\(\int \frac {x^2}{(a-b x^4)^{3/4}} \, dx\) [1251]

Optimal result

Integrand size = 16, antiderivative size = 209 \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2} b^{3/4}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2} b^{3/4}}+\frac {\log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{4 \sqrt {2} b^{3/4}}-\frac {\log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{4 \sqrt {2} b^{3/4}} \]

[Out]

1/4*arctan(-1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(3/4)*2^(1/2)+1/4*arctan(1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1
/4))/b^(3/4)*2^(1/2)+1/8*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^(3/4)*2^(1/2)
-1/8*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^(3/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {338, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2} b^{3/4}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt {2} b^{3/4}}+\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2} b^{3/4}}-\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2} b^{3/4}} \]

[In]

Int[x^2/(a - b*x^4)^(3/4),x]

[Out]

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

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 303

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

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.56 \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x \sqrt [4]{a-b x^4}}{-\sqrt {b} x^2+\sqrt {a-b x^4}}\right )-\text {arctanh}\left (\frac {\sqrt {b} x^2+\sqrt {a-b x^4}}{\sqrt {2} \sqrt [4]{b} x \sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2} b^{3/4}} \]

[In]

Integrate[x^2/(a - b*x^4)^(3/4),x]

[Out]

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

Maple [A] (verified)

Time = 4.67 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {\sqrt {2}\, \left (\ln \left (\frac {b^{\frac {1}{4}} \left (-b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2} \sqrt {b}+\sqrt {-b \,x^{4}+a}}{-b^{\frac {1}{4}} \left (-b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2} \sqrt {b}+\sqrt {-b \,x^{4}+a}}\right )+2 \arctan \left (\frac {b^{\frac {1}{4}} x +\sqrt {2}\, \left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )-2 \arctan \left (\frac {b^{\frac {1}{4}} x -\sqrt {2}\, \left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )\right )}{8 b^{\frac {3}{4}}}\)	\(148\)

[In]

int(x^2/(-b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-1/8/b^(3/4)*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)))+2*arctan((b^(1/4)*x+2^(1/2)*(-b*x^4+a)^(1/4))/b^(1/4)/x)-2*ar
ctan((b^(1/4)*x-2^(1/2)*(-b*x^4+a)^(1/4))/b^(1/4)/x))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.70 \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=-\frac {1}{4} \, \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {b x \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {b x \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} - {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} i \, \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {i \, b x \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} i \, \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, b x \left (-\frac {1}{b^{3}}\right )^{\frac {1}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate(x^2/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.19 \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=\frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(x**2/(-b*x**4+a)**(3/4),x)

[Out]

x**3*gamma(3/4)*hyper((3/4, 3/4), (7/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(3/4)*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.84 \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{4 \, b^{\frac {3}{4}}} - \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{4 \, b^{\frac {3}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {b} + \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{8 \, b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} - \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{8 \, b^{\frac {3}{4}}} \]

[In]

integrate(x^2/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(3/4) - 1/4*sqrt(2)*arct
an(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(3/4) - 1/8*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^(3/4) + 1/8*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^(3/4)

Giac [F]

\[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=\int { \frac {x^{2}}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^2/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^2/(-b*x^4 + a)^(3/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx=\int \frac {x^2}{{\left (a-b\,x^4\right )}^{3/4}} \,d x \]

[In]

int(x^2/(a - b*x^4)^(3/4),x)

[Out]

int(x^2/(a - b*x^4)^(3/4), x)