3.1906 \(\int \frac{(a+\frac{b}{x^2})^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=95 \[ \frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{16 b^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x^2}}}{16 b x}-\frac{a \sqrt{a+\frac{b}{x^2}}}{8 x^3}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3} \]

[Out]

-(a*Sqrt[a + b/x^2])/(8*x^3) - (a + b/x^2)^(3/2)/(6*x^3) - (a^2*Sqrt[a + b/x^2])/(16*b*x) + (a^3*ArcTanh[Sqrt[
b]/(Sqrt[a + b/x^2]*x)])/(16*b^(3/2))

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Rubi [A]  time = 0.045752, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {335, 279, 321, 217, 206} \[ \frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{16 b^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x^2}}}{16 b x}-\frac{a \sqrt{a+\frac{b}{x^2}}}{8 x^3}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2)/x^4,x]

[Out]

-(a*Sqrt[a + b/x^2])/(8*x^3) - (a + b/x^2)^(3/2)/(6*x^3) - (a^2*Sqrt[a + b/x^2])/(16*b*x) + (a^3*ArcTanh[Sqrt[
b]/(Sqrt[a + b/x^2]*x)])/(16*b^(3/2))

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 279

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

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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{\left (a+\frac{b}{x^2}\right )^{3/2}}{x^4} \, dx &=-\operatorname{Subst}\left (\int x^2 \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3}-\frac{1}{2} a \operatorname{Subst}\left (\int x^2 \sqrt{a+b x^2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{a \sqrt{a+\frac{b}{x^2}}}{8 x^3}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3}-\frac{1}{8} a^2 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{a \sqrt{a+\frac{b}{x^2}}}{8 x^3}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3}-\frac{a^2 \sqrt{a+\frac{b}{x^2}}}{16 b x}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{16 b}\\ &=-\frac{a \sqrt{a+\frac{b}{x^2}}}{8 x^3}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3}-\frac{a^2 \sqrt{a+\frac{b}{x^2}}}{16 b x}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^2}} x}\right )}{16 b}\\ &=-\frac{a \sqrt{a+\frac{b}{x^2}}}{8 x^3}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{6 x^3}-\frac{a^2 \sqrt{a+\frac{b}{x^2}}}{16 b x}+\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^2}} x}\right )}{16 b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0133197, size = 49, normalized size = 0.52 \[ \frac{a^3 x^3 \left (a+\frac{b}{x^2}\right )^{3/2} \left (a x^2+b\right ) \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{a x^2}{b}+1\right )}{5 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2)/x^4,x]

[Out]

(a^3*(a + b/x^2)^(3/2)*x^3*(b + a*x^2)*Hypergeometric2F1[5/2, 4, 7/2, 1 + (a*x^2)/b])/(5*b^4)

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Maple [A]  time = 0.008, size = 145, normalized size = 1.5 \begin{align*}{\frac{1}{48\,{b}^{3}{x}^{3}} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( - \left ( a{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{6}{a}^{3}+3\,{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){x}^{6}{a}^{3}+ \left ( a{x}^{2}+b \right ) ^{{\frac{5}{2}}}{x}^{4}{a}^{2}-3\,\sqrt{a{x}^{2}+b}{x}^{6}{a}^{3}b+2\, \left ( a{x}^{2}+b \right ) ^{5/2}{x}^{2}ab-8\, \left ( a{x}^{2}+b \right ) ^{5/2}{b}^{2} \right ) \left ( a{x}^{2}+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+1/x^2*b)^(3/2)/x^4,x)

[Out]

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

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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)^(3/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61426, size = 425, normalized size = 4.47 \begin{align*} \left [\frac{3 \, 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 (3 \, a^{2} b x^{4} + 14 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{96 \, b^{2} x^{5}}, -\frac{3 \, 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 (3 \, a^{2} b x^{4} + 14 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{48 \, b^{2} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(3*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*(3*a^2*b*x^4 + 14*a*b
^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b^2*x^5), -1/48*(3*a^3*sqrt(-b)*x^5*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)
/x^2)/(a*x^2 + b)) + (3*a^2*b*x^4 + 14*a*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b^2*x^5)]

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Sympy [A]  time = 5.1738, size = 119, normalized size = 1.25 \begin{align*} - \frac{a^{\frac{5}{2}}}{16 b x \sqrt{1 + \frac{b}{a x^{2}}}} - \frac{17 a^{\frac{3}{2}}}{48 x^{3} \sqrt{1 + \frac{b}{a x^{2}}}} - \frac{11 \sqrt{a} b}{24 x^{5} \sqrt{1 + \frac{b}{a x^{2}}}} + \frac{a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{16 b^{\frac{3}{2}}} - \frac{b^{2}}{6 \sqrt{a} x^{7} \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)**(3/2)/x**4,x)

[Out]

-a**(5/2)/(16*b*x*sqrt(1 + b/(a*x**2))) - 17*a**(3/2)/(48*x**3*sqrt(1 + b/(a*x**2))) - 11*sqrt(a)*b/(24*x**5*s
qrt(1 + b/(a*x**2))) + a**3*asinh(sqrt(b)/(sqrt(a)*x))/(16*b**(3/2)) - b**2/(6*sqrt(a)*x**7*sqrt(1 + b/(a*x**2
)))

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Giac [A]  time = 1.26829, size = 111, normalized size = 1.17 \begin{align*} -\frac{1}{48} \, a^{3}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{a x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{3 \,{\left (a x^{2} + b\right )}^{\frac{5}{2}} + 8 \,{\left (a x^{2} + b\right )}^{\frac{3}{2}} b - 3 \, \sqrt{a x^{2} + b} b^{2}}{a^{3} b 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)^(3/2)/x^4,x, algorithm="giac")

[Out]

-1/48*a^3*(3*arctan(sqrt(a*x^2 + b)/sqrt(-b))/(sqrt(-b)*b) + (3*(a*x^2 + b)^(5/2) + 8*(a*x^2 + b)^(3/2)*b - 3*
sqrt(a*x^2 + b)*b^2)/(a^3*b*x^6))*sgn(x)