3.1942 \(\int \frac {1}{(a+\frac {b}{x^2})^{3/2} x^6} \, dx\)

Optimal. Leaf size=71 \[ \frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 b^{5/2}}-\frac {3 \sqrt {a+\frac {b}{x^2}}}{2 b^2 x}+\frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {335, 288, 321, 217, 206} \[ -\frac {3 \sqrt {a+\frac {b}{x^2}}}{2 b^2 x}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 b^{5/2}}+\frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(3/2)*x^6),x]

[Out]

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

Rule 206

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

Rule 217

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 288

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 321

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 335

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*} \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx &=-\operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^3}-\frac {3 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{b}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^3}-\frac {3 \sqrt {a+\frac {b}{x^2}}}{2 b^2 x}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{2 b^2}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^3}-\frac {3 \sqrt {a+\frac {b}{x^2}}}{2 b^2 x}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b^2}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^3}-\frac {3 \sqrt {a+\frac {b}{x^2}}}{2 b^2 x}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 38, normalized size = 0.54 \[ -\frac {a \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {a x^2}{b}+1\right )}{b^2 x \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(3/2)*x^6),x]

[Out]

-((a*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (a*x^2)/b])/(b^2*Sqrt[a + b/x^2]*x))

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 190, normalized size = 2.68 \[ \left [\frac {3 \, {\left (a^{2} x^{3} + a b x\right )} \sqrt {b} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (3 \, a b x^{2} + b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{4 \, {\left (a b^{3} x^{3} + b^{4} x\right )}}, -\frac {3 \, {\left (a^{2} x^{3} + a b x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (3 \, a b x^{2} + b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, {\left (a b^{3} x^{3} + b^{4} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [sign(t_nostep),sign(t_nostep+sqrt(b)/a*sign(t_nostep))]sym2poly/r2sym(const gen & e,const index_
m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.01, size = 81, normalized size = 1.14 \[ \frac {\left (a \,x^{2}+b \right ) \left (3 \sqrt {a \,x^{2}+b}\, a b \,x^{2} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 a \,b^{\frac {3}{2}} x^{2}-b^{\frac {5}{2}}\right )}{2 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} b^{\frac {7}{2}} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^2)^(3/2)/x^6,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.96, size = 97, normalized size = 1.37 \[ -\frac {3 \, {\left (a + \frac {b}{x^{2}}\right )} a x^{2} - 2 \, a b}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{2} x^{3} - \sqrt {a + \frac {b}{x^{2}}} b^{3} x\right )}} - \frac {3 \, a \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{4 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

-1/2*(3*(a + b/x^2)*a*x^2 - 2*a*b)/((a + b/x^2)^(3/2)*b^2*x^3 - sqrt(a + b/x^2)*b^3*x) - 3/4*a*log((sqrt(a + b
/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(5/2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^6\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b/x^2)^(3/2)),x)

[Out]

int(1/(x^6*(a + b/x^2)^(3/2)), x)

________________________________________________________________________________________

sympy [A]  time = 4.14, size = 73, normalized size = 1.03 \[ - \frac {3 \sqrt {a}}{2 b^{2} x \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{2 b^{\frac {5}{2}}} - \frac {1}{2 \sqrt {a} b x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(3/2)/x**6,x)

[Out]

-3*sqrt(a)/(2*b**2*x*sqrt(1 + b/(a*x**2))) + 3*a*asinh(sqrt(b)/(sqrt(a)*x))/(2*b**(5/2)) - 1/(2*sqrt(a)*b*x**3
*sqrt(1 + b/(a*x**2)))

________________________________________________________________________________________