\(\int \frac {A+B x}{x^{9/2} (a^2+2 a b x+b^2 x^2)} \, dx\) [394]

Optimal result

Integrand size = 29, antiderivative size = 143 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 A}{7 a^2 x^{7/2}}+\frac {2 (2 A b-a B)}{5 a^3 x^{5/2}}-\frac {2 b (3 A b-2 a B)}{3 a^4 x^{3/2}}+\frac {2 b^2 (4 A b-3 a B)}{a^5 \sqrt {x}}+\frac {b^3 (A b-a B) \sqrt {x}}{a^5 (a+b x)}+\frac {b^{5/2} (9 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {945 A b^4 x^4+105 a b^3 x^3 (6 A-7 B x)-6 a^4 (5 A+7 B x)-14 a^2 b^2 x^2 (9 A+35 B x)+2 a^3 b x (27 A+49 B x)}{105 a^5 x^{7/2} (a+b x)}+\frac {b^{5/2} (9 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \] Input:

Integrate[(A + B*x)/(x^(9/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]
 

Output:

(945*A*b^4*x^4 + 105*a*b^3*x^3*(6*A - 7*B*x) - 6*a^4*(5*A + 7*B*x) - 14*a^ 
2*b^2*x^2*(9*A + 35*B*x) + 2*a^3*b*x*(27*A + 49*B*x))/(105*a^5*x^(7/2)*(a 
+ b*x)) + (b^(5/2)*(9*A*b - 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(1 
1/2)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.01, 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, 61, 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.

\(\displaystyle \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx\)

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

Int[(A + B*x)/(x^(9/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]
 

Output:

(A*b - a*B)/(a*b*x^(7/2)*(a + b*x)) + ((9*A*b - 7*a*B)*(-2/(7*a*x^(7/2)) - 
 (b*(-2/(5*a*x^(5/2)) - (b*(-2/(3*a*x^(3/2)) - (b*(-2/(a*Sqrt[x]) - (2*Sqr 
t[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a))/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 = 1.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(-\frac {2 A}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 \left (-2 A b +B a \right )}{5 a^{3} x^{\frac {5}{2}}}-\frac {2 b \left (3 A b -2 B a \right )}{3 a^{4} x^{\frac {3}{2}}}+\frac {2 b^{2} \left (4 A b -3 B a \right )}{a^{5} \sqrt {x}}+\frac {2 b^{3} \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (9 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{5}}\)	\(121\)
default	\(-\frac {2 A}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 \left (-2 A b +B a \right )}{5 a^{3} x^{\frac {5}{2}}}-\frac {2 b \left (3 A b -2 B a \right )}{3 a^{4} x^{\frac {3}{2}}}+\frac {2 b^{2} \left (4 A b -3 B a \right )}{a^{5} \sqrt {x}}+\frac {2 b^{3} \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (9 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{5}}\)	\(121\)
risch	\(-\frac {2 \left (-420 A \,b^{3} x^{3}+315 B a \,b^{2} x^{3}+105 A a \,b^{2} x^{2}-70 B \,a^{2} b \,x^{2}-42 A \,a^{2} b x +21 B \,a^{3} x +15 a^{3} A \right )}{105 a^{5} x^{\frac {7}{2}}}+\frac {b^{3} \left (\frac {2 \left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (9 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a^{5}}\)	\(126\)

Input:

int((B*x+A)/x^(9/2)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.57 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\left [-\frac {105 \, {\left ({\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (30 \, A a^{4} + 105 \, {\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} + 70 \, {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 14 \, {\left (7 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (7 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{210 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}, -\frac {105 \, {\left ({\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) + {\left (30 \, A a^{4} + 105 \, {\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} + 70 \, {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 14 \, {\left (7 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (7 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{105 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}\right ] \] Input:

integrate((B*x+A)/x^(9/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
 

Output:

[-1/210*(105*((7*B*a*b^3 - 9*A*b^4)*x^5 + (7*B*a^2*b^2 - 9*A*a*b^3)*x^4)*s 
qrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(30*A*a^4 
+ 105*(7*B*a*b^3 - 9*A*b^4)*x^4 + 70*(7*B*a^2*b^2 - 9*A*a*b^3)*x^3 - 14*(7 
*B*a^3*b - 9*A*a^2*b^2)*x^2 + 6*(7*B*a^4 - 9*A*a^3*b)*x)*sqrt(x))/(a^5*b*x 
^5 + a^6*x^4), -1/105*(105*((7*B*a*b^3 - 9*A*b^4)*x^5 + (7*B*a^2*b^2 - 9*A 
*a*b^3)*x^4)*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) + (30*A*a^4 + 105*(7*B*a* 
b^3 - 9*A*b^4)*x^4 + 70*(7*B*a^2*b^2 - 9*A*a*b^3)*x^3 - 14*(7*B*a^3*b - 9* 
A*a^2*b^2)*x^2 + 6*(7*B*a^4 - 9*A*a^3*b)*x)*sqrt(x))/(a^5*b*x^5 + a^6*x^4) 
]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)/x**(9/2)/(b**2*x**2+2*a*b*x+a**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {30 \, A a^{4} + 105 \, {\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} + 70 \, {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 14 \, {\left (7 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (7 \, B a^{4} - 9 \, A a^{3} b\right )} x}{105 \, {\left (a^{5} b x^{\frac {9}{2}} + a^{6} x^{\frac {7}{2}}\right )}} - \frac {{\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} \] Input:

integrate((B*x+A)/x^(9/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
 

Output:

-1/105*(30*A*a^4 + 105*(7*B*a*b^3 - 9*A*b^4)*x^4 + 70*(7*B*a^2*b^2 - 9*A*a 
*b^3)*x^3 - 14*(7*B*a^3*b - 9*A*a^2*b^2)*x^2 + 6*(7*B*a^4 - 9*A*a^3*b)*x)/ 
(a^5*b*x^(9/2) + a^6*x^(7/2)) - (7*B*a*b^3 - 9*A*b^4)*arctan(b*sqrt(x)/sqr 
t(a*b))/(sqrt(a*b)*a^5)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {{\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} - \frac {B a b^{3} \sqrt {x} - A b^{4} \sqrt {x}}{{\left (b x + a\right )} a^{5}} - \frac {2 \, {\left (315 \, B a b^{2} x^{3} - 420 \, A b^{3} x^{3} - 70 \, B a^{2} b x^{2} + 105 \, A a b^{2} x^{2} + 21 \, B a^{3} x - 42 \, A a^{2} b x + 15 \, A a^{3}\right )}}{105 \, a^{5} x^{\frac {7}{2}}} \] Input:

integrate((B*x+A)/x^(9/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
 

Output:

-(7*B*a*b^3 - 9*A*b^4)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - (B*a* 
b^3*sqrt(x) - A*b^4*sqrt(x))/((b*x + a)*a^5) - 2/105*(315*B*a*b^2*x^3 - 42 
0*A*b^3*x^3 - 70*B*a^2*b*x^2 + 105*A*a*b^2*x^2 + 21*B*a^3*x - 42*A*a^2*b*x 
 + 15*A*a^3)/(a^5*x^(7/2))
 

Mupad [B] (verification not implemented)

Time = 10.87 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {\frac {2\,x\,\left (9\,A\,b-7\,B\,a\right )}{35\,a^2}-\frac {2\,A}{7\,a}+\frac {2\,b^2\,x^3\,\left (9\,A\,b-7\,B\,a\right )}{3\,a^4}+\frac {b^3\,x^4\,\left (9\,A\,b-7\,B\,a\right )}{a^5}-\frac {2\,b\,x^2\,\left (9\,A\,b-7\,B\,a\right )}{15\,a^3}}{a\,x^{7/2}+b\,x^{9/2}}+\frac {b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (9\,A\,b-7\,B\,a\right )}{a^{11/2}} \] Input:

int((A + B*x)/(x^(9/2)*(a^2 + b^2*x^2 + 2*a*b*x)),x)
 

Output:

((2*x*(9*A*b - 7*B*a))/(35*a^2) - (2*A)/(7*a) + (2*b^2*x^3*(9*A*b - 7*B*a) 
)/(3*a^4) + (b^3*x^4*(9*A*b - 7*B*a))/a^5 - (2*b*x^2*(9*A*b - 7*B*a))/(15* 
a^3))/(a*x^(7/2) + b*x^(9/2)) + (b^(5/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*( 
9*A*b - 7*B*a))/a^(11/2)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.50 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {2 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{3}-\frac {2 a^{4}}{7}+\frac {2 a^{3} b x}{5}-\frac {2 a^{2} b^{2} x^{2}}{3}+2 a \,b^{3} x^{3}}{\sqrt {x}\, a^{5} x^{3}} \] Input:

int((B*x+A)/x^(9/2)/(b^2*x^2+2*a*b*x+a^2),x)
 

Output:

(2*(105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**3*x 
**3 - 15*a**4 + 21*a**3*b*x - 35*a**2*b**2*x**2 + 105*a*b**3*x**3))/(105*s 
qrt(x)*a**5*x**3)