3.13.10 \(\int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx\) [1210]

Optimal. Leaf size=81 \[ -\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}+\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}} \]

[Out]

-1/4*(-b*x^4+a)^(3/4)/a/x^4+1/8*b*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)-1/8*b*arctanh((-b*x^4+a)^(1/4)/a^(1
/4))/a^(5/4)

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Rubi [A]
time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {272, 44, 65, 304, 209, 212} \begin {gather*} \frac {b \text {ArcTan}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a - b*x^4)^(1/4)),x]

[Out]

-1/4*(a - b*x^4)^(3/4)/(a*x^4) + (b*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)])/(8*a^(5/4)) - (b*ArcTanh[(a - b*x^4)^(1
/4)/a^(1/4)])/(8*a^(5/4))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 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]]

Rule 304

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^5 \sqrt [4]{a-b x^4}} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a-b x}} \, dx,x,x^4\right )\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}+\frac {b \text {Subst}\left (\int \frac {1}{x \sqrt [4]{a-b x}} \, dx,x,x^4\right )}{16 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}-\frac {\text {Subst}\left (\int \frac {x^2}{\frac {a}{b}-\frac {x^4}{b}} \, dx,x,\sqrt [4]{a-b x^4}\right )}{4 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{8 a}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{8 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}+\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 81, normalized size = 1.00 \begin {gather*} -\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}+\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a - b*x^4)^(1/4)),x]

[Out]

-1/4*(a - b*x^4)^(3/4)/(a*x^4) + (b*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)])/(8*a^(5/4)) - (b*ArcTanh[(a - b*x^4)^(1
/4)/a^(1/4)])/(8*a^(5/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{5} \left (-b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-b*x^4+a)^(1/4),x)

[Out]

int(1/x^5/(-b*x^4+a)^(1/4),x)

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Maxima [A]
time = 0.50, size = 96, normalized size = 1.19 \begin {gather*} \frac {b {\left (\frac {2 \, \arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{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 )}{a^{\frac {1}{4}}}\right )}}{16 \, a} - \frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} b}{4 \, {\left ({\left (b x^{4} - a\right )} a + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

1/16*b*(2*arctan((-b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((-b*x^4 + a)^(1/4) - a^(1/4))/((-b*x^4 + a)^(1/4)
+ a^(1/4)))/a^(1/4))/a - 1/4*(-b*x^4 + a)^(3/4)*b/((b*x^4 - a)*a + a^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (61) = 122\).
time = 0.39, size = 200, normalized size = 2.47 \begin {gather*} -\frac {4 \, a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a b^{3} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} - \sqrt {a^{3} b^{4} \sqrt {\frac {b^{4}}{a^{5}}} + \sqrt {-b x^{4} + a} b^{6}} a \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}}}{b^{4}}\right ) + a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) + 4 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

-1/16*(4*a*x^4*(b^4/a^5)^(1/4)*arctan(-((-b*x^4 + a)^(1/4)*a*b^3*(b^4/a^5)^(1/4) - sqrt(a^3*b^4*sqrt(b^4/a^5)
+ sqrt(-b*x^4 + a)*b^6)*a*(b^4/a^5)^(1/4))/b^4) + a*x^4*(b^4/a^5)^(1/4)*log(a^4*(b^4/a^5)^(3/4) + (-b*x^4 + a)
^(1/4)*b^3) - a*x^4*(b^4/a^5)^(1/4)*log(-a^4*(b^4/a^5)^(3/4) + (-b*x^4 + a)^(1/4)*b^3) + 4*(-b*x^4 + a)^(3/4))
/(a*x^4)

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Sympy [C] Result contains complex when optimal does not.
time = 0.72, size = 41, normalized size = 0.51 \begin {gather*} \frac {e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{5} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-b*x**4+a)**(1/4),x)

[Out]

exp(3*I*pi/4)*gamma(5/4)*hyper((1/4, 5/4), (9/4,), a/(b*x**4))/(4*b**(1/4)*x**5*gamma(9/4))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (61) = 122\).
time = 1.86, size = 226, normalized size = 2.79 \begin {gather*} -\frac {\frac {2 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \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 )}{a^{2}} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \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 )}{a^{2}} + \frac {\sqrt {2} b^{2} \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 )}{\left (-a\right )^{\frac {1}{4}} a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \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 )}{a^{2}} + \frac {8 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} b}{a x^{4}}}{32 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

-1/32*(2*sqrt(2)*(-a)^(3/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2
+ 2*sqrt(2)*(-a)^(3/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 +
sqrt(2)*b^2*log(sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a) + sqrt(2)*
(-a)^(3/4)*b^2*log(-sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/a^2 + 8*(-b*x^4 + a)^
(3/4)*b/(a*x^4))/b

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Mupad [B]
time = 1.30, size = 61, normalized size = 0.75 \begin {gather*} \frac {b\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{5/4}}-\frac {{\left (a-b\,x^4\right )}^{3/4}}{4\,a\,x^4}-\frac {b\,\mathrm {atanh}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a - b*x^4)^(1/4)),x)

[Out]

(b*atan((a - b*x^4)^(1/4)/a^(1/4)))/(8*a^(5/4)) - (a - b*x^4)^(3/4)/(4*a*x^4) - (b*atanh((a - b*x^4)^(1/4)/a^(
1/4)))/(8*a^(5/4))

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