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

Optimal. Leaf size=71 \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}-\frac {b \sqrt {a+b x^2}}{8 a x^2}-\frac {\sqrt {a+b x^2}}{4 x^4} \]

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Rubi [A]  time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 47, 51, 63, 208} \begin {gather*} \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}-\frac {b \sqrt {a+b x^2}}{8 a x^2}-\frac {\sqrt {a+b x^2}}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/x^5,x]

[Out]

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

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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]

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]]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^2}}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+b x^2}}{4 x^4}+\frac {1}{8} b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+b x^2}}{4 x^4}-\frac {b \sqrt {a+b x^2}}{8 a x^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{16 a}\\ &=-\frac {\sqrt {a+b x^2}}{4 x^4}-\frac {b \sqrt {a+b x^2}}{8 a x^2}-\frac {b \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{8 a}\\ &=-\frac {\sqrt {a+b x^2}}{4 x^4}-\frac {b \sqrt {a+b x^2}}{8 a x^2}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.55 \begin {gather*} -\frac {b^2 \left (a+b x^2\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b x^2}{a}+1\right )}{3 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2]/x^5,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 62, normalized size = 0.87 \begin {gather*} \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {\left (-2 a-b x^2\right ) \sqrt {a+b x^2}}{8 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*x^2]/x^5,x]

[Out]

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

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fricas [A]  time = 0.64, size = 131, normalized size = 1.85 \begin {gather*} \left [\frac {\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 (a b x^{2} + 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a^{2} x^{4}}, -\frac {\sqrt {-a} b^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (a b x^{2} + 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.60, size = 72, normalized size = 1.01 \begin {gather*} -\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b x^{2} + a} a b^{3}}{a b^{2} x^{4}}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 85, normalized size = 1.20 \begin {gather*} \frac {b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, b^{2}}{8 a^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} b}{8 a^{2} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 73, normalized size = 1.03 \begin {gather*} \frac {b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} - \frac {\sqrt {b x^{2} + a} b^{2}}{8 \, a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.91, size = 54, normalized size = 0.76 \begin {gather*} \frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{3/2}}-\frac {\sqrt {b\,x^2+a}}{8\,x^4}-\frac {{\left (b\,x^2+a\right )}^{3/2}}{8\,a\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)/x^5,x)

[Out]

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

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sympy [A]  time = 3.73, size = 92, normalized size = 1.30 \begin {gather*} - \frac {a}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {b^{\frac {3}{2}}}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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