3.1946 \(\int \frac {1}{(a+\frac {b}{x^2})^{5/2} x} \, dx\)

Optimal. Leaf size=59 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {1}{a^2 \sqrt {a+\frac {b}{x^2}}}-\frac {1}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac {1}{a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {1}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(5/2)*x),x]

[Out]

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

Rule 51

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] - Dist[(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] && NeQ[b*c - a*d, 0] && 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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 91, normalized size = 1.54 \[ \frac {\frac {3 \sqrt {b} \left (a x^2+b\right ) \sqrt {\frac {a x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{x}-\sqrt {a} \left (4 a x^2+3 b\right )}{3 a^{5/2} \sqrt {a+\frac {b}{x^2}} \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(5/2)*x),x]

[Out]

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

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fricas [B]  time = 1.13, size = 232, normalized size = 3.93 \[ \left [\frac {3 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {a} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) - 2 \, {\left (4 \, a^{2} x^{4} + 3 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{6 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}}, -\frac {3 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (4 \, a^{2} x^{4} + 3 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.33, size = 154, normalized size = 2.61 \[ \frac {3 \, \log \left ({\left | b \right |}\right ) + 8}{6 \, a^{\frac {5}{2}}} - \frac {\log \left ({\left | -2 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b x^{2}}\right )} \sqrt {a} - b \right |}\right )}{2 \, a^{\frac {5}{2}}} - \frac {6 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b x^{2}}\right )}^{2} a b + 9 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b x^{2}}\right )} \sqrt {a} b^{2} + 4 \, b^{3}}{3 \, {\left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b x^{2}}\right )} \sqrt {a} + b\right )}^{3} a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x,x, algorithm="giac")

[Out]

1/6*(3*log(abs(b)) + 8)/a^(5/2) - 1/2*log(abs(-2*(sqrt(a)*x^2 - sqrt(a*x^4 + b*x^2))*sqrt(a) - b))/a^(5/2) - 1
/3*(6*(sqrt(a)*x^2 - sqrt(a*x^4 + b*x^2))^2*a*b + 9*(sqrt(a)*x^2 - sqrt(a*x^4 + b*x^2))*sqrt(a)*b^2 + 4*b^3)/(
((sqrt(a)*x^2 - sqrt(a*x^4 + b*x^2))*sqrt(a) + b)^3*a^(5/2))

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maple [A]  time = 0.01, size = 73, normalized size = 1.24 \[ -\frac {\left (a \,x^{2}+b \right ) \left (4 a^{\frac {5}{2}} x^{3}+3 a^{\frac {3}{2}} b x -3 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )\right )}{3 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} a^{\frac {7}{2}} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^2)^(5/2)/x,x)

[Out]

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

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maxima [A]  time = 1.99, size = 62, normalized size = 1.05 \[ -\frac {\log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {4 \, a + \frac {3 \, b}{x^{2}}}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.39, size = 47, normalized size = 0.80 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {a+\frac {b}{x^2}}{a^2}+\frac {1}{3\,a}}{{\left (a+\frac {b}{x^2}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x^2)^(5/2)),x)

[Out]

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

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sympy [B]  time = 3.50, size = 743, normalized size = 12.59 \[ - \frac {8 a^{7} x^{6} \sqrt {1 + \frac {b}{a x^{2}}}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} - \frac {3 a^{7} x^{6} \log {\left (\frac {b}{a x^{2}} \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} + \frac {6 a^{7} x^{6} \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} - \frac {14 a^{6} b x^{4} \sqrt {1 + \frac {b}{a x^{2}}}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} - \frac {9 a^{6} b x^{4} \log {\left (\frac {b}{a x^{2}} \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} + \frac {18 a^{6} b x^{4} \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} - \frac {6 a^{5} b^{2} x^{2} \sqrt {1 + \frac {b}{a x^{2}}}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} - \frac {9 a^{5} b^{2} x^{2} \log {\left (\frac {b}{a x^{2}} \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} + \frac {18 a^{5} b^{2} x^{2} \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} - \frac {3 a^{4} b^{3} \log {\left (\frac {b}{a x^{2}} \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} + \frac {6 a^{4} b^{3} \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )}}{6 a^{\frac {19}{2}} x^{6} + 18 a^{\frac {17}{2}} b x^{4} + 18 a^{\frac {15}{2}} b^{2} x^{2} + 6 a^{\frac {13}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(5/2)/x,x)

[Out]

-8*a**7*x**6*sqrt(1 + b/(a*x**2))/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a**(13/
2)*b**3) - 3*a**7*x**6*log(b/(a*x**2))/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a*
*(13/2)*b**3) + 6*a**7*x**6*log(sqrt(1 + b/(a*x**2)) + 1)/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/
2)*b**2*x**2 + 6*a**(13/2)*b**3) - 14*a**6*b*x**4*sqrt(1 + b/(a*x**2))/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4
+ 18*a**(15/2)*b**2*x**2 + 6*a**(13/2)*b**3) - 9*a**6*b*x**4*log(b/(a*x**2))/(6*a**(19/2)*x**6 + 18*a**(17/2)
*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a**(13/2)*b**3) + 18*a**6*b*x**4*log(sqrt(1 + b/(a*x**2)) + 1)/(6*a**(19/
2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a**(13/2)*b**3) - 6*a**5*b**2*x**2*sqrt(1 + b/(a*x*
*2))/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a**(13/2)*b**3) - 9*a**5*b**2*x**2*l
og(b/(a*x**2))/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a**(13/2)*b**3) + 18*a**5*
b**2*x**2*log(sqrt(1 + b/(a*x**2)) + 1)/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2 + 6*a
**(13/2)*b**3) - 3*a**4*b**3*log(b/(a*x**2))/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a**(15/2)*b**2*x**2
+ 6*a**(13/2)*b**3) + 6*a**4*b**3*log(sqrt(1 + b/(a*x**2)) + 1)/(6*a**(19/2)*x**6 + 18*a**(17/2)*b*x**4 + 18*a
**(15/2)*b**2*x**2 + 6*a**(13/2)*b**3)

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