Integrand size = 22, antiderivative size = 193 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {4}{5} \sqrt {1+2 x}+\frac {1}{5} \sqrt {\frac {2}{155} \left (-178+35 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {1}{5} \sqrt {\frac {2}{155} \left (-178+35 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {1}{5} \sqrt {\frac {2}{155} \left (178+35 \sqrt {35}\right )} \text {arctanh}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}}{5+\sqrt {35}+10 x}\right ) \] Output:
4/5*(1+2*x)^(1/2)+1/775*(-55180+10850*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))-1/775*(-55180+10850* 35^(1/2))^(1/2)*arctan(((20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*3 5^(1/2))^(1/2))-1/775*(55180+10850*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 = 0.42 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.59 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {2}{775} \left (310 \sqrt {1+2 x}-\sqrt {155 \left (-178+19 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )-\sqrt {155 \left (-178-19 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right ) \] Input:
Integrate[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2),x]
Output:
(2*(310*Sqrt[1 + 2*x] - Sqrt[155*(-178 + (19*I)*Sqrt[31])]*ArcTan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] - Sqrt[155*(-178 - (19*I)*Sqrt[31])]*ArcT an[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x]]))/775
Time = 0.58 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.51, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {1146, 25, 1197, 27, 1483, 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.
| \(\displaystyle \int \frac {(2 x+1)^{3/2}}{5 x^2+3 x+2} \, dx\) |
| \(\Big \downarrow \) 1146 |
| \(\displaystyle \frac {1}{5} \int -\frac {3-8 x}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx+\frac {4}{5} \sqrt {2 x+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {1}{5} \int \frac {3-8 x}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {2}{5} \int \frac {2 (7-4 (2 x+1))}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \int \frac {7-4 (2 x+1)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\) |
| \(\Big \downarrow \) 1483 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\int \frac {7 \sqrt {10 \left (2+\sqrt {35}\right )}-\left (35+4 \sqrt {35}\right ) \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\int \frac {\left (35+4 \sqrt {35}\right ) \sqrt {2 x+1}+7 \sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\frac {1}{2} \sqrt {490 \sqrt {35}-2492} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int -\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{2} \sqrt {490 \sqrt {35}-2492} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\frac {1}{2} \sqrt {490 \sqrt {35}-2492} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{2} \sqrt {490 \sqrt {35}-2492} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {490 \sqrt {35}-2492} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {490 \sqrt {35}-2492} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {490 \sqrt {35}-2492}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (35+4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {490 \sqrt {35}-2492}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {4}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\sqrt {\frac {490 \sqrt {35}-2492}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \left (35+4 \sqrt {35}\right ) \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\sqrt {\frac {490 \sqrt {35}-2492}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \left (35+4 \sqrt {35}\right ) \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\) |
Input:
Int[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2),x]
Output:
(4*Sqrt[1 + 2*x])/5 - (4*((Sqrt[(-2492 + 490*Sqrt[35])/(10*(-2 + Sqrt[35]) )]*ArcTan[(-Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt [35])]] - ((35 + 4*Sqrt[35])*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10)/(2*Sqrt[14*(2 + Sqrt[35])]) + (Sqrt[(-2492 + 4 90*Sqrt[35])/(10*(-2 + Sqrt[35]))]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sq rt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] + ((35 + 4*Sqrt[35])*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10)/(2*Sqrt[14*(2 + Sqrt[35])])))/5
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[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol ] :> Simp[e*((d + e*x)^(m - 1)/(c*(m - 1))), x] + Simp[1/c Int[(d + e*x)^ (m - 2)*(Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]
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[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r) In t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[1/(2*c*q*r) Int[(d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && N eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
Time = 2.35 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.26
| method | result | size |
| pseudoelliptic | \(\frac {1240 \sqrt {1+2 x}\, \sqrt {10 \sqrt {5}\, \sqrt {7}-20}+\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (27 \sqrt {5}+10 \sqrt {7}\right ) \left (\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )-\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )\right ) \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+620 \left (\sqrt {5}\, \sqrt {7}-4\right ) \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )}{1550 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(243\) |
| derivativedivides | \(\frac {4 \sqrt {1+2 x}}{5}+\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}-\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(389\) |
| default | \(\frac {4 \sqrt {1+2 x}}{5}+\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}-\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(389\) |
| trager | \(\frac {4 \sqrt {1+2 x}}{5}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) \ln \left (\frac {4805 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{4} x +29605 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) x +1369425 \sqrt {1+2 x}\, \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}+4712 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right )-9000 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) x -9350375 \sqrt {1+2 x}+30400 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right )}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} x -235 x -76}\right )}{775}+\frac {\operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right ) \ln \left (-\frac {-24025 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{5}+258385 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{3} x +44175 \sqrt {1+2 x}\, \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}+23560 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{3}-421716 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )+200165 \sqrt {1+2 x}-206112 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} x -121 x +76}\right )}{5}\) | \(431\) |
| risch | \(\frac {4 \sqrt {1+2 x}}{5}+\frac {27 \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{1550}+\frac {\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right ) \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{155}+\frac {27 \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 ) \left (2 \sqrt {5}\, \sqrt {7}+4\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {2 \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 {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {4 \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 {5}\, \sqrt {7}}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {27 \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{1550}-\frac {\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{155}+\frac {27 \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 ) \left (2 \sqrt {5}\, \sqrt {7}+4\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {2 \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 {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {4 \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 {5}\, \sqrt {7}}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(616\) |
Input:
int((1+2*x)^(3/2)/(5*x^2+3*x+2),x,method=_RETURNVERBOSE)
Output:
1/1550/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(1240*(1+2*x)^(1/2)*(10*5^(1/2)*7^(1/ 2)-20)^(1/2)+(10*5^(1/2)*7^(1/2)-20)^(1/2)*(27*5^(1/2)+10*7^(1/2))*(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)-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) )*(2*5^(1/2)*7^(1/2)+4)^(1/2)+620*(5^(1/2)*7^(1/2)-4)*(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))))
Time = 0.09 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.21 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=-\frac {2}{5} \, \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} - \frac {89}{155}} \arctan \left (\frac {10}{19} \, \sqrt {2 \, x + 1} {\left (5 \, \sqrt {\frac {7}{5}} + 4\right )} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} - \frac {89}{155}} + \frac {10}{19} \, {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} + \frac {89}{155}} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} - \frac {89}{155}}\right ) + \frac {2}{5} \, \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} - \frac {89}{155}} \arctan \left (-\frac {10}{19} \, \sqrt {2 \, x + 1} {\left (5 \, \sqrt {\frac {7}{5}} + 4\right )} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} - \frac {89}{155}} + \frac {10}{19} \, {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} + \frac {89}{155}} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} - \frac {89}{155}}\right ) + \frac {1}{5} \, \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} + \frac {89}{155}} \log \left (2 \, \sqrt {2 \, x + 1} {\left (10 \, \sqrt {\frac {7}{5}} - 27\right )} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} + \frac {89}{155}} + 38 \, x + 19 \, \sqrt {\frac {7}{5}} + 19\right ) - \frac {1}{5} \, \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} + \frac {89}{155}} \log \left (-2 \, \sqrt {2 \, x + 1} {\left (10 \, \sqrt {\frac {7}{5}} - 27\right )} \sqrt {\frac {35}{62} \, \sqrt {\frac {7}{5}} + \frac {89}{155}} + 38 \, x + 19 \, \sqrt {\frac {7}{5}} + 19\right ) + \frac {4}{5} \, \sqrt {2 \, x + 1} \] Input:
integrate((1+2*x)^(3/2)/(5*x^2+3*x+2),x, algorithm="fricas")
Output:
-2/5*sqrt(35/62*sqrt(7/5) - 89/155)*arctan(10/19*sqrt(2*x + 1)*(5*sqrt(7/5 ) + 4)*sqrt(35/62*sqrt(7/5) - 89/155) + 10/19*(5*sqrt(7/5) + 2)*sqrt(35/62 *sqrt(7/5) + 89/155)*sqrt(35/62*sqrt(7/5) - 89/155)) + 2/5*sqrt(35/62*sqrt (7/5) - 89/155)*arctan(-10/19*sqrt(2*x + 1)*(5*sqrt(7/5) + 4)*sqrt(35/62*s qrt(7/5) - 89/155) + 10/19*(5*sqrt(7/5) + 2)*sqrt(35/62*sqrt(7/5) + 89/155 )*sqrt(35/62*sqrt(7/5) - 89/155)) + 1/5*sqrt(35/62*sqrt(7/5) + 89/155)*log (2*sqrt(2*x + 1)*(10*sqrt(7/5) - 27)*sqrt(35/62*sqrt(7/5) + 89/155) + 38*x + 19*sqrt(7/5) + 19) - 1/5*sqrt(35/62*sqrt(7/5) + 89/155)*log(-2*sqrt(2*x + 1)*(10*sqrt(7/5) - 27)*sqrt(35/62*sqrt(7/5) + 89/155) + 38*x + 19*sqrt( 7/5) + 19) + 4/5*sqrt(2*x + 1)
\[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {3}{2}}}{5 x^{2} + 3 x + 2}\, dx \] Input:
integrate((1+2*x)**(3/2)/(5*x**2+3*x+2),x)
Output:
Integral((2*x + 1)**(3/2)/(5*x**2 + 3*x + 2), x)
\[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\int { \frac {{\left (2 \, x + 1\right )}^{\frac {3}{2}}}{5 \, x^{2} + 3 \, x + 2} \,d x } \] Input:
integrate((1+2*x)^(3/2)/(5*x^2+3*x+2),x, algorithm="maxima")
Output:
integrate((2*x + 1)^(3/2)/(5*x^2 + 3*x + 2), x)
Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (134) = 268\).
Time = 0.92 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.08 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\text {Too large to display} \] Input:
integrate((1+2*x)^(3/2)/(5*x^2+3*x+2),x, algorithm="giac")
Output:
1/46519375*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140* sqrt(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) - 8575*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) - 17150*(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*sqr t(35) + 1/2)) + 1/46519375*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2 450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*s qrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 8575*sqrt(31)*(7/5)^(1/4)*sqr t(-140*sqrt(35) + 2450) - 17150*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arc tan(-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)) + 1/93038750*sqrt(31)*(sqrt(31)*(7/5)^(3/4) *(140*sqrt(35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sq rt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) - 8575*sqrt(31 )*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) + 17150*(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) - 1/93038750*sqrt(31)*(sqrt(31)*(7/5)^(3/4)*(140*sqrt (35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)...
Time = 0.09 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.94 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {4\,\sqrt {2\,x+1}}{5}-\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {178-\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}\,2432{}\mathrm {i}}{390625\,\left (-\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}+\frac {4864\,\sqrt {31}\,\sqrt {155}\,\sqrt {178-\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}}{12109375\,\left (-\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}\right )\,\sqrt {178-\sqrt {31}\,19{}\mathrm {i}}\,2{}\mathrm {i}}{775}+\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {178+\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}\,2432{}\mathrm {i}}{390625\,\left (\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}-\frac {4864\,\sqrt {31}\,\sqrt {155}\,\sqrt {178+\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}}{12109375\,\left (\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}\right )\,\sqrt {178+\sqrt {31}\,19{}\mathrm {i}}\,2{}\mathrm {i}}{775} \] Input:
int((2*x + 1)^(3/2)/(3*x + 5*x^2 + 2),x)
Output:
(4*(2*x + 1)^(1/2))/5 - (155^(1/2)*atan((155^(1/2)*(178 - 31^(1/2)*19i)^(1 /2)*(2*x + 1)^(1/2)*2432i)/(390625*((31^(1/2)*17024i)/78125 - 34048/78125) ) + (4864*31^(1/2)*155^(1/2)*(178 - 31^(1/2)*19i)^(1/2)*(2*x + 1)^(1/2))/( 12109375*((31^(1/2)*17024i)/78125 - 34048/78125)))*(178 - 31^(1/2)*19i)^(1 /2)*2i)/775 + (155^(1/2)*atan((155^(1/2)*(31^(1/2)*19i + 178)^(1/2)*(2*x + 1)^(1/2)*2432i)/(390625*((31^(1/2)*17024i)/78125 + 34048/78125)) - (4864* 31^(1/2)*155^(1/2)*(31^(1/2)*19i + 178)^(1/2)*(2*x + 1)^(1/2))/(12109375*( (31^(1/2)*17024i)/78125 + 34048/78125)))*(31^(1/2)*19i + 178)^(1/2)*2i)/77 5
Time = 0.20 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.61 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=-\frac {2 \sqrt {\sqrt {35}-2}\, \sqrt {14}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{155}+\frac {27 \sqrt {\sqrt {35}-2}\, \sqrt {10}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{775}+\frac {2 \sqrt {\sqrt {35}-2}\, \sqrt {14}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{155}-\frac {27 \sqrt {\sqrt {35}-2}\, \sqrt {10}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{775}+\frac {\sqrt {\sqrt {35}+2}\, \sqrt {14}\, \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{155}-\frac {\sqrt {\sqrt {35}+2}\, \sqrt {14}\, \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{155}+\frac {27 \sqrt {\sqrt {35}+2}\, \sqrt {10}\, \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{1550}-\frac {27 \sqrt {\sqrt {35}+2}\, \sqrt {10}\, \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{1550}+\frac {4 \sqrt {2 x +1}}{5} \] Input:
int((1+2*x)^(3/2)/(5*x^2+3*x+2),x)
Output:
( - 20*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sq rt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 54*sqrt(sqrt(35) - 2) *sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqr t(sqrt(35) - 2)*sqrt(2))) + 20*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt (35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) - 54*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqr t(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 10*sqrt(sqrt(35) + 2)* sqrt(14)*log( - sqrt(2*x + 1)*sqrt(sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sqr t(5)*x + sqrt(5)) - 10*sqrt(sqrt(35) + 2)*sqrt(14)*log(sqrt(2*x + 1)*sqrt( sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sqrt(5)*x + sqrt(5)) + 27*sqrt(sqrt(35 ) + 2)*sqrt(10)*log( - sqrt(2*x + 1)*sqrt(sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sqrt(5)*x + sqrt(5)) - 27*sqrt(sqrt(35) + 2)*sqrt(10)*log(sqrt(2*x + 1 )*sqrt(sqrt(35) + 2)*sqrt(2) + sqrt(7) + 2*sqrt(5)*x + sqrt(5)) + 1240*sqr t(2*x + 1))/1550