\(\int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [445]

Optimal result

Integrand size = 31, antiderivative size = 190 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 A (a+b x)}{5 a x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{3 a^2 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b (A b-a B) (a+b x)}{a^3 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.56 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (a+b x) \left (\sqrt {a} \left (15 A b^2 x^2-5 a b x (A+3 B x)+a^2 (3 A+5 B x)\right )+15 b^{3/2} (A b-a B) x^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{15 a^{7/2} x^{5/2} \sqrt {(a+b x)^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.59, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {1187, 27, 87, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((a + b*x)*((-2*A)/(5*a*x^(5/2)) - ((A*b - a*B)*(-2/(3*a*x^(3/2)) - (b*(-2 
/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a)) 
/a))/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 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 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 = 0.74 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.58

method	result	size
risch	\(-\frac {2 \left (15 x^{2} b^{2} A -15 B a \,x^{2} b -5 a b A x +5 a^{2} B x +3 a^{2} A \right ) \sqrt {\left (b x +a \right )^{2}}}{15 a^{3} x^{\frac {5}{2}} \left (b x +a \right )}-\frac {2 b^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{a^{3} \sqrt {a b}\, \left (b x +a \right )}\)	\(111\)
default	\(-\frac {2 \left (b x +a \right ) \left (15 A \,x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b^{3}-15 B \,x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{2}+15 A \,x^{2} \sqrt {a b}\, b^{2}-15 B \,x^{2} \sqrt {a b}\, a b -5 A x \sqrt {a b}\, a b +5 B x \sqrt {a b}\, a^{2}+3 A \,a^{2} \sqrt {a b}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{\frac {5}{2}} \sqrt {a b}}\)	\(131\)

Input:

int((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(15*A*b^2*x^2-15*B*a*b*x^2-5*A*a*b*x+5*B*a^2*x+3*A*a^2)/a^3/x^(5/2)* 
((b*x+a)^2)^(1/2)/(b*x+a)-2*b^2*(A*b-B*a)/a^3/(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 = 0.08 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, \frac {2 \, {\left (15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) - {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \] Input:

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/15*(15*(B*a*b - A*b^2)*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a 
) - a)/(b*x + a)) + 2*(3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b 
)*x)*sqrt(x))/(a^3*x^3), 2/15*(15*(B*a*b - A*b^2)*x^3*sqrt(b/a)*arctan(sqr 
t(x)*sqrt(b/a)) - (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x) 
*sqrt(x))/(a^3*x^3)]
 

Sympy [F(-1)]

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

integrate((B*x+A)/x**(7/2)/((b*x+a)**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (126) = 252\).

Time = 0.16 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.61 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {5 \, {\left ({\left (3 \, B a b^{4} - 5 \, A b^{5}\right )} x^{2} + 3 \, {\left (5 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} x\right )} \sqrt {x} - \frac {10 \, {\left ({\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{2} - 3 \, {\left (5 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x\right )}}{\sqrt {x}} - \frac {10 \, {\left (3 \, {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{2} - {\left (5 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )}}{x^{\frac {3}{2}}} - \frac {2 \, {\left (5 \, {\left (3 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (5 \, B a^{5} - 7 \, A a^{4} b\right )} x\right )}}{x^{\frac {5}{2}}} - \frac {2 \, {\left (5 \, A a^{4} b x^{2} + 3 \, A a^{5} x\right )}}{x^{\frac {7}{2}}}}{15 \, {\left (a^{5} b x + a^{6}\right )}} + \frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {{\left (3 \, B a b^{3} - 5 \, A b^{4}\right )} x^{\frac {3}{2}} + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} \sqrt {x}}{3 \, a^{5}} \] Input:

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.64 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {2 \, {\left (15 \, B a b x^{2} \mathrm {sgn}\left (b x + a\right ) - 15 \, A b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, B a^{2} x \mathrm {sgn}\left (b x + a\right ) + 5 \, A a b x \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \] Input:

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

2*(B*a*b^2*sgn(b*x + a) - A*b^3*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/ 
(sqrt(a*b)*a^3) + 2/15*(15*B*a*b*x^2*sgn(b*x + a) - 15*A*b^2*x^2*sgn(b*x + 
 a) - 5*B*a^2*x*sgn(b*x + a) + 5*A*a*b*x*sgn(b*x + a) - 3*A*a^2*sgn(b*x + 
a))/(a^3*x^(5/2))
 

Mupad [F(-1)]

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

int((A + B*x)/(x^(7/2)*((a + b*x)^2)^(1/2)),x)
 

Output:

int((A + B*x)/(x^(7/2)*((a + b*x)^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.05 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2}{5 \sqrt {x}\, x^{2}} \] Input:

int((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x)
 

Output:

( - 2)/(5*sqrt(x)*x**2)