\(\int \frac {x^{7/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [404]

Optimal result

Integrand size = 29, antiderivative size = 182 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {(A b-a B) x^{7/2}}{5 b^2 (a+b x)^5}-\frac {(7 A b-17 a B) x^{5/2}}{40 b^3 (a+b x)^4}-\frac {(7 A b-33 a B) x^{3/2}}{48 b^4 (a+b x)^3}-\frac {(7 A b-65 a B) \sqrt {x}}{64 b^5 (a+b x)^2}+\frac {(7 A b-193 a B) \sqrt {x}}{128 a b^5 (a+b x)}+\frac {7 (A b+9 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 a^{3/2} b^{11/2}} \] Output:

-1/5*(A*b-B*a)*x^(7/2)/b^2/(b*x+a)^5-1/40*(7*A*b-17*B*a)*x^(5/2)/b^3/(b*x+ 
a)^4-1/48*(7*A*b-33*B*a)*x^(3/2)/b^4/(b*x+a)^3-1/64*(7*A*b-65*B*a)*x^(1/2) 
/b^5/(b*x+a)^2+1/128*(7*A*b-193*B*a)*x^(1/2)/a/b^5/(b*x+a)+7/128*(A*b+9*B* 
a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(3/2)/b^(11/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.79 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {\sqrt {x} \left (945 a^5 B-105 A b^5 x^4+105 a^4 b (A+42 B x)+14 a^3 b^2 x (35 A+576 B x)+5 a b^4 x^3 (158 A+579 B x)+2 a^2 b^3 x^2 (448 A+3555 B x)\right )}{1920 a b^5 (a+b x)^5}+\frac {7 (A b+9 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 a^{3/2} b^{11/2}} \] Input:

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

Output:

-1/1920*(Sqrt[x]*(945*a^5*B - 105*A*b^5*x^4 + 105*a^4*b*(A + 42*B*x) + 14* 
a^3*b^2*x*(35*A + 576*B*x) + 5*a*b^4*x^3*(158*A + 579*B*x) + 2*a^2*b^3*x^2 
*(448*A + 3555*B*x)))/(a*b^5*(a + b*x)^5) + (7*(A*b + 9*a*B)*ArcTan[(Sqrt[ 
b]*Sqrt[x])/Sqrt[a]])/(128*a^(3/2)*b^(11/2))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.97, 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, 51, 51, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^6 \int \frac {x^{7/2} (A+B x)}{b^6 (a+b x)^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^{7/2} (A+B x)}{(a+b x)^6}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(9 a B+A b) \int \frac {x^{7/2}}{(a+b x)^5}dx}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(9 a B+A b) \left (\frac {7 \int \frac {x^{5/2}}{(a+b x)^4}dx}{8 b}-\frac {x^{7/2}}{4 b (a+b x)^4}\right )}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(9 a B+A b) \left (\frac {7 \left (\frac {5 \int \frac {x^{3/2}}{(a+b x)^3}dx}{6 b}-\frac {x^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{7/2}}{4 b (a+b x)^4}\right )}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(9 a B+A b) \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{(a+b x)^2}dx}{4 b}-\frac {x^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{7/2}}{4 b (a+b x)^4}\right )}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(9 a B+A b) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {x} (a+b x)}dx}{2 b}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 b}-\frac {x^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{7/2}}{4 b (a+b x)^4}\right )}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(9 a B+A b) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{a+b x}d\sqrt {x}}{b}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 b}-\frac {x^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{7/2}}{4 b (a+b x)^4}\right )}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(9 a B+A b) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 b}-\frac {x^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {x^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{7/2}}{4 b (a+b x)^4}\right )}{10 a b}+\frac {x^{9/2} (A b-a B)}{5 a b (a+b x)^5}\)

Input:

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

Output:

((A*b - a*B)*x^(9/2))/(5*a*b*(a + b*x)^5) + ((A*b + 9*a*B)*(-1/4*x^(7/2)/( 
b*(a + b*x)^4) + (7*(-1/3*x^(5/2)/(b*(a + b*x)^3) + (5*(-1/2*x^(3/2)/(b*(a 
 + b*x)^2) + (3*(-(Sqrt[x]/(b*(a + b*x))) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[ 
a]]/(Sqrt[a]*b^(3/2))))/(4*b)))/(6*b)))/(8*b)))/(10*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 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 = 2.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\frac {\left (7 A b -193 B a \right ) x^{\frac {9}{2}}}{128 a b}-\frac {79 \left (A b +9 B a \right ) x^{\frac {7}{2}}}{192 b^{2}}-\frac {7 a \left (A b +9 B a \right ) x^{\frac {5}{2}}}{15 b^{3}}-\frac {49 a^{2} \left (A b +9 B a \right ) x^{\frac {3}{2}}}{192 b^{4}}-\frac {7 \left (A b +9 B a \right ) a^{3} \sqrt {x}}{128 b^{5}}}{\left (b x +a \right )^{5}}+\frac {7 \left (A b +9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 a \,b^{5} \sqrt {a b}}\)	\(135\)
default	\(\frac {\frac {\left (7 A b -193 B a \right ) x^{\frac {9}{2}}}{128 a b}-\frac {79 \left (A b +9 B a \right ) x^{\frac {7}{2}}}{192 b^{2}}-\frac {7 a \left (A b +9 B a \right ) x^{\frac {5}{2}}}{15 b^{3}}-\frac {49 a^{2} \left (A b +9 B a \right ) x^{\frac {3}{2}}}{192 b^{4}}-\frac {7 \left (A b +9 B a \right ) a^{3} \sqrt {x}}{128 b^{5}}}{\left (b x +a \right )^{5}}+\frac {7 \left (A b +9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 a \,b^{5} \sqrt {a b}}\)	\(135\)

Input:

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

Output:

2*(1/256*(7*A*b-193*B*a)/a/b*x^(9/2)-79/384*(A*b+9*B*a)/b^2*x^(7/2)-7/30*a 
*(A*b+9*B*a)/b^3*x^(5/2)-49/384*a^2*(A*b+9*B*a)/b^4*x^(3/2)-7/256*(A*b+9*B 
*a)*a^3/b^5*x^(1/2))/(b*x+a)^5+7/128*(A*b+9*B*a)/a/b^5/(a*b)^(1/2)*arctan( 
b*x^(1/2)/(a*b)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (150) = 300\).

Time = 0.10 (sec) , antiderivative size = 639, normalized size of antiderivative = 3.51 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\left [-\frac {105 \, {\left (9 \, B a^{6} + A a^{5} b + {\left (9 \, B a b^{5} + A b^{6}\right )} x^{5} + 5 \, {\left (9 \, B a^{2} b^{4} + A a b^{5}\right )} x^{4} + 10 \, {\left (9 \, B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{3} + 10 \, {\left (9 \, B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 5 \, {\left (9 \, B a^{5} b + A a^{4} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (945 \, B a^{6} b + 105 \, A a^{5} b^{2} + 15 \, {\left (193 \, B a^{2} b^{5} - 7 \, A a b^{6}\right )} x^{4} + 790 \, {\left (9 \, B a^{3} b^{4} + A a^{2} b^{5}\right )} x^{3} + 896 \, {\left (9 \, B a^{4} b^{3} + A a^{3} b^{4}\right )} x^{2} + 490 \, {\left (9 \, B a^{5} b^{2} + A a^{4} b^{3}\right )} x\right )} \sqrt {x}}{3840 \, {\left (a^{2} b^{11} x^{5} + 5 \, a^{3} b^{10} x^{4} + 10 \, a^{4} b^{9} x^{3} + 10 \, a^{5} b^{8} x^{2} + 5 \, a^{6} b^{7} x + a^{7} b^{6}\right )}}, -\frac {105 \, {\left (9 \, B a^{6} + A a^{5} b + {\left (9 \, B a b^{5} + A b^{6}\right )} x^{5} + 5 \, {\left (9 \, B a^{2} b^{4} + A a b^{5}\right )} x^{4} + 10 \, {\left (9 \, B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{3} + 10 \, {\left (9 \, B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 5 \, {\left (9 \, B a^{5} b + A a^{4} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (945 \, B a^{6} b + 105 \, A a^{5} b^{2} + 15 \, {\left (193 \, B a^{2} b^{5} - 7 \, A a b^{6}\right )} x^{4} + 790 \, {\left (9 \, B a^{3} b^{4} + A a^{2} b^{5}\right )} x^{3} + 896 \, {\left (9 \, B a^{4} b^{3} + A a^{3} b^{4}\right )} x^{2} + 490 \, {\left (9 \, B a^{5} b^{2} + A a^{4} b^{3}\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{2} b^{11} x^{5} + 5 \, a^{3} b^{10} x^{4} + 10 \, a^{4} b^{9} x^{3} + 10 \, a^{5} b^{8} x^{2} + 5 \, a^{6} b^{7} x + a^{7} b^{6}\right )}}\right ] \] Input:

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

Output:

[-1/3840*(105*(9*B*a^6 + A*a^5*b + (9*B*a*b^5 + A*b^6)*x^5 + 5*(9*B*a^2*b^ 
4 + A*a*b^5)*x^4 + 10*(9*B*a^3*b^3 + A*a^2*b^4)*x^3 + 10*(9*B*a^4*b^2 + A* 
a^3*b^3)*x^2 + 5*(9*B*a^5*b + A*a^4*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sq 
rt(-a*b)*sqrt(x))/(b*x + a)) + 2*(945*B*a^6*b + 105*A*a^5*b^2 + 15*(193*B* 
a^2*b^5 - 7*A*a*b^6)*x^4 + 790*(9*B*a^3*b^4 + A*a^2*b^5)*x^3 + 896*(9*B*a^ 
4*b^3 + A*a^3*b^4)*x^2 + 490*(9*B*a^5*b^2 + A*a^4*b^3)*x)*sqrt(x))/(a^2*b^ 
11*x^5 + 5*a^3*b^10*x^4 + 10*a^4*b^9*x^3 + 10*a^5*b^8*x^2 + 5*a^6*b^7*x + 
a^7*b^6), -1/1920*(105*(9*B*a^6 + A*a^5*b + (9*B*a*b^5 + A*b^6)*x^5 + 5*(9 
*B*a^2*b^4 + A*a*b^5)*x^4 + 10*(9*B*a^3*b^3 + A*a^2*b^4)*x^3 + 10*(9*B*a^4 
*b^2 + A*a^3*b^3)*x^2 + 5*(9*B*a^5*b + A*a^4*b^2)*x)*sqrt(a*b)*arctan(sqrt 
(a*b)/(b*sqrt(x))) + (945*B*a^6*b + 105*A*a^5*b^2 + 15*(193*B*a^2*b^5 - 7* 
A*a*b^6)*x^4 + 790*(9*B*a^3*b^4 + A*a^2*b^5)*x^3 + 896*(9*B*a^4*b^3 + A*a^ 
3*b^4)*x^2 + 490*(9*B*a^5*b^2 + A*a^4*b^3)*x)*sqrt(x))/(a^2*b^11*x^5 + 5*a 
^3*b^10*x^4 + 10*a^4*b^9*x^3 + 10*a^5*b^8*x^2 + 5*a^6*b^7*x + a^7*b^6)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.09 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {15 \, {\left (193 \, B a b^{4} - 7 \, A b^{5}\right )} x^{\frac {9}{2}} + 790 \, {\left (9 \, B a^{2} b^{3} + A a b^{4}\right )} x^{\frac {7}{2}} + 896 \, {\left (9 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{\frac {5}{2}} + 490 \, {\left (9 \, B a^{4} b + A a^{3} b^{2}\right )} x^{\frac {3}{2}} + 105 \, {\left (9 \, B a^{5} + A a^{4} b\right )} \sqrt {x}}{1920 \, {\left (a b^{10} x^{5} + 5 \, a^{2} b^{9} x^{4} + 10 \, a^{3} b^{8} x^{3} + 10 \, a^{4} b^{7} x^{2} + 5 \, a^{5} b^{6} x + a^{6} b^{5}\right )}} + \frac {7 \, {\left (9 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a b^{5}} \] Input:

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

Output:

-1/1920*(15*(193*B*a*b^4 - 7*A*b^5)*x^(9/2) + 790*(9*B*a^2*b^3 + A*a*b^4)* 
x^(7/2) + 896*(9*B*a^3*b^2 + A*a^2*b^3)*x^(5/2) + 490*(9*B*a^4*b + A*a^3*b 
^2)*x^(3/2) + 105*(9*B*a^5 + A*a^4*b)*sqrt(x))/(a*b^10*x^5 + 5*a^2*b^9*x^4 
 + 10*a^3*b^8*x^3 + 10*a^4*b^7*x^2 + 5*a^5*b^6*x + a^6*b^5) + 7/128*(9*B*a 
 + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^5)
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.85 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {7 \, {\left (9 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a b^{5}} - \frac {2895 \, B a b^{4} x^{\frac {9}{2}} - 105 \, A b^{5} x^{\frac {9}{2}} + 7110 \, B a^{2} b^{3} x^{\frac {7}{2}} + 790 \, A a b^{4} x^{\frac {7}{2}} + 8064 \, B a^{3} b^{2} x^{\frac {5}{2}} + 896 \, A a^{2} b^{3} x^{\frac {5}{2}} + 4410 \, B a^{4} b x^{\frac {3}{2}} + 490 \, A a^{3} b^{2} x^{\frac {3}{2}} + 945 \, B a^{5} \sqrt {x} + 105 \, A a^{4} b \sqrt {x}}{1920 \, {\left (b x + a\right )}^{5} a b^{5}} \] Input:

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

Output:

7/128*(9*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^5) - 1/1920 
*(2895*B*a*b^4*x^(9/2) - 105*A*b^5*x^(9/2) + 7110*B*a^2*b^3*x^(7/2) + 790* 
A*a*b^4*x^(7/2) + 8064*B*a^3*b^2*x^(5/2) + 896*A*a^2*b^3*x^(5/2) + 4410*B* 
a^4*b*x^(3/2) + 490*A*a^3*b^2*x^(3/2) + 945*B*a^5*sqrt(x) + 105*A*a^4*b*sq 
rt(x))/((b*x + a)^5*a*b^5)
 

Mupad [B] (verification not implemented)

Time = 10.70 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {7\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b+9\,B\,a\right )}{128\,a^{3/2}\,b^{11/2}}-\frac {\frac {79\,x^{7/2}\,\left (A\,b+9\,B\,a\right )}{192\,b^2}+\frac {49\,a^2\,x^{3/2}\,\left (A\,b+9\,B\,a\right )}{192\,b^4}+\frac {7\,a^3\,\sqrt {x}\,\left (A\,b+9\,B\,a\right )}{128\,b^5}-\frac {x^{9/2}\,\left (7\,A\,b-193\,B\,a\right )}{128\,a\,b}+\frac {7\,a\,x^{5/2}\,\left (A\,b+9\,B\,a\right )}{15\,b^3}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5} \] Input:

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

Output:

(7*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b + 9*B*a))/(128*a^(3/2)*b^(11/2)) - 
 ((79*x^(7/2)*(A*b + 9*B*a))/(192*b^2) + (49*a^2*x^(3/2)*(A*b + 9*B*a))/(1 
92*b^4) + (7*a^3*x^(1/2)*(A*b + 9*B*a))/(128*b^5) - (x^(9/2)*(7*A*b - 193* 
B*a))/(128*a*b) + (7*a*x^(5/2)*(A*b + 9*B*a))/(15*b^3))/(a^5 + b^5*x^5 + 5 
*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.19 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}+420 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b x +630 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}+420 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{3}+105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{4}-105 \sqrt {x}\, a^{4} b -385 \sqrt {x}\, a^{3} b^{2} x -511 \sqrt {x}\, a^{2} b^{3} x^{2}-279 \sqrt {x}\, a \,b^{4} x^{3}}{192 a \,b^{5} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:

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

Output:

(105*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**4 + 420*sqrt(b 
)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3*b*x + 630*sqrt(b)*sqrt( 
a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**2 + 420*sqrt(b)*sqrt(a 
)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**3*x**3 + 105*sqrt(b)*sqrt(a)*at 
an((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**4*x**4 - 105*sqrt(x)*a**4*b - 385*sqr 
t(x)*a**3*b**2*x - 511*sqrt(x)*a**2*b**3*x**2 - 279*sqrt(x)*a*b**4*x**3)/( 
192*a*b**5*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3*x**3 + b**4*x* 
*4))