3.12.15 \(\int \frac {1}{x (a+b x^4)^{3/4}} \, dx\) [1115]

Optimal. Leaf size=55 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}} \]

[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 = 0.02, 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 = {272, 65, 218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\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 {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^4)^(3/4)),x]

[Out]

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

Rule 65

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 272

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} \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/4}} \, dx,x,x^4\right )\\ &=\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{2 \sqrt {a}}-\frac {\text {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 = 0.05, size = 46, normalized size = 0.84 \begin {gather*} -\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}} \end {gather*}

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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (b \,x^{4}+a \right )^{\frac {3}{4}}}\, dx\]

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 = 0.51, size = 57, normalized size = 1.04 \begin {gather*} -\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}}} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (39) = 78\).
time = 0.39, size = 109, normalized size = 1.98 \begin {gather*} \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 ) \end {gather*}

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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Sympy [C] Result contains complex when optimal does not.
time = 0.47, size = 39, normalized size = 0.71 \begin {gather*} - \frac {\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 e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{3} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (39) = 78\).
time = 1.60, size = 186, normalized size = 3.38 \begin {gather*} -\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} \end {gather*}

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

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Mupad [B]
time = 1.16, size = 34, normalized size = 0.62 \begin {gather*} -\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^{3/4}} \end {gather*}

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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