Integrand size = 22, antiderivative size = 172 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=-\sqrt {\frac {2}{7 \left (-2+\sqrt {35}\right )}} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\sqrt {\frac {2}{7 \left (-2+\sqrt {35}\right )}} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\sqrt {\frac {2}{7 \left (2+\sqrt {35}\right )}} \text {arctanh}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}}{5+\sqrt {35}+10 x}\right ) \] Output:
-2^(1/2)/(-14+7*35^(1/2))^(1/2)*arctan(((20+10*35^(1/2))^(1/2)-10*(1+2*x)^ (1/2))/(-20+10*35^(1/2))^(1/2))+2^(1/2)/(-14+7*35^(1/2))^(1/2)*arctan(((20 +10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*35^(1/2))^(1/2))+2^(1/2)/(14 +7*35^(1/2))^(1/2)*arctanh((20+10*35^(1/2))^(1/2)*(1+2*x)^(1/2)/(5+35^(1/2 )+10*x))
Result contains complex when optimal does not.
Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {2 \left (\sqrt {2+i \sqrt {31}} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )+\sqrt {2-i \sqrt {31}} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right )}{\sqrt {217}} \] Input:
Integrate[1/(Sqrt[1 + 2*x]*(2 + 3*x + 5*x^2)),x]
Output:
(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 = 0.47 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.48, 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 |
| \(\displaystyle 4 \left (\frac {5 \left (\frac {1}{5} \sqrt {\frac {2+\sqrt {35}}{\sqrt {35}-2}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{5} \sqrt {\frac {2+\sqrt {35}}{\sqrt {35}-2}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
Input:
Int[1/(Sqrt[1 + 2*x]*(2 + 3*x + 5*x^2)),x]
Output:
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])]))
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 = 2.11 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.27
| method | result | size |
| pseudoelliptic | \(\frac {\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (7 \sqrt {5}-2 \sqrt {7}\right ) \left (\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )-\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )\right ) \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-124 \sqrt {5}\, \sqrt {7}\, \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )}{434 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(219\) |
| derivativedivides | \(\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}+\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}-\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(380\) |
| default | \(\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}+\frac {\left (-35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{2170}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}-\frac {\left (35 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{217 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(380\) |
| trager | \(\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 +20181 \sqrt {1+2 x}\, \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2}-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 )-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 -4123 \sqrt {1+2 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 )}{217 \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )^{2} x -x -4}\right )}{217}+\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 x \operatorname {RootOf}\left (6727 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+5\right )+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 )\) | \(421\) |
Input:
int(1/(1+2*x)^(1/2)/(5*x^2+3*x+2),x,method=_RETURNVERBOSE)
Output:
1/434/(10*5^(1/2)*7^(1/2)-20)^(1/2)*((10*5^(1/2)*7^(1/2)-20)^(1/2)*(7*5^(1 /2)-2*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)-ln(-(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5 ^(1/2)*7^(1/2)+5+10*x))*(2*5^(1/2)*7^(1/2)+4)^(1/2)-124*5^(1/2)*7^(1/2)*(a rctan((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))))
Time = 0.09 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.16 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=2 \, \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} + \frac {1}{217}} \arctan \left (14 \, {\left ({\left (7 \, \sqrt {\frac {5}{7}} + 2\right )} \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} - \frac {1}{217}} + \sqrt {\frac {5}{7}} \sqrt {2 \, x + 1}\right )} \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} + \frac {1}{217}}\right ) - 2 \, \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} + \frac {1}{217}} \arctan \left (14 \, {\left ({\left (7 \, \sqrt {\frac {5}{7}} + 2\right )} \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} - \frac {1}{217}} - \sqrt {\frac {5}{7}} \sqrt {2 \, x + 1}\right )} \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} + \frac {1}{217}}\right ) + \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} - \frac {1}{217}} \log \left (14 \, \sqrt {2 \, x + 1} {\left (2 \, \sqrt {\frac {5}{7}} + 5\right )} \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} - \frac {1}{217}} + 10 \, x + 7 \, \sqrt {\frac {5}{7}} + 5\right ) - \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} - \frac {1}{217}} \log \left (-14 \, \sqrt {2 \, x + 1} {\left (2 \, \sqrt {\frac {5}{7}} + 5\right )} \sqrt {\frac {1}{62} \, \sqrt {\frac {5}{7}} - \frac {1}{217}} + 10 \, x + 7 \, \sqrt {\frac {5}{7}} + 5\right ) \] Input:
integrate(1/(1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="fricas")
Output:
2*sqrt(1/62*sqrt(5/7) + 1/217)*arctan(14*((7*sqrt(5/7) + 2)*sqrt(1/62*sqrt (5/7) - 1/217) + sqrt(5/7)*sqrt(2*x + 1))*sqrt(1/62*sqrt(5/7) + 1/217)) - 2*sqrt(1/62*sqrt(5/7) + 1/217)*arctan(14*((7*sqrt(5/7) + 2)*sqrt(1/62*sqrt (5/7) - 1/217) - sqrt(5/7)*sqrt(2*x + 1))*sqrt(1/62*sqrt(5/7) + 1/217)) + sqrt(1/62*sqrt(5/7) - 1/217)*log(14*sqrt(2*x + 1)*(2*sqrt(5/7) + 5)*sqrt(1 /62*sqrt(5/7) - 1/217) + 10*x + 7*sqrt(5/7) + 5) - sqrt(1/62*sqrt(5/7) - 1 /217)*log(-14*sqrt(2*x + 1)*(2*sqrt(5/7) + 5)*sqrt(1/62*sqrt(5/7) - 1/217) + 10*x + 7*sqrt(5/7) + 5)
\[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\int \frac {1}{\sqrt {2 x + 1} \cdot \left (5 x^{2} + 3 x + 2\right )}\, dx \] Input:
integrate(1/(1+2*x)**(1/2)/(5*x**2+3*x+2),x)
Output:
Integral(1/(sqrt(2*x + 1)*(5*x**2 + 3*x + 2)), x)
\[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\int { \frac {1}{{\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {2 \, x + 1}} \,d x } \] Input:
integrate(1/(1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="maxima")
Output:
integrate(1/((5*x^2 + 3*x + 2)*sqrt(2*x + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (132) = 264\).
Time = 0.57 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.62 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {1}{7595} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{7595} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (-\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} - \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{15190} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450} - 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450}\right )} \log \left (2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + 2 \, x + \sqrt {\frac {7}{5}} + 1\right ) - \frac {1}{15190} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450} - 2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450}\right )} \log \left (-2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + 2 \, x + \sqrt {\frac {7}{5}} + 1\right ) \] Input:
integrate(1/(1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="giac")
Output:
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 = 0.08 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {256\,\sqrt {7}\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}}{6125\,\left (\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}+\frac {\sqrt {217}\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}\,128{}\mathrm {i}}{6125\,\left (\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}\right )\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,2{}\mathrm {i}}{217}+\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {256\,\sqrt {7}\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}}{6125\,\left (-\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}-\frac {\sqrt {217}\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,\sqrt {2\,x+1}\,128{}\mathrm {i}}{6125\,\left (-\frac {256}{875}+\frac {\sqrt {31}\,128{}\mathrm {i}}{875}\right )}\right )\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,2{}\mathrm {i}}{217} \] Input:
int(1/((2*x + 1)^(1/2)*(3*x + 5*x^2 + 2)),x)
Output:
(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
Time = 0.22 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.78 \[ \int \frac {1}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx=\frac {\sqrt {2}\, \left (-4 \sqrt {\sqrt {35}-2}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )-14 \sqrt {\sqrt {35}-2}\, \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )+4 \sqrt {\sqrt {35}-2}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )+14 \sqrt {\sqrt {35}-2}\, \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )+2 \sqrt {\sqrt {35}+2}\, \sqrt {7}\, \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )-2 \sqrt {\sqrt {35}+2}\, \sqrt {7}\, \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )-7 \sqrt {\sqrt {35}+2}\, \sqrt {5}\, \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )+7 \sqrt {\sqrt {35}+2}\, \sqrt {5}\, \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )\right )}{434} \] Input:
int(1/(1+2*x)^(1/2)/(5*x^2+3*x+2),x)
Output:
(sqrt(2)*( - 4*sqrt(sqrt(35) - 2)*sqrt(7)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) - 14*sqrt(sqrt(3 5) - 2)*sqrt(5)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5) )/(sqrt(sqrt(35) - 2)*sqrt(2))) + 4*sqrt(sqrt(35) - 2)*sqrt(7)*atan((sqrt( sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt( 2))) + 14*sqrt(sqrt(35) - 2)*sqrt(5)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2* sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 2*sqrt(sqrt(35) + 2 )*sqrt(7)*log( - sqrt(2*x + 1)*sqrt(sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sq rt(5)*x + sqrt(5)) - 2*sqrt(sqrt(35) + 2)*sqrt(7)*log(sqrt(2*x + 1)*sqrt(s qrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sqrt(5)*x + sqrt(5)) - 7*sqrt(sqrt(35) + 2)*sqrt(5)*log( - sqrt(2*x + 1)*sqrt(sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2 *sqrt(5)*x + sqrt(5)) + 7*sqrt(sqrt(35) + 2)*sqrt(5)*log(sqrt(2*x + 1)*sqr t(sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sqrt(5)*x + sqrt(5))))/434