\(\int \frac {1}{(a+\frac {b}{x^2})^{3/2} x^8} \, dx\) [1943]

Optimal result

Integrand size = 15, antiderivative size = 95 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {342, 294, 327, 223, 212} \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=-\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{8 b^{7/2}}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {1}{b x^5 \sqrt {a+\frac {b}{x^2}}} \]

[In]

Int[1/((a + b/x^2)^(3/2)*x^8),x]

[Out]

1/(b*Sqrt[a + b/x^2]*x^5) - (5*Sqrt[a + b/x^2])/(4*b^2*x^3) + (15*a*Sqrt[a + b/x^2])/(8*b^3*x) - (15*a^2*ArcTa
nh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(8*b^(7/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x^6}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \text {Subst}\left (\int \frac {x^4}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{b} \\ & = \frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {(15 a) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{4 b^2} \\ & = \frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{8 b^3} \\ & = \frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^3} \\ & = \frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\frac {\sqrt {b} \left (-2 b^2+5 a b x^2+15 a^2 x^4\right )-15 a^2 x^4 \sqrt {b+a x^2} \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {a+\frac {b}{x^2}} x^5} \]

[In]

Integrate[1/((a + b/x^2)^(3/2)*x^8),x]

[Out]

(Sqrt[b]*(-2*b^2 + 5*a*b*x^2 + 15*a^2*x^4) - 15*a^2*x^4*Sqrt[b + a*x^2]*ArcTanh[Sqrt[b + a*x^2]/Sqrt[b]])/(8*b
^(7/2)*Sqrt[a + b/x^2]*x^5)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {\left (a \,x^{2}+b \right ) \left (-15 b^{\frac {3}{2}} a^{2} x^{4}+15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \sqrt {a \,x^{2}+b}\, a^{2} b \,x^{4}-5 b^{\frac {5}{2}} a \,x^{2}+2 b^{\frac {7}{2}}\right )}{8 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{7} b^{\frac {9}{2}}}\)	\(94\)
risch	\(\frac {\left (a \,x^{2}+b \right ) \left (7 a \,x^{2}-2 b \right )}{8 b^{3} x^{5} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (\frac {a^{2}}{b^{3} \sqrt {a \,x^{2}+b}}-\frac {15 a^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right )}{8 b^{\frac {7}{2}}}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\)	\(114\)

[In]

int(1/(a+b/x^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/8*(a*x^2+b)*(-15*b^(3/2)*a^2*x^4+15*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*(a*x^2+b)^(1/2)*a^2*b*x^4-5*b^(5/2)
*a*x^2+2*b^(7/2))/((a*x^2+b)/x^2)^(3/2)/x^7/b^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.40 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\left [\frac {15 \, {\left (a^{3} x^{5} + a^{2} b x^{3}\right )} \sqrt {b} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} - 2 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, {\left (a b^{4} x^{5} + b^{5} x^{3}\right )}}, \frac {15 \, {\left (a^{3} x^{5} + a^{2} b x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} - 2 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, {\left (a b^{4} x^{5} + b^{5} x^{3}\right )}}\right ] \]

[In]

integrate(1/(a+b/x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

[1/16*(15*(a^3*x^5 + a^2*b*x^3)*sqrt(b)*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) + 2*(15*a^
2*b*x^4 + 5*a*b^2*x^2 - 2*b^3)*sqrt((a*x^2 + b)/x^2))/(a*b^4*x^5 + b^5*x^3), 1/8*(15*(a^3*x^5 + a^2*b*x^3)*sqr
t(-b)*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (15*a^2*b*x^4 + 5*a*b^2*x^2 - 2*b^3)*sqrt((a*x^2
+ b)/x^2))/(a*b^4*x^5 + b^5*x^3)]

Sympy [A] (verification not implemented)

Time = 3.94 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\frac {15 a^{\frac {3}{2}}}{8 b^{3} x \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {5 \sqrt {a}}{8 b^{2} x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{8 b^{\frac {7}{2}}} - \frac {1}{4 \sqrt {a} b x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} \]

[In]

integrate(1/(a+b/x**2)**(3/2)/x**8,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\frac {15 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} a^{2} x^{4} - 25 \, {\left (a + \frac {b}{x^{2}}\right )} a^{2} b x^{2} + 8 \, a^{2} b^{2}}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} b^{3} x^{5} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{4} x^{3} + \sqrt {a + \frac {b}{x^{2}}} b^{5} x\right )}} + \frac {15 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{16 \, b^{\frac {7}{2}}} \]

[In]

integrate(1/(a+b/x^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

1/8*(15*(a + b/x^2)^2*a^2*x^4 - 25*(a + b/x^2)*a^2*b*x^2 + 8*a^2*b^2)/((a + b/x^2)^(5/2)*b^3*x^5 - 2*(a + b/x^
2)^(3/2)*b^4*x^3 + sqrt(a + b/x^2)*b^5*x) + 15/16*a^2*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + s
qrt(b)))/b^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\frac {15 \, a^{2} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{3} \mathrm {sgn}\left (x\right )} + \frac {a^{2}}{\sqrt {a x^{2} + b} b^{3} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{2} - 9 \, \sqrt {a x^{2} + b} a^{2} b}{8 \, a^{2} b^{3} x^{4} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/(a+b/x^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

15/8*a^2*arctan(sqrt(a*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^3*sgn(x)) + a^2/(sqrt(a*x^2 + b)*b^3*sgn(x)) + 1/8*(7*(a
*x^2 + b)^(3/2)*a^2 - 9*sqrt(a*x^2 + b)*a^2*b)/(a^2*b^3*x^4*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx=\int \frac {1}{x^8\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \,d x \]

[In]

int(1/(x^8*(a + b/x^2)^(3/2)),x)

[Out]

int(1/(x^8*(a + b/x^2)^(3/2)), x)