\(\int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx\) [553]

Optimal result

Integrand size = 22, antiderivative size = 206 \[ \int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx=\frac {16}{25} \sqrt {1+2 x}+\frac {4}{15} (1+2 x)^{3/2}+\frac {1}{25} \sqrt {\frac {2}{155} \left (7162+1225 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {1}{25} \sqrt {\frac {2}{155} \left (7162+1225 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {1}{25} \sqrt {\frac {2}{155} \left (-7162+1225 \sqrt {35}\right )} \text {arctanh}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}}{5+\sqrt {35}+10 x}\right ) \] Output:

16/25*(1+2*x)^(1/2)+4/15*(1+2*x)^(3/2)+1/3875*(2220220+379750*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))-1/3875*(2220220+379750*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))-1/3875*(-2220220+379750*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 = 0.47 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx=\frac {620 \sqrt {1+2 x} (17+10 x)-6 \sqrt {155 \left (7162+199 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )-6 \sqrt {155 \left (7162-199 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )}{11625} \] Input:

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

Output:

(620*Sqrt[1 + 2*x]*(17 + 10*x) - 6*Sqrt[155*(7162 + (199*I)*Sqrt[31])]*Arc 
Tan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] - 6*Sqrt[155*(7162 - (199*I)* 
Sqrt[31])]*ArcTan[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x]])/11625
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1146, 25, 1196, 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.

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

\(\Big \downarrow \) 1146

\(\displaystyle \frac {1}{5} \int -\frac {(3-8 x) \sqrt {2 x+1}}{5 x^2+3 x+2}dx+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4}{15} (2 x+1)^{3/2}-\frac {1}{5} \int \frac {(3-8 x) \sqrt {2 x+1}}{5 x^2+3 x+2}dx\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {1}{5} \int \frac {38 x+47}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {2}{5} \int \frac {2 (19 (2 x+1)+28)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \int \frac {19 (2 x+1)+28}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 1483

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\int \frac {28 \sqrt {10 \left (2+\sqrt {35}\right )}-\left (140-19 \sqrt {35}\right ) \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\int \frac {\left (140-19 \sqrt {35}\right ) \sqrt {2 x+1}+28 \sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\sqrt {\frac {7}{2} \left (7162+1225 \sqrt {35}\right )} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int -\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\sqrt {\frac {7}{2} \left (7162+1225 \sqrt {35}\right )} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\sqrt {\frac {7}{2} \left (7162+1225 \sqrt {35}\right )} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\sqrt {\frac {7}{2} \left (7162+1225 \sqrt {35}\right )} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {14 \left (7162+1225 \sqrt {35}\right )} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {14 \left (7162+1225 \sqrt {35}\right )} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {7 \left (7162+1225 \sqrt {35}\right )}{5 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (140-19 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {7 \left (7162+1225 \sqrt {35}\right )}{5 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\right )+\frac {4}{15} (2 x+1)^{3/2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{5} \left (\frac {16}{5} \sqrt {2 x+1}-\frac {4}{5} \left (\frac {\sqrt {\frac {7 \left (7162+1225 \sqrt {35}\right )}{5 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \left (140-19 \sqrt {35}\right ) \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\sqrt {\frac {7 \left (7162+1225 \sqrt {35}\right )}{5 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \left (140-19 \sqrt {35}\right ) \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )\right )+\frac {4}{15} (2 x+1)^{3/2}\)

Input:

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

Output:

(4*(1 + 2*x)^(3/2))/15 + ((16*Sqrt[1 + 2*x])/5 - (4*((Sqrt[(7*(7162 + 1225 
*Sqrt[35]))/(5*(-2 + Sqrt[35]))]*ArcTan[(-Sqrt[10*(2 + Sqrt[35])] + 10*Sqr 
t[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] - ((140 - 19*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[(7*(7162 + 1225*Sqrt[35]))/(5*(-2 + Sqrt[35]))]*ArcT 
an[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] 
+ ((140 - 19*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)/5
 

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 1146

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]
 

rule 1196

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c   Int 
[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + 
 b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & 
& GtQ[m, 0]
 

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 = 2.43 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.20

method	result	size
pseudoelliptic	\(\frac {12400 \left (x +\frac {17}{10}\right ) \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {1+2 x}+3 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (178 \sqrt {5}-135 \sqrt {7}\right ) \left (\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )-\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )\right ) \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+7440 \left (\sqrt {5}\, \sqrt {7}+\frac {19}{4}\right ) \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )}{23250 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(247\)
derivativedivides	\(\frac {4 \left (1+2 x \right )^{\frac {3}{2}}}{15}+\frac {16 \sqrt {1+2 x}}{25}+\frac {\left (178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{7750}+\frac {2 \left (-248 \sqrt {5}\, \sqrt {7}+\frac {\left (178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (-178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{7750}+\frac {2 \left (-248 \sqrt {5}\, \sqrt {7}-\frac {\left (-178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(398\)
default	\(\frac {4 \left (1+2 x \right )^{\frac {3}{2}}}{15}+\frac {16 \sqrt {1+2 x}}{25}+\frac {\left (178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{7750}+\frac {2 \left (-248 \sqrt {5}\, \sqrt {7}+\frac {\left (178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (-178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{7750}+\frac {2 \left (-248 \sqrt {5}\, \sqrt {7}-\frac {\left (-178 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+135 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(398\)
trager	\(\left (\frac {8 x}{15}+\frac {68}{75}\right ) \sqrt {1+2 x}-\frac {\operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right ) \ln \left (-\frac {1537600 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{5}[phone] \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{3} x [phone] \sqrt {1+2 x}\, \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}-15792640 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{3}[phone] \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right ) x [phone] \sqrt {1+2 x}-700258712 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2} x +7759 x +796}\right )}{25}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}+2220220\right ) \ln \left (-\frac {307520 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}+2220220\right ) \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{4} x [phone] \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}+2220220\right ) x +3158528 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}+2220220\right )-502002375 \sqrt {1+2 x}\, \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}+2220220\right ) x [phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2}+2220220\right )[phone] \sqrt {1+2 x}}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+444044 \textit {\_Z}^{2}[phone]\right )^{2} x +6565 x -796}\right )}{3875}\)	\(436\)
risch	\(\frac {4 \left (10 x +17\right ) \sqrt {1+2 x}}{75}+\frac {89 \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}}{3875}-\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 {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{1550}+\frac {178 \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 )}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\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 ) \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {16 \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}}{25 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {89 \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}}{3875}+\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 {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{1550}+\frac {178 \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 )}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\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 ) \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {16 \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}}{25 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(621\)

Input:

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

Output:

1/23250/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(12400*(x+17/10)*(10*5^(1/2)*7^(1/2) 
-20)^(1/2)*(1+2*x)^(1/2)+3*(10*5^(1/2)*7^(1/2)-20)^(1/2)*(178*5^(1/2)-135* 
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)+7440*(5^(1/2)*7^(1/2)+19/4)* 
(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))))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.08 \[ \int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx=\frac {4}{75} \, {\left (10 \, x + 17\right )} \sqrt {2 \, x + 1} - \frac {2}{25} \, \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} + \frac {3581}{155}} \arctan \left (\frac {10}{199} \, {\left (\sqrt {2 \, x + 1} {\left (20 \, \sqrt {\frac {7}{5}} - 19\right )} + \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} - \frac {3581}{155}} {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )}\right )} \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} + \frac {3581}{155}}\right ) + \frac {2}{25} \, \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} + \frac {3581}{155}} \arctan \left (-\frac {10}{199} \, {\left (\sqrt {2 \, x + 1} {\left (20 \, \sqrt {\frac {7}{5}} - 19\right )} - \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} - \frac {3581}{155}} {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )}\right )} \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} + \frac {3581}{155}}\right ) - \frac {1}{25} \, \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} - \frac {3581}{155}} \log \left (2 \, \sqrt {2 \, x + 1} {\left (135 \, \sqrt {\frac {7}{5}} + 178\right )} \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} - \frac {3581}{155}} + 398 \, x + 199 \, \sqrt {\frac {7}{5}} + 199\right ) + \frac {1}{25} \, \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} - \frac {3581}{155}} \log \left (-2 \, \sqrt {2 \, x + 1} {\left (135 \, \sqrt {\frac {7}{5}} + 178\right )} \sqrt {\frac {1225}{62} \, \sqrt {\frac {7}{5}} - \frac {3581}{155}} + 398 \, x + 199 \, \sqrt {\frac {7}{5}} + 199\right ) \] Input:

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

Output:

4/75*(10*x + 17)*sqrt(2*x + 1) - 2/25*sqrt(1225/62*sqrt(7/5) + 3581/155)*a 
rctan(10/199*(sqrt(2*x + 1)*(20*sqrt(7/5) - 19) + sqrt(1225/62*sqrt(7/5) - 
 3581/155)*(5*sqrt(7/5) + 2))*sqrt(1225/62*sqrt(7/5) + 3581/155)) + 2/25*s 
qrt(1225/62*sqrt(7/5) + 3581/155)*arctan(-10/199*(sqrt(2*x + 1)*(20*sqrt(7 
/5) - 19) - sqrt(1225/62*sqrt(7/5) - 3581/155)*(5*sqrt(7/5) + 2))*sqrt(122 
5/62*sqrt(7/5) + 3581/155)) - 1/25*sqrt(1225/62*sqrt(7/5) - 3581/155)*log( 
2*sqrt(2*x + 1)*(135*sqrt(7/5) + 178)*sqrt(1225/62*sqrt(7/5) - 3581/155) + 
 398*x + 199*sqrt(7/5) + 199) + 1/25*sqrt(1225/62*sqrt(7/5) - 3581/155)*lo 
g(-2*sqrt(2*x + 1)*(135*sqrt(7/5) + 178)*sqrt(1225/62*sqrt(7/5) - 3581/155 
) + 398*x + 199*sqrt(7/5) + 199)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 605 vs. \(2 (143) = 286\).

Time = 0.89 (sec) , antiderivative size = 605, normalized size of antiderivative = 2.94 \[ \int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/930387500*sqrt(31)*(3990*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-1 
40*sqrt(35) + 2450) - 19*sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) 
 + 38*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 7980*(7/5)^(3/4)*sqrt(140* 
sqrt(35) + 2450)*(2*sqrt(35) - 35) + 137200*sqrt(31)*(7/5)^(1/4)*sqrt(-140 
*sqrt(35) + 2450) + 274400*(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))/sqr 
t(-1/35*sqrt(35) + 1/2)) - 1/930387500*sqrt(31)*(3990*sqrt(31)*(7/5)^(3/4) 
*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 19*sqrt(31)*(7/5)^(3/4)*(- 
140*sqrt(35) + 2450)^(3/2) + 38*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 
7980*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 137200*sqrt 
(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 274400*(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)) - 1/1860775000*sqrt(31 
)*(19*sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 3990*sqrt(31)*(7/ 
5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 7980*(7/5)^(3/4)*(2 
*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) + 38*(7/5)^(3/4)*(-140*sqrt(35) 
 + 2450)^(3/2) + 137200*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) - 2 
74400*(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) + 1/1860775000*sqrt(3 
1)*(19*sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 3990*sqrt(31)...
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.93 \[ \int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx=\frac {16\,\sqrt {2\,x+1}}{25}+\frac {4\,{\left (2\,x+1\right )}^{3/2}}{15}-\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-7162-\sqrt {31}\,199{}\mathrm {i}}\,\sqrt {2\,x+1}\,25472{}\mathrm {i}}{48828125\,\left (\frac {4814208}{9765625}+\frac {\sqrt {31}\,713216{}\mathrm {i}}{9765625}\right )}+\frac {50944\,\sqrt {31}\,\sqrt {155}\,\sqrt {-7162-\sqrt {31}\,199{}\mathrm {i}}\,\sqrt {2\,x+1}}{1513671875\,\left (\frac {4814208}{9765625}+\frac {\sqrt {31}\,713216{}\mathrm {i}}{9765625}\right )}\right )\,\sqrt {-7162-\sqrt {31}\,199{}\mathrm {i}}\,2{}\mathrm {i}}{3875}+\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-7162+\sqrt {31}\,199{}\mathrm {i}}\,\sqrt {2\,x+1}\,25472{}\mathrm {i}}{48828125\,\left (-\frac {4814208}{9765625}+\frac {\sqrt {31}\,713216{}\mathrm {i}}{9765625}\right )}-\frac {50944\,\sqrt {31}\,\sqrt {155}\,\sqrt {-7162+\sqrt {31}\,199{}\mathrm {i}}\,\sqrt {2\,x+1}}{1513671875\,\left (-\frac {4814208}{9765625}+\frac {\sqrt {31}\,713216{}\mathrm {i}}{9765625}\right )}\right )\,\sqrt {-7162+\sqrt {31}\,199{}\mathrm {i}}\,2{}\mathrm {i}}{3875} \] Input:

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

Output:

(16*(2*x + 1)^(1/2))/25 + (4*(2*x + 1)^(3/2))/15 - (155^(1/2)*atan((155^(1 
/2)*(- 31^(1/2)*199i - 7162)^(1/2)*(2*x + 1)^(1/2)*25472i)/(48828125*((31^ 
(1/2)*713216i)/9765625 + 4814208/9765625)) + (50944*31^(1/2)*155^(1/2)*(- 
31^(1/2)*199i - 7162)^(1/2)*(2*x + 1)^(1/2))/(1513671875*((31^(1/2)*713216 
i)/9765625 + 4814208/9765625)))*(- 31^(1/2)*199i - 7162)^(1/2)*2i)/3875 + 
(155^(1/2)*atan((155^(1/2)*(31^(1/2)*199i - 7162)^(1/2)*(2*x + 1)^(1/2)*25 
472i)/(48828125*((31^(1/2)*713216i)/9765625 - 4814208/9765625)) - (50944*3 
1^(1/2)*155^(1/2)*(31^(1/2)*199i - 7162)^(1/2)*(2*x + 1)^(1/2))/(151367187 
5*((31^(1/2)*713216i)/9765625 - 4814208/9765625)))*(31^(1/2)*199i - 7162)^ 
(1/2)*2i)/3875
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.55 \[ \int \frac {(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx=\frac {27 \sqrt {\sqrt {35}-2}\, \sqrt {14}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{775}+\frac {178 \sqrt {\sqrt {35}-2}\, \sqrt {10}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}-2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{3875}-\frac {27 \sqrt {\sqrt {35}-2}\, \sqrt {14}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{775}-\frac {178 \sqrt {\sqrt {35}-2}\, \sqrt {10}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {35}+2}\, \sqrt {2}+2 \sqrt {2 x +1}\, \sqrt {5}}{\sqrt {\sqrt {35}-2}\, \sqrt {2}}\right )}{3875}-\frac {27 \sqrt {\sqrt {35}+2}\, \sqrt {14}\, \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{1550}+\frac {27 \sqrt {\sqrt {35}+2}\, \sqrt {14}\, \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{1550}+\frac {89 \sqrt {\sqrt {35}+2}\, \sqrt {10}\, \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{3875}-\frac {89 \sqrt {\sqrt {35}+2}\, \sqrt {10}\, \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {\sqrt {35}+2}\, \sqrt {2}+\sqrt {7}+2 \sqrt {5}\, x +\sqrt {5}\right )}{3875}+\frac {8 \sqrt {2 x +1}\, x}{15}+\frac {68 \sqrt {2 x +1}}{75} \] Input:

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

Output:

(810*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))) + 1068*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))) - 810*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqr 
t(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)) 
) - 1068*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))) - 405*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)) + 405*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)) + 534*sqrt(s 
qrt(35) + 2)*sqrt(10)*log( - sqrt(2*x + 1)*sqrt(sqrt(35) + 2)*sqrt(2) + sq 
rt(7) + 2*sqrt(5)*x + sqrt(5)) - 534*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)) + 1 
2400*sqrt(2*x + 1)*x + 21080*sqrt(2*x + 1))/23250