\(\int \frac {x^{7/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [457]

Optimal result

Integrand size = 31, antiderivative size = 285 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(35 A b-187 a B) \sqrt {x}}{64 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) x^{7/2}}{4 b^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(7 A b-15 a B) x^{5/2}}{24 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(35 A b-123 a B) x^{3/2}}{96 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B \sqrt {x} (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 (A b-9 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 \sqrt {a} b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

-1/64*(35*A*b-187*B*a)*x^(1/2)/b^5/((b*x+a)^2)^(1/2)-1/4*(A*b-B*a)*x^(7/2) 
/b^2/(b*x+a)^3/((b*x+a)^2)^(1/2)-1/24*(7*A*b-15*B*a)*x^(5/2)/b^3/(b*x+a)^2 
/((b*x+a)^2)^(1/2)-1/96*(35*A*b-123*B*a)*x^(3/2)/b^4/(b*x+a)/((b*x+a)^2)^( 
1/2)+2*B*x^(1/2)*(b*x+a)/b^5/((b*x+a)^2)^(1/2)+35/64*(A*b-9*B*a)*(b*x+a)*a 
rctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(1/2)/b^(11/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.51 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {b} \sqrt {x} \left (945 a^4 B-105 a^3 b (A-33 B x)+3 b^4 x^3 (-93 A+128 B x)+7 a^2 b^2 x (-55 A+657 B x)+a b^3 x^2 (-511 A+2511 B x)\right )+\frac {105 (A b-9 a B) (a+b x)^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}}}{192 b^{11/2} (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

Integrate[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(Sqrt[b]*Sqrt[x]*(945*a^4*B - 105*a^3*b*(A - 33*B*x) + 3*b^4*x^3*(-93*A + 
128*B*x) + 7*a^2*b^2*x*(-55*A + 657*B*x) + a*b^3*x^2*(-511*A + 2511*B*x)) 
+ (105*(A*b - 9*a*B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[a 
])/(192*b^(11/2)*(a + b*x)^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.68, 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, 51, 51, 51, 60, 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.

\(\displaystyle \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^5 (a+b x) \int \frac {x^{7/2} (A+B x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {x^{7/2} (A+B x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \int \frac {x^{7/2}}{(a+b x)^4}dx}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \left (\frac {7 \int \frac {x^{5/2}}{(a+b x)^3}dx}{6 b}-\frac {x^{7/2}}{3 b (a+b x)^3}\right )}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \left (\frac {7 \left (\frac {5 \int \frac {x^{3/2}}{(a+b x)^2}dx}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{7/2}}{3 b (a+b x)^3}\right )}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{a+b x}dx}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{7/2}}{3 b (a+b x)^3}\right )}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{7/2}}{3 b (a+b x)^3}\right )}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{7/2}}{3 b (a+b x)^3}\right )}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{4 a b (a+b x)^4}-\frac {(A b-9 a B) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{7/2}}{3 b (a+b x)^3}\right )}{8 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

Int[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

((a + b*x)*(((A*b - a*B)*x^(9/2))/(4*a*b*(a + b*x)^4) - ((A*b - 9*a*B)*(-1 
/3*x^(7/2)/(b*(a + b*x)^3) + (7*(-1/2*x^(5/2)/(b*(a + b*x)^2) + (5*(-(x^(3 
/2)/(b*(a + b*x))) + (3*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x 
])/Sqrt[a]])/b^(3/2)))/(2*b)))/(4*b)))/(6*b)))/(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 51

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

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 = 1.14 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.56

method	result	size
risch	\(\frac {2 B \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{b^{5} \left (b x +a \right )}+\frac {\left (\frac {2 \left (-\frac {93}{128} A \,b^{4}+\frac {325}{128} B a \,b^{3}\right ) x^{\frac {7}{2}}+2 \left (\frac {765}{128} B \,a^{2} b^{2}-\frac {511}{384} A a \,b^{3}\right ) x^{\frac {5}{2}}-\frac {a^{2} b \left (385 A b -1929 B a \right ) x^{\frac {3}{2}}}{192}+2 \left (\frac {187}{128} a^{4} B -\frac {35}{128} A \,a^{3} b \right ) \sqrt {x}}{\left (b x +a \right )^{4}}+\frac {35 \left (A b -9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{5} \left (b x +a \right )}\)	\(159\)
default	\(-\frac {\left (279 A \sqrt {a b}\, x^{\frac {7}{2}} b^{4}-2511 B \sqrt {a b}\, x^{\frac {7}{2}} a \,b^{3}+511 A \sqrt {a b}\, x^{\frac {5}{2}} a \,b^{3}-105 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b^{5} x^{4}-4599 B \sqrt {a b}\, x^{\frac {5}{2}} a^{2} b^{2}-384 B \sqrt {a b}\, x^{\frac {9}{2}} b^{4}+945 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{4}-420 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{3}+3780 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{3}+385 A \sqrt {a b}\, x^{\frac {3}{2}} a^{2} b^{2}-630 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{2}-3465 B \sqrt {a b}\, x^{\frac {3}{2}} a^{3} b +5670 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x^{2}-420 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x +3780 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b x +105 A \sqrt {a b}\, \sqrt {x}\, a^{3} b -105 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b -945 B \sqrt {a b}\, \sqrt {x}\, a^{4}+945 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{5}\right ) \left (b x +a \right )}{192 \sqrt {a b}\, b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(368\)

Input:

int(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2*B/b^5*x^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)+1/b^5*(2*((-93/128*A*b^4+325/128 
*B*a*b^3)*x^(7/2)+(765/128*B*a^2*b^2-511/384*A*a*b^3)*x^(5/2)-1/384*a^2*b* 
(385*A*b-1929*B*a)*x^(3/2)+(187/128*a^4*B-35/128*A*a^3*b)*x^(1/2))/(b*x+a) 
^4+35/64*(A*b-9*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 = 555, normalized size of antiderivative = 1.95 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [\frac {105 \, {\left (9 \, B a^{5} - A a^{4} b + {\left (9 \, B a b^{4} - A b^{5}\right )} x^{4} + 4 \, {\left (9 \, B a^{2} b^{3} - A a b^{4}\right )} x^{3} + 6 \, {\left (9 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (9 \, B a^{4} b - A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (384 \, B a b^{5} x^{4} + 945 \, B a^{5} b - 105 \, A a^{4} b^{2} + 279 \, {\left (9 \, B a^{2} b^{4} - A a b^{5}\right )} x^{3} + 511 \, {\left (9 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} x^{2} + 385 \, {\left (9 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a b^{10} x^{4} + 4 \, a^{2} b^{9} x^{3} + 6 \, a^{3} b^{8} x^{2} + 4 \, a^{4} b^{7} x + a^{5} b^{6}\right )}}, \frac {105 \, {\left (9 \, B a^{5} - A a^{4} b + {\left (9 \, B a b^{4} - A b^{5}\right )} x^{4} + 4 \, {\left (9 \, B a^{2} b^{3} - A a b^{4}\right )} x^{3} + 6 \, {\left (9 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (9 \, B a^{4} b - A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (384 \, B a b^{5} x^{4} + 945 \, B a^{5} b - 105 \, A a^{4} b^{2} + 279 \, {\left (9 \, B a^{2} b^{4} - A a b^{5}\right )} x^{3} + 511 \, {\left (9 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} x^{2} + 385 \, {\left (9 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a b^{10} x^{4} + 4 \, a^{2} b^{9} x^{3} + 6 \, a^{3} b^{8} x^{2} + 4 \, a^{4} b^{7} x + a^{5} b^{6}\right )}}\right ] \] Input:

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas 
")
 

Output:

[1/384*(105*(9*B*a^5 - A*a^4*b + (9*B*a*b^4 - A*b^5)*x^4 + 4*(9*B*a^2*b^3 
- A*a*b^4)*x^3 + 6*(9*B*a^3*b^2 - A*a^2*b^3)*x^2 + 4*(9*B*a^4*b - A*a^3*b^ 
2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(384* 
B*a*b^5*x^4 + 945*B*a^5*b - 105*A*a^4*b^2 + 279*(9*B*a^2*b^4 - A*a*b^5)*x^ 
3 + 511*(9*B*a^3*b^3 - A*a^2*b^4)*x^2 + 385*(9*B*a^4*b^2 - A*a^3*b^3)*x)*s 
qrt(x))/(a*b^10*x^4 + 4*a^2*b^9*x^3 + 6*a^3*b^8*x^2 + 4*a^4*b^7*x + a^5*b^ 
6), 1/192*(105*(9*B*a^5 - A*a^4*b + (9*B*a*b^4 - A*b^5)*x^4 + 4*(9*B*a^2*b 
^3 - A*a*b^4)*x^3 + 6*(9*B*a^3*b^2 - A*a^2*b^3)*x^2 + 4*(9*B*a^4*b - A*a^3 
*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (384*B*a*b^5*x^4 + 945* 
B*a^5*b - 105*A*a^4*b^2 + 279*(9*B*a^2*b^4 - A*a*b^5)*x^3 + 511*(9*B*a^3*b 
^3 - A*a^2*b^4)*x^2 + 385*(9*B*a^4*b^2 - A*a^3*b^3)*x)*sqrt(x))/(a*b^10*x^ 
4 + 4*a^2*b^9*x^3 + 6*a^3*b^8*x^2 + 4*a^4*b^7*x + a^5*b^6)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(7/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.31 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {105 \, {\left (3 \, {\left (11 \, B a b^{5} - A b^{6}\right )} x^{2} + {\left (9 \, B a^{2} b^{4} + A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 30 \, {\left ({\left (359 \, B a^{2} b^{4} - 21 \, A a b^{5}\right )} x^{2} + {\left (61 \, B a^{3} b^{3} + 21 \, A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} + 20 \, {\left (66 \, {\left (11 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} x^{2} + 13 \, {\left (9 \, B a^{4} b^{2} + A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} + 2 \, {\left (405 \, {\left (11 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} x^{2} + 77 \, {\left (9 \, B a^{5} b + A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 7 \, {\left (27 \, {\left (11 \, B a^{5} b - A a^{4} b^{2}\right )} x^{2} + 5 \, {\left (9 \, B a^{6} + A a^{5} b\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{2} b^{9} x^{5} + 5 \, a^{3} b^{8} x^{4} + 10 \, a^{4} b^{7} x^{3} + 10 \, a^{5} b^{6} x^{2} + 5 \, a^{6} b^{5} x + a^{7} b^{4}\right )}} - \frac {35 \, {\left (9 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} b^{5}} - \frac {7 \, {\left (3 \, {\left (11 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} - 10 \, {\left (9 \, B a^{2} - A a b\right )} \sqrt {x}\right )}}{128 \, a^{2} b^{5}} \] Input:

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima 
")
 

Output:

1/1920*(105*(3*(11*B*a*b^5 - A*b^6)*x^2 + (9*B*a^2*b^4 + A*a*b^5)*x)*x^(9/ 
2) + 30*((359*B*a^2*b^4 - 21*A*a*b^5)*x^2 + (61*B*a^3*b^3 + 21*A*a^2*b^4)* 
x)*x^(7/2) + 20*(66*(11*B*a^3*b^3 - A*a^2*b^4)*x^2 + 13*(9*B*a^4*b^2 + A*a 
^3*b^3)*x)*x^(5/2) + 2*(405*(11*B*a^4*b^2 - A*a^3*b^3)*x^2 + 77*(9*B*a^5*b 
 + A*a^4*b^2)*x)*x^(3/2) + 7*(27*(11*B*a^5*b - A*a^4*b^2)*x^2 + 5*(9*B*a^6 
 + A*a^5*b)*x)*sqrt(x))/(a^2*b^9*x^5 + 5*a^3*b^8*x^4 + 10*a^4*b^7*x^3 + 10 
*a^5*b^6*x^2 + 5*a^6*b^5*x + a^7*b^4) - 35/64*(9*B*a - A*b)*arctan(b*sqrt( 
x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 7/128*(3*(11*B*a*b - A*b^2)*x^(3/2) - 10*( 
9*B*a^2 - A*a*b)*sqrt(x))/(a^2*b^5)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.56 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {2 \, B \sqrt {x}}{b^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {35 \, {\left (9 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {975 \, B a b^{3} x^{\frac {7}{2}} - 279 \, A b^{4} x^{\frac {7}{2}} + 2295 \, B a^{2} b^{2} x^{\frac {5}{2}} - 511 \, A a b^{3} x^{\frac {5}{2}} + 1929 \, B a^{3} b x^{\frac {3}{2}} - 385 \, A a^{2} b^{2} x^{\frac {3}{2}} + 561 \, B a^{4} \sqrt {x} - 105 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} b^{5} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
 

Output:

2*B*sqrt(x)/(b^5*sgn(b*x + a)) - 35/64*(9*B*a - A*b)*arctan(b*sqrt(x)/sqrt 
(a*b))/(sqrt(a*b)*b^5*sgn(b*x + a)) + 1/192*(975*B*a*b^3*x^(7/2) - 279*A*b 
^4*x^(7/2) + 2295*B*a^2*b^2*x^(5/2) - 511*A*a*b^3*x^(5/2) + 1929*B*a^3*b*x 
^(3/2) - 385*A*a^2*b^2*x^(3/2) + 561*B*a^4*sqrt(x) - 105*A*a^3*b*sqrt(x))/ 
((b*x + a)^4*b^5*sgn(b*x + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:

int((x^(7/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int((x^(7/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.60 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b x -315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}-105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{3}+105 \sqrt {x}\, a^{3} b +280 \sqrt {x}\, a^{2} b^{2} x +231 \sqrt {x}\, a \,b^{3} x^{2}+48 \sqrt {x}\, b^{4} x^{3}}{24 b^{5} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

int(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

( - 105*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3 - 315*sqr 
t(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b*x - 315*sqrt(b)*sq 
rt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**2*x**2 - 105*sqrt(b)*sqrt(a 
)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**3*x**3 + 105*sqrt(x)*a**3*b + 280 
*sqrt(x)*a**2*b**2*x + 231*sqrt(x)*a*b**3*x**2 + 48*sqrt(x)*b**4*x**3)/(24 
*b**5*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))