3.1239 \(\int \frac {1}{x (a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=57 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}} \]

[Out]

-1/2*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-1/2*arctanh((-b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)

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Rubi [A]  time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {266, 63, 212, 206, 203} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a - b*x^4)^(3/4)),x]

[Out]

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

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 212

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 48, normalized size = 0.84 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a - b*x^4)^(3/4)),x]

[Out]

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

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fricas [B]  time = 0.66, size = 113, normalized size = 1.98 \[ \frac {1}{a^{3}}^{\frac {1}{4}} \arctan \left (\sqrt {a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {-b x^{4} + a}} a^{2} \frac {1}{a^{3}}^{\frac {3}{4}} - {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{2} \frac {1}{a^{3}}^{\frac {3}{4}}\right ) - \frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (a \frac {1}{a^{3}}^{\frac {1}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (-a \frac {1}{a^{3}}^{\frac {1}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 192, normalized size = 3.37 \[ -\frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{8 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*(-a)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a - 1/4*sqr
t(2)*(-a)^(1/4)*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)^(1/4)*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)^(1/4)*l
og(-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 = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {3}{4}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(-b*x^4+a)^(3/4),x)

[Out]

int(1/x/(-b*x^4+a)^(3/4),x)

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maxima [A]  time = 2.28, size = 60, normalized size = 1.05 \[ -\frac {\arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2 \, a^{\frac {3}{4}}} + \frac {\log \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{4 \, a^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.16, size = 36, normalized size = 0.63 \[ -\frac {\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )+\mathrm {atanh}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a - b*x^4)^(3/4)),x)

[Out]

-(atan((a - b*x^4)^(1/4)/a^(1/4)) + atanh((a - b*x^4)^(1/4)/a^(1/4)))/(2*a^(3/4))

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sympy [C]  time = 2.39, size = 42, normalized size = 0.74 \[ - \frac {e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{3} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(-3*I*pi/4)*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a/(b*x**4))/(4*b**(3/4)*x**3*gamma(7/4))

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