rubi rules used
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[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Sym bol] :> Simp[2*e Subst[Int[1/(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] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/ c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r) Int[(r - x)/(q - r* x + x^2), x], x] + Simp[1/(2*c*q*r) Int[(r + x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]
Time = 0.48 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.14
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \left (\sqrt {5}-\frac {2 \sqrt {7}}{7}\right ) \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{62}+\frac {\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \left (\sqrt {5}-\frac {2 \sqrt {7}}{7}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{62}+\frac {2 \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right ) \sqrt {7}\, \sqrt {5}}{7}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(249\) |
| derivativedivides | \(\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}+\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}-\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(380\) |
| default | \(\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}+\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}-\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(380\) |
| trager | \(\operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right ) \ln \left (-\frac {517979 x \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{5}-20615 x \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{3}+651 \sqrt {1+2 x}\, \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}-19096 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{3}+200 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right ) x +145 \sqrt {1+2 x}+320 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )}{217 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2} x +5 x +4}\right )-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+47089 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+868\right ) \ln \left (\frac {73997 \operatorname {RootOf}\left (\textit {\_Z}^{2}+47089 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+868\right ) \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{4} x +5673 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+47089 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+868\right ) x +2728 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+47089 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+868\right )+20181 \sqrt {1+2 x}\, \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+108 \operatorname {RootOf}\left (\textit {\_Z}^{2}+47089 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+868\right ) x +96 \operatorname {RootOf}\left (\textit {\_Z}^{2}+47089 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}+868\right )-4123 \sqrt {1+2 x}}{217 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2} x -x -4}\right )}{217}\) | \(421\) |
2/7/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(-7/124*(10*5^(1/2)*7^(1/2)-20)^(1/2)*(2 *5^(1/2)*7^(1/2)+4)^(1/2)*(5^(1/2)-2/7*7^(1/2))*ln(5^(1/2)*7^(1/2)-(2*5^(1 /2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5+10*x)+7/124*(10*5^(1/2)*7^(1/ 2)-20)^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*(5^(1/2)-2/7*7^(1/2))*ln(5^(1/2)* 7^(1/2)+(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5+10*x)+(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)))*7^(1/2)*5^(1/2))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=-\frac {1}{434} \, \sqrt {217} \sqrt {4 i \, \sqrt {31} - 8} \log \left (\sqrt {217} \sqrt {4 i \, \sqrt {31} - 8} {\left (2 i \, \sqrt {31} - 31\right )} + 2170 \, \sqrt {2 \, x + 1}\right ) + \frac {1}{434} \, \sqrt {217} \sqrt {4 i \, \sqrt {31} - 8} \log \left (\sqrt {217} \sqrt {4 i \, \sqrt {31} - 8} {\left (-2 i \, \sqrt {31} + 31\right )} + 2170 \, \sqrt {2 \, x + 1}\right ) + \frac {1}{434} \, \sqrt {217} \sqrt {-4 i \, \sqrt {31} - 8} \log \left (\sqrt {217} {\left (2 i \, \sqrt {31} + 31\right )} \sqrt {-4 i \, \sqrt {31} - 8} + 2170 \, \sqrt {2 \, x + 1}\right ) - \frac {1}{434} \, \sqrt {217} \sqrt {-4 i \, \sqrt {31} - 8} \log \left (\sqrt {217} {\left (-2 i \, \sqrt {31} - 31\right )} \sqrt {-4 i \, \sqrt {31} - 8} + 2170 \, \sqrt {2 \, x + 1}\right ) \]
-1/434*sqrt(217)*sqrt(4*I*sqrt(31) - 8)*log(sqrt(217)*sqrt(4*I*sqrt(31) - 8)*(2*I*sqrt(31) - 31) + 2170*sqrt(2*x + 1)) + 1/434*sqrt(217)*sqrt(4*I*sq rt(31) - 8)*log(sqrt(217)*sqrt(4*I*sqrt(31) - 8)*(-2*I*sqrt(31) + 31) + 21 70*sqrt(2*x + 1)) + 1/434*sqrt(217)*sqrt(-4*I*sqrt(31) - 8)*log(sqrt(217)* (2*I*sqrt(31) + 31)*sqrt(-4*I*sqrt(31) - 8) + 2170*sqrt(2*x + 1)) - 1/434* sqrt(217)*sqrt(-4*I*sqrt(31) - 8)*log(sqrt(217)*(-2*I*sqrt(31) - 31)*sqrt( -4*I*sqrt(31) - 8) + 2170*sqrt(2*x + 1))
\[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\int \frac {1}{\sqrt {2 x + 1} \cdot \left (5 x^{2} + 3 x + 2\right )}\, dx \]
\[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\int { \frac {1}{{\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {2 \, x + 1}} \,d x } \]
Time = 0.49 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {1}{7595} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{7595} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (-\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} - \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{15190} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450} - 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450}\right )} \log \left (2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + 2 \, x + \sqrt {\frac {7}{5}} + 1\right ) - \frac {1}{15190} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450} - 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450}\right )} \log \left (-2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + 2 \, x + \sqrt {\frac {7}{5}} + 1\right ) \]
1/7595*sqrt(31)*(sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 2*(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)) + 1/759 5*sqrt(31)*(sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 2*(7/5)^(1/4 )*sqrt(140*sqrt(35) + 2450))*arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/3 5*sqrt(35) + 1/2) - sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2)) + 1/15190*s qrt(31)*(sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) - 2*(7/5)^(1/4)*sq rt(-140*sqrt(35) + 2450))*log(2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(3 5) + 1/2) + 2*x + sqrt(7/5) + 1) - 1/15190*sqrt(31)*(sqrt(31)*(7/5)^(1/4)* sqrt(140*sqrt(35) + 2450) - 2*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450))*log( -2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(35) + 1/2) + 2*x + sqrt(7/5) + 1)
Time = 9.82 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {256\,\sqrt {7}\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}}{6125\,\left (\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}+\frac {\sqrt {217}\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}\,128{}\mathrm {i}}{6125\,\left (\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}\right )\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,2{}\mathrm {i}}{217}+\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {256\,\sqrt {7}\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}}{6125\,\left (-\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}-\frac {\sqrt {217}\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}\,128{}\mathrm {i}}{6125\,\left (-\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}\right )\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,2{}\mathrm {i}}{217} \]
(217^(1/2)*atan((256*7^(1/2)*(- 31^(1/2)*1i - 2)^(1/2)*(2*x + 1)^(1/2))/(6 125*((31^(1/2)*128i)/875 + 256/875)) + (217^(1/2)*(- 31^(1/2)*1i - 2)^(1/2 )*(2*x + 1)^(1/2)*128i)/(6125*((31^(1/2)*128i)/875 + 256/875)))*(- 31^(1/2 )*1i - 2)^(1/2)*2i)/217 + (217^(1/2)*atan((256*7^(1/2)*(31^(1/2)*1i - 2)^( 1/2)*(2*x + 1)^(1/2))/(6125*((31^(1/2)*128i)/875 - 256/875)) - (217^(1/2)* (31^(1/2)*1i - 2)^(1/2)*(2*x + 1)^(1/2)*128i)/(6125*((31^(1/2)*128i)/875 - 256/875)))*(31^(1/2)*1i - 2)^(1/2)*2i)/217