\(\int \frac {x^2}{(-b+a x^4)^{3/4} (b+a x^4)} \, dx\) [1184]

Optimal result

Integrand size = 26, antiderivative size = 87 \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {508, 304, 209, 212} \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4} b}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4} b} \]

[In]

Int[x^2/((-b + a*x^4)^(3/4)*(b + a*x^4)),x]

[Out]

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

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 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 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\frac {-\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]

[In]

Integrate[x^2/((-b + a*x^4)^(3/4)*(b + a*x^4)),x]

[Out]

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

Maple [A] (verified)

Time = 3.91 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-2^{\frac {1}{4}} a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2^{\frac {1}{4}} a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right )\right )}{8 a^{\frac {3}{4}} b}\)	\(87\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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