\(\int \frac {x^{5/2} (A+B x)}{(a+b x+c x^2)^2} \, dx\) [92]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 403 \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {2 B \sqrt {x}}{c^2}+\frac {\sqrt {x} \left (a \left (b^2 B-A b c-2 a B c\right )+\left (b^3 B-A b^2 c-3 a b B c+2 a A c^2\right ) x\right )}{c^2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\left (3 b^3 B-A b^2 c-13 a b B c+6 a A c^2-\frac {3 b^4 B-A b^3 c-19 a b^2 B c+8 a A b c^2+20 a^2 B c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} c^{5/2} \left (b^2-4 a c\right ) \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (3 b^3 B-A b^2 c-13 a b B c+6 a A c^2+\frac {3 b^4 B-A b^3 c-19 a b^2 B c+8 a A b c^2+20 a^2 B c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} c^{5/2} \left (b^2-4 a c\right ) \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

2*B*x^(1/2)/c^2+x^(1/2)*(a*(-A*b*c-2*B*a*c+B*b^2)+(2*A*a*c^2-A*b^2*c-3*B*a 
*b*c+B*b^3)*x)/c^2/(-4*a*c+b^2)/(c*x^2+b*x+a)-1/2*(3*B*b^3-A*b^2*c-13*B*a* 
b*c+6*A*a*c^2-(8*A*a*b*c^2-A*b^3*c+20*B*a^2*c^2-19*B*a*b^2*c+3*B*b^4)/(-4* 
a*c+b^2)^(1/2))*arctan(2^(1/2)*c^(1/2)*x^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2 
))*2^(1/2)/c^(5/2)/(-4*a*c+b^2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-1/2*(3*B*b^3- 
A*b^2*c-13*B*a*b*c+6*A*a*c^2+(8*A*a*b*c^2-A*b^3*c+20*B*a^2*c^2-19*B*a*b^2* 
c+3*B*b^4)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/2)*c^(1/2)*x^(1/2)/(b+(-4*a*c+b 
^2)^(1/2))^(1/2))*2^(1/2)/c^(5/2)/(-4*a*c+b^2)/(b+(-4*a*c+b^2)^(1/2))^(1/2 
)
 

Mathematica [A] (verified)

Time = 4.69 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.14 \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {x} \left (10 a^2 B c+b^2 x (-3 b B+c (A-2 B x))+a \left (-3 b^2 B+2 c^2 x (-A+4 B x)+b c (A+11 B x)\right )\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}-\frac {\sqrt {2} \left (-3 b^4 B+b^2 c \left (19 a B-A \sqrt {b^2-4 a c}\right )+2 a c^2 \left (-10 a B+3 A \sqrt {b^2-4 a c}\right )+b^3 \left (A c+3 B \sqrt {b^2-4 a c}\right )-a b c \left (8 A c+13 B \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \left (3 b^4 B-b^2 c \left (19 a B+A \sqrt {b^2-4 a c}\right )+2 a c^2 \left (10 a B+3 A \sqrt {b^2-4 a c}\right )+a b c \left (8 A c-13 B \sqrt {b^2-4 a c}\right )+b^3 \left (-A c+3 B \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{2 c^{5/2}} \] Input:

Integrate[(x^(5/2)*(A + B*x))/(a + b*x + c*x^2)^2,x]
 

Output:

((-2*Sqrt[c]*Sqrt[x]*(10*a^2*B*c + b^2*x*(-3*b*B + c*(A - 2*B*x)) + a*(-3* 
b^2*B + 2*c^2*x*(-A + 4*B*x) + b*c*(A + 11*B*x))))/((b^2 - 4*a*c)*(a + x*( 
b + c*x))) - (Sqrt[2]*(-3*b^4*B + b^2*c*(19*a*B - A*Sqrt[b^2 - 4*a*c]) + 2 
*a*c^2*(-10*a*B + 3*A*Sqrt[b^2 - 4*a*c]) + b^3*(A*c + 3*B*Sqrt[b^2 - 4*a*c 
]) - a*b*c*(8*A*c + 13*B*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[ 
x])/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 
 4*a*c]]) - (Sqrt[2]*(3*b^4*B - b^2*c*(19*a*B + A*Sqrt[b^2 - 4*a*c]) + 2*a 
*c^2*(10*a*B + 3*A*Sqrt[b^2 - 4*a*c]) + a*b*c*(8*A*c - 13*B*Sqrt[b^2 - 4*a 
*c]) + b^3*(-(A*c) + 3*B*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[ 
x])/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 
 4*a*c]]))/(2*c^(5/2))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1233, 27, 1196, 25, 1197, 1480, 218}

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.

\(\displaystyle \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1233

\(\displaystyle \frac {\int \frac {\sqrt {x} \left (3 a (b B-2 A c)+\left (3 B b^2-A c b-10 a B c\right ) x\right )}{2 \left (c x^2+b x+a\right )}dx}{c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {x} \left (3 a (b B-2 A c)+\left (3 B b^2-A c b-10 a B c\right ) x\right )}{c x^2+b x+a}dx}{2 c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {\frac {\int -\frac {a \left (3 B b^2-A c b-10 a B c\right )+\left (3 B b^3-A c b^2-13 a B c b+6 a A c^2\right ) x}{\sqrt {x} \left (c x^2+b x+a\right )}dx}{c}+\frac {2 \sqrt {x} \left (-10 a B c-A b c+3 b^2 B\right )}{c}}{2 c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2 \sqrt {x} \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {\int \frac {a \left (3 B b^2-A c b-10 a B c\right )+\left (3 B b^3-A c b^2-13 a B c b+6 a A c^2\right ) x}{\sqrt {x} \left (c x^2+b x+a\right )}dx}{c}}{2 c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {\frac {2 \sqrt {x} \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {2 \int \frac {a \left (3 B b^2-A c b-10 a B c\right )+\left (3 B b^3-A c b^2-13 a B c b+6 a A c^2\right ) x}{c x^2+b x+a}d\sqrt {x}}{c}}{2 c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {\frac {2 \sqrt {x} \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {2 \left (\frac {1}{2} \left (-\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \int \frac {1}{\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )+c x}d\sqrt {x}+\frac {1}{2} \left (\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \int \frac {1}{\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )+c x}d\sqrt {x}\right )}{c}}{2 c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {2 \sqrt {x} \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {2 \left (\frac {\left (-\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )}{c}}{2 c \left (b^2-4 a c\right )}-\frac {x^{3/2} \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

Input:

Int[(x^(5/2)*(A + B*x))/(a + b*x + c*x^2)^2,x]
 

Output:

-((x^(3/2)*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(c*(b^2 - 4*a* 
c)*(a + b*x + c*x^2))) + ((2*(3*b^2*B - A*b*c - 10*a*B*c)*Sqrt[x])/c - (2* 
(((3*b^3*B - A*b^2*c - 13*a*b*B*c + 6*a*A*c^2 - (3*b^4*B - A*b^3*c - 19*a* 
b^2*B*c + 8*a*A*b*c^2 + 20*a^2*B*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*S 
qrt[c]*Sqrt[x])/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sq 
rt[b^2 - 4*a*c]]) + ((3*b^3*B - A*b^2*c - 13*a*b*B*c + 6*a*A*c^2 + (3*b^4* 
B - A*b^3*c - 19*a*b^2*B*c + 8*a*A*b*c^2 + 20*a^2*B*c^2)/Sqrt[b^2 - 4*a*c] 
)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]* 
Sqrt[c]*Sqrt[b + Sqrt[b^2 - 4*a*c]])))/c)/(2*c*(b^2 - 4*a*c))
 

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 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 1233
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) 
^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c 
*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f 
*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( 
m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* 
p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && 
GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | 
|  !ILtQ[m + 2*p + 3, 0])
 

rule 1480
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 
Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.13

method result size
derivativedivides \(\frac {\frac {2 \left (-\frac {\left (2 A a \,c^{2}-A \,b^{2} c -3 B a b c +B \,b^{3}\right ) x^{\frac {3}{2}}}{2 \left (4 a c -b^{2}\right )}+\frac {a \left (A b c +2 a B c -B \,b^{2}\right ) \sqrt {x}}{8 a c -2 b^{2}}\right )}{c \,x^{2}+b x +a}+\frac {4 c \left (-\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}-8 A a b \,c^{2}+A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}-20 B \,a^{2} c^{2}+19 B a \,b^{2} c -3 B \,b^{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}+8 A a b \,c^{2}-A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}+20 B \,a^{2} c^{2}-19 B a \,b^{2} c +3 B \,b^{4}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {x}\, c \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{c^{2}}+\frac {2 B \sqrt {x}}{c^{2}}\) \(455\)
default \(\frac {\frac {2 \left (-\frac {\left (2 A a \,c^{2}-A \,b^{2} c -3 B a b c +B \,b^{3}\right ) x^{\frac {3}{2}}}{2 \left (4 a c -b^{2}\right )}+\frac {a \left (A b c +2 a B c -B \,b^{2}\right ) \sqrt {x}}{8 a c -2 b^{2}}\right )}{c \,x^{2}+b x +a}+\frac {4 c \left (-\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}-8 A a b \,c^{2}+A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}-20 B \,a^{2} c^{2}+19 B a \,b^{2} c -3 B \,b^{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}+8 A a b \,c^{2}-A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}+20 B \,a^{2} c^{2}-19 B a \,b^{2} c +3 B \,b^{4}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {x}\, c \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{c^{2}}+\frac {2 B \sqrt {x}}{c^{2}}\) \(455\)
risch \(\frac {2 B \sqrt {x}}{c^{2}}+\frac {\frac {-\frac {\left (2 A a \,c^{2}-A \,b^{2} c -3 B a b c +B \,b^{3}\right ) x^{\frac {3}{2}}}{4 a c -b^{2}}+\frac {2 a \left (A b c +2 a B c -B \,b^{2}\right ) \sqrt {x}}{8 a c -2 b^{2}}}{c \,x^{2}+b x +a}+\frac {4 c \left (-\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}-8 A a b \,c^{2}+A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}-20 B \,a^{2} c^{2}+19 B a \,b^{2} c -3 B \,b^{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}+8 A a b \,c^{2}-A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}+20 B \,a^{2} c^{2}-19 B a \,b^{2} c +3 B \,b^{4}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {x}\, c \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{c^{2}}\) \(455\)

Input:

int(x^(5/2)*(B*x+A)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

2/c^2*((-1/2*(2*A*a*c^2-A*b^2*c-3*B*a*b*c+B*b^3)/(4*a*c-b^2)*x^(3/2)+1/2*a 
*(A*b*c+2*B*a*c-B*b^2)/(4*a*c-b^2)*x^(1/2))/(c*x^2+b*x+a)+2/(4*a*c-b^2)*c* 
(-1/8*(6*A*a*c^2*(-4*a*c+b^2)^(1/2)-A*b^2*c*(-4*a*c+b^2)^(1/2)-8*A*a*b*c^2 
+A*b^3*c-13*B*a*b*c*(-4*a*c+b^2)^(1/2)+3*B*b^3*(-4*a*c+b^2)^(1/2)-20*B*a^2 
*c^2+19*B*a*b^2*c-3*B*b^4)/c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^ 
(1/2))*c)^(1/2)*arctanh(c*x^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2 
))+1/8*(6*A*a*c^2*(-4*a*c+b^2)^(1/2)-A*b^2*c*(-4*a*c+b^2)^(1/2)+8*A*a*b*c^ 
2-A*b^3*c-13*B*a*b*c*(-4*a*c+b^2)^(1/2)+3*B*b^3*(-4*a*c+b^2)^(1/2)+20*B*a^ 
2*c^2-19*B*a*b^2*c+3*B*b^4)/c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^ 
(1/2))*c)^(1/2)*arctan(x^(1/2)*c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)) 
))+2*B*x^(1/2)/c^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7251 vs. \(2 (359) = 718\).

Time = 15.67 (sec) , antiderivative size = 7251, normalized size of antiderivative = 17.99 \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} x^{\frac {5}{2}}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \] Input:

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x+a)^2,x, algorithm="maxima")
 

Output:

((A*b*c - (b^2 - 2*a*c)*B)*x^(5/2) - (B*a*b - 2*A*a*c)*x^(3/2))/(a*b^2*c - 
 4*a^2*c^2 + (b^2*c^2 - 4*a*c^3)*x^2 + (b^3*c - 4*a*b*c^2)*x) + integrate( 
-1/2*((A*b*c - (3*b^2 - 10*a*c)*B)*x^(3/2) - 3*(B*a*b - 2*A*a*c)*sqrt(x))/ 
(a*b^2*c - 4*a^2*c^2 + (b^2*c^2 - 4*a*c^3)*x^2 + (b^3*c - 4*a*b*c^2)*x), x 
)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5685 vs. \(2 (359) = 718\).

Time = 1.16 (sec) , antiderivative size = 5685, normalized size of antiderivative = 14.11 \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x+a)^2,x, algorithm="giac")
 

Output:

2*B*sqrt(x)/c^2 + (B*b^3*x^(3/2) - 3*B*a*b*c*x^(3/2) - A*b^2*c*x^(3/2) + 2 
*A*a*c^2*x^(3/2) + B*a*b^2*sqrt(x) - 2*B*a^2*c*sqrt(x) - A*a*b*c*sqrt(x))/ 
((b^2*c^2 - 4*a*c^3)*(c*x^2 + b*x + a)) + 1/8*((2*b^4*c^3 - 20*a*b^2*c^4 + 
 48*a^2*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^ 
4*c + 10*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c 
^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 - 
 24*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^3 - 12 
*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 - sqrt( 
2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^3 + 6*sqrt(2)*s 
qrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^4 - 2*(b^2 - 4*a*c)*b 
^2*c^3 + 12*(b^2 - 4*a*c)*a*c^4)*(b^2*c^2 - 4*a*c^3)^2*A - (6*b^5*c^2 - 50 
*a*b^3*c^3 + 104*a^2*b*c^4 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b 
^2 - 4*a*c)*c)*b^5 + 25*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4* 
a*c)*c)*a*b^3*c + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c) 
*c)*b^4*c - 52*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a 
^2*b*c^2 - 26*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a* 
b^2*c^2 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3* 
c^2 + 13*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 
 - 6*(b^2 - 4*a*c)*b^3*c^2 + 26*(b^2 - 4*a*c)*a*b*c^3)*(b^2*c^2 - 4*a*c^3) 
^2*B + 2*(sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^5*c^4 - 8*sqrt(2)...
 

Mupad [B] (verification not implemented)

Time = 15.15 (sec) , antiderivative size = 16631, normalized size of antiderivative = 41.27 \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

int((x^(5/2)*(A + B*x))/(a + b*x + c*x^2)^2,x)
 

Output:

(2*B*x^(1/2))/c^2 - atan(((((2560*B*a^5*c^7 - 4*A*a*b^7*c^4 + 256*A*a^4*b* 
c^7 + 12*B*a*b^8*c^3 + 48*A*a^2*b^5*c^5 - 192*A*a^3*b^3*c^6 - 184*B*a^2*b^ 
6*c^4 + 1056*B*a^3*b^4*c^5 - 2688*B*a^4*b^2*c^6)/(64*a^3*c^6 - b^6*c^3 + 1 
2*a*b^4*c^4 - 48*a^2*b^2*c^5) - (2*x^(1/2)*((9*B^2*b^4*(-(4*a*c - b^2)^9)^ 
(1/2) - A^2*b^11*c^2 - 9*B^2*b^13 + 6*A*B*b^12*c - 288*A^2*a^2*b^7*c^4 + 1 
504*A^2*a^3*b^5*c^5 - 3840*A^2*a^4*b^3*c^6 - 2077*B^2*a^2*b^9*c^2 + 10656* 
B^2*a^3*b^7*c^3 - 30240*B^2*a^4*b^5*c^4 + 44800*B^2*a^5*b^3*c^5 + A^2*b^2* 
c^2*(-(4*a*c - b^2)^9)^(1/2) + 25*B^2*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) + 1 
5360*A*B*a^6*c^7 + 213*B^2*a*b^11*c + 27*A^2*a*b^9*c^3 + 3840*A^2*a^5*b*c^ 
7 - 9*A^2*a*c^3*(-(4*a*c - b^2)^9)^(1/2) - 26880*B^2*a^6*b*c^6 + 1548*A*B* 
a^2*b^8*c^3 - 8064*A*B*a^3*b^6*c^4 + 22400*A*B*a^4*b^4*c^5 - 30720*A*B*a^5 
*b^2*c^6 - 51*B^2*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2) - 152*A*B*a*b^10*c^2 - 
6*A*B*b^3*c*(-(4*a*c - b^2)^9)^(1/2) + 44*A*B*a*b*c^2*(-(4*a*c - b^2)^9)^( 
1/2))/(8*(4096*a^6*c^11 + b^12*c^5 - 24*a*b^10*c^6 + 240*a^2*b^8*c^7 - 128 
0*a^3*b^6*c^8 + 3840*a^4*b^4*c^9 - 6144*a^5*b^2*c^10)))^(1/2)*(4*b^7*c^5 - 
 48*a*b^5*c^6 - 256*a^3*b*c^8 + 192*a^2*b^3*c^7))/(16*a^2*c^5 + b^4*c^3 - 
8*a*b^2*c^4))*((9*B^2*b^4*(-(4*a*c - b^2)^9)^(1/2) - A^2*b^11*c^2 - 9*B^2* 
b^13 + 6*A*B*b^12*c - 288*A^2*a^2*b^7*c^4 + 1504*A^2*a^3*b^5*c^5 - 3840*A^ 
2*a^4*b^3*c^6 - 2077*B^2*a^2*b^9*c^2 + 10656*B^2*a^3*b^7*c^3 - 30240*B^2*a 
^4*b^5*c^4 + 44800*B^2*a^5*b^3*c^5 + A^2*b^2*c^2*(-(4*a*c - b^2)^9)^(1/...
 

Reduce [B] (verification not implemented)

Time = 3.93 (sec) , antiderivative size = 3675, normalized size of antiderivative = 9.12 \[ \int \frac {x^{5/2} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^(5/2)*(B*x+A)/(c*x^2+b*x+a)^2,x)
 

Output:

( - 24*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - 
b) - 2*sqrt(x)*sqrt(c))/sqrt(2*sqrt(c)*sqrt(a) + b))*a**3*c**3 + 34*sqrt(a 
)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x 
)*sqrt(c))/sqrt(2*sqrt(c)*sqrt(a) + b))*a**2*b**2*c**2 - 24*sqrt(a)*sqrt(2 
*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x)*sqrt(c 
))/sqrt(2*sqrt(c)*sqrt(a) + b))*a**2*b*c**3*x - 24*sqrt(a)*sqrt(2*sqrt(c)* 
sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x)*sqrt(c))/sqrt(2 
*sqrt(c)*sqrt(a) + b))*a**2*c**4*x**2 - 6*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + 
 b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x)*sqrt(c))/sqrt(2*sqrt(c)* 
sqrt(a) + b))*a*b**4*c + 34*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt 
(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x)*sqrt(c))/sqrt(2*sqrt(c)*sqrt(a) + b))* 
a*b**3*c**2*x + 34*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c 
)*sqrt(a) - b) - 2*sqrt(x)*sqrt(c))/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b**2*c* 
*3*x**2 - 6*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt( 
a) - b) - 2*sqrt(x)*sqrt(c))/sqrt(2*sqrt(c)*sqrt(a) + b))*b**5*c*x - 6*sqr 
t(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqr 
t(x)*sqrt(c))/sqrt(2*sqrt(c)*sqrt(a) + b))*b**4*c**2*x**2 + 56*sqrt(c)*sqr 
t(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x)*sqr 
t(c))/sqrt(2*sqrt(c)*sqrt(a) + b))*a**3*b*c**2 - 40*sqrt(c)*sqrt(2*sqrt(c) 
*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(x)*sqrt(c))/sq...