3.1955 \(\int \frac{1}{(a+\frac{b}{x^2})^{5/2} x^6} \, dx\)

Optimal. Leaf size=68 \[ \frac{1}{b^2 x \sqrt{a+\frac{b}{x^2}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{b^{5/2}}+\frac{1}{3 b x^3 \left (a+\frac{b}{x^2}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0347532, antiderivative size = 68, 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 = {335, 288, 217, 206} \[ \frac{1}{b^2 x \sqrt{a+\frac{b}{x^2}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{b^{5/2}}+\frac{1}{3 b x^3 \left (a+\frac{b}{x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(5/2)*x^6),x]

[Out]

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

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]

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 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 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])

Rubi steps

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

Mathematica [C]  time = 0.0095532, size = 46, normalized size = 0.68 \[ \frac{\left (a x^2+b\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{a x^2}{b}+1\right )}{3 b x^5 \left (a+\frac{b}{x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(5/2)*x^6),x]

[Out]

((b + a*x^2)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (a*x^2)/b])/(3*b*(a + b/x^2)^(5/2)*x^5)

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Maple [A]  time = 0.006, size = 77, normalized size = 1.1 \begin{align*} -{\frac{a{x}^{2}+b}{3\,{x}^{5}} \left ( -3\,{b}^{3/2}{x}^{2}a+3\, \left ( a{x}^{2}+b \right ) ^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ) b-4\,{b}^{5/2} \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{5}{2}}}{b}^{-{\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^6,x)

[Out]

-1/3*(a*x^2+b)*(-3*b^(3/2)*x^2*a+3*(a*x^2+b)^(3/2)*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*b-4*b^(5/2))/((a*x^2+b)
/x^2)^(5/2)/x^5/b^(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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92627, size = 500, normalized size = 7.35 \begin{align*} \left [\frac{3 \,{\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\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^{3} + 4 \, b^{2} x\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{6 \,{\left (a^{2} b^{3} x^{4} + 2 \, a b^{4} x^{2} + b^{5}\right )}}, \frac{3 \,{\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\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^{3} + 4 \, b^{2} x\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{3 \,{\left (a^{2} b^{3} x^{4} + 2 \, a b^{4} x^{2} + b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^6,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 4.51955, size = 740, normalized size = 10.88 \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**6,x)

[Out]

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

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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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^6,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(5/2)*x^6), x)