\(\int \frac {A+B x}{x^{7/2} (a^2+2 a b x+b^2 x^2)} \, dx\) [393]

Optimal result

Integrand size = 29, antiderivative size = 121 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 A}{5 a^2 x^{5/2}}+\frac {2 (2 A b-a B)}{3 a^3 x^{3/2}}-\frac {2 b (3 A b-2 a B)}{a^4 \sqrt {x}}-\frac {b^2 (A b-a B) \sqrt {x}}{a^4 (a+b x)}-\frac {b^{3/2} (7 A b-5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {-105 A b^3 x^3-2 a^3 (3 A+5 B x)+5 a b^2 x^2 (-14 A+15 B x)+2 a^2 b x (7 A+25 B x)}{15 a^4 x^{5/2} (a+b x)}+\frac {b^{3/2} (-7 A b+5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2}} \] Input:

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

Output:

(-105*A*b^3*x^3 - 2*a^3*(3*A + 5*B*x) + 5*a*b^2*x^2*(-14*A + 15*B*x) + 2*a 
^2*b*x*(7*A + 25*B*x))/(15*a^4*x^(5/2)*(a + b*x)) + (b^(3/2)*(-7*A*b + 5*a 
*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(9/2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1184, 27, 87, 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.

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

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(A*b - a*B)/(a*b*x^(5/2)*(a + b*x)) + ((7*A*b - 5*a*B)*(-2/(5*a*x^(5/2)) - 
 (b*(-2/(3*a*x^(3/2)) - (b*(-2/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sq 
rt[x])/Sqrt[a]])/a^(3/2)))/a))/a))/(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 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 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 = 1.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {2 A}{5 a^{2} x^{\frac {5}{2}}}-\frac {2 \left (-2 A b +B a \right )}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 b \left (3 A b -2 B a \right )}{a^{4} \sqrt {x}}-\frac {2 b^{2} \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (7 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}\)	\(101\)
default	\(-\frac {2 A}{5 a^{2} x^{\frac {5}{2}}}-\frac {2 \left (-2 A b +B a \right )}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 b \left (3 A b -2 B a \right )}{a^{4} \sqrt {x}}-\frac {2 b^{2} \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (7 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}\)	\(101\)
risch	\(-\frac {2 \left (45 x^{2} b^{2} A -30 B a \,x^{2} b -10 a b A x +5 a^{2} B x +3 a^{2} A \right )}{15 a^{4} x^{\frac {5}{2}}}-\frac {b^{2} \left (\frac {2 \left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (7 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a^{4}}\)	\(103\)

Input:

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

Output:

-2/5*A/a^2/x^(5/2)-2/3*(-2*A*b+B*a)/a^3/x^(3/2)-2*b*(3*A*b-2*B*a)/a^4/x^(1 
/2)-2*b^2/a^4*((1/2*A*b-1/2*B*a)*x^(1/2)/(b*x+a)+1/2*(7*A*b-5*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 = 316, normalized size of antiderivative = 2.61 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\left [-\frac {15 \, {\left ({\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (6 \, A a^{3} - 15 \, {\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 10 \, {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 2 \, {\left (5 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{30 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac {15 \, {\left ({\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) - {\left (6 \, A a^{3} - 15 \, {\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 10 \, {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 2 \, {\left (5 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{15 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \] Input:

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

Output:

[-1/30*(15*((5*B*a*b^2 - 7*A*b^3)*x^4 + (5*B*a^2*b - 7*A*a*b^2)*x^3)*sqrt( 
-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(6*A*a^3 - 15* 
(5*B*a*b^2 - 7*A*b^3)*x^3 - 10*(5*B*a^2*b - 7*A*a*b^2)*x^2 + 2*(5*B*a^3 - 
7*A*a^2*b)*x)*sqrt(x))/(a^4*b*x^4 + a^5*x^3), 1/15*(15*((5*B*a*b^2 - 7*A*b 
^3)*x^4 + (5*B*a^2*b - 7*A*a*b^2)*x^3)*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) 
 - (6*A*a^3 - 15*(5*B*a*b^2 - 7*A*b^3)*x^3 - 10*(5*B*a^2*b - 7*A*a*b^2)*x^ 
2 + 2*(5*B*a^3 - 7*A*a^2*b)*x)*sqrt(x))/(a^4*b*x^4 + a^5*x^3)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1017 vs. \(2 (117) = 234\).

Time = 113.34 (sec) , antiderivative size = 1017, normalized size of antiderivative = 8.40 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0) 
), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2)))/a**2, Eq(b, 0)), ((-2*A/(9*x**( 
9/2)) - 2*B/(7*x**(7/2)))/b**2, Eq(a, 0)), (-12*A*a**3*sqrt(-a/b)/(30*a**5 
*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) + 28*A*a**2*b*x*sqrt 
(-a/b)/(30*a**5*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) - 105 
*A*a*b**2*x**(5/2)*log(sqrt(x) - sqrt(-a/b))/(30*a**5*x**(5/2)*sqrt(-a/b) 
+ 30*a**4*b*x**(7/2)*sqrt(-a/b)) + 105*A*a*b**2*x**(5/2)*log(sqrt(x) + sqr 
t(-a/b))/(30*a**5*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) - 1 
40*A*a*b**2*x**2*sqrt(-a/b)/(30*a**5*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7 
/2)*sqrt(-a/b)) - 105*A*b**3*x**(7/2)*log(sqrt(x) - sqrt(-a/b))/(30*a**5*x 
**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) + 105*A*b**3*x**(7/2)* 
log(sqrt(x) + sqrt(-a/b))/(30*a**5*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2 
)*sqrt(-a/b)) - 210*A*b**3*x**3*sqrt(-a/b)/(30*a**5*x**(5/2)*sqrt(-a/b) + 
30*a**4*b*x**(7/2)*sqrt(-a/b)) - 20*B*a**3*x*sqrt(-a/b)/(30*a**5*x**(5/2)* 
sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) + 75*B*a**2*b*x**(5/2)*log(sqr 
t(x) - sqrt(-a/b))/(30*a**5*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt( 
-a/b)) - 75*B*a**2*b*x**(5/2)*log(sqrt(x) + sqrt(-a/b))/(30*a**5*x**(5/2)* 
sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) + 100*B*a**2*b*x**2*sqrt(-a/b) 
/(30*a**5*x**(5/2)*sqrt(-a/b) + 30*a**4*b*x**(7/2)*sqrt(-a/b)) + 75*B*a*b* 
*2*x**(7/2)*log(sqrt(x) - sqrt(-a/b))/(30*a**5*x**(5/2)*sqrt(-a/b) + 30...
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {6 \, A a^{3} - 15 \, {\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 10 \, {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 2 \, {\left (5 \, B a^{3} - 7 \, A a^{2} b\right )} x}{15 \, {\left (a^{4} b x^{\frac {7}{2}} + a^{5} x^{\frac {5}{2}}\right )}} + \frac {{\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} \] Input:

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

Output:

-1/15*(6*A*a^3 - 15*(5*B*a*b^2 - 7*A*b^3)*x^3 - 10*(5*B*a^2*b - 7*A*a*b^2) 
*x^2 + 2*(5*B*a^3 - 7*A*a^2*b)*x)/(a^4*b*x^(7/2) + a^5*x^(5/2)) + (5*B*a*b 
^2 - 7*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {{\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {B a b^{2} \sqrt {x} - A b^{3} \sqrt {x}}{{\left (b x + a\right )} a^{4}} + \frac {2 \, {\left (30 \, B a b x^{2} - 45 \, A b^{2} x^{2} - 5 \, B a^{2} x + 10 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{4} x^{\frac {5}{2}}} \] Input:

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

Output:

(5*B*a*b^2 - 7*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + (B*a*b 
^2*sqrt(x) - A*b^3*sqrt(x))/((b*x + a)*a^4) + 2/15*(30*B*a*b*x^2 - 45*A*b^ 
2*x^2 - 5*B*a^2*x + 10*A*a*b*x - 3*A*a^2)/(a^4*x^(5/2))
 

Mupad [B] (verification not implemented)

Time = 10.79 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {\frac {2\,A}{5\,a}-\frac {2\,x\,\left (7\,A\,b-5\,B\,a\right )}{15\,a^2}+\frac {b^2\,x^3\,\left (7\,A\,b-5\,B\,a\right )}{a^4}+\frac {2\,b\,x^2\,\left (7\,A\,b-5\,B\,a\right )}{3\,a^3}}{a\,x^{5/2}+b\,x^{7/2}}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (7\,A\,b-5\,B\,a\right )}{a^{9/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.50 \[ \int \frac {A+B x}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {-2 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{2}-\frac {2 a^{3}}{5}+\frac {2 a^{2} b x}{3}-2 a \,b^{2} x^{2}}{\sqrt {x}\, a^{4} x^{2}} \] Input:

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

Output:

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