\(\int \frac {x^{7/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [395]

Optimal result

Integrand size = 29, antiderivative size = 156 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {35 (A b-3 a B) \sqrt {x}}{8 b^5}+\frac {2 B x^{3/2}}{3 b^4}-\frac {(A b-a B) x^{7/2}}{3 b^2 (a+b x)^3}-\frac {(7 A b-13 a B) x^{5/2}}{12 b^3 (a+b x)^2}-\frac {(35 A b-89 a B) x^{3/2}}{24 b^4 (a+b x)}-\frac {35 \sqrt {a} (A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 b^{11/2}} \] Output:

35/8*(A*b-3*B*a)*x^(1/2)/b^5+2/3*B*x^(3/2)/b^4-1/3*(A*b-B*a)*x^(7/2)/b^2/( 
b*x+a)^3-1/12*(7*A*b-13*B*a)*x^(5/2)/b^3/(b*x+a)^2-1/24*(35*A*b-89*B*a)*x^ 
(3/2)/b^4/(b*x+a)-35/8*a^(1/2)*(A*b-3*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2)) 
/b^(11/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.82 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {x} \left (-315 a^4 B+7 a^2 b^2 x (40 A-99 B x)+3 a b^3 x^2 (77 A-48 B x)+105 a^3 b (A-8 B x)+16 b^4 x^3 (3 A+B x)\right )}{24 b^5 (a+b x)^3}+\frac {35 \sqrt {a} (-A b+3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 b^{11/2}} \] Input:

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

Output:

(Sqrt[x]*(-315*a^4*B + 7*a^2*b^2*x*(40*A - 99*B*x) + 3*a*b^3*x^2*(77*A - 4 
8*B*x) + 105*a^3*b*(A - 8*B*x) + 16*b^4*x^3*(3*A + B*x)))/(24*b^5*(a + b*x 
)^3) + (35*Sqrt[a]*(-(A*b) + 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8* 
b^(11/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {1184, 27, 87, 51, 51, 60, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^4 \int \frac {x^{7/2} (A+B x)}{b^4 (a+b x)^4}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^{7/2} (A+B x)}{(a+b x)^4}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {x^{9/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-3 a B) \int \frac {x^{7/2}}{(a+b x)^3}dx}{2 a b}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^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}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {2 \left (B b x +3 A b -12 B a \right ) \sqrt {x}}{3 b^{5}}-\frac {a \left (\frac {2 \left (-\frac {29}{16} A \,b^{3}+\frac {55}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}-\frac {a b \left (17 A b -35 B a \right ) x^{\frac {3}{2}}}{3}+2 \left (-\frac {19}{16} A \,a^{2} b +\frac {41}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {35 \left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}\)	\(118\)
derivativedivides	\(\frac {\frac {2 B b \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-8 B a \sqrt {x}}{b^{5}}-\frac {2 a \left (\frac {\left (-\frac {29}{16} A \,b^{3}+\frac {55}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}-\frac {a b \left (17 A b -35 B a \right ) x^{\frac {3}{2}}}{6}+\left (-\frac {19}{16} A \,a^{2} b +\frac {41}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {35 \left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{b^{5}}\)	\(122\)
default	\(\frac {\frac {2 B b \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-8 B a \sqrt {x}}{b^{5}}-\frac {2 a \left (\frac {\left (-\frac {29}{16} A \,b^{3}+\frac {55}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}-\frac {a b \left (17 A b -35 B a \right ) x^{\frac {3}{2}}}{6}+\left (-\frac {19}{16} A \,a^{2} b +\frac {41}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {35 \left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{b^{5}}\)	\(122\)

Input:

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

Output:

2/3*(B*b*x+3*A*b-12*B*a)*x^(1/2)/b^5-a/b^5*(2*((-29/16*A*b^3+55/16*B*a*b^2 
)*x^(5/2)-1/6*a*b*(17*A*b-35*B*a)*x^(3/2)+(-19/16*A*a^2*b+41/16*B*a^3)*x^( 
1/2))/(b*x+a)^3+35/8*(A*b-3*B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)) 
)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.99 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [-\frac {105 \, {\left (3 \, B a^{4} - A a^{3} b + {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (16 \, B b^{4} x^{4} - 315 \, B a^{4} + 105 \, A a^{3} b - 48 \, {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} - 231 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 280 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{48 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, \frac {105 \, {\left (3 \, B a^{4} - A a^{3} b + {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (16 \, B b^{4} x^{4} - 315 \, B a^{4} + 105 \, A a^{3} b - 48 \, {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} - 231 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 280 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \] Input:

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

Output:

[-1/48*(105*(3*B*a^4 - A*a^3*b + (3*B*a*b^3 - A*b^4)*x^3 + 3*(3*B*a^2*b^2 
- A*a*b^3)*x^2 + 3*(3*B*a^3*b - A*a^2*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sq 
rt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(16*B*b^4*x^4 - 315*B*a^4 + 105*A*a^3 
*b - 48*(3*B*a*b^3 - A*b^4)*x^3 - 231*(3*B*a^2*b^2 - A*a*b^3)*x^2 - 280*(3 
*B*a^3*b - A*a^2*b^2)*x)*sqrt(x))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a 
^3*b^5), 1/24*(105*(3*B*a^4 - A*a^3*b + (3*B*a*b^3 - A*b^4)*x^3 + 3*(3*B*a 
^2*b^2 - A*a*b^3)*x^2 + 3*(3*B*a^3*b - A*a^2*b^2)*x)*sqrt(a/b)*arctan(b*sq 
rt(x)*sqrt(a/b)/a) + (16*B*b^4*x^4 - 315*B*a^4 + 105*A*a^3*b - 48*(3*B*a*b 
^3 - A*b^4)*x^3 - 231*(3*B*a^2*b^2 - A*a*b^3)*x^2 - 280*(3*B*a^3*b - A*a^2 
*b^2)*x)*sqrt(x))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {3 \, {\left (55 \, B a^{2} b^{2} - 29 \, A a b^{3}\right )} x^{\frac {5}{2}} + 8 \, {\left (35 \, B a^{3} b - 17 \, A a^{2} b^{2}\right )} x^{\frac {3}{2}} + 3 \, {\left (41 \, B a^{4} - 19 \, A a^{3} b\right )} \sqrt {x}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} + \frac {35 \, {\left (3 \, B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (4 \, B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{5}} \] Input:

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

Output:

-1/24*(3*(55*B*a^2*b^2 - 29*A*a*b^3)*x^(5/2) + 8*(35*B*a^3*b - 17*A*a^2*b^ 
2)*x^(3/2) + 3*(41*B*a^4 - 19*A*a^3*b)*sqrt(x))/(b^8*x^3 + 3*a*b^7*x^2 + 3 
*a^2*b^6*x + a^3*b^5) + 35/8*(3*B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b)) 
/(sqrt(a*b)*b^5) + 2/3*(B*b*x^(3/2) - 3*(4*B*a - A*b)*sqrt(x))/b^5
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {35 \, {\left (3 \, B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} - \frac {165 \, B a^{2} b^{2} x^{\frac {5}{2}} - 87 \, A a b^{3} x^{\frac {5}{2}} + 280 \, B a^{3} b x^{\frac {3}{2}} - 136 \, A a^{2} b^{2} x^{\frac {3}{2}} + 123 \, B a^{4} \sqrt {x} - 57 \, A a^{3} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} b^{5}} + \frac {2 \, {\left (B b^{8} x^{\frac {3}{2}} - 12 \, B a b^{7} \sqrt {x} + 3 \, A b^{8} \sqrt {x}\right )}}{3 \, b^{12}} \] Input:

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

Output:

35/8*(3*B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/24* 
(165*B*a^2*b^2*x^(5/2) - 87*A*a*b^3*x^(5/2) + 280*B*a^3*b*x^(3/2) - 136*A* 
a^2*b^2*x^(3/2) + 123*B*a^4*sqrt(x) - 57*A*a^3*b*sqrt(x))/((b*x + a)^3*b^5 
) + 2/3*(B*b^8*x^(3/2) - 12*B*a*b^7*sqrt(x) + 3*A*b^8*sqrt(x))/b^12
 

Mupad [B] (verification not implemented)

Time = 10.75 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.13 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\sqrt {x}\,\left (\frac {2\,A}{b^4}-\frac {8\,B\,a}{b^5}\right )-\frac {x^{5/2}\,\left (\frac {55\,B\,a^2\,b^2}{8}-\frac {29\,A\,a\,b^3}{8}\right )-x^{3/2}\,\left (\frac {17\,A\,a^2\,b^2}{3}-\frac {35\,B\,a^3\,b}{3}\right )+\sqrt {x}\,\left (\frac {41\,B\,a^4}{8}-\frac {19\,A\,a^3\,b}{8}\right )}{a^3\,b^5+3\,a^2\,b^6\,x+3\,a\,b^7\,x^2+b^8\,x^3}+\frac {2\,B\,x^{3/2}}{3\,b^4}+\frac {35\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-3\,B\,a\right )}{3\,B\,a^2-A\,a\,b}\right )\,\left (A\,b-3\,B\,a\right )}{8\,b^{11/2}} \] Input:

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

Output:

x^(1/2)*((2*A)/b^4 - (8*B*a)/b^5) - (x^(5/2)*((55*B*a^2*b^2)/8 - (29*A*a*b 
^3)/8) - x^(3/2)*((17*A*a^2*b^2)/3 - (35*B*a^3*b)/3) + x^(1/2)*((41*B*a^4) 
/8 - (19*A*a^3*b)/8))/(a^3*b^5 + b^8*x^3 + 3*a^2*b^6*x + 3*a*b^7*x^2) + (2 
*B*x^(3/2))/(3*b^4) + (35*a^(1/2)*atan((a^(1/2)*b^(1/2)*x^(1/2)*(A*b - 3*B 
*a))/(3*B*a^2 - A*a*b))*(A*b - 3*B*a))/(8*b^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}+210 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b x +105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}-105 \sqrt {x}\, a^{3} b -175 \sqrt {x}\, a^{2} b^{2} x -56 \sqrt {x}\, a \,b^{3} x^{2}+8 \sqrt {x}\, b^{4} x^{3}}{12 b^{5} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:

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

Output:

(105*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3 + 210*sqrt(b 
)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b*x + 105*sqrt(b)*sqrt( 
a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**2*x**2 - 105*sqrt(x)*a**3*b - 
175*sqrt(x)*a**2*b**2*x - 56*sqrt(x)*a*b**3*x**2 + 8*sqrt(x)*b**4*x**3)/(1 
2*b**5*(a**2 + 2*a*b*x + b**2*x**2))