Integrand size = 19, antiderivative size = 80 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}-\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )+\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right ) \]
-2*(a*x^4+b*x^2)^(1/4)/x-a^(1/4)*arctan(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))+a^( 1/4)*arctanh(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))
Time = 0.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {x \left (b+a x^2\right )^{3/4} \left (-2 \sqrt [4]{b+a x^2}-\sqrt [4]{a} \sqrt {x} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\sqrt [4]{a} \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{\left (x^2 \left (b+a x^2\right )\right )^{3/4}} \]
(x*(b + a*x^2)^(3/4)*(-2*(b + a*x^2)^(1/4) - a^(1/4)*Sqrt[x]*ArcTan[(a^(1/ 4)*Sqrt[x])/(b + a*x^2)^(1/4)] + a^(1/4)*Sqrt[x]*ArcTanh[(a^(1/4)*Sqrt[x]) /(b + a*x^2)^(1/4)]))/(x^2*(b + a*x^2))^(3/4)
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.50, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1425, 1431, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4+b x^2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle a \int \frac {x^2}{\left (a x^4+b x^2\right )^{3/4}}dx-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {a x^{3/2} \left (a x^2+b\right )^{3/4} \int \frac {\sqrt {x}}{\left (a x^2+b\right )^{3/4}}dx}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 a x^{3/2} \left (a x^2+b\right )^{3/4} \int \frac {x}{\left (a x^2+b\right )^{3/4}}d\sqrt {x}}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {2 a x^{3/2} \left (a x^2+b\right )^{3/4} \int \frac {x}{1-a x^2}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 a x^{3/2} \left (a x^2+b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} x+1}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}\right )}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 a x^{3/2} \left (a x^2+b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 a x^{3/2} \left (a x^2+b\right )^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}\) |
(-2*(b*x^2 + a*x^4)^(1/4))/x + (2*a*x^(3/2)*(b + a*x^2)^(3/4)*(-1/2*ArcTan [(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]/a^(3/4) + ArcTanh[(a^(1/4)*Sqrt[x])/ (b + a*x^2)^(1/4)]/(2*a^(3/4))))/(b*x^2 + a*x^4)^(3/4)
3.11.66.3.1 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] :> 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[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Time = 2.88 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.21
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right ) a^{\frac {1}{4}} x +2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a^{\frac {1}{4}} x -4 \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{2 x}\) | \(97\) |
1/2*(ln((a^(1/4)*x+(x^2*(a*x^2+b))^(1/4))/(-a^(1/4)*x+(x^2*(a*x^2+b))^(1/4 )))*a^(1/4)*x+2*arctan(1/a^(1/4)/x*(x^2*(a*x^2+b))^(1/4))*a^(1/4)*x-4*(x^2 *(a*x^2+b))^(1/4))/x
Timed out. \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )}}{x^{2}}\, dx \]
\[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}}}{x^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (66) = 132\).
Time = 0.28 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.31 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} \]
1/2*sqrt(2)*(-a)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x ^2)^(1/4))/(-a)^(1/4)) + 1/2*sqrt(2)*(-a)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt( 2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4)) + 1/4*sqrt(2)*(-a)^(1/4)* log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) - 1 /4*sqrt(2)*(-a)^(1/4)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) - 2*(a + b/x^2)^(1/4)
Time = 5.79 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx=-\frac {2\,{\left (a\,x^4+b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a\,x^2}{b}\right )}{x\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}} \]