3.854 \(\int \frac{1}{x (a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=46 \[ \frac{1}{2 a \sqrt{a+b x^4}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0266344, antiderivative size = 46, 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} \[ \frac{1}{2 a \sqrt{a+b x^4}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^4)^(3/2)),x]

[Out]

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

Mathematica [C]  time = 0.0067697, size = 36, normalized size = 0.78 \[ \frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b x^4}{a}+1\right )}{2 a \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^4)^(3/2)),x]

[Out]

Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x^4)/a]/(2*a*Sqrt[a + b*x^4])

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Maple [A]  time = 0.01, size = 44, normalized size = 1. \begin{align*}{\frac{1}{2\,a}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{1}{2}\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/(b*x^4+a)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51133, size = 301, normalized size = 6.54 \begin{align*} \left [\frac{{\left (b x^{4} + a\right )} \sqrt{a} \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}{4 \,{\left (a^{2} b x^{4} + a^{3}\right )}}, \frac{{\left (b x^{4} + a\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{4} + a} \sqrt{-a}}{a}\right ) + \sqrt{b x^{4} + a} a}{2 \,{\left (a^{2} b x^{4} + a^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.73768, size = 184, normalized size = 4. \begin{align*} \frac{2 a^{3} \sqrt{1 + \frac{b x^{4}}{a}}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} + \frac{a^{3} \log{\left (\frac{b x^{4}}{a} \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} - \frac{2 a^{3} \log{\left (\sqrt{1 + \frac{b x^{4}}{a}} + 1 \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} + \frac{a^{2} b x^{4} \log{\left (\frac{b x^{4}}{a} \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} - \frac{2 a^{2} b x^{4} \log{\left (\sqrt{1 + \frac{b x^{4}}{a}} + 1 \right )}}{4 a^{\frac{9}{2}} + 4 a^{\frac{7}{2}} b x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**4+a)**(3/2),x)

[Out]

2*a**3*sqrt(1 + b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**3*log(b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**
4) - 2*a**3*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**2*b*x**4*log(b*x**4/a)/(4*a**(9/
2) + 4*a**(7/2)*b*x**4) - 2*a**2*b*x**4*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4)

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Giac [A]  time = 1.10521, size = 55, normalized size = 1.2 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b x^{4} + a}}{\sqrt{-a}}\right )}{2 \, \sqrt{-a} a} + \frac{1}{2 \, \sqrt{b x^{4} + a} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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