3.1127 \(\int \frac{x^2}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=57 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0214927, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {331, 298, 203, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^4)^(3/4),x]

[Out]

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

Rule 331

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 298

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 203

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{x^2}{\left (a+b x^4\right )^{3/4}} \, dx &=\operatorname{Subst}\left (\int \frac{x^2}{1-b x^4} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac{\operatorname{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{\operatorname{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]  time = 0.007414, size = 50, normalized size = 0.88 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( b{x}^{4}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.56737, size = 377, normalized size = 6.61 \begin{align*} -\frac{1}{b^{3}}^{\frac{1}{4}} \arctan \left (\frac{b^{2} \frac{1}{b^{3}}^{\frac{3}{4}} x \sqrt{\frac{b^{2} \sqrt{\frac{1}{b^{3}}} x^{2} + \sqrt{b x^{4} + a}}{x^{2}}} -{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{2} \frac{1}{b^{3}}^{\frac{3}{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} \, \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.38612, size = 37, normalized size = 0.65 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*x^4 + a)^(3/4), x)