3.1003 \(\int x^2 \sqrt [4]{a+b x^4} \, dx\)

Optimal. Leaf size=77 \[ -\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{3/4}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a+b x^4} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {279, 331, 298, 203, 206} \[ -\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{3/4}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a+b x^4} \]

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[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(3/4)) + (a*ArcTanh[(b^(1/4)*x)/(a
+ b*x^4)^(1/4)])/(8*b^(3/4))

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])

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 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 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]

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

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Mathematica [C]  time = 0.05, size = 51, normalized size = 0.66 \[ \frac {x^3 \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{3 \sqrt [4]{\frac {b x^4}{a}+1}} \]

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 [B]  time = 0.79, size = 191, normalized size = 2.48 \[ \frac {1}{4} \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{3} - \frac {1}{4} \, \left (\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {\left (\frac {a^{4}}{b^{3}}\right )^{\frac {3}{4}} b^{2} x \sqrt {\frac {\sqrt {\frac {a^{4}}{b^{3}}} b^{2} x^{2} + \sqrt {b x^{4} + a} a^{2}}{x^{2}}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a \left (\frac {a^{4}}{b^{3}}\right )^{\frac {3}{4}} b^{2}}{a^{4} x}\right ) + \frac {1}{16} \, \left (\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {\left (\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b x + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{x}\right ) - \frac {1}{16} \, \left (\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {\left (\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b x - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{x}\right ) \]

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 + sq
rt(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 = 0.14, 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 = 3.00, size = 101, normalized size = 1.31 \[ \frac {a \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{8 \, b^{\frac {3}{4}}} - \frac {a \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{16 \, b^{\frac {3}{4}}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{4 \, {\left (b - \frac {b x^{4} + a}{x^{4}}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - 1/16*a*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (
b*x^4 + a)^(1/4)/x))/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 = -0.01 \[ \int x^2\,{\left (b\,x^4+a\right )}^{1/4} \,d x \]

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 = 1.65, size = 39, normalized size = 0.51 \[ \frac {\sqrt [4]{a} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

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

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