\(\int \frac {A+B x}{x^{7/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [454]

Optimal result

Integrand size = 31, antiderivative size = 289 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b^2 (15 A b-11 a B) \sqrt {x}}{4 a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (A b-a B) \sqrt {x}}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A (a+b x)}{5 a^3 x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (3 A b-a B) (a+b x)}{3 a^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b (2 A b-a B) (a+b x)}{a^5 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^{3/2} (9 A b-5 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

-1/4*b^2*(15*A*b-11*B*a)*x^(1/2)/a^5/((b*x+a)^2)^(1/2)-1/2*b^2*(A*b-B*a)*x 
^(1/2)/a^4/(b*x+a)/((b*x+a)^2)^(1/2)-2/5*A*(b*x+a)/a^3/x^(5/2)/((b*x+a)^2) 
^(1/2)+2/3*(3*A*b-B*a)*(b*x+a)/a^4/x^(3/2)/((b*x+a)^2)^(1/2)-6*b*(2*A*b-B* 
a)*(b*x+a)/a^5/x^(1/2)/((b*x+a)^2)^(1/2)-7/4*b^(3/2)*(9*A*b-5*B*a)*(b*x+a) 
*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(11/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.54 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-945 A b^4 x^4+525 a b^3 x^3 (-3 A+B x)-8 a^4 (3 A+5 B x)+8 a^3 b x (9 A+35 B x)+7 a^2 b^2 x^2 (-72 A+125 B x)\right )+105 b^{3/2} (-9 A b+5 a B) x^{5/2} (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{60 a^{11/2} x^{5/2} (a+b x) \sqrt {(a+b x)^2}} \] Input:

Integrate[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
 

Output:

(Sqrt[a]*(-945*A*b^4*x^4 + 525*a*b^3*x^3*(-3*A + B*x) - 8*a^4*(3*A + 5*B*x 
) + 8*a^3*b*x*(9*A + 35*B*x) + 7*a^2*b^2*x^2*(-72*A + 125*B*x)) + 105*b^(3 
/2)*(-9*A*b + 5*a*B)*x^(5/2)*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] 
)/(60*a^(11/2)*x^(5/2)*(a + b*x)*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.62, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {1187, 27, 87, 52, 61, 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.

\(\displaystyle \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^3 (a+b x) \int \frac {A+B x}{b^3 x^{7/2} (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^{7/2} (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \int \frac {1}{x^{7/2} (a+b x)^2}dx}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \left (\frac {7 \int \frac {1}{x^{7/2} (a+b x)}dx}{2 a}+\frac {1}{a x^{5/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \left (\frac {7 \left (-\frac {b \int \frac {1}{x^{5/2} (a+b x)}dx}{a}-\frac {2}{5 a x^{5/2}}\right )}{2 a}+\frac {1}{a x^{5/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \left (\frac {7 \left (-\frac {b \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{2 a}+\frac {1}{a x^{5/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \left (\frac {7 \left (-\frac {b \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{2 a}+\frac {1}{a x^{5/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \left (\frac {7 \left (-\frac {b \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{2 a}+\frac {1}{a x^{5/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a+b x) \left (\frac {(9 A b-5 a B) \left (\frac {7 \left (-\frac {b \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{2 a}+\frac {1}{a x^{5/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

Int[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
 

Output:

((a + b*x)*((A*b - a*B)/(2*a*b*x^(5/2)*(a + b*x)^2) + ((9*A*b - 5*a*B)*(1/ 
(a*x^(5/2)*(a + b*x)) + (7*(-2/(5*a*x^(5/2)) - (b*(-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 
))/a))/(2*a)))/(4*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 = 1.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.54

method	result	size
risch	\(-\frac {2 \left (90 x^{2} b^{2} A -45 B a \,x^{2} b -15 a b A x +5 a^{2} B x +3 a^{2} A \right ) \sqrt {\left (b x +a \right )^{2}}}{15 a^{5} x^{\frac {5}{2}} \left (b x +a \right )}-\frac {b^{2} \left (\frac {2 \left (\frac {15}{8} b^{2} A -\frac {11}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (17 A b -13 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {7 \left (9 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{a^{5} \left (b x +a \right )}\)	\(156\)
default	\(-\frac {\left (945 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {9}{2}} b^{5}-525 B \sqrt {a b}\, x^{4} a \,b^{3}+1575 A \sqrt {a b}\, x^{3} a \,b^{3}+1890 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {7}{2}} a \,b^{4}-1050 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {7}{2}} a^{2} b^{3}-525 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {5}{2}} a^{3} b^{2}+504 A \sqrt {a b}\, x^{2} a^{2} b^{2}-280 B \sqrt {a b}\, x^{2} a^{3} b -72 A \sqrt {a b}\, x \,a^{3} b +945 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {5}{2}} a^{2} b^{3}-875 B \sqrt {a b}\, x^{3} a^{2} b^{2}+945 A \sqrt {a b}\, x^{4} b^{4}+40 B \sqrt {a b}\, x \,a^{4}-525 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) x^{\frac {9}{2}} a \,b^{4}+24 A \sqrt {a b}\, a^{4}\right ) \left (b x +a \right )}{60 \sqrt {a b}\, x^{\frac {5}{2}} a^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(289\)

Input:

int((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(90*A*b^2*x^2-45*B*a*b*x^2-15*A*a*b*x+5*B*a^2*x+3*A*a^2)/a^5/x^(5/2) 
*((b*x+a)^2)^(1/2)/(b*x+a)-1/a^5*b^2*(2*((15/8*b^2*A-11/8*a*b*B)*x^(3/2)+1 
/8*a*(17*A*b-13*B*a)*x^(1/2))/(b*x+a)^2+7/4*(9*A*b-5*B*a)/(a*b)^(1/2)*arct 
an(b*x^(1/2)/(a*b)^(1/2)))*((b*x+a)^2)^(1/2)/(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.50 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{120 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, \frac {105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) - {\left (24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{60 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \] Input:

integrate((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas 
")
 

Output:

[-1/120*(105*((5*B*a*b^3 - 9*A*b^4)*x^5 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*x^4 
+ (5*B*a^3*b - 9*A*a^2*b^2)*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b 
/a) - a)/(b*x + a)) + 2*(24*A*a^4 - 105*(5*B*a*b^3 - 9*A*b^4)*x^4 - 175*(5 
*B*a^2*b^2 - 9*A*a*b^3)*x^3 - 56*(5*B*a^3*b - 9*A*a^2*b^2)*x^2 + 8*(5*B*a^ 
4 - 9*A*a^3*b)*x)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3), 1/60*(10 
5*((5*B*a*b^3 - 9*A*b^4)*x^5 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*x^4 + (5*B*a^3* 
b - 9*A*a^2*b^2)*x^3)*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) - (24*A*a^4 - 10 
5*(5*B*a*b^3 - 9*A*b^4)*x^4 - 175*(5*B*a^2*b^2 - 9*A*a*b^3)*x^3 - 56*(5*B* 
a^3*b - 9*A*a^2*b^2)*x^2 + 8*(5*B*a^4 - 9*A*a^3*b)*x)*sqrt(x))/(a^5*b^2*x^ 
5 + 2*a^6*b*x^4 + a^7*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)/x**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (191) = 382\).

Time = 0.18 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.40 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {1260 \, {\left (5 \, B a^{3} b^{4} - 11 \, A a^{2} b^{5}\right )} x^{\frac {5}{2}} + 35 \, {\left (5 \, {\left (B a b^{6} - 3 \, A b^{7}\right )} x^{2} + 9 \, {\left (5 \, B a^{2} b^{5} - 11 \, A a b^{6}\right )} x\right )} x^{\frac {5}{2}} - 105 \, {\left (15 \, {\left (B a^{3} b^{4} - 3 \, A a^{2} b^{5}\right )} x^{2} - 17 \, {\left (5 \, B a^{4} b^{3} - 11 \, A a^{3} b^{4}\right )} x\right )} \sqrt {x} - \frac {112 \, {\left (25 \, {\left (B a^{4} b^{3} - 3 \, A a^{3} b^{4}\right )} x^{2} - 9 \, {\left (5 \, B a^{5} b^{2} - 11 \, A a^{4} b^{3}\right )} x\right )}}{\sqrt {x}} - \frac {48 \, {\left (35 \, {\left (B a^{5} b^{2} - 3 \, A a^{4} b^{3}\right )} x^{2} - 3 \, {\left (5 \, B a^{6} b - 11 \, A a^{5} b^{2}\right )} x\right )}}{x^{\frac {3}{2}}} - \frac {16 \, {\left (15 \, {\left (B a^{6} b - 3 \, A a^{5} b^{2}\right )} x^{2} + {\left (5 \, B a^{7} - 11 \, A a^{6} b\right )} x\right )}}{x^{\frac {5}{2}}} - \frac {16 \, {\left (5 \, A a^{6} b x^{2} + 3 \, A a^{7} x\right )}}{x^{\frac {7}{2}}}}{120 \, {\left (a^{7} b^{3} x^{3} + 3 \, a^{8} b^{2} x^{2} + 3 \, a^{9} b x + a^{10}\right )}} + \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} - \frac {7 \, {\left (5 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{\frac {3}{2}} + 6 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} \sqrt {x}\right )}}{24 \, a^{7}} \] Input:

integrate((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima 
")
 

Output:

1/120*(1260*(5*B*a^3*b^4 - 11*A*a^2*b^5)*x^(5/2) + 35*(5*(B*a*b^6 - 3*A*b^ 
7)*x^2 + 9*(5*B*a^2*b^5 - 11*A*a*b^6)*x)*x^(5/2) - 105*(15*(B*a^3*b^4 - 3* 
A*a^2*b^5)*x^2 - 17*(5*B*a^4*b^3 - 11*A*a^3*b^4)*x)*sqrt(x) - 112*(25*(B*a 
^4*b^3 - 3*A*a^3*b^4)*x^2 - 9*(5*B*a^5*b^2 - 11*A*a^4*b^3)*x)/sqrt(x) - 48 
*(35*(B*a^5*b^2 - 3*A*a^4*b^3)*x^2 - 3*(5*B*a^6*b - 11*A*a^5*b^2)*x)/x^(3/ 
2) - 16*(15*(B*a^6*b - 3*A*a^5*b^2)*x^2 + (5*B*a^7 - 11*A*a^6*b)*x)/x^(5/2 
) - 16*(5*A*a^6*b*x^2 + 3*A*a^7*x)/x^(7/2))/(a^7*b^3*x^3 + 3*a^8*b^2*x^2 + 
 3*a^9*b*x + a^10) + 7/4*(5*B*a*b^2 - 9*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b)) 
/(sqrt(a*b)*a^5) - 7/24*(5*(B*a*b^3 - 3*A*b^4)*x^(3/2) + 6*(5*B*a^2*b^2 - 
9*A*a*b^3)*sqrt(x))/a^7
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.55 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {11 \, B a b^{3} x^{\frac {3}{2}} - 15 \, A b^{4} x^{\frac {3}{2}} + 13 \, B a^{2} b^{2} \sqrt {x} - 17 \, A a b^{3} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (45 \, B a b x^{2} - 90 \, A b^{2} x^{2} - 5 \, B a^{2} x + 15 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
 

Output:

7/4*(5*B*a*b^2 - 9*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5*sgn(b 
*x + a)) + 1/4*(11*B*a*b^3*x^(3/2) - 15*A*b^4*x^(3/2) + 13*B*a^2*b^2*sqrt( 
x) - 17*A*a*b^3*sqrt(x))/((b*x + a)^2*a^5*sgn(b*x + a)) + 2/15*(45*B*a*b*x 
^2 - 90*A*b^2*x^2 - 5*B*a^2*x + 15*A*a*b*x - 3*A*a^2)/(a^5*x^(5/2)*sgn(b*x 
 + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{x^{7/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \] Input:

int((A + B*x)/(x^(7/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)
 

Output:

int((A + B*x)/(x^(7/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.37 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}-105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{3}-6 a^{4}+14 a^{3} b x -70 a^{2} b^{2} x^{2}-105 a \,b^{3} x^{3}}{15 \sqrt {x}\, a^{5} x^{2} \left (b x +a \right )} \] Input:

int((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
 

Output:

( - 105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**2 
*x**2 - 105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b* 
*3*x**3 - 6*a**4 + 14*a**3*b*x - 70*a**2*b**2*x**2 - 105*a*b**3*x**3)/(15* 
sqrt(x)*a**5*x**2*(a + b*x))