\(\int \frac {(1+2 x)^{3/2}}{(2+3 x+5 x^2)^2} \, dx\) [561]

Optimal result

Integrand size = 22, antiderivative size = 210 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=-\frac {(5-4 x) \sqrt {1+2 x}}{31 \left (2+3 x+5 x^2\right )}-\frac {1}{31} \sqrt {\frac {2}{155} \left (218+47 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {1}{31} \sqrt {\frac {2}{155} \left (218+47 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {1}{31} \sqrt {\frac {2}{155} \left (-218+47 \sqrt {35}\right )} \text {arctanh}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}}{5+\sqrt {35}+10 x}\right ) \] Output:

-1/31*(5-4*x)*(1+2*x)^(1/2)/(5*x^2+3*x+2)-1/4805*(67580+14570*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))+1/4805*(67580+14570*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))+1/4805*(-67580+14570*35^(1/2))^(1/ 
2)*arctanh((20+10*35^(1/2))^(1/2)*(1+2*x)^(1/2)/(5+35^(1/2)+10*x))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.01 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.61 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\frac {2 \left (\frac {155 \sqrt {1+2 x} (-5+4 x)}{4+6 x+10 x^2}+\sqrt {155 \left (218+31 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )+\sqrt {155 \left (218-31 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right )}{4805} \] Input:

Integrate[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2)^2,x]
 

Output:

(2*((155*Sqrt[1 + 2*x]*(-5 + 4*x))/(4 + 6*x + 10*x^2) + Sqrt[155*(218 + (3 
1*I)*Sqrt[31])]*ArcTan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] + Sqrt[155 
*(218 - (31*I)*Sqrt[31])]*ArcTan[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x 
]]))/4805
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.47, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1164, 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.

\(\displaystyle \int \frac {(2 x+1)^{3/2}}{\left (5 x^2+3 x+2\right )^2} \, dx\)

\(\Big \downarrow \) 1164

\(\displaystyle \frac {1}{31} \int \frac {4 x+9}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {2}{31} \int \frac {2 (2 (2 x+1)+7)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4}{31} \int \frac {2 (2 x+1)+7}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1483

\(\displaystyle \frac {4}{31} \left (\frac {\int \frac {7 \sqrt {10 \left (2+\sqrt {35}\right )}-\left (35-2 \sqrt {35}\right ) \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\int \frac {\left (35-2 \sqrt {35}\right ) \sqrt {2 x+1}+7 \sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {4}{31} \left (\frac {\frac {1}{2} \sqrt {3052+658 \sqrt {35}} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int -\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{2} \sqrt {3052+658 \sqrt {35}} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4}{31} \left (\frac {\frac {1}{2} \sqrt {3052+658 \sqrt {35}} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{2} \sqrt {3052+658 \sqrt {35}} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {4}{31} \left (\frac {\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {3052+658 \sqrt {35}} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {3052+658 \sqrt {35}} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {4}{31} \left (\frac {\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {3052+658 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (35-2 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {3052+658 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {4}{31} \left (\frac {\sqrt {\frac {3052+658 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \left (35-2 \sqrt {35}\right ) \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\sqrt {\frac {3052+658 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \left (35-2 \sqrt {35}\right ) \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{31 \left (5 x^2+3 x+2\right )}\)

Input:

Int[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2)^2,x]
 

Output:

-1/31*((5 - 4*x)*Sqrt[1 + 2*x])/(2 + 3*x + 5*x^2) + (4*((Sqrt[(3052 + 658* 
Sqrt[35])/(10*(-2 + Sqrt[35]))]*ArcTan[(-Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt 
[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] - ((35 - 2*Sqrt[35])*Log[Sqrt[35] - S 
qrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10)/(2*Sqrt[14*(2 + S 
qrt[35])]) + (Sqrt[(3052 + 658*Sqrt[35])/(10*(-2 + Sqrt[35]))]*ArcTan[(Sqr 
t[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] + ((35 
- 2*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])])))/31
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 1083

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]
 

rule 1103

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]
 

rule 1142

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]
 

rule 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1197

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]
 

rule 1483

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]
 

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.29

method	result	size
pseudoelliptic	\(-\frac {-1240 \left (x -\frac {5}{4}\right ) \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {1+2 x}+\left (5 x^{2}+3 x +2\right ) \left (\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (39 \sqrt {5}-20 \sqrt {7}\right ) \left (\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )-\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )\right ) \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+620 \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right ) \left (\sqrt {5}\, \sqrt {7}+2\right )\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (48050 x^{2}+28830 x +19220\right )}\)	\(270\)
derivativedivides	\(\frac {\frac {8 \left (1+2 x \right )^{\frac {3}{2}}}{155}-\frac {28 \sqrt {1+2 x}}{155}}{\left (1+2 x \right )^{2}+\frac {3}{5}-\frac {8 x}{5}}+\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{9610}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}-\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{961 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{9610}-\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}+\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{961 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(415\)
default	\(\frac {\frac {8 \left (1+2 x \right )^{\frac {3}{2}}}{155}-\frac {28 \sqrt {1+2 x}}{155}}{\left (1+2 x \right )^{2}+\frac {3}{5}-\frac {8 x}{5}}+\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{9610}+\frac {2 \left (62 \sqrt {5}\, \sqrt {7}-\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{961 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{9610}-\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}+\frac {\left (39 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-20 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{961 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(415\)
trager	\(\frac {\left (-5+4 x \right ) \sqrt {1+2 x}}{155 x^{2}+93 x +62}-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+16895\right ) \ln \left (-\frac {12147040 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+16895\right ) \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{4} x [phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+16895\right ) x [phone] \sqrt {1+2 x}\, \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+4858816 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+16895\right )+4334340 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+16895\right ) x -165085385 \sqrt {1+2 x}+2471072 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}+16895\right )}{620 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2} x +125 x -124}\right )}{4805}+\frac {2 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right ) \ln \left (\frac {60735200 x \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{5}[phone] x \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{3}+2710020 \sqrt {1+2 x}\, \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2}-24294080 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{3}+19068 x \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )+7231091 \sqrt {1+2 x}-4728864 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )}{620 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+54064 \textit {\_Z}^{2}+15463\right )^{2} x +311 x +124}\right )}{31}\)	\(448\)
risch	\(\frac {\left (-5+4 x \right ) \sqrt {1+2 x}}{155 x^{2}+93 x +62}-\frac {39 \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}}{9610}+\frac {2 \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}}{961}-\frac {39 \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 )}{961 \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}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{961 \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}}{31 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {39 \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}}{9610}-\frac {2 \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}}{961}-\frac {39 \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 )}{961 \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}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{961 \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}}{31 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(633\)

Input:

int((1+2*x)^(3/2)/(5*x^2+3*x+2)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(-1240*(x-5/4)*(10*5^(1/2)*7^(1/2)-20)^(1 
/2)*(1+2*x)^(1/2)+(5*x^2+3*x+2)*((10*5^(1/2)*7^(1/2)-20)^(1/2)*(39*5^(1/2) 
-20*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*(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)))*(5^(1/2)*7^(1/2)+2)))/(48050*x^2+28830*x+19220)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.30 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\frac {2 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} + \frac {109}{155}} \arctan \left (\frac {10}{31} \, {\left (\sqrt {2 \, x + 1} {\left (5 \, \sqrt {\frac {7}{5}} - 2\right )} + {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} - \frac {109}{155}}\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} + \frac {109}{155}}\right ) - 2 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} + \frac {109}{155}} \arctan \left (-\frac {10}{31} \, {\left (\sqrt {2 \, x + 1} {\left (5 \, \sqrt {\frac {7}{5}} - 2\right )} - {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} - \frac {109}{155}}\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} + \frac {109}{155}}\right ) + {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} - \frac {109}{155}} \log \left (2 \, \sqrt {2 \, x + 1} {\left (20 \, \sqrt {\frac {7}{5}} + 39\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} - \frac {109}{155}} + 62 \, x + 31 \, \sqrt {\frac {7}{5}} + 31\right ) - {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} - \frac {109}{155}} \log \left (-2 \, \sqrt {2 \, x + 1} {\left (20 \, \sqrt {\frac {7}{5}} + 39\right )} \sqrt {\frac {47}{62} \, \sqrt {\frac {7}{5}} - \frac {109}{155}} + 62 \, x + 31 \, \sqrt {\frac {7}{5}} + 31\right ) + {\left (4 \, x - 5\right )} \sqrt {2 \, x + 1}}{31 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \] Input:

integrate((1+2*x)^(3/2)/(5*x^2+3*x+2)^2,x, algorithm="fricas")
 

Output:

1/31*(2*(5*x^2 + 3*x + 2)*sqrt(47/62*sqrt(7/5) + 109/155)*arctan(10/31*(sq 
rt(2*x + 1)*(5*sqrt(7/5) - 2) + (5*sqrt(7/5) + 2)*sqrt(47/62*sqrt(7/5) - 1 
09/155))*sqrt(47/62*sqrt(7/5) + 109/155)) - 2*(5*x^2 + 3*x + 2)*sqrt(47/62 
*sqrt(7/5) + 109/155)*arctan(-10/31*(sqrt(2*x + 1)*(5*sqrt(7/5) - 2) - (5* 
sqrt(7/5) + 2)*sqrt(47/62*sqrt(7/5) - 109/155))*sqrt(47/62*sqrt(7/5) + 109 
/155)) + (5*x^2 + 3*x + 2)*sqrt(47/62*sqrt(7/5) - 109/155)*log(2*sqrt(2*x 
+ 1)*(20*sqrt(7/5) + 39)*sqrt(47/62*sqrt(7/5) - 109/155) + 62*x + 31*sqrt( 
7/5) + 31) - (5*x^2 + 3*x + 2)*sqrt(47/62*sqrt(7/5) - 109/155)*log(-2*sqrt 
(2*x + 1)*(20*sqrt(7/5) + 39)*sqrt(47/62*sqrt(7/5) - 109/155) + 62*x + 31* 
sqrt(7/5) + 31) + (4*x - 5)*sqrt(2*x + 1))/(5*x^2 + 3*x + 2)
 

Sympy [F]

\[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {3}{2}}}{\left (5 x^{2} + 3 x + 2\right )^{2}}\, dx \] Input:

integrate((1+2*x)**(3/2)/(5*x**2+3*x+2)**2,x)
 

Output:

Integral((2*x + 1)**(3/2)/(5*x**2 + 3*x + 2)**2, x)
 

Maxima [F]

\[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\int { \frac {{\left (2 \, x + 1\right )}^{\frac {3}{2}}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{2}} \,d x } \] Input:

integrate((1+2*x)^(3/2)/(5*x^2+3*x+2)^2,x, algorithm="maxima")
 

Output:

integrate((2*x + 1)^(3/2)/(5*x^2 + 3*x + 2)^2, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 622 vs. \(2 (151) = 302\).

Time = 0.80 (sec) , antiderivative size = 622, normalized size of antiderivative = 2.96 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((1+2*x)^(3/2)/(5*x^2+3*x+2)^2,x, algorithm="giac")
 

Output:

1/576840250*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) + 17150*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35 
) + 2450) + 34300*(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*s 
qrt(35) + 1/2)) + 1/576840250*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(3 
5) + 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) + 17150*sqrt(31)*(7/5)^(1/4) 
*sqrt(-140*sqrt(35) + 2450) + 34300*(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)) + 1/1153680500*sqrt(31)*(sqrt(31)*(7/5) 
^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqr 
t(35) + 2450)*(2*sqrt(35) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(- 
140*sqrt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 17150* 
sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) - 34300*(7/5)^(1/4)*sqrt(-1 
40*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) - 1/1153680500*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...
 

Mupad [B] (verification not implemented)

Time = 5.62 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.99 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=-\frac {\frac {28\,\sqrt {2\,x+1}}{155}-\frac {8\,{\left (2\,x+1\right )}^{3/2}}{155}}{{\left (2\,x+1\right )}^2-\frac {8\,x}{5}+\frac {3}{5}}+\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-218-\sqrt {31}\,31{}\mathrm {i}}\,\sqrt {2\,x+1}\,128{}\mathrm {i}}{3003125\,\left (\frac {3584}{600625}+\frac {\sqrt {31}\,896{}\mathrm {i}}{600625}\right )}+\frac {256\,\sqrt {31}\,\sqrt {155}\,\sqrt {-218-\sqrt {31}\,31{}\mathrm {i}}\,\sqrt {2\,x+1}}{93096875\,\left (\frac {3584}{600625}+\frac {\sqrt {31}\,896{}\mathrm {i}}{600625}\right )}\right )\,\sqrt {-218-\sqrt {31}\,31{}\mathrm {i}}\,2{}\mathrm {i}}{4805}-\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-218+\sqrt {31}\,31{}\mathrm {i}}\,\sqrt {2\,x+1}\,128{}\mathrm {i}}{3003125\,\left (-\frac {3584}{600625}+\frac {\sqrt {31}\,896{}\mathrm {i}}{600625}\right )}-\frac {256\,\sqrt {31}\,\sqrt {155}\,\sqrt {-218+\sqrt {31}\,31{}\mathrm {i}}\,\sqrt {2\,x+1}}{93096875\,\left (-\frac {3584}{600625}+\frac {\sqrt {31}\,896{}\mathrm {i}}{600625}\right )}\right )\,\sqrt {-218+\sqrt {31}\,31{}\mathrm {i}}\,2{}\mathrm {i}}{4805} \] Input:

int((2*x + 1)^(3/2)/(3*x + 5*x^2 + 2)^2,x)
 

Output:

(155^(1/2)*atan((155^(1/2)*(- 31^(1/2)*31i - 218)^(1/2)*(2*x + 1)^(1/2)*12 
8i)/(3003125*((31^(1/2)*896i)/600625 + 3584/600625)) + (256*31^(1/2)*155^( 
1/2)*(- 31^(1/2)*31i - 218)^(1/2)*(2*x + 1)^(1/2))/(93096875*((31^(1/2)*89 
6i)/600625 + 3584/600625)))*(- 31^(1/2)*31i - 218)^(1/2)*2i)/4805 - ((28*( 
2*x + 1)^(1/2))/155 - (8*(2*x + 1)^(3/2))/155)/((2*x + 1)^2 - (8*x)/5 + 3/ 
5) - (155^(1/2)*atan((155^(1/2)*(31^(1/2)*31i - 218)^(1/2)*(2*x + 1)^(1/2) 
*128i)/(3003125*((31^(1/2)*896i)/600625 - 3584/600625)) - (256*31^(1/2)*15 
5^(1/2)*(31^(1/2)*31i - 218)^(1/2)*(2*x + 1)^(1/2))/(93096875*((31^(1/2)*8 
96i)/600625 - 3584/600625)))*(31^(1/2)*31i - 218)^(1/2)*2i)/4805
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 969, normalized size of antiderivative = 4.61 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((1+2*x)^(3/2)/(5*x^2+3*x+2)^2,x)
 

Output:

( - 200*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*s 
qrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x**2 - 120*sqrt(sqrt(3 
5) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5 
))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x - 80*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))) - 390*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt( 
2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x**2 - 234*sqr 
t(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1 
)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x - 156*sqrt(sqrt(35) - 2)*sqrt(1 
0)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt( 
35) - 2)*sqrt(2))) + 200*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)))*x**2 
+ 120*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqr 
t(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x + 80*sqrt(sqrt(35) - 2 
)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sq 
rt(sqrt(35) - 2)*sqrt(2))) + 390*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sq 
rt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2) 
))*x**2 + 234*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) 
 + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x + 156*sqrt(sqr 
t(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*...