Integrand size = 31, antiderivative size = 109 \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b^2 x^3}-\frac {a \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]}{4 b} \]
Time = 0.00 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (-8 \left (-b+a x^2\right )^{5/4}-5 a b x^{5/2} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]\right )}{20 b^2 x^3 \sqrt [4]{-b+a x^2}} \]
((-(b*x^2) + a*x^4)^(1/4)*(-8*(-b + a*x^2)^(5/4) - 5*a*b*x^(5/2)*RootSum[a ^2 - a*b - 2*a*#1^4 + #1^8 & , (-(Log[Sqrt[x]]*#1) + Log[(-b + a*x^2)^(1/4 ) - Sqrt[x]*#1]*#1)/(-a + #1^4) & ]))/(20*b^2*x^3*(-b + a*x^2)^(1/4))
Leaf count is larger than twice the leaf count of optimal. \(260\) vs. \(2(109)=218\).
Time = 1.31 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.39, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2467, 25, 1593, 1845, 27, 803, 796, 7276, 2009}
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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4-b x^2}}{x^4 \left (a x^4-b\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \int -\frac {\sqrt [4]{a x^2-b}}{x^{7/2} \left (b-a x^4\right )}dx}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{a x^4-b x^2} \int \frac {\sqrt [4]{a x^2-b}}{x^{7/2} \left (b-a x^4\right )}dx}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 1593 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \int \frac {\sqrt [4]{a x^2-b}}{x^3 \left (b-a x^4\right )}d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 1845 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \left (\frac {\int \frac {a b \left (1-x^2\right )}{x \left (a x^2-b\right )^{3/4} \left (b-a x^4\right )}d\sqrt {x}}{b}-\int \frac {1}{x^3 \left (a x^2-b\right )^{3/4}}d\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \left (a \int \frac {1-x^2}{x \left (a x^2-b\right )^{3/4} \left (b-a x^4\right )}d\sqrt {x}-\int \frac {1}{x^3 \left (a x^2-b\right )^{3/4}}d\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \left (-\frac {4 a \int \frac {1}{x \left (a x^2-b\right )^{3/4}}d\sqrt {x}}{5 b}+a \int \frac {1-x^2}{x \left (a x^2-b\right )^{3/4} \left (b-a x^4\right )}d\sqrt {x}-\frac {\sqrt [4]{a x^2-b}}{5 b x^{5/2}}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \left (a \int \frac {1-x^2}{x \left (a x^2-b\right )^{3/4} \left (b-a x^4\right )}d\sqrt {x}-\frac {4 a \sqrt [4]{a x^2-b}}{5 b^2 \sqrt {x}}-\frac {\sqrt [4]{a x^2-b}}{5 b x^{5/2}}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \left (a \int \left (\frac {\sqrt [4]{a x^2-b} x}{b \left (b-a x^4\right )}+\frac {1}{b \left (a x^2-b\right )^{3/4} x}\right )d\sqrt {x}-\frac {4 a \sqrt [4]{a x^2-b}}{5 b^2 \sqrt {x}}-\frac {\sqrt [4]{a x^2-b}}{5 b x^{5/2}}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt [4]{a x^4-b x^2} \left (a \left (\frac {x^{3/2} \sqrt [4]{a x^2-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {\sqrt {a} x^2}{\sqrt {b}},\frac {a x^2}{b}\right )}{6 b^2 \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {x^{3/2} \sqrt [4]{a x^2-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {\sqrt {a} x^2}{\sqrt {b}},\frac {a x^2}{b}\right )}{6 b^2 \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {\sqrt [4]{a x^2-b}}{b^2 \sqrt {x}}\right )-\frac {4 a \sqrt [4]{a x^2-b}}{5 b^2 \sqrt {x}}-\frac {\sqrt [4]{a x^2-b}}{5 b x^{5/2}}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\) |
(-2*(-(b*x^2) + a*x^4)^(1/4)*(-1/5*(-b + a*x^2)^(1/4)/(b*x^(5/2)) - (4*a*( -b + a*x^2)^(1/4))/(5*b^2*Sqrt[x]) + a*((-b + a*x^2)^(1/4)/(b^2*Sqrt[x]) + (x^(3/2)*(-b + a*x^2)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, -((Sqrt[a]*x^2)/S qrt[b]), (a*x^2)/b])/(6*b^2*(1 - (a*x^2)/b)^(1/4)) + (x^(3/2)*(-b + a*x^2) ^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (Sqrt[a]*x^2)/Sqrt[b], (a*x^2)/b])/(6*b ^2*(1 - (a*x^2)/b)^(1/4)))))/(Sqrt[x]*(-b + a*x^2)^(1/4))
3.16.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_ ), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/f Subst[Int[x^(k*(m + 1 ) - 1)*(d + e*(x^(2*k)/f))^q*(a + c*(x^(4*k)/f))^p, x], x, (f*x)^(1/k)], x] ] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]
Int[(((f_.)*(x_))^(m_)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( n2_.)), x_Symbol] :> Simp[d/a Int[(f*x)^m*(d + e*x^n)^(q - 1), x], x] + S imp[1/(a*f^n) Int[(f*x)^(m + n)*(d + e*x^n)^(q - 1)*(Simp[a*e - c*d*x^n, x]/(a + c*x^(2*n))), x], x] /; FreeQ[{a, c, d, e, f}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && !IntegerQ[q] && GtQ[q, 0] && LtQ[m, 0]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.00 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92
method | result | size |
pseudoelliptic | \(\frac {-5 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}-a b \right )}{\sum }\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{4}-a}\right ) b \,x^{3}-8 \left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \left (a \,x^{2}-b \right )}{20 x^{3} b^{2}}\) | \(100\) |
1/20*(-5*a*sum(_R*ln((-_R*x+(x^2*(a*x^2-b))^(1/4))/x)/(_R^4-a),_R=RootOf(_ Z^8-2*_Z^4*a+a^2-a*b))*b*x^3-8*(x^2*(a*x^2-b))^(1/4)*(a*x^2-b))/x^3/b^2
Timed out. \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=\text {Timed out} \]
Not integrable
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.28 \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}}}{{\left (a x^{4} - b\right )} x^{4}} \,d x } \]
Not integrable
Time = 0.90 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.28 \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}}}{{\left (a x^{4} - b\right )} x^{4}} \,d x } \]
Not integrable
Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.29 \[ \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^4 \left (-b+a x^4\right )} \, dx=-\int \frac {{\left (a\,x^4-b\,x^2\right )}^{1/4}}{x^4\,\left (b-a\,x^4\right )} \,d x \]