3.835 \(\int \frac{1}{x^5 \sqrt{a-b x^4}} \, dx\)

Optimal. Leaf size=52 \[ -\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b x^4}}{\sqrt{a}}\right )}{4 a^{3/2}}-\frac{\sqrt{a-b x^4}}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.0283322, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {266, 51, 63, 208} \[ -\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b x^4}}{\sqrt{a}}\right )}{4 a^{3/2}}-\frac{\sqrt{a-b x^4}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[a - b*x^4]),x]

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0700338, size = 61, normalized size = 1.17 \[ -\frac{b \sqrt{a-b x^4} \left (\frac{a}{b x^4}+\frac{\tanh ^{-1}\left (\sqrt{1-\frac{b x^4}{a}}\right )}{\sqrt{1-\frac{b x^4}{a}}}\right )}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[a - b*x^4]),x]

[Out]

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

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Maple [A]  time = 0.013, size = 50, normalized size = 1. \begin{align*} -{\frac{1}{4\,a{x}^{4}}\sqrt{-b{x}^{4}+a}}-{\frac{b}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{-b{x}^{4}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-b*x^4+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51011, size = 269, normalized size = 5.17 \begin{align*} \left [\frac{\sqrt{a} b x^{4} \log \left (\frac{b x^{4} + 2 \, \sqrt{-b x^{4} + a} \sqrt{a} - 2 \, a}{x^{4}}\right ) - 2 \, \sqrt{-b x^{4} + a} a}{8 \, a^{2} x^{4}}, \frac{\sqrt{-a} b x^{4} \arctan \left (\frac{\sqrt{-b x^{4} + a} \sqrt{-a}}{a}\right ) - \sqrt{-b x^{4} + a} a}{4 \, a^{2} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.94393, size = 133, normalized size = 2.56 \begin{align*} \begin{cases} - \frac{\sqrt{b} \sqrt{\frac{a}{b x^{4}} - 1}}{4 a x^{2}} - \frac{b \operatorname{acosh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 a^{\frac{3}{2}}} & \text{for}\: \frac{\left |{a}\right |}{\left |{b}\right | \left |{x^{4}}\right |} > 1 \\\frac{i}{4 \sqrt{b} x^{6} \sqrt{- \frac{a}{b x^{4}} + 1}} - \frac{i \sqrt{b}}{4 a x^{2} \sqrt{- \frac{a}{b x^{4}} + 1}} + \frac{i b \operatorname{asin}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 a^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-b*x**4+a)**(1/2),x)

[Out]

Piecewise((-sqrt(b)*sqrt(a/(b*x**4) - 1)/(4*a*x**2) - b*acosh(sqrt(a)/(sqrt(b)*x**2))/(4*a**(3/2)), Abs(a)/(Ab
s(b)*Abs(x**4)) > 1), (I/(4*sqrt(b)*x**6*sqrt(-a/(b*x**4) + 1)) - I*sqrt(b)/(4*a*x**2*sqrt(-a/(b*x**4) + 1)) +
I*b*asin(sqrt(a)/(sqrt(b)*x**2))/(4*a**(3/2)), True))

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Giac [A]  time = 1.11545, size = 69, normalized size = 1.33 \begin{align*} \frac{1}{4} \, b{\left (\frac{\arctan \left (\frac{\sqrt{-b x^{4} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{\sqrt{-b x^{4} + a}}{a b x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

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