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

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

[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)

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Rubi [A]  time = 0.0327098, antiderivative size = 59, normalized size of antiderivative = 1., 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 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]]

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]

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*}

Mathematica [A]  time = 0.156926, 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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Maple [A]  time = 0.007, size = 73, normalized size = 1.2 \begin{align*} -{\frac{a{x}^{2}+b}{3\,{x}^{5}} \left ( 4\,{x}^{3}{a}^{5/2}+3\,{a}^{3/2}xb-3\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) \left ( a{x}^{2}+b \right ) ^{3/2}a \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{5}{2}}}{a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a*x^2+b)*(4*x^3*a^(5/2)+3*a^(3/2)*x*b-3*ln(x*a^(1/2)+(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.70268, size = 504, normalized size = 8.54 \begin{align*} \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 ] \end{align*}

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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Sympy [B]  time = 3.17367, size = 743, normalized size = 12.59 \begin{align*} \text{result too large to display} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x^{2}}\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a + b/x^2)^(5/2)*x), x)