3.1910 \(\int (a+\frac{b}{x^2})^{5/2} \, dx\)

Optimal. Leaf size=86 \[ -\frac{15}{8} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )+x \left (a+\frac{b}{x^2}\right )^{5/2}-\frac{5 b \left (a+\frac{b}{x^2}\right )^{3/2}}{4 x}-\frac{15 a b \sqrt{a+\frac{b}{x^2}}}{8 x} \]

[Out]

(-15*a*b*Sqrt[a + b/x^2])/(8*x) - (5*b*(a + b/x^2)^(3/2))/(4*x) + (a + b/x^2)^(5/2)*x - (15*a^2*Sqrt[b]*ArcTan
h[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/8

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Rubi [A]  time = 0.0363724, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {242, 277, 195, 217, 206} \[ -\frac{15}{8} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )+x \left (a+\frac{b}{x^2}\right )^{5/2}-\frac{5 b \left (a+\frac{b}{x^2}\right )^{3/2}}{4 x}-\frac{15 a b \sqrt{a+\frac{b}{x^2}}}{8 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(5/2),x]

[Out]

(-15*a*b*Sqrt[a + b/x^2])/(8*x) - (5*b*(a + b/x^2)^(3/2))/(4*x) + (a + b/x^2)^(5/2)*x - (15*a^2*Sqrt[b]*ArcTan
h[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/8

Rule 242

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

Rule 277

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [C]  time = 0.0121806, size = 49, normalized size = 0.57 \[ -\frac{a^2 x^5 \left (a+\frac{b}{x^2}\right )^{5/2} \left (a x^2+b\right ) \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{a x^2}{b}+1\right )}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(5/2),x]

[Out]

-(a^2*(a + b/x^2)^(5/2)*x^5*(b + a*x^2)*Hypergeometric2F1[3, 7/2, 9/2, 1 + (a*x^2)/b])/(7*b^3)

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Maple [B]  time = 0.006, size = 144, normalized size = 1.7 \begin{align*} -{\frac{x}{8\,{b}^{2}} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{5}{2}}} \left ( -3\, \left ( a{x}^{2}+b \right ) ^{5/2}{x}^{4}{a}^{2}+15\,{b}^{5/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){x}^{4}{a}^{2}+3\, \left ( a{x}^{2}+b \right ) ^{7/2}{x}^{2}a-5\, \left ( a{x}^{2}+b \right ) ^{3/2}{x}^{4}{a}^{2}b-15\,\sqrt{a{x}^{2}+b}{x}^{4}{a}^{2}{b}^{2}+2\, \left ( a{x}^{2}+b \right ) ^{7/2}b \right ) \left ( a{x}^{2}+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((a*x^2+b)/x^2)^(5/2)*x*(-3*(a*x^2+b)^(5/2)*x^4*a^2+15*b^(5/2)*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*x^4*a^
2+3*(a*x^2+b)^(7/2)*x^2*a-5*(a*x^2+b)^(3/2)*x^4*a^2*b-15*(a*x^2+b)^(1/2)*x^4*a^2*b^2+2*(a*x^2+b)^(7/2)*b)/(a*x
^2+b)^(5/2)/b^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((a+b/x^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55633, size = 396, normalized size = 4.6 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x^{3} \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 (8 \, a^{2} x^{4} - 9 \, a b x^{2} - 2 \, b^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{16 \, x^{3}}, \frac{15 \, a^{2} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) +{\left (8 \, a^{2} x^{4} - 9 \, a b x^{2} - 2 \, b^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{8 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(15*a^2*sqrt(b)*x^3*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) + 2*(8*a^2*x^4 - 9*a*b*x
^2 - 2*b^2)*sqrt((a*x^2 + b)/x^2))/x^3, 1/8*(15*a^2*sqrt(-b)*x^3*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^
2 + b)) + (8*a^2*x^4 - 9*a*b*x^2 - 2*b^2)*sqrt((a*x^2 + b)/x^2))/x^3]

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Sympy [A]  time = 4.16591, size = 117, normalized size = 1.36 \begin{align*} \frac{a^{\frac{5}{2}} x}{\sqrt{1 + \frac{b}{a x^{2}}}} - \frac{a^{\frac{3}{2}} b}{8 x \sqrt{1 + \frac{b}{a x^{2}}}} - \frac{11 \sqrt{a} b^{2}}{8 x^{3} \sqrt{1 + \frac{b}{a x^{2}}}} - \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{8} - \frac{b^{3}}{4 \sqrt{a} x^{5} \sqrt{1 + \frac{b}{a x^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*x/sqrt(1 + b/(a*x**2)) - a**(3/2)*b/(8*x*sqrt(1 + b/(a*x**2))) - 11*sqrt(a)*b**2/(8*x**3*sqrt(1 + b/(
a*x**2))) - 15*a**2*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x))/8 - b**3/(4*sqrt(a)*x**5*sqrt(1 + b/(a*x**2)))

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Giac [A]  time = 1.25049, size = 105, normalized size = 1.22 \begin{align*} \frac{1}{8} \,{\left (\frac{15 \, b \arctan \left (\frac{\sqrt{a x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 8 \, \sqrt{a x^{2} + b} - \frac{9 \,{\left (a x^{2} + b\right )}^{\frac{3}{2}} b - 7 \, \sqrt{a x^{2} + b} b^{2}}{a^{2} x^{4}}\right )} a^{2} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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