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

Optimal. Leaf size=29 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a x^4-b}}{\sqrt {b}}\right )}{2 \sqrt {b}} \]

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {266, 63, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {a x^4-b}}{\sqrt {b}}\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-b + a*x^4]),x]

[Out]

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

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 {1}{x \sqrt {-b+a x^4}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-b+a x}} \, dx,x,x^4\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {b}{a}+\frac {x^2}{a}} \, dx,x,\sqrt {-b+a x^4}\right )}{2 a}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {-b+a x^4}}{\sqrt {b}}\right )}{2 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {a x^4-b}}{\sqrt {b}}\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[-b + a*x^4]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 29, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {-b+a x^4}}{\sqrt {b}}\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*Sqrt[-b + a*x^4]),x]

[Out]

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

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fricas [A]  time = 0.46, size = 64, normalized size = 2.21 \begin {gather*} \left [-\frac {\sqrt {-b} \log \left (\frac {a x^{4} - 2 \, \sqrt {a x^{4} - b} \sqrt {-b} - 2 \, b}{x^{4}}\right )}{4 \, b}, \frac {\arctan \left (\frac {\sqrt {a x^{4} - b}}{\sqrt {b}}\right )}{2 \, \sqrt {b}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 21, normalized size = 0.72 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {a x^{4} - b}}{\sqrt {b}}\right )}{2 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 35, normalized size = 1.21

method	result	size
default	\(-\frac {\ln \left (\frac {-2 b +2 \sqrt {-b}\, \sqrt {a \,x^{4}-b}}{x^{2}}\right )}{2 \sqrt {-b}}\)	\(35\)
elliptic	\(-\frac {\ln \left (\frac {-2 b +2 \sqrt {-b}\, \sqrt {a \,x^{4}-b}}{x^{2}}\right )}{2 \sqrt {-b}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a*x^4-b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.70, size = 21, normalized size = 0.72 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {a x^{4} - b}}{\sqrt {b}}\right )}{2 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.38, size = 21, normalized size = 0.72 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {a\,x^4-b}}{\sqrt {b}}\right )}{2\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^4 - b)^(1/2)),x)

[Out]

atan((a*x^4 - b)^(1/2)/b^(1/2))/(2*b^(1/2))

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sympy [A]  time = 0.93, size = 53, normalized size = 1.83 \begin {gather*} \begin {cases} \frac {i \operatorname {acosh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{2 \sqrt {b}} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {\operatorname {asin}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{2 \sqrt {b}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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