3.1182 \(\int \frac {\sqrt [4]{a-b x^4}}{x^9} \, dx\)

Optimal. Leaf size=105 \[ \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{7/4}}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{7/4}}+\frac {b \sqrt [4]{a-b x^4}}{32 a x^4}-\frac {\sqrt [4]{a-b x^4}}{8 x^8} \]

[Out]

-1/8*(-b*x^4+a)^(1/4)/x^8+1/32*b*(-b*x^4+a)^(1/4)/a/x^4+3/64*b^2*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(7/4)+3/64
*b^2*arctanh((-b*x^4+a)^(1/4)/a^(1/4))/a^(7/4)

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Rubi [A]  time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {266, 47, 51, 63, 212, 206, 203} \[ \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{7/4}}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{7/4}}+\frac {b \sqrt [4]{a-b x^4}}{32 a x^4}-\frac {\sqrt [4]{a-b x^4}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4)/x^9,x]

[Out]

-(a - b*x^4)^(1/4)/(8*x^8) + (b*(a - b*x^4)^(1/4))/(32*a*x^4) + (3*b^2*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)])/(64*
a^(7/4)) + (3*b^2*ArcTanh[(a - b*x^4)^(1/4)/a^(1/4)])/(64*a^(7/4))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 {\sqrt [4]{a-b x^4}}{x^9} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt [4]{a-b x}}{x^3} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 x^8}-\frac {1}{32} b \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}}{8 x^8}+\frac {b \sqrt [4]{a-b x^4}}{32 a x^4}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a-b x)^{3/4}} \, dx,x,x^4\right )}{128 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 x^8}+\frac {b \sqrt [4]{a-b x^4}}{32 a x^4}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {x^4}{b}} \, dx,x,\sqrt [4]{a-b x^4}\right )}{32 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 x^8}+\frac {b \sqrt [4]{a-b x^4}}{32 a x^4}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{64 a^{3/2}}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{64 a^{3/2}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 x^8}+\frac {b \sqrt [4]{a-b x^4}}{32 a x^4}+\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{7/4}}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.39 \[ -\frac {b^2 \left (a-b x^4\right )^{5/4} \, _2F_1\left (\frac {5}{4},3;\frac {9}{4};1-\frac {b x^4}{a}\right )}{5 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4)/x^9,x]

[Out]

-1/5*(b^2*(a - b*x^4)^(5/4)*Hypergeometric2F1[5/4, 3, 9/4, 1 - (b*x^4)/a])/a^3

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fricas [B]  time = 1.00, size = 214, normalized size = 2.04 \[ -\frac {12 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x^{8} \arctan \left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{5} b^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {3}{4}} - \sqrt {\sqrt {-b x^{4} + a} b^{4} + a^{4} \sqrt {\frac {b^{8}}{a^{7}}}} a^{5} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {3}{4}}}{b^{8}}\right ) - 3 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x^{8} \log \left (3 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{2} + 3 \, a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}}\right ) + 3 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x^{8} \log \left (3 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{2} - 3 \, a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (b x^{4} - 4 \, a\right )} {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, a x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^9,x, algorithm="fricas")

[Out]

-1/128*(12*a*(b^8/a^7)^(1/4)*x^8*arctan(-((-b*x^4 + a)^(1/4)*a^5*b^2*(b^8/a^7)^(3/4) - sqrt(sqrt(-b*x^4 + a)*b
^4 + a^4*sqrt(b^8/a^7))*a^5*(b^8/a^7)^(3/4))/b^8) - 3*a*(b^8/a^7)^(1/4)*x^8*log(3*(-b*x^4 + a)^(1/4)*b^2 + 3*a
^2*(b^8/a^7)^(1/4)) + 3*a*(b^8/a^7)^(1/4)*x^8*log(3*(-b*x^4 + a)^(1/4)*b^2 - 3*a^2*(b^8/a^7)^(1/4)) - 4*(b*x^4
- 4*a)*(-b*x^4 + a)^(1/4))/(a*x^8)

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giac [B]  time = 0.21, size = 251, normalized size = 2.39 \[ \frac {\frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{a^{2}} + \frac {3 \, \sqrt {2} b^{3} \log \left (-\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a} - \frac {8 \, {\left ({\left (-b x^{4} + a\right )}^{\frac {5}{4}} b^{3} + 3 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a b^{3}\right )}}{a b^{2} x^{8}}}{256 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^9,x, algorithm="giac")

[Out]

1/256*(6*sqrt(2)*(-a)^(1/4)*b^3*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^3*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^3*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^3*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)^(5/4)*b^3 + 3*(-b*x^4 + a)^(1/4)*a*b^3)/(a*b^2*x^8))/b

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^4+a)^(1/4)/x^9,x)

[Out]

int((-b*x^4+a)^(1/4)/x^9,x)

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maxima [A]  time = 2.97, size = 138, normalized size = 1.31 \[ -\frac {{\left (-b x^{4} + a\right )}^{\frac {5}{4}} b^{2} + 3 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} - a\right )}^{2} a + 2 \, {\left (b x^{4} - a\right )} a^{2} + a^{3}\right )}} + \frac {3 \, {\left (\frac {2 \, b^{2} \arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b^{2} \log \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{128 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^9,x, algorithm="maxima")

[Out]

-1/32*((-b*x^4 + a)^(5/4)*b^2 + 3*(-b*x^4 + a)^(1/4)*a*b^2)/((b*x^4 - a)^2*a + 2*(b*x^4 - a)*a^2 + a^3) + 3/12
8*(2*b^2*arctan((-b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) - b^2*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 = 1.45, size = 83, normalized size = 0.79 \[ \frac {3\,b^2\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{7/4}}-\frac {3\,{\left (a-b\,x^4\right )}^{1/4}}{32\,x^8}-\frac {{\left (a-b\,x^4\right )}^{5/4}}{32\,a\,x^8}-\frac {b^2\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,3{}\mathrm {i}}{64\,a^{7/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^4)^(1/4)/x^9,x)

[Out]

(3*b^2*atan((a - b*x^4)^(1/4)/a^(1/4)))/(64*a^(7/4)) - (3*(a - b*x^4)^(1/4))/(32*x^8) - (b^2*atan(((a - b*x^4)
^(1/4)*1i)/a^(1/4))*3i)/(64*a^(7/4)) - (a - b*x^4)^(5/4)/(32*a*x^8)

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sympy [C]  time = 2.09, size = 42, normalized size = 0.40 \[ - \frac {\sqrt [4]{b} e^{\frac {i \pi }{4}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 x^{7} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**4+a)**(1/4)/x**9,x)

[Out]

-b**(1/4)*exp(I*pi/4)*gamma(7/4)*hyper((-1/4, 7/4), (11/4,), a/(b*x**4))/(4*x**7*gamma(11/4))

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