\(\int \frac {x^2}{(a+b x^4)^{3/4}} \, dx\) [1127]

Optimal result

Integrand size = 15, antiderivative size = 57 \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {338, 304, 209, 212} \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \]

[In]

Int[x^2/(a + b*x^4)^(3/4),x]

[Out]

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

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 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]

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {-\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \]

[In]

Integrate[x^2/(a + b*x^4)^(3/4),x]

[Out]

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

Maple [A] (verified)

Time = 4.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{4 b^{\frac {3}{4}}}\)	\(65\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.21 \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {1}{4} \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (\frac {b \frac {1}{b^{3}}^{\frac {1}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (-\frac {b \frac {1}{b^{3}}^{\frac {1}{4}} x - {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} i \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (\frac {i \, b \frac {1}{b^{3}}^{\frac {1}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} i \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (\frac {-i \, b \frac {1}{b^{3}}^{\frac {1}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.65 \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} \]

[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(I*pi)/a)/(4*a**(3/4)*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{2 \, b^{\frac {3}{4}}} - \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{4 \, b^{\frac {3}{4}}} \]

[In]

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

[Out]

1/2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - 1/4*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4
+ a)^(1/4)/x))/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. \[ \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx=\int \frac {x^2}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \]

[In]

int(x^2/(a + b*x^4)^(3/4),x)

[Out]

int(x^2/(a + b*x^4)^(3/4), x)