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

Optimal. Leaf size=92 \[ -\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{64 b^{11/4}}-\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{64 b^{11/4}}+\frac {\sqrt [4]{a x^4+b} \left (7 a x^4-4 b\right )}{32 b^2 x^8} \]

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Rubi [A]  time = 0.08, antiderivative size = 104, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 51, 63, 212, 206, 203} \begin {gather*} -\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{64 b^{11/4}}-\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{64 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]

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.40 \begin {gather*} -\frac {a^2 \sqrt [4]{a x^4+b} \, _2F_1\left (\frac {1}{4},3;\frac {5}{4};\frac {a x^4}{b}+1\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 = 0.08, size = 92, normalized size = 1.00 \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 {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{64 b^{11/4}}-\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{64 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 + a*x^4)^(1/4)/b^(1/4)])/(64*b^(11/4)) -
(21*a^2*ArcTanh[(b + a*x^4)^(1/4)/b^(1/4)])/(64*b^(11/4))

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fricas [B]  time = 0.47, size = 216, normalized size = 2.35 \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^1
1) + 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 [B]  time = 0.18, size = 244, normalized size = 2.65 \begin {gather*} \frac {\frac {42 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {3}{4}} b^{2}} + \frac {42 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {3}{4}} b^{2}} + \frac {21 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{\left (-b\right )^{\frac {3}{4}} b^{2}} + \frac {21 \, \sqrt {2} a^{3} \left (-b\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{b^{3}} + \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)^(3/4)*b^
2) + 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)^(3/4)*b^2
) + 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)^(3/4)*b^2) + 2
1*sqrt(2)*a^3*(-b)^(1/4)*log(-sqrt(2)*(a*x^4 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^4 + b) + sqrt(-b))/b^3 + 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 = 0.55, size = 132, normalized size = 1.43 \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 \, a^{2} \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} - \frac {a^{2} \log \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, 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
/128*(2*a^2*arctan((a*x^4 + b)^(1/4)/b^(1/4))/b^(3/4) - a^2*log(((a*x^4 + b)^(1/4) - b^(1/4))/((a*x^4 + b)^(1/
4) + b^(1/4)))/b^(3/4))/b^2

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mupad [B]  time = 1.19, size = 82, normalized size = 0.89 \begin {gather*} \frac {7\,{\left (a\,x^4+b\right )}^{5/4}}{32\,b^2\,x^8}-\frac {11\,{\left (a\,x^4+b\right )}^{1/4}}{32\,b\,x^8}-\frac {21\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{64\,b^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}\,1{}\mathrm {i}}{b^{1/4}}\right )\,21{}\mathrm {i}}{64\,b^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(b + a*x^4)^(3/4)),x)

[Out]

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

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sympy [C]  time = 1.60, size = 39, normalized size = 0.42 \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^{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(I*pi)/(a*x**4))/(4*a**(3/4)*x**11*gamma(15/4))

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