3.8.59 \(\int x^2 \sqrt [4]{x^2+x^4} \, dx\) [759]

Optimal. Leaf size=59 \[ \frac {1}{16} \left (x+4 x^3\right ) \sqrt [4]{x^2+x^4}+\frac {3}{32} \text {ArcTan}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {3}{32} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right ) \]

[Out]

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

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(125\) vs. \(2(59)=118\).
time = 0.08, antiderivative size = 125, normalized size of antiderivative = 2.12, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2046, 2049, 2057, 335, 338, 304, 209, 212} \begin {gather*} \frac {3 \left (x^2+1\right )^{3/4} x^{3/2} \text {ArcTan}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{32 \left (x^4+x^2\right )^{3/4}}+\frac {1}{16} \sqrt [4]{x^4+x^2} x-\frac {3 \left (x^2+1\right )^{3/4} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{32 \left (x^4+x^2\right )^{3/4}}+\frac {1}{4} \sqrt [4]{x^4+x^2} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(x^2 + x^4)^(1/4),x]

[Out]

(x*(x^2 + x^4)^(1/4))/16 + (x^3*(x^2 + x^4)^(1/4))/4 + (3*x^(3/2)*(1 + x^2)^(3/4)*ArcTan[Sqrt[x]/(1 + x^2)^(1/
4)])/(32*(x^2 + x^4)^(3/4)) - (3*x^(3/2)*(1 + x^2)^(3/4)*ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)])/(32*(x^2 + x^4)^(3/
4))

Rule 209

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

Rule 212

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

Rule 304

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 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 338

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 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int x^2 \sqrt [4]{x^2+x^4} \, dx &=\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{8} \int \frac {x^4}{\left (x^2+x^4\right )^{3/4}} \, dx\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {3}{32} \int \frac {x^2}{\left (x^2+x^4\right )^{3/4}} \, dx\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 x^{3/2} \left (1+x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (1+x^2\right )^{3/4}} \, dx}{32 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 x^{3/2} \left (1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{16 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 x^{3/2} \left (1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{16 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 x^{3/2} \left (1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \left (x^2+x^4\right )^{3/4}}+\frac {\left (3 x^{3/2} \left (1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}+\frac {3 x^{3/2} \left (1+x^2\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \left (x^2+x^4\right )^{3/4}}-\frac {3 x^{3/2} \left (1+x^2\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \left (x^2+x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 89, normalized size = 1.51 \begin {gather*} \frac {\sqrt [4]{x^2+x^4} \left (2 x^{3/2} \sqrt [4]{1+x^2} \left (1+4 x^2\right )+3 \text {ArcTan}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )-3 \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{32 \sqrt {x} \sqrt [4]{1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(x^2 + x^4)^(1/4),x]

[Out]

((x^2 + x^4)^(1/4)*(2*x^(3/2)*(1 + x^2)^(1/4)*(1 + 4*x^2) + 3*ArcTan[Sqrt[x]/(1 + x^2)^(1/4)] - 3*ArcTanh[Sqrt
[x]/(1 + x^2)^(1/4)]))/(32*Sqrt[x]*(1 + x^2)^(1/4))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + x^2)^(1/4)*x^2, x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (47) = 94\).
time = 0.89, size = 102, normalized size = 1.73 \begin {gather*} \frac {1}{16} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (4 \, x^{3} + x\right )} + \frac {3}{64} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {3}{64} \, \log \left (-\frac {2 \, x^{3} - 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} + x^{2}} x + x - 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(x^4 + x^2)^(1/4)*(4*x^3 + x) + 3/64*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x^4 + x^2)^(3/4))/x) + 3/64*log(-
(2*x^3 - 2*(x^4 + x^2)^(1/4)*x^2 + 2*sqrt(x^4 + x^2)*x + x - 2*(x^4 + x^2)^(3/4))/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt [4]{x^{2} \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 57, normalized size = 0.97 \begin {gather*} \frac {1}{16} \, {\left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right )} x^{4} - \frac {3}{32} \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{64} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{64} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((1/x^2 + 1)^(5/4) + 3*(1/x^2 + 1)^(1/4))*x^4 - 3/32*arctan((1/x^2 + 1)^(1/4)) - 3/64*log((1/x^2 + 1)^(1/
4) + 1) + 3/64*log((1/x^2 + 1)^(1/4) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^2 + x^4)^(1/4),x)

[Out]

int(x^2*(x^2 + x^4)^(1/4), x)

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