3.1241 \(\int \frac {1}{x^9 (a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {21 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {21 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {7 b \sqrt [4]{a-b x^4}}{32 a^2 x^4}-\frac {\sqrt [4]{a-b x^4}}{8 a x^8} \]

[Out]

-1/8*(-b*x^4+a)^(1/4)/a/x^8-7/32*b*(-b*x^4+a)^(1/4)/a^2/x^4-21/64*b^2*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(11/4
)-21/64*b^2*arctanh((-b*x^4+a)^(1/4)/a^(1/4))/a^(11/4)

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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, 212, 206, 203} \[ -\frac {21 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {21 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {7 b \sqrt [4]{a-b x^4}}{32 a^2 x^4}-\frac {\sqrt [4]{a-b x^4}}{8 a x^8} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(a - b*x^4)^(3/4)),x]

[Out]

-(a - b*x^4)^(1/4)/(8*a*x^8) - (7*b*(a - b*x^4)^(1/4))/(32*a^2*x^4) - (21*b^2*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)
])/(64*a^(11/4)) - (21*b^2*ArcTanh[(a - b*x^4)^(1/4)/a^(1/4)])/(64*a^(11/4))

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^9 \left (a-b x^4\right )^{3/4}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x^3 (a-b x)^{3/4}} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 a x^8}+\frac {(7 b) \operatorname {Subst}\left (\int \frac {1}{x^2 (a-b x)^{3/4}} \, dx,x,x^4\right )}{32 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 a x^8}-\frac {7 b \sqrt [4]{a-b x^4}}{32 a^2 x^4}+\frac {\left (21 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a-b x)^{3/4}} \, dx,x,x^4\right )}{128 a^2}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 a x^8}-\frac {7 b \sqrt [4]{a-b x^4}}{32 a^2 x^4}-\frac {(21 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^2}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 a x^8}-\frac {7 b \sqrt [4]{a-b x^4}}{32 a^2 x^4}-\frac {\left (21 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{64 a^{5/2}}-\frac {\left (21 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a-b x^4}\right )}{64 a^{5/2}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{8 a x^8}-\frac {7 b \sqrt [4]{a-b x^4}}{32 a^2 x^4}-\frac {21 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {21 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 39, normalized size = 0.36 \[ -\frac {b^2 \sqrt [4]{a-b x^4} \, _2F_1\left (\frac {1}{4},3;\frac {5}{4};1-\frac {b x^4}{a}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.70, size = 221, normalized size = 2.05 \[ \frac {84 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{8} b^{2} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {3}{4}} - \sqrt {a^{6} \sqrt {\frac {b^{8}}{a^{11}}} + \sqrt {-b x^{4} + a} b^{4}} a^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {3}{4}}}{b^{8}}\right ) - 21 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \log \left (21 \, a^{3} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{2}\right ) + 21 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-21 \, a^{3} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{2}\right ) - 4 \, {\left (7 \, b x^{4} + 4 \, a\right )} {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, a^{2} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/128*(84*a^2*x^8*(b^8/a^11)^(1/4)*arctan(-((-b*x^4 + a)^(1/4)*a^8*b^2*(b^8/a^11)^(3/4) - sqrt(a^6*sqrt(b^8/a^
11) + sqrt(-b*x^4 + a)*b^4)*a^8*(b^8/a^11)^(3/4))/b^8) - 21*a^2*x^8*(b^8/a^11)^(1/4)*log(21*a^3*(b^8/a^11)^(1/
4) + 21*(-b*x^4 + a)^(1/4)*b^2) + 21*a^2*x^8*(b^8/a^11)^(1/4)*log(-21*a^3*(b^8/a^11)^(1/4) + 21*(-b*x^4 + a)^(
1/4)*b^2) - 4*(7*b*x^4 + 4*a)*(-b*x^4 + a)^(1/4))/(a^2*x^8)

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giac [B]  time = 0.17, size = 252, normalized size = 2.33 \[ -\frac {\frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {21 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{a^{3}} + \frac {21 \, \sqrt {2} b^{3} \log \left (-\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a^{2}} - \frac {8 \, {\left (7 \, {\left (-b x^{4} + a\right )}^{\frac {5}{4}} b^{3} - 11 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a b^{3}\right )}}{a^{2} b^{2} x^{8}}}{256 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

-1/256*(42*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
^3 + 42*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^3
+ 21*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^3 + 21
*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^2) - 8*(7
*(-b*x^4 + a)^(5/4)*b^3 - 11*(-b*x^4 + a)^(1/4)*a*b^3)/(a^2*b^2*x^8))/b

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {3}{4}} x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^9/(-b*x^4+a)^(3/4),x)

[Out]

int(1/x^9/(-b*x^4+a)^(3/4),x)

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maxima [A]  time = 2.38, size = 141, normalized size = 1.31 \[ \frac {7 \, {\left (-b x^{4} + a\right )}^{\frac {5}{4}} b^{2} - 11 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} - a\right )}^{2} a^{2} + 2 \, {\left (b x^{4} - a\right )} a^{3} + a^{4}\right )}} - \frac {21 \, {\left (\frac {2 \, b^{2} \arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b^{2} \log \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{128 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

1/32*(7*(-b*x^4 + a)^(5/4)*b^2 - 11*(-b*x^4 + a)^(1/4)*a*b^2)/((b*x^4 - a)^2*a^2 + 2*(b*x^4 - a)*a^3 + a^4) -
21/128*(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^2

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mupad [B]  time = 1.39, size = 86, normalized size = 0.80 \[ \frac {7\,{\left (a-b\,x^4\right )}^{5/4}}{32\,a^2\,x^8}-\frac {11\,{\left (a-b\,x^4\right )}^{1/4}}{32\,a\,x^8}-\frac {21\,b^2\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{11/4}}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,21{}\mathrm {i}}{64\,a^{11/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(a - b*x^4)^(3/4)),x)

[Out]

(b^2*atan(((a - b*x^4)^(1/4)*1i)/a^(1/4))*21i)/(64*a^(11/4)) - (21*b^2*atan((a - b*x^4)^(1/4)/a^(1/4)))/(64*a^
(11/4)) - (11*(a - b*x^4)^(1/4))/(32*a*x^8) + (7*(a - b*x^4)^(5/4))/(32*a^2*x^8)

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sympy [C]  time = 3.26, size = 42, normalized size = 0.39 \[ - \frac {e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{11} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**9/(-b*x**4+a)**(3/4),x)

[Out]

-exp(-3*I*pi/4)*gamma(11/4)*hyper((3/4, 11/4), (15/4,), a/(b*x**4))/(4*b**(3/4)*x**11*gamma(15/4))

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