\(\int \frac {A+B x}{x^{5/2} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [463]

Optimal result

Integrand size = 31, antiderivative size = 336 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {b (515 A b-187 a B) \sqrt {x}}{64 a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (A b-a B) \sqrt {x}}{4 a^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (23 A b-15 a B) \sqrt {x}}{24 a^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (259 A b-123 a B) \sqrt {x}}{96 a^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A (a+b x)}{3 a^5 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (5 A b-a B) (a+b x)}{a^6 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {105 \sqrt {b} (11 A b-3 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

1/64*b*(515*A*b-187*B*a)*x^(1/2)/a^6/((b*x+a)^2)^(1/2)+1/4*b*(A*b-B*a)*x^( 
1/2)/a^3/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/24*b*(23*A*b-15*B*a)*x^(1/2)/a^4/(b 
*x+a)^2/((b*x+a)^2)^(1/2)+1/96*b*(259*A*b-123*B*a)*x^(1/2)/a^5/(b*x+a)/((b 
*x+a)^2)^(1/2)-2/3*A*(b*x+a)/a^5/x^(3/2)/((b*x+a)^2)^(1/2)+2*(5*A*b-B*a)*( 
b*x+a)/a^6/x^(1/2)/((b*x+a)^2)^(1/2)+105/64*b^(1/2)*(11*A*b-3*B*a)*(b*x+a) 
*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(13/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.51 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {\sqrt {a} \left (-3465 A b^5 x^5+128 a^5 (A+3 B x)+105 a b^4 x^4 (-121 A+9 B x)+231 a^2 b^3 x^3 (-73 A+15 B x)+9 a^3 b^2 x^2 (-1023 A+511 B x)+a^4 b x (-1408 A+2511 B x)\right )}{x^{3/2}}+315 \sqrt {b} (11 A b-3 a B) (a+b x)^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{192 a^{13/2} (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

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

Output:

(-((Sqrt[a]*(-3465*A*b^5*x^5 + 128*a^5*(A + 3*B*x) + 105*a*b^4*x^4*(-121*A 
 + 9*B*x) + 231*a^2*b^3*x^3*(-73*A + 15*B*x) + 9*a^3*b^2*x^2*(-1023*A + 51 
1*B*x) + a^4*b*x*(-1408*A + 2511*B*x)))/x^(3/2)) + 315*Sqrt[b]*(11*A*b - 3 
*a*B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(192*a^(13/2)*(a + b* 
x)^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.63, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {1187, 27, 87, 52, 52, 52, 61, 61, 73, 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.

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

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^5 (a+b x) \int \frac {A+B x}{b^5 x^{5/2} (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^{5/2} (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \int \frac {1}{x^{5/2} (a+b x)^4}dx}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \int \frac {1}{x^{5/2} (a+b x)^3}dx}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \left (\frac {7 \int \frac {1}{x^{5/2} (a+b x)^2}dx}{4 a}+\frac {1}{2 a x^{3/2} (a+b x)^2}\right )}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \left (\frac {7 \left (\frac {5 \int \frac {1}{x^{5/2} (a+b x)}dx}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a}+\frac {1}{2 a x^{3/2} (a+b x)^2}\right )}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a}+\frac {1}{2 a x^{3/2} (a+b x)^2}\right )}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a}+\frac {1}{2 a x^{3/2} (a+b x)^2}\right )}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a}+\frac {1}{2 a x^{3/2} (a+b x)^2}\right )}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a+b x) \left (\frac {(11 A b-3 a B) \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a}+\frac {1}{2 a x^{3/2} (a+b x)^2}\right )}{2 a}+\frac {1}{3 a x^{3/2} (a+b x)^3}\right )}{8 a b}+\frac {A b-a B}{4 a b x^{3/2} (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

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

Output:

((a + b*x)*((A*b - a*B)/(4*a*b*x^(3/2)*(a + b*x)^4) + ((11*A*b - 3*a*B)*(1 
/(3*a*x^(3/2)*(a + b*x)^3) + (3*(1/(2*a*x^(3/2)*(a + b*x)^2) + (7*(1/(a*x^ 
(3/2)*(a + b*x)) + (5*(-2/(3*a*x^(3/2)) - (b*(-2/(a*Sqrt[x]) - (2*Sqrt[b]* 
ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a))/(2*a)))/(4*a)))/(2*a)))/( 
8*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

Defintions of rubi rules used

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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 1187

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ 
IntPart[p]*(b/2 + c*x)^(2*FracPart[p]))   Int[(d + e*x)^m*(f + g*x)^n*(b/2 
+ c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 
 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.52

method	result	size
risch	\(-\frac {2 \left (-15 A b x +3 B a x +A a \right ) \sqrt {\left (b x +a \right )^{2}}}{3 a^{6} x^{\frac {3}{2}} \left (b x +a \right )}+\frac {b \left (\frac {2 \left (\frac {515}{128} A \,b^{4}-\frac {187}{128} B a \,b^{3}\right ) x^{\frac {7}{2}}+\frac {a \,b^{2} \left (5153 A b -1929 B a \right ) x^{\frac {5}{2}}}{192}+2 \left (\frac {5855}{384} a^{2} A \,b^{2}-\frac {765}{128} B \,a^{3} b \right ) x^{\frac {3}{2}}+2 \left (\frac {765}{128} A \,a^{3} b -\frac {325}{128} a^{4} B \right ) \sqrt {x}}{\left (b x +a \right )^{4}}+\frac {105 \left (11 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{a^{6} \left (b x +a \right )}\)	\(174\)
default	\(-\frac {\left (-3465 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {11}{2}} b^{6}+945 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {11}{2}} a \,b^{5}-13860 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {9}{2}} a \,b^{5}+3780 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {9}{2}} a^{2} b^{4}-20790 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {7}{2}} a^{2} b^{4}+5670 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {7}{2}} a^{3} b^{3}-13860 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {5}{2}} a^{3} b^{3}+3780 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {5}{2}} a^{4} b^{2}-3465 A \sqrt {a b}\, x^{5} b^{5}+945 B \sqrt {a b}\, x^{5} a \,b^{4}-12705 A \sqrt {a b}\, x^{4} a \,b^{4}-3465 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {3}{2}} a^{4} b^{2}+3465 B \sqrt {a b}\, x^{4} a^{2} b^{3}+945 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {3}{2}} a^{5} b -16863 A \sqrt {a b}\, x^{3} a^{2} b^{3}+4599 B \sqrt {a b}\, x^{3} a^{3} b^{2}-9207 A \sqrt {a b}\, x^{2} a^{3} b^{2}+2511 B \sqrt {a b}\, x^{2} a^{4} b -1408 A \sqrt {a b}\, x \,a^{4} b +384 B \sqrt {a b}\, x \,a^{5}+128 A \sqrt {a b}\, a^{5}\right ) \left (b x +a \right )}{192 \sqrt {a b}\, x^{\frac {3}{2}} a^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(413\)

Input:

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

Output:

-2/3*(-15*A*b*x+3*B*a*x+A*a)/a^6/x^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+1/a^6*b 
*(2*((515/128*A*b^4-187/128*B*a*b^3)*x^(7/2)+1/384*a*b^2*(5153*A*b-1929*B* 
a)*x^(5/2)+(5855/384*a^2*A*b^2-765/128*B*a^3*b)*x^(3/2)+(765/128*A*a^3*b-3 
25/128*a^4*B)*x^(1/2))/(b*x+a)^4+105/64*(11*A*b-3*B*a)/(a*b)^(1/2)*arctan( 
b*x^(1/2)/(a*b)^(1/2)))*((b*x+a)^2)^(1/2)/(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 611, normalized size of antiderivative = 1.82 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {315 \, {\left ({\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{6} + 4 \, {\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{5} + 6 \, {\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{4} + 4 \, {\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{3} + {\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (128 \, A a^{5} + 315 \, {\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{5} + 1155 \, {\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{4} + 1533 \, {\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{3} + 837 \, {\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2} + 128 \, {\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{6} b^{4} x^{6} + 4 \, a^{7} b^{3} x^{5} + 6 \, a^{8} b^{2} x^{4} + 4 \, a^{9} b x^{3} + a^{10} x^{2}\right )}}, -\frac {315 \, {\left ({\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{6} + 4 \, {\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{5} + 6 \, {\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{4} + 4 \, {\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{3} + {\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) + {\left (128 \, A a^{5} + 315 \, {\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{5} + 1155 \, {\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{4} + 1533 \, {\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{3} + 837 \, {\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2} + 128 \, {\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{6} b^{4} x^{6} + 4 \, a^{7} b^{3} x^{5} + 6 \, a^{8} b^{2} x^{4} + 4 \, a^{9} b x^{3} + a^{10} x^{2}\right )}}\right ] \] Input:

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

Output:

[-1/384*(315*((3*B*a*b^4 - 11*A*b^5)*x^6 + 4*(3*B*a^2*b^3 - 11*A*a*b^4)*x^ 
5 + 6*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^4 + 4*(3*B*a^4*b - 11*A*a^3*b^2)*x^3 
+ (3*B*a^5 - 11*A*a^4*b)*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) 
 - a)/(b*x + a)) + 2*(128*A*a^5 + 315*(3*B*a*b^4 - 11*A*b^5)*x^5 + 1155*(3 
*B*a^2*b^3 - 11*A*a*b^4)*x^4 + 1533*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^3 + 837 
*(3*B*a^4*b - 11*A*a^3*b^2)*x^2 + 128*(3*B*a^5 - 11*A*a^4*b)*x)*sqrt(x))/( 
a^6*b^4*x^6 + 4*a^7*b^3*x^5 + 6*a^8*b^2*x^4 + 4*a^9*b*x^3 + a^10*x^2), -1/ 
192*(315*((3*B*a*b^4 - 11*A*b^5)*x^6 + 4*(3*B*a^2*b^3 - 11*A*a*b^4)*x^5 + 
6*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^4 + 4*(3*B*a^4*b - 11*A*a^3*b^2)*x^3 + (3 
*B*a^5 - 11*A*a^4*b)*x^2)*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) + (128*A*a^5 
 + 315*(3*B*a*b^4 - 11*A*b^5)*x^5 + 1155*(3*B*a^2*b^3 - 11*A*a*b^4)*x^4 + 
1533*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^3 + 837*(3*B*a^4*b - 11*A*a^3*b^2)*x^2 
 + 128*(3*B*a^5 - 11*A*a^4*b)*x)*sqrt(x))/(a^6*b^4*x^6 + 4*a^7*b^3*x^5 + 6 
*a^8*b^2*x^4 + 4*a^9*b*x^3 + a^10*x^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (224) = 448\).

Time = 0.21 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.47 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {315 \, {\left ({\left (B a b^{7} - 11 \, A b^{8}\right )} x^{2} + 11 \, {\left (3 \, B a^{2} b^{6} - 13 \, A a b^{7}\right )} x\right )} x^{\frac {9}{2}} + 630 \, {\left ({\left (B a^{2} b^{6} - 11 \, A a b^{7}\right )} x^{2} + 33 \, {\left (3 \, B a^{3} b^{5} - 13 \, A a^{2} b^{6}\right )} x\right )} x^{\frac {7}{2}} - 420 \, {\left (6 \, {\left (B a^{3} b^{5} - 11 \, A a^{2} b^{6}\right )} x^{2} - 121 \, {\left (3 \, B a^{4} b^{4} - 13 \, A a^{3} b^{5}\right )} x\right )} x^{\frac {5}{2}} - 42 \, {\left (255 \, {\left (B a^{4} b^{4} - 11 \, A a^{3} b^{5}\right )} x^{2} - 1529 \, {\left (3 \, B a^{5} b^{3} - 13 \, A a^{4} b^{4}\right )} x\right )} x^{\frac {3}{2}} - 33 \, {\left (483 \, {\left (B a^{5} b^{3} - 11 \, A a^{4} b^{4}\right )} x^{2} - 1315 \, {\left (3 \, B a^{6} b^{2} - 13 \, A a^{5} b^{3}\right )} x\right )} \sqrt {x} - \frac {1280 \, {\left (9 \, {\left (B a^{6} b^{2} - 11 \, A a^{5} b^{3}\right )} x^{2} - 11 \, {\left (3 \, B a^{7} b - 13 \, A a^{6} b^{2}\right )} x\right )}}{\sqrt {x}} - \frac {1280 \, {\left (3 \, {\left (B a^{7} b - 11 \, A a^{6} b^{2}\right )} x^{2} - {\left (3 \, B a^{8} - 13 \, A a^{7} b\right )} x\right )}}{x^{\frac {3}{2}}} + \frac {1280 \, {\left (3 \, A a^{7} b x^{2} + A a^{8} x\right )}}{x^{\frac {5}{2}}}}{1920 \, {\left (a^{8} b^{5} x^{5} + 5 \, a^{9} b^{4} x^{4} + 10 \, a^{10} b^{3} x^{3} + 10 \, a^{11} b^{2} x^{2} + 5 \, a^{12} b x + a^{13}\right )}} - \frac {105 \, {\left (3 \, B a b - 11 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{6}} + \frac {21 \, {\left ({\left (B a b^{2} - 11 \, A b^{3}\right )} x^{\frac {3}{2}} + 10 \, {\left (3 \, B a^{2} b - 11 \, A a b^{2}\right )} \sqrt {x}\right )}}{128 \, a^{8}} \] Input:

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

Output:

-1/1920*(315*((B*a*b^7 - 11*A*b^8)*x^2 + 11*(3*B*a^2*b^6 - 13*A*a*b^7)*x)* 
x^(9/2) + 630*((B*a^2*b^6 - 11*A*a*b^7)*x^2 + 33*(3*B*a^3*b^5 - 13*A*a^2*b 
^6)*x)*x^(7/2) - 420*(6*(B*a^3*b^5 - 11*A*a^2*b^6)*x^2 - 121*(3*B*a^4*b^4 
- 13*A*a^3*b^5)*x)*x^(5/2) - 42*(255*(B*a^4*b^4 - 11*A*a^3*b^5)*x^2 - 1529 
*(3*B*a^5*b^3 - 13*A*a^4*b^4)*x)*x^(3/2) - 33*(483*(B*a^5*b^3 - 11*A*a^4*b 
^4)*x^2 - 1315*(3*B*a^6*b^2 - 13*A*a^5*b^3)*x)*sqrt(x) - 1280*(9*(B*a^6*b^ 
2 - 11*A*a^5*b^3)*x^2 - 11*(3*B*a^7*b - 13*A*a^6*b^2)*x)/sqrt(x) - 1280*(3 
*(B*a^7*b - 11*A*a^6*b^2)*x^2 - (3*B*a^8 - 13*A*a^7*b)*x)/x^(3/2) + 1280*( 
3*A*a^7*b*x^2 + A*a^8*x)/x^(5/2))/(a^8*b^5*x^5 + 5*a^9*b^4*x^4 + 10*a^10*b 
^3*x^3 + 10*a^11*b^2*x^2 + 5*a^12*b*x + a^13) - 105/64*(3*B*a*b - 11*A*b^2 
)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^6) + 21/128*((B*a*b^2 - 11*A*b^ 
3)*x^(3/2) + 10*(3*B*a^2*b - 11*A*a*b^2)*sqrt(x))/a^8
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.54 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {105 \, {\left (3 \, B a b - 11 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, {\left (3 \, B a x - 15 \, A b x + A a\right )}}{3 \, a^{6} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right )} - \frac {561 \, B a b^{4} x^{\frac {7}{2}} - 1545 \, A b^{5} x^{\frac {7}{2}} + 1929 \, B a^{2} b^{3} x^{\frac {5}{2}} - 5153 \, A a b^{4} x^{\frac {5}{2}} + 2295 \, B a^{3} b^{2} x^{\frac {3}{2}} - 5855 \, A a^{2} b^{3} x^{\frac {3}{2}} + 975 \, B a^{4} b \sqrt {x} - 2295 \, A a^{3} b^{2} \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a^{6} \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

-105/64*(3*B*a*b - 11*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^6*sg 
n(b*x + a)) - 2/3*(3*B*a*x - 15*A*b*x + A*a)/(a^6*x^(3/2)*sgn(b*x + a)) - 
1/192*(561*B*a*b^4*x^(7/2) - 1545*A*b^5*x^(7/2) + 1929*B*a^2*b^3*x^(5/2) - 
 5153*A*a*b^4*x^(5/2) + 2295*B*a^3*b^2*x^(3/2) - 5855*A*a^2*b^3*x^(3/2) + 
975*B*a^4*b*sqrt(x) - 2295*A*a^3*b^2*sqrt(x))/((b*x + a)^4*a^6*sgn(b*x + a 
))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.58 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {315 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b x +945 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}+945 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{3}+315 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{4}-16 a^{5}+144 a^{4} b x +693 a^{3} b^{2} x^{2}+840 a^{2} b^{3} x^{3}+315 a \,b^{4} x^{4}}{24 \sqrt {x}\, a^{6} x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

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

Output:

(315*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3*b*x 
+ 945*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b** 
2*x**2 + 945*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a 
*b**3*x**3 + 315*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a) 
))*b**4*x**4 - 16*a**5 + 144*a**4*b*x + 693*a**3*b**2*x**2 + 840*a**2*b**3 
*x**3 + 315*a*b**4*x**4)/(24*sqrt(x)*a**6*x*(a**3 + 3*a**2*b*x + 3*a*b**2* 
x**2 + b**3*x**3))