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

Optimal. Leaf size=167 \[ -\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}{\sqrt {a x^4-b}-\sqrt {b}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^4-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt [4]{a x^4-b}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {\sqrt [4]{a x^4-b} \left (7 a x^4+4 b\right )}{32 b^2 x^8} \]

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Rubi [A]  time = 0.25, antiderivative size = 260, normalized size of antiderivative = 1.56, number of steps used = 13, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {266, 51, 63, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {21 a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}+\sqrt {a x^4-b}+\sqrt {b}\right )}{128 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}+\sqrt {a x^4-b}+\sqrt {b}\right )}{128 \sqrt {2} b^{11/4}}-\frac {21 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{64 \sqrt {2} b^{11/4}}+\frac {7 a \sqrt [4]{a x^4-b}}{32 b^2 x^4}+\frac {\sqrt [4]{a x^4-b}}{8 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(-b + a*x^4)^(3/4)),x]

[Out]

(-b + a*x^4)^(1/4)/(8*b*x^8) + (7*a*(-b + a*x^4)^(1/4))/(32*b^2*x^4) - (21*a^2*ArcTan[1 - (Sqrt[2]*(-b + a*x^4
)^(1/4))/b^(1/4)])/(64*Sqrt[2]*b^(11/4)) + (21*a^2*ArcTan[1 + (Sqrt[2]*(-b + a*x^4)^(1/4))/b^(1/4)])/(64*Sqrt[
2]*b^(11/4)) - (21*a^2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4) + Sqrt[-b + a*x^4]])/(128*Sqrt[2]*b^(1
1/4)) + (21*a^2*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4) + Sqrt[-b + a*x^4]])/(128*Sqrt[2]*b^(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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.23 \begin {gather*} \frac {a^2 \sqrt [4]{a x^4-b} \, _2F_1\left (\frac {1}{4},3;\frac {5}{4};1-\frac {a x^4}{b}\right )}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 166, normalized size = 0.99 \begin {gather*} \frac {\sqrt [4]{-b+a x^4} \left (4 b+7 a x^4\right )}{32 b^2 x^8}+\frac {21 a^2 \tan ^{-1}\left (\frac {-\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^4}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}{\sqrt {b}+\sqrt {-b+a x^4}}\right )}{64 \sqrt {2} b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 234, normalized size = 1.40 \begin {gather*} \frac {84 \, b^{2} x^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}} a^{2} b^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}} - \sqrt {b^{6} \sqrt {-\frac {a^{8}}{b^{11}}} + \sqrt {a x^{4} - b} a^{4}} b^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}}}{a^{8}}\right ) + 21 \, b^{2} x^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (21 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 \, b^{2} x^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-21 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 4 \, {\left (7 \, a x^{4} + 4 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{128 \, b^{2} x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 224, normalized size = 1.34 \begin {gather*} \frac {\frac {42 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {42 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {21 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} - \frac {21 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} + \frac {8 \, {\left (7 \, {\left (a x^{4} - b\right )}^{\frac {5}{4}} a^{3} + 11 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{8}}}{256 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{9} \left (a \,x^{4}-b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 250, normalized size = 1.50 \begin {gather*} \frac {7 \, {\left (a x^{4} - b\right )}^{\frac {5}{4}} a^{2} + 11 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a^{2} b}{32 \, {\left ({\left (a x^{4} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{4} - b\right )} b^{3} + b^{4}\right )}} + \frac {21 \, {\left (\frac {2 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{256 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(7*(a*x^4 - b)^(5/4)*a^2 + 11*(a*x^4 - b)^(1/4)*a^2*b)/((a*x^4 - b)^2*b^2 + 2*(a*x^4 - b)*b^3 + b^4) + 21
/256*(2*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^4 - b)^(1/4))/b^(1/4))/b^(3/4) + 2*sqrt(2)*a^
2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^4 - b)^(1/4))/b^(1/4))/b^(3/4) + sqrt(2)*a^2*log(sqrt(2)*(a*x^
4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(3/4) - sqrt(2)*a^2*log(-sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4)
+ sqrt(a*x^4 - b) + sqrt(b))/b^(3/4))/b^2

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mupad [B]  time = 1.39, size = 98, normalized size = 0.59 \begin {gather*} \frac {11\,{\left (a\,x^4-b\right )}^{1/4}}{32\,b\,x^8}+\frac {7\,{\left (a\,x^4-b\right )}^{5/4}}{32\,b^2\,x^8}-\frac {21\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^4-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{64\,{\left (-b\right )}^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^4-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,21{}\mathrm {i}}{64\,{\left (-b\right )}^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(a*x^4 - b)^(3/4)),x)

[Out]

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

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sympy [C]  time = 1.65, size = 41, normalized size = 0.25 \begin {gather*} - \frac {\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 {b e^{2 i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{4}} x^{11} \Gamma \left (\frac {15}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*exp_polar(2*I*pi)/(a*x**4))/(4*a**(3/4)*x**11*gamma(15/4))

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