3.1912 \(\int \frac{(a+\frac{b}{x^2})^{5/2}}{x^2} \, dx\)

Optimal. Leaf size=92 \[ -\frac{5 a^2 \sqrt{a+\frac{b}{x^2}}}{16 x}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{16 \sqrt{b}}-\frac{5 a \left (a+\frac{b}{x^2}\right )^{3/2}}{24 x}-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x} \]

[Out]

(-5*a^2*Sqrt[a + b/x^2])/(16*x) - (5*a*(a + b/x^2)^(3/2))/(24*x) - (a + b/x^2)^(5/2)/(6*x) - (5*a^3*ArcTanh[Sq
rt[b]/(Sqrt[a + b/x^2]*x)])/(16*Sqrt[b])

________________________________________________________________________________________

Rubi [A]  time = 0.0373123, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {335, 195, 217, 206} \[ -\frac{5 a^2 \sqrt{a+\frac{b}{x^2}}}{16 x}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{16 \sqrt{b}}-\frac{5 a \left (a+\frac{b}{x^2}\right )^{3/2}}{24 x}-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(5/2)/x^2,x]

[Out]

(-5*a^2*Sqrt[a + b/x^2])/(16*x) - (5*a*(a + b/x^2)^(3/2))/(24*x) - (a + b/x^2)^(5/2)/(6*x) - (5*a^3*ArcTanh[Sq
rt[b]/(Sqrt[a + b/x^2]*x)])/(16*Sqrt[b])

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 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 \frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{x^2} \, dx &=-\operatorname{Subst}\left (\int \left (a+b x^2\right )^{5/2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x}-\frac{1}{6} (5 a) \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{5 a \left (a+\frac{b}{x^2}\right )^{3/2}}{24 x}-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x}-\frac{1}{8} \left (5 a^2\right ) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x^2}}}{16 x}-\frac{5 a \left (a+\frac{b}{x^2}\right )^{3/2}}{24 x}-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x}-\frac{1}{16} \left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x^2}}}{16 x}-\frac{5 a \left (a+\frac{b}{x^2}\right )^{3/2}}{24 x}-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x}-\frac{1}{16} \left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^2}} x}\right )\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x^2}}}{16 x}-\frac{5 a \left (a+\frac{b}{x^2}\right )^{3/2}}{24 x}-\frac{\left (a+\frac{b}{x^2}\right )^{5/2}}{6 x}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^2}} x}\right )}{16 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0386905, size = 96, normalized size = 1.04 \[ -\frac{\sqrt{a+\frac{b}{x^2}} \left (59 a^2 b x^4+15 a^3 x^6 \sqrt{\frac{a x^2}{b}+1} \tanh ^{-1}\left (\sqrt{\frac{a x^2}{b}+1}\right )+33 a^3 x^6+34 a b^2 x^2+8 b^3\right )}{48 x^5 \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(5/2)/x^2,x]

[Out]

-(Sqrt[a + b/x^2]*(8*b^3 + 34*a*b^2*x^2 + 59*a^2*b*x^4 + 33*a^3*x^6 + 15*a^3*x^6*Sqrt[1 + (a*x^2)/b]*ArcTanh[S
qrt[1 + (a*x^2)/b]]))/(48*x^5*(b + a*x^2))

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 166, normalized size = 1.8 \begin{align*} -{\frac{1}{48\,{b}^{3}x} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{5}{2}}} \left ( -3\, \left ( a{x}^{2}+b \right ) ^{5/2}{x}^{6}{a}^{3}+3\, \left ( a{x}^{2}+b \right ) ^{7/2}{x}^{4}{a}^{2}-5\, \left ( a{x}^{2}+b \right ) ^{3/2}{x}^{6}{a}^{3}b+15\,{b}^{5/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){x}^{6}{a}^{3}-15\,\sqrt{a{x}^{2}+b}{x}^{6}{a}^{3}{b}^{2}+2\, \left ( a{x}^{2}+b \right ) ^{7/2}{x}^{2}ab+8\, \left ( a{x}^{2}+b \right ) ^{7/2}{b}^{2} \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^2,x)

[Out]

-1/48*((a*x^2+b)/x^2)^(5/2)/x*(-3*(a*x^2+b)^(5/2)*x^6*a^3+3*(a*x^2+b)^(7/2)*x^4*a^2-5*(a*x^2+b)^(3/2)*x^6*a^3*
b+15*b^(5/2)*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*x^6*a^3-15*(a*x^2+b)^(1/2)*x^6*a^3*b^2+2*(a*x^2+b)^(7/2)*x^2*
a*b+8*(a*x^2+b)^(7/2)*b^2)/(a*x^2+b)^(5/2)/b^3

________________________________________________________________________________________

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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.62368, size = 424, normalized size = 4.61 \begin{align*} \left [\frac{15 \, a^{3} \sqrt{b} x^{5} \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 (33 \, a^{2} b x^{4} + 26 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{96 \, b x^{5}}, \frac{15 \, a^{3} \sqrt{-b} x^{5} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) -{\left (33 \, a^{2} b x^{4} + 26 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{48 \, b x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(15*a^3*sqrt(b)*x^5*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(33*a^2*b*x^4 + 26*a
*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b*x^5), 1/48*(15*a^3*sqrt(-b)*x^5*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)
/x^2)/(a*x^2 + b)) - (33*a^2*b*x^4 + 26*a*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b*x^5)]

________________________________________________________________________________________

Sympy [A]  time = 4.82063, size = 99, normalized size = 1.08 \begin{align*} - \frac{11 a^{\frac{5}{2}} \sqrt{1 + \frac{b}{a x^{2}}}}{16 x} - \frac{13 a^{\frac{3}{2}} b \sqrt{1 + \frac{b}{a x^{2}}}}{24 x^{3}} - \frac{\sqrt{a} b^{2} \sqrt{1 + \frac{b}{a x^{2}}}}{6 x^{5}} - \frac{5 a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{16 \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11*a**(5/2)*sqrt(1 + b/(a*x**2))/(16*x) - 13*a**(3/2)*b*sqrt(1 + b/(a*x**2))/(24*x**3) - sqrt(a)*b**2*sqrt(1
+ b/(a*x**2))/(6*x**5) - 5*a**3*asinh(sqrt(b)/(sqrt(a)*x))/(16*sqrt(b))

________________________________________________________________________________________

Giac [A]  time = 1.29464, size = 104, normalized size = 1.13 \begin{align*} \frac{1}{48} \, a^{3}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{a x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{33 \,{\left (a x^{2} + b\right )}^{\frac{5}{2}} - 40 \,{\left (a x^{2} + b\right )}^{\frac{3}{2}} b + 15 \, \sqrt{a x^{2} + b} b^{2}}{a^{3} x^{6}}\right )} \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^2,x, algorithm="giac")

[Out]

1/48*a^3*(15*arctan(sqrt(a*x^2 + b)/sqrt(-b))/sqrt(-b) - (33*(a*x^2 + b)^(5/2) - 40*(a*x^2 + b)^(3/2)*b + 15*s
qrt(a*x^2 + b)*b^2)/(a^3*x^6))*sgn(x)