3.1085 \(\int \frac {1}{x \sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=55 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, 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, 63, 298, 203, 206} \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^4)^(1/4)),x]

[Out]

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

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 48, normalized size = 0.87 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^4)^(1/4)),x]

[Out]

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

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fricas [B]  time = 0.97, size = 81, normalized size = 1.47 \[ -\frac {\arctan \left (\frac {\sqrt {\sqrt {b x^{4} + a} + \sqrt {a}}}{a^{\frac {1}{4}}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} - \frac {\log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}\right )}{4 \, a^{\frac {1}{4}}} + \frac {\log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}\right )}{4 \, a^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 186, normalized size = 3.38 \[ -\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.89, size = 57, normalized size = 1.04 \[ \frac {\arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2 \, a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{4 \, a^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.16, size = 36, normalized size = 0.65 \[ \frac {\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )-\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^4)^(1/4)),x)

[Out]

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

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sympy [C]  time = 1.25, size = 37, normalized size = 0.67 \[ - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**(1/4)*x*gamma(5/4))

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