3.1251 \(\int \frac{x^2}{(a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=209 \[ \frac{\log \left (\frac{\sqrt{b} x^2}{\sqrt{a-b x^4}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt{2} b^{3/4}}-\frac{\log \left (\frac{\sqrt{b} x^2}{\sqrt{a-b x^4}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt{2} b^{3/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt{2} b^{3/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt{2} b^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0877794, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {331, 297, 1162, 617, 204, 1165, 628} \[ \frac{\log \left (\frac{\sqrt{b} x^2}{\sqrt{a-b x^4}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt{2} b^{3/4}}-\frac{\log \left (\frac{\sqrt{b} x^2}{\sqrt{a-b x^4}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt{2} b^{3/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt{2} b^{3/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt{2} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a - b*x^4)^(3/4),x]

[Out]

-ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)]/(2*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*Sq
rt[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 297

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 1162

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 617

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 204

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

Rule 1165

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]

Rule 628

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]

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-\sqrt{b} x^2}{1+b x^4} \, dx,x,\frac{x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt{b}}+\frac{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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]  time = 0.0125413, size = 58, normalized size = 0.28 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt [4]{-b} x}{\sqrt [4]{a-b x^4}}\right )-\tan ^{-1}\left (\frac{\sqrt [4]{-b} x}{\sqrt [4]{a-b x^4}}\right )}{2 (-b)^{3/4}} \]

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 = 0.026, 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 [A]  time = 1.82359, size = 382, normalized size = 1.83 \begin{align*} -\left (-\frac{1}{b^{3}}\right )^{\frac{1}{4}} \arctan \left (\frac{b^{2} x \sqrt{\frac{b^{2} x^{2} \sqrt{-\frac{1}{b^{3}}} + \sqrt{-b x^{4} + a}}{x^{2}}} \left (-\frac{1}{b^{3}}\right )^{\frac{3}{4}} -{\left (-b x^{4} + a\right )}^{\frac{1}{4}} b^{2} \left (-\frac{1}{b^{3}}\right )^{\frac{3}{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} \, \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 ) \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]

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

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Sympy [C]  time = 1.54875, size = 39, normalized size = 0.19 \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^{2 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(2*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)