\(\int \frac {(1+2 x)^{5/2}}{(2+3 x+5 x^2)^2} \, dx\) [560]

Optimal result

Integrand size = 22, antiderivative size = 223 \[ \int \frac {(1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=-\frac {8}{155} \sqrt {1+2 x}-\frac {(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}-\frac {1}{155} \sqrt {\frac {2}{155} \left (32678+10325 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {1}{155} \sqrt {\frac {2}{155} \left (32678+10325 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {1}{155} \sqrt {\frac {2}{155} \left (-32678+10325 \sqrt {35}\right )} \text {arctanh}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}}{5+\sqrt {35}+10 x}\right ) \] Output:

-8/155*(1+2*x)^(1/2)-(5-4*x)*(1+2*x)^(3/2)/(155*x^2+93*x+62)-1/24025*(1013 
0180+3200750*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))+1/24025*(10130180+3200750*35^(1/2))^(1/2)*arc 
tan(((20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*35^(1/2))^(1/2))-1/2 
4025*(-10130180+3200750*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 = 1.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.58 \[ \int \frac {(1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\frac {2 \left (-\frac {155 \sqrt {1+2 x} (41+54 x)}{4+6 x+10 x^2}+\sqrt {155 \left (32678-9269 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )+\sqrt {155 \left (32678+9269 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right )}{24025} \] Input:

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

Output:

(2*((-155*Sqrt[1 + 2*x]*(41 + 54*x))/(4 + 6*x + 10*x^2) + Sqrt[155*(32678 
- (9269*I)*Sqrt[31])]*ArcTan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] + Sq 
rt[155*(32678 + (9269*I)*Sqrt[31])]*ArcTan[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sq 
rt[1 + 2*x]]))/24025
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.47, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {1164, 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.

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

\(\Big \downarrow \) 1164

\(\displaystyle \frac {1}{31} \int \frac {(19-4 x) \sqrt {2 x+1}}{5 x^2+3 x+2}dx-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {1}{31} \left (\frac {1}{5} \int \frac {194 x+111}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {1}{31} \left (\frac {2}{5} \int \frac {2 (97 (2 x+1)+14)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \int \frac {97 (2 x+1)+14}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1483

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \left (\frac {\int \frac {14 \sqrt {10 \left (2+\sqrt {35}\right )}-\left (70-97 \sqrt {35}\right ) \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\int \frac {\left (70-97 \sqrt {35}\right ) \sqrt {2 x+1}+14 \sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \left (\frac {\frac {1}{2} \sqrt {457492+144550 \sqrt {35}} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int -\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{2} \sqrt {457492+144550 \sqrt {35}} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \left (\frac {\frac {1}{2} \sqrt {457492+144550 \sqrt {35}} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{2} \sqrt {457492+144550 \sqrt {35}} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \left (\frac {\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {457492+144550 \sqrt {35}} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\sqrt {457492+144550 \sqrt {35}} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \left (\frac {\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {457492+144550 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\frac {1}{10} \left (70-97 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\sqrt {\frac {457492+144550 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{31} \left (\frac {4}{5} \left (\frac {\sqrt {\frac {457492+144550 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \left (70-97 \sqrt {35}\right ) \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\sqrt {\frac {457492+144550 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \left (70-97 \sqrt {35}\right ) \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )-\frac {8}{5} \sqrt {2 x+1}\right )-\frac {(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}\)

Input:

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

Output:

-1/31*((5 - 4*x)*(1 + 2*x)^(3/2))/(2 + 3*x + 5*x^2) + ((-8*Sqrt[1 + 2*x])/ 
5 + (4*((Sqrt[(457492 + 144550*Sqrt[35])/(10*(-2 + Sqrt[35]))]*ArcTan[(-Sq 
rt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] - ((70 
 - 97*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[(457492 + 144550*Sqrt[3 
5])/(10*(-2 + Sqrt[35]))]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2* 
x])/Sqrt[10*(-2 + Sqrt[35])]] + ((70 - 97*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)/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 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 = 2.46 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.21

method	result	size
pseudoelliptic	\(-\frac {16740 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (x +\frac {41}{54}\right ) \sqrt {1+2 x}+\left (5 x^{2}+3 x +2\right ) \left (\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (264 \sqrt {5}-505 \sqrt {7}\right ) \left (\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )-\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )\right ) \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+1240 \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right ) \left (\sqrt {5}\, \sqrt {7}+\frac {97}{2}\right )\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (240250 x^{2}+144150 x +96100\right )}\)	\(270\)
derivativedivides	\(\frac {-\frac {108 \left (1+2 x \right )^{\frac {3}{2}}}{775}-\frac {56 \sqrt {1+2 x}}{775}}{\left (1+2 x \right )^{2}+\frac {3}{5}-\frac {8 x}{5}}-\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{48050}-\frac {2 \left (-124 \sqrt {5}\, \sqrt {7}+\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{48050}+\frac {2 \left (124 \sqrt {5}\, \sqrt {7}-\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(415\)
default	\(\frac {-\frac {108 \left (1+2 x \right )^{\frac {3}{2}}}{775}-\frac {56 \sqrt {1+2 x}}{775}}{\left (1+2 x \right )^{2}+\frac {3}{5}-\frac {8 x}{5}}-\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{48050}-\frac {2 \left (-124 \sqrt {5}\, \sqrt {7}+\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{48050}+\frac {2 \left (124 \sqrt {5}\, \sqrt {7}-\frac {\left (264 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-505 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(415\)
trager	\(-\frac {\left (41+54 x \right ) \sqrt {1+2 x}}{155 \left (5 x^{2}+3 x +2\right )}+\frac {2 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right ) \ln \left (-\frac {10763200 x \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{5}[phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{3} x [phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{3}-1956115500 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2} \sqrt {1+2 x}[phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right ) x [phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )[phone] \sqrt {1+2 x}}{620 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2} x +4871 x -37076}\right )}{155}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2}+2532545\right ) \ln \left (-\frac {2152640 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2}+2532545\right ) \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{4} x [phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2}+2532545\right ) x -257455744 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2}+2532545\right )[phone] \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2} \sqrt {1+2 x}-9025451250 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2}+2532545\right ) x -20781098000 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2}+2532545\right )[phone] \sqrt {1+2 x}}{620 \operatorname {RootOf}\left (76880 \textit {\_Z}^{4}+8104144 \textit {\_Z}^{2}[phone]\right )^{2} x +60485 x +37076}\right )}{24025}\)	\(449\)
risch	\(-\frac {\left (41+54 x \right ) \sqrt {1+2 x}}{155 \left (5 x^{2}+3 x +2\right )}-\frac {132 \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}}{24025}+\frac {101 \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}}{9610}-\frac {264 \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 )}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {101 \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}}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {8 \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}}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {132 \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}}{24025}-\frac {101 \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}}{9610}-\frac {264 \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 )}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {101 \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}}{4805 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {8 \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}}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(633\)

Input:

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

Output:

-1/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(16740*(10*5^(1/2)*7^(1/2)-20)^(1/2)*(x+4 
1/54)*(1+2*x)^(1/2)+(5*x^2+3*x+2)*((10*5^(1/2)*7^(1/2)-20)^(1/2)*(264*5^(1 
/2)-505*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)+1240*(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)+97/2)))/(240250*x^2+144150*x+961 
00)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.23 \[ \int \frac {(1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=-\frac {2 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} + \frac {16339}{155}} \arctan \left (\frac {10}{9269} \, {\left (\sqrt {2 \, x + 1} {\left (10 \, \sqrt {\frac {7}{5}} - 97\right )} + \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} - \frac {16339}{155}} {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )}\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} + \frac {16339}{155}}\right ) - 2 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} + \frac {16339}{155}} \arctan \left (-\frac {10}{9269} \, {\left (\sqrt {2 \, x + 1} {\left (10 \, \sqrt {\frac {7}{5}} - 97\right )} - \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} - \frac {16339}{155}} {\left (5 \, \sqrt {\frac {7}{5}} + 2\right )}\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} + \frac {16339}{155}}\right ) + {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} - \frac {16339}{155}} \log \left (2 \, \sqrt {2 \, x + 1} {\left (505 \, \sqrt {\frac {7}{5}} + 264\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} - \frac {16339}{155}} + 18538 \, x + 9269 \, \sqrt {\frac {7}{5}} + 9269\right ) - {\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} - \frac {16339}{155}} \log \left (-2 \, \sqrt {2 \, x + 1} {\left (505 \, \sqrt {\frac {7}{5}} + 264\right )} \sqrt {\frac {10325}{62} \, \sqrt {\frac {7}{5}} - \frac {16339}{155}} + 18538 \, x + 9269 \, \sqrt {\frac {7}{5}} + 9269\right ) + {\left (54 \, x + 41\right )} \sqrt {2 \, x + 1}}{155 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \] Input:

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

Output:

-1/155*(2*(5*x^2 + 3*x + 2)*sqrt(10325/62*sqrt(7/5) + 16339/155)*arctan(10 
/9269*(sqrt(2*x + 1)*(10*sqrt(7/5) - 97) + sqrt(10325/62*sqrt(7/5) - 16339 
/155)*(5*sqrt(7/5) + 2))*sqrt(10325/62*sqrt(7/5) + 16339/155)) - 2*(5*x^2 
+ 3*x + 2)*sqrt(10325/62*sqrt(7/5) + 16339/155)*arctan(-10/9269*(sqrt(2*x 
+ 1)*(10*sqrt(7/5) - 97) - sqrt(10325/62*sqrt(7/5) - 16339/155)*(5*sqrt(7/ 
5) + 2))*sqrt(10325/62*sqrt(7/5) + 16339/155)) + (5*x^2 + 3*x + 2)*sqrt(10 
325/62*sqrt(7/5) - 16339/155)*log(2*sqrt(2*x + 1)*(505*sqrt(7/5) + 264)*sq 
rt(10325/62*sqrt(7/5) - 16339/155) + 18538*x + 9269*sqrt(7/5) + 9269) - (5 
*x^2 + 3*x + 2)*sqrt(10325/62*sqrt(7/5) - 16339/155)*log(-2*sqrt(2*x + 1)* 
(505*sqrt(7/5) + 264)*sqrt(10325/62*sqrt(7/5) - 16339/155) + 18538*x + 926 
9*sqrt(7/5) + 9269) + (54*x + 41)*sqrt(2*x + 1))/(5*x^2 + 3*x + 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((2*x + 1)^(5/2)/(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. 624 vs. \(2 (160) = 320\).

Time = 0.85 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.80 \[ \int \frac {(1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/5768402500*sqrt(31)*(20370*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(- 
140*sqrt(35) + 2450) - 97*sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2 
) + 194*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 40740*(7/5)^(3/4)*sqrt(1 
40*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 68600*sqrt(31)*(7/5)^(1/4)*sqrt(-1 
40*sqrt(35) + 2450) + 137200*(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))/s 
qrt(-1/35*sqrt(35) + 1/2)) + 1/5768402500*sqrt(31)*(20370*sqrt(31)*(7/5)^( 
3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 97*sqrt(31)*(7/5)^(3/4 
)*(-140*sqrt(35) + 2450)^(3/2) + 194*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/ 
2) + 40740*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 68600 
*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 137200*(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)) + 1/11536805000*s 
qrt(31)*(97*sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 20370*sqrt( 
31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 40740*(7/5)^ 
(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) + 194*(7/5)^(3/4)*(-140 
*sqrt(35) + 2450)^(3/2) + 68600*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2 
450) - 137200*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450))*log(2*(7/5)^(1/4)*sq 
rt(2*x + 1)*sqrt(1/35*sqrt(35) + 1/2) + 2*x + sqrt(7/5) + 1) - 1/115368050 
00*sqrt(31)*(97*sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 2037...
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.93 \[ \int \frac {(1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx=-\frac {\frac {56\,\sqrt {2\,x+1}}{775}+\frac {108\,{\left (2\,x+1\right )}^{3/2}}{775}}{{\left (2\,x+1\right )}^2-\frac {8\,x}{5}+\frac {3}{5}}-\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-32678-\sqrt {31}\,9269{}\mathrm {i}}\,\sqrt {2\,x+1}\,38272{}\mathrm {i}}{375390625\,\left (-\frac {27058304}{75078125}+\frac {\sqrt {31}\,535808{}\mathrm {i}}{75078125}\right )}-\frac {76544\,\sqrt {31}\,\sqrt {155}\,\sqrt {-32678-\sqrt {31}\,9269{}\mathrm {i}}\,\sqrt {2\,x+1}}{11637109375\,\left (-\frac {27058304}{75078125}+\frac {\sqrt {31}\,535808{}\mathrm {i}}{75078125}\right )}\right )\,\sqrt {-32678-\sqrt {31}\,9269{}\mathrm {i}}\,2{}\mathrm {i}}{24025}+\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-32678+\sqrt {31}\,9269{}\mathrm {i}}\,\sqrt {2\,x+1}\,38272{}\mathrm {i}}{375390625\,\left (\frac {27058304}{75078125}+\frac {\sqrt {31}\,535808{}\mathrm {i}}{75078125}\right )}+\frac {76544\,\sqrt {31}\,\sqrt {155}\,\sqrt {-32678+\sqrt {31}\,9269{}\mathrm {i}}\,\sqrt {2\,x+1}}{11637109375\,\left (\frac {27058304}{75078125}+\frac {\sqrt {31}\,535808{}\mathrm {i}}{75078125}\right )}\right )\,\sqrt {-32678+\sqrt {31}\,9269{}\mathrm {i}}\,2{}\mathrm {i}}{24025} \] Input:

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

Output:

(155^(1/2)*atan((155^(1/2)*(31^(1/2)*9269i - 32678)^(1/2)*(2*x + 1)^(1/2)* 
38272i)/(375390625*((31^(1/2)*535808i)/75078125 + 27058304/75078125)) + (7 
6544*31^(1/2)*155^(1/2)*(31^(1/2)*9269i - 32678)^(1/2)*(2*x + 1)^(1/2))/(1 
1637109375*((31^(1/2)*535808i)/75078125 + 27058304/75078125)))*(31^(1/2)*9 
269i - 32678)^(1/2)*2i)/24025 - (155^(1/2)*atan((155^(1/2)*(- 31^(1/2)*926 
9i - 32678)^(1/2)*(2*x + 1)^(1/2)*38272i)/(375390625*((31^(1/2)*535808i)/7 
5078125 - 27058304/75078125)) - (76544*31^(1/2)*155^(1/2)*(- 31^(1/2)*9269 
i - 32678)^(1/2)*(2*x + 1)^(1/2))/(11637109375*((31^(1/2)*535808i)/7507812 
5 - 27058304/75078125)))*(- 31^(1/2)*9269i - 32678)^(1/2)*2i)/24025 - ((56 
*(2*x + 1)^(1/2))/775 + (108*(2*x + 1)^(3/2))/775)/((2*x + 1)^2 - (8*x)/5 
+ 3/5)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 969, normalized size of antiderivative = 4.35 \[ \int \frac {(1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 5050*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 - 3030*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 - 2020*sqrt(sqrt(35) - 2)*sqrt(14)*at 
an((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 
 2)*sqrt(2))) - 2640*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 - 15 
84*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 - 1056*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))) + 5050*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sq 
rt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2) 
))*x**2 + 3030*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 + 2020*sqrt(s 
qrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*s 
qrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 2640*sqrt(sqrt(35) - 2)*sqrt(10)*a 
tan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) 
- 2)*sqrt(2)))*x**2 + 1584*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 + 
 1056*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) + 2*...