3.1211 \(\int \frac {1}{x^9 \sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=108 \[ \frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b \left (a-b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8} \]

[Out]

-1/8*(-b*x^4+a)^(3/4)/a/x^8-5/32*b*(-b*x^4+a)^(3/4)/a^2/x^4+5/64*b^2*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)-
5/64*b^2*arctanh((-b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)

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Rubi [A]  time = 0.06, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {266, 51, 63, 298, 203, 206} \[ \frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b \left (a-b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(a - b*x^4)^(1/4)),x]

[Out]

-(a - b*x^4)^(3/4)/(8*a*x^8) - (5*b*(a - b*x^4)^(3/4))/(32*a^2*x^4) + (5*b^2*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)]
)/(64*a^(9/4)) - (5*b^2*ArcTanh[(a - b*x^4)^(1/4)/a^(1/4)])/(64*a^(9/4))

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 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]]

Rule 298

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^9 \sqrt [4]{a-b x^4}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt [4]{a-b x}} \, dx,x,x^4\right )\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8}+\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a-b x}} \, dx,x,x^4\right )}{32 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8}-\frac {5 b \left (a-b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{a-b x}} \, dx,x,x^4\right )}{128 a^2}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8}-\frac {5 b \left (a-b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {x^2}{\frac {a}{b}-\frac {x^4}{b}} \, dx,x,\sqrt [4]{a-b x^4}\right )}{32 a^2}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8}-\frac {5 b \left (a-b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{64 a^2}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{64 a^2}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{8 a x^8}-\frac {5 b \left (a-b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.38 \[ -\frac {b^2 \left (a-b x^4\right )^{3/4} \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};1-\frac {b x^4}{a}\right )}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^9*(a - b*x^4)^(1/4)),x]

[Out]

-1/3*(b^2*(a - b*x^4)^(3/4)*Hypergeometric2F1[3/4, 3, 7/4, 1 - (b*x^4)/a])/a^3

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fricas [B]  time = 0.57, size = 224, normalized size = 2.07 \[ -\frac {20 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b^{6} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} - \sqrt {a^{5} b^{8} \sqrt {\frac {b^{8}}{a^{9}}} + \sqrt {-b x^{4} + a} b^{12}} a^{2} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}}}{b^{8}}\right ) + 5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) + 4 \, {\left (5 \, b x^{4} + 4 \, a\right )} {\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, a^{2} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(20*a^2*x^8*(b^8/a^9)^(1/4)*arctan(-((-b*x^4 + a)^(1/4)*a^2*b^6*(b^8/a^9)^(1/4) - sqrt(a^5*b^8*sqrt(b^8
/a^9) + sqrt(-b*x^4 + a)*b^12)*a^2*(b^8/a^9)^(1/4))/b^8) + 5*a^2*x^8*(b^8/a^9)^(1/4)*log(125*a^7*(b^8/a^9)^(3/
4) + 125*(-b*x^4 + a)^(1/4)*b^6) - 5*a^2*x^8*(b^8/a^9)^(1/4)*log(-125*a^7*(b^8/a^9)^(3/4) + 125*(-b*x^4 + a)^(
1/4)*b^6) + 4*(5*b*x^4 + 4*a)*(-b*x^4 + a)^(3/4))/(a^2*x^8)

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giac [B]  time = 0.18, size = 252, normalized size = 2.33 \[ -\frac {\frac {10 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {10 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {5 \, \sqrt {2} b^{3} \log \left (\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \log \left (-\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{a^{3}} - \frac {8 \, {\left (5 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} b^{3} - 9 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a b^{3}\right )}}{a^{2} b^{2} x^{8}}}{256 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(10*sqrt(2)*(-a)^(3/4)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a
^3 + 10*sqrt(2)*(-a)^(3/4)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^3
+ 5*sqrt(2)*b^3*log(sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^2) + 5
*sqrt(2)*(-a)^(3/4)*b^3*log(-sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/a^3 - 8*(5*(
-b*x^4 + a)^(7/4)*b^3 - 9*(-b*x^4 + a)^(3/4)*a*b^3)/(a^2*b^2*x^8))/b

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.41, size = 137, normalized size = 1.27 \[ \frac {5 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a^{2}} + \frac {5 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} b^{2} - 9 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} - a\right )}^{2} a^{2} + 2 \, {\left (b x^{4} - a\right )} a^{3} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/128*b^2*(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^2 + 1/32*(5*(-b*x^4 + a)^(7/4)*b^2 - 9*(-b*x^4 + a)^(3/4)*a*b^2)/((b*x^4 - a)^2*a^2
+ 2*(b*x^4 - a)*a^3 + a^4)

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mupad [B]  time = 1.39, size = 86, normalized size = 0.80 \[ \frac {5\,b^2\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{9/4}}-\frac {9\,{\left (a-b\,x^4\right )}^{3/4}}{32\,a\,x^8}+\frac {5\,{\left (a-b\,x^4\right )}^{7/4}}{32\,a^2\,x^8}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{64\,a^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(a - b*x^4)^(1/4)),x)

[Out]

(5*b^2*atan((a - b*x^4)^(1/4)/a^(1/4)))/(64*a^(9/4)) + (b^2*atan(((a - b*x^4)^(1/4)*1i)/a^(1/4))*5i)/(64*a^(9/
4)) - (9*(a - b*x^4)^(3/4))/(32*a*x^8) + (5*(a - b*x^4)^(7/4))/(32*a^2*x^8)

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sympy [C]  time = 2.12, size = 41, normalized size = 0.38 \[ - \frac {e^{- \frac {i \pi }{4}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{9} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(-I*pi/4)*gamma(9/4)*hyper((1/4, 9/4), (13/4,), a/(b*x**4))/(4*b**(1/4)*x**9*gamma(13/4))

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