3.375 \(\int \frac{(a+b x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=68 \[ -\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{3 b \sqrt{a+b x^2}}{8 x^2}-\frac{\left (a+b x^2\right )^{3/2}}{4 x^4} \]

[Out]

(-3*b*Sqrt[a + b*x^2])/(8*x^2) - (a + b*x^2)^(3/2)/(4*x^4) - (3*b^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*Sqrt[
a])

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Rubi [A]  time = 0.0395221, 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 = {266, 47, 63, 208} \[ -\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{3 b \sqrt{a+b x^2}}{8 x^2}-\frac{\left (a+b x^2\right )^{3/2}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^5,x]

[Out]

(-3*b*Sqrt[a + b*x^2])/(8*x^2) - (a + b*x^2)^(3/2)/(4*x^4) - (3*b^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*Sqrt[
a])

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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{\left (a+b x^2\right )^{3/2}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{3/2}}{4 x^4}+\frac{1}{8} (3 b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{3 b \sqrt{a+b x^2}}{8 x^2}-\frac{\left (a+b x^2\right )^{3/2}}{4 x^4}+\frac{1}{16} \left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=-\frac{3 b \sqrt{a+b x^2}}{8 x^2}-\frac{\left (a+b x^2\right )^{3/2}}{4 x^4}+\frac{1}{8} (3 b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=-\frac{3 b \sqrt{a+b x^2}}{8 x^2}-\frac{\left (a+b x^2\right )^{3/2}}{4 x^4}-\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0354422, size = 76, normalized size = 1.12 \[ -\frac{2 a^2+3 b^2 x^4 \sqrt{\frac{b x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )+7 a b x^2+5 b^2 x^4}{8 x^4 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^5,x]

[Out]

-(2*a^2 + 7*a*b*x^2 + 5*b^2*x^4 + 3*b^2*x^4*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/(8*x^4*Sqrt[a +
b*x^2])

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Maple [A]  time = 0.004, size = 102, normalized size = 1.5 \begin{align*} -{\frac{1}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{b}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{2}}{8\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{3\,{b}^{2}}{8\,a}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/x^5,x)

[Out]

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

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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((b*x^2+a)^(3/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54177, size = 317, normalized size = 4.66 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (5 \, a b x^{2} + 2 \, a^{2}\right )} \sqrt{b x^{2} + a}}{16 \, a x^{4}}, \frac{3 \, \sqrt{-a} b^{2} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (5 \, a b x^{2} + 2 \, a^{2}\right )} \sqrt{b x^{2} + a}}{8 \, a x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/16*(3*sqrt(a)*b^2*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(5*a*b*x^2 + 2*a^2)*sqrt(b*x^
2 + a))/(a*x^4), 1/8*(3*sqrt(-a)*b^2*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (5*a*b*x^2 + 2*a^2)*sqrt(b*x^2 + a
))/(a*x^4)]

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Sympy [A]  time = 3.00593, size = 71, normalized size = 1.04 \begin{align*} - \frac{a \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{4 x^{3}} - \frac{5 b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{8 x} - \frac{3 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8 \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/x**5,x)

[Out]

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

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Giac [A]  time = 2.6188, size = 82, normalized size = 1.21 \begin{align*} \frac{1}{8} \, b^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} - 3 \, \sqrt{b x^{2} + a} a}{b^{2} x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="giac")

[Out]

1/8*b^2*(3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (5*(b*x^2 + a)^(3/2) - 3*sqrt(b*x^2 + a)*a)/(b^2*x^4))