3.1190 \(\int x^2 \sqrt [4]{a-b x^4} \, dx\)

Optimal. Leaf size=232 \[ -\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{3/4}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{8 \sqrt {2} b^{3/4}}+\frac {a \log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{16 \sqrt {2} b^{3/4}}-\frac {a \log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{16 \sqrt {2} b^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a-b x^4} \]

[Out]

1/4*x^3*(-b*x^4+a)^(1/4)+1/16*a*arctan(-1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(3/4)*2^(1/2)+1/16*a*arctan(1+
b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(3/4)*2^(1/2)+1/32*a*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)-1/32*a*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)

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Rubi [A]  time = 0.11, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {279, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {a \log \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{16 \sqrt {2} b^{3/4}}-\frac {a \log \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{16 \sqrt {2} b^{3/4}}-\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{3/4}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{8 \sqrt {2} b^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a-b x^4} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a - b*x^4)^(1/4),x]

[Out]

(x^3*(a - b*x^4)^(1/4))/4 - (a*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(8*Sqrt[2]*b^(3/4)) + (a*Arc
Tan[1 + (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(8*Sqrt[2]*b^(3/4)) + (a*Log[1 + (Sqrt[b]*x^2)/Sqrt[a - b*x^4]
- (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(16*Sqrt[2]*b^(3/4)) - (a*Log[1 + (Sqrt[b]*x^2)/Sqrt[a - b*x^4] + (
Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(16*Sqrt[2]*b^(3/4))

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 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 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 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]

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 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]

Rubi steps

\begin {align*} \int x^2 \sqrt [4]{a-b x^4} \, dx &=\frac {1}{4} x^3 \sqrt [4]{a-b x^4}+\frac {1}{4} a \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx\\ &=\frac {1}{4} x^3 \sqrt [4]{a-b x^4}+\frac {1}{4} a \operatorname {Subst}\left (\int \frac {x^2}{1+b x^4} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )\\ &=\frac {1}{4} x^3 \sqrt [4]{a-b x^4}-\frac {a \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 )}{8 \sqrt {b}}+\frac {a \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 )}{8 \sqrt {b}}\\ &=\frac {1}{4} x^3 \sqrt [4]{a-b x^4}+\frac {a \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 )}{16 b}+\frac {a \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 )}{16 b}+\frac {a \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 )}{16 \sqrt {2} b^{3/4}}+\frac {a \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 )}{16 \sqrt {2} b^{3/4}}\\ &=\frac {1}{4} x^3 \sqrt [4]{a-b x^4}+\frac {a \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 )}{16 \sqrt {2} b^{3/4}}-\frac {a \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 )}{16 \sqrt {2} b^{3/4}}+\frac {a \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 )}{8 \sqrt {2} b^{3/4}}-\frac {a \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 )}{8 \sqrt {2} b^{3/4}}\\ &=\frac {1}{4} x^3 \sqrt [4]{a-b x^4}-\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{3/4}}+\frac {a \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{8 \sqrt {2} b^{3/4}}+\frac {a \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 )}{16 \sqrt {2} b^{3/4}}-\frac {a \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 )}{16 \sqrt {2} b^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.22 \[ \frac {x^3 \sqrt [4]{a-b x^4} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {b x^4}{a}\right )}{3 \sqrt [4]{1-\frac {b x^4}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a - b*x^4)^(1/4),x]

[Out]

(x^3*(a - b*x^4)^(1/4)*Hypergeometric2F1[-1/4, 3/4, 7/4, (b*x^4)/a])/(3*(1 - (b*x^4)/a)^(1/4))

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fricas [A]  time = 0.85, size = 204, normalized size = 0.88 \[ \frac {1}{4} \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{3} - \frac {1}{4} \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {\left (-\frac {a^{4}}{b^{3}}\right )^{\frac {3}{4}} b^{2} x \sqrt {\frac {\sqrt {-\frac {a^{4}}{b^{3}}} b^{2} x^{2} + \sqrt {-b x^{4} + a} a^{2}}{x^{2}}} - {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {3}{4}} b^{2}}{a^{4} x}\right ) - \frac {1}{16} \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {\left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b x + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a}{x}\right ) + \frac {1}{16} \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {\left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b x - {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

1/4*(-b*x^4 + a)^(1/4)*x^3 - 1/4*(-a^4/b^3)^(1/4)*arctan(((-a^4/b^3)^(3/4)*b^2*x*sqrt((sqrt(-a^4/b^3)*b^2*x^2
+ sqrt(-b*x^4 + a)*a^2)/x^2) - (-b*x^4 + a)^(1/4)*a*(-a^4/b^3)^(3/4)*b^2)/(a^4*x)) - 1/16*(-a^4/b^3)^(1/4)*log
(((-a^4/b^3)^(1/4)*b*x + (-b*x^4 + a)^(1/4)*a)/x) + 1/16*(-a^4/b^3)^(1/4)*log(-((-a^4/b^3)^(1/4)*b*x - (-b*x^4
+ a)^(1/4)*a)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)*x^2, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-b*x^4+a)^(1/4),x)

[Out]

int(x^2*(-b*x^4+a)^(1/4),x)

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maxima [A]  time = 2.96, size = 214, normalized size = 0.92 \[ -\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{16 \, b^{\frac {3}{4}}} - \frac {\sqrt {2} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{16 \, b^{\frac {3}{4}}} - \frac {\sqrt {2} a \log \left (\sqrt {b} + \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{32 \, b^{\frac {3}{4}}} + \frac {\sqrt {2} a \log \left (\sqrt {b} - \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{32 \, b^{\frac {3}{4}}} + \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a}{4 \, {\left (b - \frac {b x^{4} - a}{x^{4}}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

-1/16*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(3/4) - 1/16*sqrt(2)*
a*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(3/4) - 1/32*sqrt(2)*a*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) + 1/32*sqrt(2)*a*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) + 1/4*(-b*x^4 + a)^(1/4)*a/((b - (b*x^4 - a)/x^4
)*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (a-b\,x^4\right )}^{1/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a - b*x^4)^(1/4),x)

[Out]

int(x^2*(a - b*x^4)^(1/4), x)

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sympy [C]  time = 1.85, size = 41, normalized size = 0.18 \[ \frac {\sqrt [4]{a} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-b*x**4+a)**(1/4),x)

[Out]

a**(1/4)*x**3*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*gamma(7/4))

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