\(\int \frac {x}{(a+\frac {b}{x^2})^{5/2}} \, dx\) [405]

Optimal result

Integrand size = 13, antiderivative size = 88 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {5 b}{6 a^2 \left (a+\frac {b}{x^2}\right )^{3/2}}+\frac {5 b}{2 a^3 \sqrt {a+\frac {b}{x^2}}}+\frac {x^2}{2 a \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {5 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2 a^{7/2}} \] Output:

5/6*b/a^2/(a+b/x^2)^(3/2)+5/2*b/a^3/(a+b/x^2)^(1/2)+1/2*x^2/a/(a+b/x^2)^(3 
/2)-5/2*b*arctanh((a+b/x^2)^(1/2)/a^(1/2))/a^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {\sqrt {a} x \left (15 b^2+20 a b x^2+3 a^2 x^4\right )+30 b \left (b+a x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {b}-\sqrt {b+a x^2}}\right )}{6 a^{7/2} \sqrt {a+\frac {b}{x^2}} x \left (b+a x^2\right )} \] Input:

Integrate[x/(a + b/x^2)^(5/2),x]
 

Output:

(Sqrt[a]*x*(15*b^2 + 20*a*b*x^2 + 3*a^2*x^4) + 30*b*(b + a*x^2)^(3/2)*ArcT 
anh[(Sqrt[a]*x)/(Sqrt[b] - Sqrt[b + a*x^2])])/(6*a^(7/2)*Sqrt[a + b/x^2]*x 
*(b + a*x^2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {798, 52, 61, 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.

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

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[x/(a + b/x^2)^(5/2),x]
 

Output:

(x^2/(a*(a + b/x^2)^(3/2)) + (5*b*(2/(3*a*(a + b/x^2)^(3/2)) + (2/(a*Sqrt[ 
a + b/x^2]) - (2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/a^(3/2))/a))/(2*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 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 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\left (a \,x^{2}+b \right ) \left (3 x^{5} a^{\frac {7}{2}}+20 a^{\frac {5}{2}} b \,x^{3}+15 a^{\frac {3}{2}} b^{2} x -15 \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) \left (a \,x^{2}+b \right )^{\frac {3}{2}} a b \right )}{6 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} x^{5} a^{\frac {9}{2}}}\)	\(85\)
risch	\(\frac {a \,x^{2}+b}{2 a^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (-\frac {5 b \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )}{2 a^{\frac {7}{2}}}-\frac {b^{2} \sqrt {a \left (x -\frac {\sqrt {-a b}}{a}\right )^{2}+2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{a}\right )}}{12 a^{4} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{a}\right )^{2}}+\frac {7 b \sqrt {a \left (x -\frac {\sqrt {-a b}}{a}\right )^{2}+2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{a}\right )}}{6 a^{4} \left (x -\frac {\sqrt {-a b}}{a}\right )}+\frac {b^{2} \sqrt {a \left (x +\frac {\sqrt {-a b}}{a}\right )^{2}-2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{a}\right )}}{12 a^{4} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{a}\right )^{2}}+\frac {7 b \sqrt {a \left (x +\frac {\sqrt {-a b}}{a}\right )^{2}-2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{a}\right )}}{6 a^{4} \left (x +\frac {\sqrt {-a b}}{a}\right )}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\)	\(334\)

Input:

int(x/(a+b/x^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(a*x^2+b)*(3*x^5*a^(7/2)+20*a^(5/2)*b*x^3+15*a^(3/2)*b^2*x-15*ln(a^(1/ 
2)*x+(a*x^2+b)^(1/2))*(a*x^2+b)^(3/2)*a*b)/((a*x^2+b)/x^2)^(5/2)/x^5/a^(9/ 
2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.95 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt {a} \log \left (-2 \, a x^{2} + 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + 2 \, {\left (3 \, a^{3} x^{6} + 20 \, a^{2} b x^{4} + 15 \, a b^{2} x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{12 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}, \frac {15 \, {\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (3 \, a^{3} x^{6} + 20 \, a^{2} b x^{4} + 15 \, a b^{2} x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{6 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}\right ] \] Input:

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

Output:

[1/12*(15*(a^2*b*x^4 + 2*a*b^2*x^2 + b^3)*sqrt(a)*log(-2*a*x^2 + 2*sqrt(a) 
*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(3*a^3*x^6 + 20*a^2*b*x^4 + 15*a*b^2*x 
^2)*sqrt((a*x^2 + b)/x^2))/(a^6*x^4 + 2*a^5*b*x^2 + a^4*b^2), 1/6*(15*(a^2 
*b*x^4 + 2*a*b^2*x^2 + b^3)*sqrt(-a)*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/ 
x^2)/(a*x^2 + b)) + (3*a^3*x^6 + 20*a^2*b*x^4 + 15*a*b^2*x^2)*sqrt((a*x^2 
+ b)/x^2))/(a^6*x^4 + 2*a^5*b*x^2 + a^4*b^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 819 vs. \(2 (80) = 160\).

Time = 3.16 (sec) , antiderivative size = 819, normalized size of antiderivative = 9.31 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

6*a**17*x**8*sqrt(1 + b/(a*x**2))/(12*a**(39/2)*x**6 + 36*a**(37/2)*b*x**4 
 + 36*a**(35/2)*b**2*x**2 + 12*a**(33/2)*b**3) + 46*a**16*b*x**6*sqrt(1 + 
b/(a*x**2))/(12*a**(39/2)*x**6 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x 
**2 + 12*a**(33/2)*b**3) + 15*a**16*b*x**6*log(b/(a*x**2))/(12*a**(39/2)*x 
**6 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**2 + 12*a**(33/2)*b**3) - 
30*a**16*b*x**6*log(sqrt(1 + b/(a*x**2)) + 1)/(12*a**(39/2)*x**6 + 36*a**( 
37/2)*b*x**4 + 36*a**(35/2)*b**2*x**2 + 12*a**(33/2)*b**3) + 70*a**15*b**2 
*x**4*sqrt(1 + b/(a*x**2))/(12*a**(39/2)*x**6 + 36*a**(37/2)*b*x**4 + 36*a 
**(35/2)*b**2*x**2 + 12*a**(33/2)*b**3) + 45*a**15*b**2*x**4*log(b/(a*x**2 
))/(12*a**(39/2)*x**6 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**2 + 12* 
a**(33/2)*b**3) - 90*a**15*b**2*x**4*log(sqrt(1 + b/(a*x**2)) + 1)/(12*a** 
(39/2)*x**6 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**2 + 12*a**(33/2)* 
b**3) + 30*a**14*b**3*x**2*sqrt(1 + b/(a*x**2))/(12*a**(39/2)*x**6 + 36*a* 
*(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**2 + 12*a**(33/2)*b**3) + 45*a**14*b* 
*3*x**2*log(b/(a*x**2))/(12*a**(39/2)*x**6 + 36*a**(37/2)*b*x**4 + 36*a**( 
35/2)*b**2*x**2 + 12*a**(33/2)*b**3) - 90*a**14*b**3*x**2*log(sqrt(1 + b/( 
a*x**2)) + 1)/(12*a**(39/2)*x**6 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2 
*x**2 + 12*a**(33/2)*b**3) + 15*a**13*b**4*log(b/(a*x**2))/(12*a**(39/2)*x 
**6 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**2 + 12*a**(33/2)*b**3) - 
30*a**13*b**4*log(sqrt(1 + b/(a*x**2)) + 1)/(12*a**(39/2)*x**6 + 36*a**...
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {15 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b - 10 \, {\left (a + \frac {b}{x^{2}}\right )} a b - 2 \, a^{2} b}{6 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{3} - {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{4}\right )}} + \frac {5 \, b \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{4 \, a^{\frac {7}{2}}} \] Input:

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

Output:

1/6*(15*(a + b/x^2)^2*b - 10*(a + b/x^2)*a*b - 2*a^2*b)/((a + b/x^2)^(5/2) 
*a^3 - (a + b/x^2)^(3/2)*a^4) + 5/4*b*log((sqrt(a + b/x^2) - sqrt(a))/(sqr 
t(a + b/x^2) + sqrt(a)))/a^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {3 \, x^{2}}{a \mathrm {sgn}\left (x\right )} + \frac {20 \, b}{a^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {15 \, b^{2}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} x}{6 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}}} - \frac {5 \, b \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{4 \, a^{\frac {7}{2}}} + \frac {5 \, b \log \left ({\left | -\sqrt {a} x + \sqrt {a x^{2} + b} \right |}\right )}{2 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

1/6*(x^2*(3*x^2/(a*sgn(x)) + 20*b/(a^2*sgn(x))) + 15*b^2/(a^3*sgn(x)))*x/( 
a*x^2 + b)^(3/2) - 5/4*b*log(abs(b))*sgn(x)/a^(7/2) + 5/2*b*log(abs(-sqrt( 
a)*x + sqrt(a*x^2 + b)))/(a^(7/2)*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {10\,b}{3\,a^2\,{\left (a+\frac {b}{x^2}\right )}^{3/2}}+\frac {x^2}{2\,a\,{\left (a+\frac {b}{x^2}\right )}^{3/2}}-\frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2\,a^{7/2}}+\frac {5\,b^2}{2\,a^3\,x^2\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \] Input:

int(x/(a + b/x^2)^(5/2),x)
 

Output:

(10*b)/(3*a^2*(a + b/x^2)^(3/2)) + x^2/(2*a*(a + b/x^2)^(3/2)) - (5*b*atan 
h((a + b/x^2)^(1/2)/a^(1/2)))/(2*a^(7/2)) + (5*b^2)/(2*a^3*x^2*(a + b/x^2) 
^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.15 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {6 \sqrt {a \,x^{2}+b}\, a^{3} x^{5}+40 \sqrt {a \,x^{2}+b}\, a^{2} b \,x^{3}+30 \sqrt {a \,x^{2}+b}\, a \,b^{2} x -30 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x}{\sqrt {b}}\right ) a^{2} b \,x^{4}-60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x}{\sqrt {b}}\right ) a \,b^{2} x^{2}-30 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x}{\sqrt {b}}\right ) b^{3}-5 \sqrt {a}\, a^{2} b \,x^{4}-10 \sqrt {a}\, a \,b^{2} x^{2}-5 \sqrt {a}\, b^{3}}{12 a^{4} \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right )} \] Input:

int(x/(a+b/x^2)^(5/2),x)
 

Output:

(6*sqrt(a*x**2 + b)*a**3*x**5 + 40*sqrt(a*x**2 + b)*a**2*b*x**3 + 30*sqrt( 
a*x**2 + b)*a*b**2*x - 30*sqrt(a)*log((sqrt(a*x**2 + b) + sqrt(a)*x)/sqrt( 
b))*a**2*b*x**4 - 60*sqrt(a)*log((sqrt(a*x**2 + b) + sqrt(a)*x)/sqrt(b))*a 
*b**2*x**2 - 30*sqrt(a)*log((sqrt(a*x**2 + b) + sqrt(a)*x)/sqrt(b))*b**3 - 
 5*sqrt(a)*a**2*b*x**4 - 10*sqrt(a)*a*b**2*x**2 - 5*sqrt(a)*b**3)/(12*a**4 
*(a**2*x**4 + 2*a*b*x**2 + b**2))