3.1240 \(\int \frac {1}{x^5 (a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=81 \[ -\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}-\frac {\sqrt [4]{a-b x^4}}{4 a x^4} \]

[Out]

-1/4*(-b*x^4+a)^(1/4)/a/x^4-3/8*b*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(7/4)-3/8*b*arctanh((-b*x^4+a)^(1/4)/a^(1
/4))/a^(7/4)

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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 = {266, 51, 63, 212, 206, 203} \[ -\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}-\frac {\sqrt [4]{a-b x^4}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a - b*x^4)^(3/4)),x]

[Out]

-(a - b*x^4)^(1/4)/(4*a*x^4) - (3*b*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)])/(8*a^(7/4)) - (3*b*ArcTanh[(a - b*x^4)^
(1/4)/a^(1/4)])/(8*a^(7/4))

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 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^5 \left (a-b x^4\right )^{3/4}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x^2 (a-b x)^{3/4}} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{a-b x^4}}{4 a x^4}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x (a-b x)^{3/4}} \, dx,x,x^4\right )}{16 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{4 a x^4}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {x^4}{b}} \, dx,x,\sqrt [4]{a-b x^4}\right )}{4 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{4 a x^4}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{8 a^{3/2}}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{8 a^{3/2}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{4 a x^4}-\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.46 \[ -\frac {b \sqrt [4]{a-b x^4} \, _2F_1\left (\frac {1}{4},2;\frac {5}{4};1-\frac {b x^4}{a}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a - b*x^4)^(3/4)),x]

[Out]

-((b*(a - b*x^4)^(1/4)*Hypergeometric2F1[1/4, 2, 5/4, 1 - (b*x^4)/a])/a^2)

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fricas [B]  time = 0.73, size = 199, normalized size = 2.46 \[ \frac {12 \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{5} b \left (\frac {b^{4}}{a^{7}}\right )^{\frac {3}{4}} - \sqrt {a^{4} \sqrt {\frac {b^{4}}{a^{7}}} + \sqrt {-b x^{4} + a} b^{2}} a^{5} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {3}{4}}}{b^{4}}\right ) - 3 \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (3 \, a^{2} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b\right ) + 3 \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-3 \, a^{2} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b\right ) - 4 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{16 \, a x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/16*(12*a*x^4*(b^4/a^7)^(1/4)*arctan(-((-b*x^4 + a)^(1/4)*a^5*b*(b^4/a^7)^(3/4) - sqrt(a^4*sqrt(b^4/a^7) + sq
rt(-b*x^4 + a)*b^2)*a^5*(b^4/a^7)^(3/4))/b^4) - 3*a*x^4*(b^4/a^7)^(1/4)*log(3*a^2*(b^4/a^7)^(1/4) + 3*(-b*x^4
+ a)^(1/4)*b) + 3*a*x^4*(b^4/a^7)^(1/4)*log(-3*a^2*(b^4/a^7)^(1/4) + 3*(-b*x^4 + a)^(1/4)*b) - 4*(-b*x^4 + a)^
(1/4))/(a*x^4)

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giac [B]  time = 0.17, size = 228, normalized size = 2.81 \[ -\frac {\frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{a^{2}} + \frac {3 \, \sqrt {2} b^{2} \log \left (-\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {8 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b}{a x^{4}}}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

-1/32*(6*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2
+ 6*sqrt(2)*(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 +
3*sqrt(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/a^2 + 3*sqrt
(2)*b^2*log(-sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/((-a)^(3/4)*a) + 8*(-b*x^4 +
a)^(1/4)*b/(a*x^4))/b

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {3}{4}} x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-b*x^4+a)^(3/4),x)

[Out]

int(1/x^5/(-b*x^4+a)^(3/4),x)

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maxima [A]  time = 2.42, size = 98, normalized size = 1.21 \[ -\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} b}{4 \, {\left ({\left (b x^{4} - a\right )} a + a^{2}\right )}} - \frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{16 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.30, size = 61, normalized size = 0.75 \[ -\frac {{\left (a-b\,x^4\right )}^{1/4}}{4\,a\,x^4}-\frac {3\,b\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{7/4}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{7/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a - b*x^4)^(3/4)),x)

[Out]

- (a - b*x^4)^(1/4)/(4*a*x^4) - (3*b*atan((a - b*x^4)^(1/4)/a^(1/4)))/(8*a^(7/4)) - (3*b*atanh((a - b*x^4)^(1/
4)/a^(1/4)))/(8*a^(7/4))

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sympy [C]  time = 2.31, size = 39, normalized size = 0.48 \[ \frac {e^{\frac {i \pi }{4}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{7} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-b*x**4+a)**(3/4),x)

[Out]

exp(I*pi/4)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a/(b*x**4))/(4*b**(3/4)*x**7*gamma(11/4))

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