Integrand size = 11, antiderivative size = 53 \[ \int \sqrt [4]{x^2+x^4} \, dx=\frac {1}{2} x \sqrt [4]{x^2+x^4}-\frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )+\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right ) \]
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int \sqrt [4]{x^2+x^4} \, dx=\frac {\sqrt [4]{x^2+x^4} \left (2 x^{3/2} \sqrt [4]{1+x^2}-\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{4 \sqrt {x} \sqrt [4]{1+x^2}} \]
((x^2 + x^4)^(1/4)*(2*x^(3/2)*(1 + x^2)^(1/4) - ArcTan[Sqrt[x]/(1 + x^2)^( 1/4)] + ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)]))/(4*Sqrt[x]*(1 + x^2)^(1/4))
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.70, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {1402, 248, 266, 854, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt [4]{x^4+x^2} \, dx\) |
\(\Big \downarrow \) 1402 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \int \sqrt {x} \sqrt [4]{x^2+1}dx}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \left (\frac {1}{4} \int \frac {\sqrt {x}}{\left (x^2+1\right )^{3/4}}dx+\frac {1}{2} \sqrt [4]{x^2+1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \left (\frac {1}{2} \int \frac {x}{\left (x^2+1\right )^{3/4}}d\sqrt {x}+\frac {1}{2} \sqrt [4]{x^2+1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \left (\frac {1}{2} \int \frac {x}{1-x^2}d\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}+\frac {1}{2} \sqrt [4]{x^2+1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}-\frac {1}{2} \int \frac {1}{x+1}d\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )+\frac {1}{2} \sqrt [4]{x^2+1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}-\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )\right )+\frac {1}{2} \sqrt [4]{x^2+1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^2} \left (\frac {1}{2} \left (\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )\right )+\frac {1}{2} \sqrt [4]{x^2+1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2+1}}\) |
((x^2 + x^4)^(1/4)*((x^(3/2)*(1 + x^2)^(1/4))/2 + (-1/2*ArcTan[Sqrt[x]/(1 + x^2)^(1/4)] + ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)]/2)/2))/(Sqrt[x]*(1 + x^2) ^(1/4))
3.7.70.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(b*x^2 + c*x^4)^p /(x^(2*p)*(b + c*x^2)^p) Int[x^(2*p)*(b + c*x^2)^p, x], x] /; FreeQ[{b, c , p}, x] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 3.58 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.32
method | result | size |
meijerg | \(\frac {2 x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{2}\right )}{3}\) | \(17\) |
pseudoelliptic | \(\frac {x \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2}-\frac {\ln \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right )}{8}+\frac {\arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right )}{4}+\frac {\ln \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+x}{x}\right )}{8}\) | \(76\) |
trager | \(\frac {x \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{2}-\frac {\ln \left (\frac {2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}+x^{2}}\, x +2 x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}-2 x^{3}-x}{x}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x}{x}\right )}{8}\) | \(144\) |
risch | \(\frac {x \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2}+\frac {\left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}+2 x^{6}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}+5 x^{4}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}+4 x^{2}+1}{\left (x^{2}+1\right )^{2}}\right )}{8}-\frac {\ln \left (-\frac {-2 x^{6}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}-5 x^{4}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}}+4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}-4 x^{2}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}}-1}{\left (x^{2}+1\right )^{2}}\right )}{8}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{2} \left (x^{2}+1\right )^{3}\right )^{\frac {1}{4}}}{\left (x^{2}+1\right ) x}\) | \(410\) |
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (41) = 82\).
Time = 0.60 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.79 \[ \int \sqrt [4]{x^2+x^4} \, dx=\frac {1}{2} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x - \frac {1}{8} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {1}{8} \, \log \left (\frac {2 \, x^{3} + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} + x^{2}} x + x + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \]
1/2*(x^4 + x^2)^(1/4)*x - 1/8*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x^4 + x^2 )^(3/4))/x) + 1/8*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)
\[ \int \sqrt [4]{x^2+x^4} \, dx=\int \sqrt [4]{x^{4} + x^{2}}\, dx \]
\[ \int \sqrt [4]{x^2+x^4} \, dx=\int { {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} \,d x } \]
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \sqrt [4]{x^2+x^4} \, dx=\frac {1}{2} \, x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + \frac {1}{4} \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{8} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
1/2*x^2*(1/x^2 + 1)^(1/4) + 1/4*arctan((1/x^2 + 1)^(1/4)) + 1/8*log((1/x^2 + 1)^(1/4) + 1) - 1/8*log((1/x^2 + 1)^(1/4) - 1)
Time = 5.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.55 \[ \int \sqrt [4]{x^2+x^4} \, dx=\frac {2\,x\,{\left (x^4+x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -x^2\right )}{3\,{\left (x^2+1\right )}^{1/4}} \]