\(\int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx\) [1220]

Optimal result

Integrand size = 16, antiderivative size = 233 \[ \int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx=-\frac {x \left (a-b x^4\right )^{3/4}}{4 b}-\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}}+\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}}-\frac {a \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}}+\frac {a \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}} \]

[Out]

-1/4*x*(-b*x^4+a)^(3/4)/b+1/16*a*arctan(-1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(5/4)*2^(1/2)+1/16*a*arctan(1
+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(5/4)*2^(1/2)-1/32*a*ln(1-b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4)+x^2*b^(1/2
)/(-b*x^4+a)^(1/2))/b^(5/4)*2^(1/2)+1/32*a*ln(1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4)+x^2*b^(1/2)/(-b*x^4+a)^(1/2
))/b^(5/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {327, 246, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx=-\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}}+\frac {a \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{8 \sqrt {2} b^{5/4}}-\frac {a \log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{16 \sqrt {2} b^{5/4}}+\frac {a \log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{16 \sqrt {2} b^{5/4}}-\frac {x \left (a-b x^4\right )^{3/4}}{4 b} \]

[In]

Int[x^4/(a - b*x^4)^(1/4),x]

[Out]

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

Rule 210

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

Rule 217

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 327

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

Rule 631

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 642

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 1176

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 1179

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*} \text {integral}& = -\frac {x \left (a-b x^4\right )^{3/4}}{4 b}+\frac {a \int \frac {1}{\sqrt [4]{a-b x^4}} \, dx}{4 b} \\ & = -\frac {x \left (a-b x^4\right )^{3/4}}{4 b}+\frac {a \text {Subst}\left (\int \frac {1}{1+b x^4} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{4 b} \\ & = -\frac {x \left (a-b x^4\right )^{3/4}}{4 b}+\frac {a \text {Subst}\left (\int \frac {1-\sqrt {b} x^2}{1+b x^4} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{8 b}+\frac {a \text {Subst}\left (\int \frac {1+\sqrt {b} x^2}{1+b x^4} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{8 b} \\ & = -\frac {x \left (a-b x^4\right )^{3/4}}{4 b}+\frac {a \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {b}}-\frac {\sqrt {2} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{16 b^{3/2}}+\frac {a \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {b}}+\frac {\sqrt {2} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{16 b^{3/2}}-\frac {a \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{b}}+2 x}{-\frac {1}{\sqrt {b}}-\frac {\sqrt {2} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}}-\frac {a \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{b}}-2 x}{-\frac {1}{\sqrt {b}}+\frac {\sqrt {2} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}} \\ & = -\frac {x \left (a-b x^4\right )^{3/4}}{4 b}-\frac {a \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}}+\frac {a \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}}+\frac {a \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}}-\frac {a \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}} \\ & = -\frac {x \left (a-b x^4\right )^{3/4}}{4 b}-\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}}+\frac {a \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{5/4}}-\frac {a \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}}+\frac {a \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{16 \sqrt {2} b^{5/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.62 \[ \int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx=-\frac {4 \sqrt [4]{b} x \left (a-b x^4\right )^{3/4}+\sqrt {2} a \arctan \left (\frac {-\sqrt {b} x^2+\sqrt {a-b x^4}}{\sqrt {2} \sqrt [4]{b} x \sqrt [4]{a-b x^4}}\right )-\sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} x \sqrt [4]{a-b x^4}}{\sqrt {b} x^2+\sqrt {a-b x^4}}\right )}{16 b^{5/4}} \]

[In]

Integrate[x^4/(a - b*x^4)^(1/4),x]

[Out]

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

Maple [A] (verified)

Time = 4.59 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\arctan \left (\frac {b^{\frac {1}{4}} x -\sqrt {2}\, \left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) \sqrt {2}\, a -\arctan \left (\frac {b^{\frac {1}{4}} x +\sqrt {2}\, \left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) \sqrt {2}\, a -\frac {\sqrt {2}\, \ln \left (\frac {-b^{\frac {1}{4}} \left (-b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2} \sqrt {b}+\sqrt {-b \,x^{4}+a}}{b^{\frac {1}{4}} \left (-b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2} \sqrt {b}+\sqrt {-b \,x^{4}+a}}\right ) a}{2}-4 x \left (-b \,x^{4}+a \right )^{\frac {3}{4}} b^{\frac {1}{4}}}{16 b^{\frac {5}{4}}}\)	\(174\)

[In]

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

[Out]

1/16/b^(5/4)*(arctan((b^(1/4)*x-2^(1/2)*(-b*x^4+a)^(1/4))/b^(1/4)/x)*2^(1/2)*a-arctan((b^(1/4)*x+2^(1/2)*(-b*x
^4+a)^(1/4))/b^(1/4)/x)*2^(1/2)*a-1/2*2^(1/2)*ln((-b^(1/4)*(-b*x^4+a)^(1/4)*2^(1/2)*x+x^2*b^(1/2)+(-b*x^4+a)^(
1/2))/(b^(1/4)*(-b*x^4+a)^(1/4)*2^(1/2)*x+x^2*b^(1/2)+(-b*x^4+a)^(1/2)))*a-4*x*(-b*x^4+a)^(3/4)*b^(1/4))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx=-\frac {b \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {b^{4} x \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - b \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {b^{4} x \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} - {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - i \, b \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {i \, b^{4} x \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) + i \, b \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, b^{4} x \left (-\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) + 4 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} x}{16 \, b} \]

[In]

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

[Out]

-1/16*(b*(-a^4/b^5)^(1/4)*log((b^4*x*(-a^4/b^5)^(3/4) + (-b*x^4 + a)^(1/4)*a^3)/x) - b*(-a^4/b^5)^(1/4)*log(-(
b^4*x*(-a^4/b^5)^(3/4) - (-b*x^4 + a)^(1/4)*a^3)/x) - I*b*(-a^4/b^5)^(1/4)*log((I*b^4*x*(-a^4/b^5)^(3/4) + (-b
*x^4 + a)^(1/4)*a^3)/x) + I*b*(-a^4/b^5)^(1/4)*log((-I*b^4*x*(-a^4/b^5)^(3/4) + (-b*x^4 + a)^(1/4)*a^3)/x) + 4
*(-b*x^4 + a)^(3/4)*x)/b

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

x**5*gamma(5/4)*hyper((1/4, 5/4), (9/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(1/4)*gamma(9/4))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{\sqrt [4]{a-b x^4}} \, dx=-\frac {{\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {b} + \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} - \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{b^{\frac {1}{4}}}\right )} a}{32 \, b} - \frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} a}{4 \, {\left (b^{2} - \frac {{\left (b x^{4} - a\right )} b}{x^{4}}\right )} x^{3}} \]

[In]

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

[Out]

-1/32*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(1/4) + 2*sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(1/4) - sqrt(2)*log(sqrt(b) + sqrt(2)*
(-b*x^4 + a)^(1/4)*b^(1/4)/x + sqrt(-b*x^4 + a)/x^2)/b^(1/4) + sqrt(2)*log(sqrt(b) - sqrt(2)*(-b*x^4 + a)^(1/4
)*b^(1/4)/x + sqrt(-b*x^4 + a)/x^2)/b^(1/4))*a/b - 1/4*(-b*x^4 + a)^(3/4)*a/((b^2 - (b*x^4 - a)*b/x^4)*x^3)

Giac [F]

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

[In]

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

[Out]

integrate(x^4/(-b*x^4 + a)^(1/4), x)

Mupad [F(-1)]

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

[In]

int(x^4/(a - b*x^4)^(1/4),x)

[Out]

int(x^4/(a - b*x^4)^(1/4), x)