\(\int \frac {\sqrt [4]{-b+a x^4} (-8 b+a x^8)}{x^{10} (b+a x^4)} \, dx\) [2134]

Optimal result

Integrand size = 35, antiderivative size = 155 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\frac {\sqrt [4]{-b+a x^4} \left (8 b^2-16 a b x^4+80 a^2 x^8-9 a b x^8\right )}{9 b^2 x^9}+\frac {\left (8 a^{9/4}-a^{5/4} b\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2^{3/4} b^2}-\frac {\left (8 a^{9/4}-a^{5/4} b\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2^{3/4} b^2} \]

[Out]

1/9*(a*x^4-b)^(1/4)*(80*a^2*x^8-9*a*b*x^8-16*a*b*x^4+8*b^2)/b^2/x^9+1/2*(8*a^(9/4)-a^(5/4)*b)*arctan(2^(1/4)*a
^(1/4)*x/(a*x^4-b)^(1/4))*2^(1/4)/b^2-1/2*(8*a^(9/4)-a^(5/4)*b)*arctanh(2^(1/4)*a^(1/4)*x/(a*x^4-b)^(1/4))*2^(
1/4)/b^2

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.45 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6857, 277, 270, 283, 338, 304, 209, 212, 525, 524} \[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\frac {a^{5/4} (8 a-b) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 b^2}-\frac {a^{5/4} (8 a-b) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 b^2}+\frac {a^2 x^3 (8 a-b) \sqrt [4]{a x^4-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}+\frac {a (8 a-b) \sqrt [4]{a x^4-b}}{b^2 x}+\frac {8 a \left (a x^4-b\right )^{5/4}}{9 b^2 x^5}-\frac {8 \left (a x^4-b\right )^{5/4}}{9 b x^9} \]

[In]

Int[((-b + a*x^4)^(1/4)*(-8*b + a*x^8))/(x^10*(b + a*x^4)),x]

[Out]

(a*(8*a - b)*(-b + a*x^4)^(1/4))/(b^2*x) - (8*(-b + a*x^4)^(5/4))/(9*b*x^9) + (8*a*(-b + a*x^4)^(5/4))/(9*b^2*
x^5) + (a^2*(8*a - b)*x^3*(-b + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, -((a*x^4)/b), (a*x^4)/b])/(3*b^3*(1 -
(a*x^4)/b)^(1/4)) + (a^(5/4)*(8*a - b)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2*b^2) - (a^(5/4)*(8*a - b)*A
rcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2*b^2)

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8 \sqrt [4]{-b+a x^4}}{x^{10}}+\frac {8 a \sqrt [4]{-b+a x^4}}{b x^6}-\frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x^2}+\frac {a^2 (8 a-b) x^2 \sqrt [4]{-b+a x^4}}{b^2 \left (b+a x^4\right )}\right ) \, dx \\ & = -\left (8 \int \frac {\sqrt [4]{-b+a x^4}}{x^{10}} \, dx\right )-\frac {(a (8 a-b)) \int \frac {\sqrt [4]{-b+a x^4}}{x^2} \, dx}{b^2}+\frac {\left (a^2 (8 a-b)\right ) \int \frac {x^2 \sqrt [4]{-b+a x^4}}{b+a x^4} \, dx}{b^2}+\frac {(8 a) \int \frac {\sqrt [4]{-b+a x^4}}{x^6} \, dx}{b} \\ & = \frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{5 b^2 x^5}-\frac {\left (a^2 (8 a-b)\right ) \int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx}{b^2}-\frac {(32 a) \int \frac {\sqrt [4]{-b+a x^4}}{x^6} \, dx}{9 b}+\frac {\left (a^2 (8 a-b) \sqrt [4]{-b+a x^4}\right ) \int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{b+a x^4} \, dx}{b^2 \sqrt [4]{1-\frac {a x^4}{b}}} \\ & = \frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{9 b^2 x^5}+\frac {a^2 (8 a-b) x^3 \sqrt [4]{-b+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}-\frac {\left (a^2 (8 a-b)\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{b^2} \\ & = \frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{9 b^2 x^5}+\frac {a^2 (8 a-b) x^3 \sqrt [4]{-b+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}-\frac {\left (a^{3/2} (8 a-b)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2}+\frac {\left (a^{3/2} (8 a-b)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2} \\ & = \frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{9 b^2 x^5}+\frac {a^2 (8 a-b) x^3 \sqrt [4]{-b+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}+\frac {a^{5/4} (8 a-b) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2}-\frac {a^{5/4} (8 a-b) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\frac {\sqrt [4]{-b+a x^4} \left (8 b^2-16 a b x^4+80 a^2 x^8-9 a b x^8\right )}{9 b^2 x^9}+\frac {a^{5/4} (8 a-b) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2^{3/4} b^2}-\frac {a^{5/4} (8 a-b) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2^{3/4} b^2} \]

[In]

Integrate[((-b + a*x^4)^(1/4)*(-8*b + a*x^8))/(x^10*(b + a*x^4)),x]

[Out]

((-b + a*x^4)^(1/4)*(8*b^2 - 16*a*b*x^4 + 80*a^2*x^8 - 9*a*b*x^8))/(9*b^2*x^9) + (a^(5/4)*(8*a - b)*ArcTan[(2^
(1/4)*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2^(3/4)*b^2) - (a^(5/4)*(8*a - b)*ArcTanh[(2^(1/4)*a^(1/4)*x)/(-b + a*x
^4)^(1/4)])/(2^(3/4)*b^2)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {9 x^{9} 2^{\frac {1}{4}} \left (a^{\frac {5}{4}} b -8 a^{\frac {9}{4}}\right ) \ln \left (\frac {-x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )+18 x^{9} 2^{\frac {1}{4}} \left (a^{\frac {5}{4}} b -8 a^{\frac {9}{4}}\right ) \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right )+320 \left (a \,x^{4}-b \right )^{\frac {1}{4}} \left (a \left (a -\frac {9 b}{80}\right ) x^{8}-\frac {a b \,x^{4}}{5}+\frac {b^{2}}{10}\right )}{36 b^{2} x^{9}}\)	\(156\)

[In]

int((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x,method=_RETURNVERBOSE)

[Out]

1/36*(9*x^9*2^(1/4)*(a^(5/4)*b-8*a^(9/4))*ln((-x*2^(1/4)*a^(1/4)-(a*x^4-b)^(1/4))/(x*2^(1/4)*a^(1/4)-(a*x^4-b)
^(1/4)))+18*x^9*2^(1/4)*(a^(5/4)*b-8*a^(9/4))*arctan(1/2*2^(3/4)/a^(1/4)/x*(a*x^4-b)^(1/4))+320*(a*x^4-b)^(1/4
)*(a*(a-9/80*b)*x^8-1/5*a*b*x^4+1/10*b^2))/b^2/x^9

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\int \frac {\sqrt [4]{a x^{4} - b} \left (a x^{8} - 8 b\right )}{x^{10} \left (a x^{4} + b\right )}\, dx \]

[In]

integrate((a*x**4-b)**(1/4)*(a*x**8-8*b)/x**10/(a*x**4+b),x)

[Out]

Integral((a*x**4 - b)**(1/4)*(a*x**8 - 8*b)/(x**10*(a*x**4 + b)), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{8} - 8 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{{\left (a x^{4} + b\right )} x^{10}} \,d x } \]

[In]

integrate((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x, algorithm="maxima")

[Out]

integrate((a*x^8 - 8*b)*(a*x^4 - b)^(1/4)/((a*x^4 + b)*x^10), x)

Giac [F]

\[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{8} - 8 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{{\left (a x^{4} + b\right )} x^{10}} \,d x } \]

[In]

integrate((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x, algorithm="giac")

[Out]

integrate((a*x^8 - 8*b)*(a*x^4 - b)^(1/4)/((a*x^4 + b)*x^10), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx=\int -\frac {{\left (a\,x^4-b\right )}^{1/4}\,\left (8\,b-a\,x^8\right )}{x^{10}\,\left (a\,x^4+b\right )} \,d x \]

[In]

int(-((a*x^4 - b)^(1/4)*(8*b - a*x^8))/(x^10*(b + a*x^4)),x)

[Out]

int(-((a*x^4 - b)^(1/4)*(8*b - a*x^8))/(x^10*(b + a*x^4)), x)