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

Optimal. Leaf size=50 \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 a^{3/2}}-\frac{\sqrt{a+b x^4}}{4 a x^4} \]

[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 = 0.0259341, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, 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 = 0.0550041, size = 64, normalized size = 1.28 \[ \frac{b \sqrt{a+b x^4} \left (\frac{\tanh ^{-1}\left (\sqrt{\frac{b x^4}{a}+1}\right )}{2 \sqrt{\frac{b x^4}{a}+1}}-\frac{a}{2 b x^4}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[a + b*x^4]),x]

[Out]

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

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Maple [A]  time = 0.011, size = 48, 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 = 1.48439, size = 265, normalized size = 5.3 \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 = 2.46112, size = 46, normalized size = 0.92 \begin{align*} - \frac{\sqrt{b} \sqrt{\frac{a}{b x^{4}} + 1}}{4 a x^{2}} + \frac{b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 a^{\frac{3}{2}}} \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]

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

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Giac [A]  time = 1.11564, size = 65, normalized size = 1.3 \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))