3.7.22 \(\int x^2 \sqrt [4]{1+x^4} \, dx\)

Optimal. Leaf size=49 \[ -\frac {1}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{4} \sqrt [4]{x^4+1} x^3 \]

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {279, 331, 298, 203, 206} \begin {gather*} -\frac {1}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{4} \sqrt [4]{x^4+1} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(1 + x^4)^(1/4),x]

[Out]

(x^3*(1 + x^4)^(1/4))/4 - ArcTan[x/(1 + x^4)^(1/4)]/8 + ArcTanh[x/(1 + x^4)^(1/4)]/8

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

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Mathematica [C]  time = 0.00, size = 22, normalized size = 0.45 \begin {gather*} \frac {1}{3} x^3 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(1 + x^4)^(1/4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.15, size = 49, normalized size = 1.00 \begin {gather*} \frac {1}{4} x^3 \sqrt [4]{1+x^4}-\frac {1}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*(1 + x^4)^(1/4),x]

[Out]

(x^3*(1 + x^4)^(1/4))/4 - ArcTan[x/(1 + x^4)^(1/4)]/8 + ArcTanh[x/(1 + x^4)^(1/4)]/8

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fricas [A]  time = 0.47, size = 62, normalized size = 1.27 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{16} \, \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{16} \, \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^4 + 1)^(1/4)*x^3 + 1/8*arctan((x^4 + 1)^(1/4)/x) + 1/16*log((x + (x^4 + 1)^(1/4))/x) - 1/16*log(-(x - (
x^4 + 1)^(1/4))/x)

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giac [A]  time = 0.51, size = 59, normalized size = 1.20 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^4 + 1)^(1/4)*x^3 + 1/8*arctan((x^4 + 1)^(1/4)/x) + 1/16*log((x^4 + 1)^(1/4)/x + 1) - 1/16*log((x^4 + 1)
^(1/4)/x - 1)

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maple [C]  time = 1.78, size = 17, normalized size = 0.35

method	result	size
meijerg	\(\frac {\hypergeom \left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right ) x^{3}}{3}\)	\(17\)
risch	\(\frac {x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}{4}+\frac {\hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right ) x^{3}}{12}\)	\(30\)
trager	\(\frac {x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x^{2}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{16}-\frac {\ln \left (2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{2} \sqrt {x^{4}+1}+2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}-2 x^{4}-1\right )}{16}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^4+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/3*hypergeom([-1/4,3/4],[7/4],-x^4)*x^3

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maxima [A]  time = 0.42, size = 72, normalized size = 1.47 \begin {gather*} \frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{4 \, x {\left (\frac {x^{4} + 1}{x^{4}} - 1\right )}} + \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^4 + 1)^(1/4)/(x*((x^4 + 1)/x^4 - 1)) + 1/8*arctan((x^4 + 1)^(1/4)/x) + 1/16*log((x^4 + 1)^(1/4)/x + 1)
- 1/16*log((x^4 + 1)^(1/4)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,{\left (x^4+1\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^4 + 1)^(1/4),x)

[Out]

int(x^2*(x^4 + 1)^(1/4), x)

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sympy [C]  time = 0.82, size = 31, normalized size = 0.63 \begin {gather*} \frac {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 | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(x**4+1)**(1/4),x)

[Out]

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

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