\(\int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx\) [1087]

Optimal result

Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=-\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \]

[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)-5/6
4*b^2*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 44, 65, 304, 209, 212} \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8} \]

[In]

Int[1/(x^9*(a + b*x^4)^(1/4)),x]

[Out]

-1/8*(a + b*x^4)^(3/4)/(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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 209

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

Rule 212

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

Rule 272

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 304

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*} \text {integral}& = \frac {1}{4} \text {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) \text {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 ) \text {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) \text {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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (-4 a+5 b x^4\right )}{32 a^2 x^8}+\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \]

[In]

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

[Out]

((a + b*x^4)^(3/4)*(-4*a + 5*b*x^4))/(32*a^2*x^8) + (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))

Maple [A] (verified)

Time = 4.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(\frac {10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b^{2} x^{8}-5 \ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b^{2} x^{8}+20 b \,x^{4} a^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}-16 a^{\frac {5}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{128 a^{\frac {9}{4}} x^{8}}\)	\(108\)

[In]

int(1/x^9/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/128/a^(9/4)*(10*arctan((b*x^4+a)^(1/4)/a^(1/4))*b^2*x^8-5*ln((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^
(1/4)))*b^2*x^8+20*b*x^4*a^(1/4)*(b*x^4+a)^(3/4)-16*a^(5/4)*(b*x^4+a)^(3/4))/x^8

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.10 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=-\frac {5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) + 5 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 4 \, {\left (5 \, b x^{4} - 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, a^{2} x^{8}} \]

[In]

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

[Out]

-1/128*(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*I*a^2*x^8*(b^8/
a^9)^(1/4)*log(125*I*a^7*(b^8/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^6) + 5*I*a^2*x^8*(b^8/a^9)^(1/4)*log(-125*I
*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) + 1
25*(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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.63 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=- \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{9} \Gamma \left (\frac {13}{4}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a^{2}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{2} - 9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{4} + a\right )} a^{3} + a^{4}\right )}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (80) = 160\).

Time = 0.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.35 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {\frac {10 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {10 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{3}} + \frac {5 \, \sqrt {2} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {8 \, {\left (5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{3} - 9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{3}\right )}}{a^{2} b^{2} x^{8}}}{256 \, b} \]

[In]

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

[Out]

1/256*(10*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a^
2) + 10*sqrt(2)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-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 + 5*sq
rt(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) + 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

Mupad [B] (verification not implemented)

Time = 5.95 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{9/4}}-\frac {9\,{\left (b\,x^4+a\right )}^{3/4}}{32\,a\,x^8}+\frac {5\,{\left (b\,x^4+a\right )}^{7/4}}{32\,a^2\,x^8}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{64\,a^{9/4}} \]

[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)