\(\int \frac {A+B x^2}{x^5 (a+b x^2)^{3/2}} \, dx\) [264]

Optimal result

Integrand size = 22, antiderivative size = 114 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=\frac {b (A b-a B)}{a^3 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{4 a^2 x^4}+\frac {(7 A b-4 a B) \sqrt {a+b x^2}}{8 a^3 x^2}-\frac {3 b (5 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{7/2}} \] Output:

b*(A*b-B*a)/a^3/(b*x^2+a)^(1/2)-1/4*A*(b*x^2+a)^(1/2)/a^2/x^4+1/8*(7*A*b-4 
*B*a)*(b*x^2+a)^(1/2)/a^3/x^2-3/8*b*(5*A*b-4*B*a)*arctanh((b*x^2+a)^(1/2)/ 
a^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 a^2 A+5 a A b x^2-4 a^2 B x^2+15 A b^2 x^4-12 a b B x^4}{8 a^3 x^4 \sqrt {a+b x^2}}+\frac {3 b (-5 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{7/2}} \] Input:

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x]
 

Output:

(-2*a^2*A + 5*a*A*b*x^2 - 4*a^2*B*x^2 + 15*A*b^2*x^4 - 12*a*b*B*x^4)/(8*a^ 
3*x^4*Sqrt[a + b*x^2]) + (3*b*(-5*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqr 
t[a]])/(8*a^(7/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {354, 87, 52, 61, 73, 221}

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.

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

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \left (b x^2+a\right )^{3/2}}dx^2\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x]
 

Output:

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

Defintions of rubi rules used

rule 52

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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {b \left (\frac {A b -B a}{\sqrt {b \,x^{2}+a}}-\frac {\sqrt {b \,x^{2}+a}\, \left (-7 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 b \,x^{4}}-\frac {3 \left (5 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )}{a^{3}}\)	\(89\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-7 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 a^{3} x^{4}}+\frac {b \left (-\frac {7 A b -4 B a}{\sqrt {b \,x^{2}+a}}+3 a \left (5 A b -4 B a \right ) \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )\right )}{8 a^{3}}\)	\(119\)
default	\(A \left (-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{2}+a}}-\frac {5 b \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )}{4 a}\right )+B \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )\)	\(162\)

Input:

int((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

b/a^3*((A*b-B*a)/(b*x^2+a)^(1/2)-1/8*(b*x^2+a)^(1/2)/b*(-7*A*b*x^2+4*B*a*x 
^2+2*A*a)/x^4-3/8*(5*A*b-4*B*a)/a^(1/2)*arctanh((b*x^2+a)^(1/2)/a^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.54 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4}\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + 2 \, A a^{3} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}, -\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + 2 \, A a^{3} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}\right ] \] Input:

integrate((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[-1/16*(3*((4*B*a*b^2 - 5*A*b^3)*x^6 + (4*B*a^2*b - 5*A*a*b^2)*x^4)*sqrt(a 
)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*(4*B*a^2*b - 
5*A*a*b^2)*x^4 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^ 
4*b*x^6 + a^5*x^4), -1/8*(3*((4*B*a*b^2 - 5*A*b^3)*x^6 + (4*B*a^2*b - 5*A* 
a*b^2)*x^4)*sqrt(-a)*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (3*(4*B*a^2*b - 
5*A*a*b^2)*x^4 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^ 
4*b*x^6 + a^5*x^4)]
 

Sympy [A] (verification not implemented)

Time = 30.73 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=A \left (- \frac {1}{4 a \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {5 \sqrt {b}}{8 a^{2} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {15 b^{\frac {3}{2}}}{8 a^{3} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {7}{2}}}\right ) + B \left (- \frac {1}{2 a \sqrt {b} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b}}{2 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {5}{2}}}\right ) \] Input:

integrate((B*x**2+A)/x**5/(b*x**2+a)**(3/2),x)
 

Output:

A*(-1/(4*a*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + 5*sqrt(b)/(8*a**2*x**3*sqr 
t(a/(b*x**2) + 1)) + 15*b**(3/2)/(8*a**3*x*sqrt(a/(b*x**2) + 1)) - 15*b**2 
*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(7/2))) + B*(-1/(2*a*sqrt(b)*x**3*sqrt(a 
/(b*x**2) + 1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sq 
rt(a)/(sqrt(b)*x))/(2*a**(5/2)))
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=\frac {3 \, B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {15 \, A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {7}{2}}} - \frac {3 \, B b}{2 \, \sqrt {b x^{2} + a} a^{2}} + \frac {15 \, A b^{2}}{8 \, \sqrt {b x^{2} + a} a^{3}} - \frac {B}{2 \, \sqrt {b x^{2} + a} a x^{2}} + \frac {5 \, A b}{8 \, \sqrt {b x^{2} + a} a^{2} x^{2}} - \frac {A}{4 \, \sqrt {b x^{2} + a} a x^{4}} \] Input:

integrate((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

3/2*B*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) - 15/8*A*b^2*arcsinh(a/(sqrt 
(a*b)*abs(x)))/a^(7/2) - 3/2*B*b/(sqrt(b*x^2 + a)*a^2) + 15/8*A*b^2/(sqrt( 
b*x^2 + a)*a^3) - 1/2*B/(sqrt(b*x^2 + a)*a*x^2) + 5/8*A*b/(sqrt(b*x^2 + a) 
*a^2*x^2) - 1/4*A/(sqrt(b*x^2 + a)*a*x^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{3}} - \frac {B a b - A b^{2}}{\sqrt {b x^{2} + a} a^{3}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x^{2} + a} B a^{2} b - 7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2} + 9 \, \sqrt {b x^{2} + a} A a b^{2}}{8 \, a^{3} b^{2} x^{4}} \] Input:

integrate((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

-3/8*(4*B*a*b - 5*A*b^2)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) - 
 (B*a*b - A*b^2)/(sqrt(b*x^2 + a)*a^3) - 1/8*(4*(b*x^2 + a)^(3/2)*B*a*b - 
4*sqrt(b*x^2 + a)*B*a^2*b - 7*(b*x^2 + a)^(3/2)*A*b^2 + 9*sqrt(b*x^2 + a)* 
A*a*b^2)/(a^3*b^2*x^4)
 

Mupad [B] (verification not implemented)

Time = 1.72 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=\frac {15\,A\,b^2}{8\,a^3\,\sqrt {b\,x^2+a}}-\frac {3\,B\,b}{2\,a^2\,\sqrt {b\,x^2+a}}-\frac {15\,A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{7/2}}-\frac {A}{4\,a\,x^4\,\sqrt {b\,x^2+a}}-\frac {B}{2\,a\,x^2\,\sqrt {b\,x^2+a}}+\frac {3\,B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}}+\frac {5\,A\,b}{8\,a^2\,x^2\,\sqrt {b\,x^2+a}} \] Input:

int((A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x)
 

Output:

(15*A*b^2)/(8*a^3*(a + b*x^2)^(1/2)) - (3*B*b)/(2*a^2*(a + b*x^2)^(1/2)) - 
 (15*A*b^2*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(8*a^(7/2)) - A/(4*a*x^4*(a + 
 b*x^2)^(1/2)) - B/(2*a*x^2*(a + b*x^2)^(1/2)) + (3*B*b*atanh((a + b*x^2)^ 
(1/2)/a^(1/2)))/(2*a^(5/2)) + (5*A*b)/(8*a^2*x^2*(a + b*x^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a^{2}+3 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}}{8 a^{3} x^{4}} \] Input:

int((B*x^2+A)/x^5/(b*x^2+a)^(3/2),x)
 

Output:

( - 2*sqrt(a + b*x**2)*a**2 + 3*sqrt(a + b*x**2)*a*b*x**2 + 3*sqrt(a)*log( 
(sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4 - 3*sqrt(a)*lo 
g((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4)/(8*a**3*x** 
4)