Integrand size = 28, antiderivative size = 122 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {\sqrt [4]{b x^2+a x^4} \left (32 b+b x^2+4 a x^4\right )}{16 x}+\frac {\left (32 a b+3 b^2\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{32 a^{3/4}}+\frac {\left (-32 a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{32 a^{3/4}} \]
1/16*(a*x^4+b*x^2)^(1/4)*(4*a*x^4+b*x^2+32*b)/x+1/32*(32*a*b+3*b^2)*arctan (a^(1/4)*x/(a*x^4+b*x^2)^(1/4))/a^(3/4)+1/32*(-32*a*b-3*b^2)*arctanh(a^(1/ 4)*x/(a*x^4+b*x^2)^(1/4))/a^(3/4)
Time = 0.49 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {x \left (b+a x^2\right )^{3/4} \left (2 a^{3/4} \sqrt [4]{b+a x^2} \left (4 a x^4+b \left (32+x^2\right )\right )+b (32 a+3 b) \sqrt {x} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )-b (32 a+3 b) \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{32 a^{3/4} \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \]
(x*(b + a*x^2)^(3/4)*(2*a^(3/4)*(b + a*x^2)^(1/4)*(4*a*x^4 + b*(32 + x^2)) + b*(32*a + 3*b)*Sqrt[x]*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] - b* (32*a + 3*b)*Sqrt[x]*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]))/(32*a^ (3/4)*(x^2*(b + a*x^2))^(3/4))
Leaf count is larger than twice the leaf count of optimal. \(321\) vs. \(2(122)=244\).
Time = 0.54 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.63, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2449, 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.
\(\displaystyle \int \frac {\left (a x^4-b\right ) \sqrt [4]{a x^4+b x^2}}{x^2} \, dx\) |
\(\Big \downarrow \) 2449 |
\(\displaystyle \int \left (a x^2 \sqrt [4]{a x^4+b x^2}-\frac {b \sqrt [4]{a x^4+b x^2}}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 b^2 x^{3/2} \left (a x^2+b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{32 a^{3/4} \left (a x^4+b x^2\right )^{3/4}}-\frac {3 b^2 x^{3/2} \left (a x^2+b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{32 a^{3/4} \left (a x^4+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\left (a x^4+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\left (a x^4+b x^2\right )^{3/4}}+\frac {1}{16} b x \sqrt [4]{a x^4+b x^2}+\frac {2 b \sqrt [4]{a x^4+b x^2}}{x}+\frac {1}{4} a x^3 \sqrt [4]{a x^4+b x^2}\) |
(2*b*(b*x^2 + a*x^4)^(1/4))/x + (b*x*(b*x^2 + a*x^4)^(1/4))/16 + (a*x^3*(b *x^2 + a*x^4)^(1/4))/4 + (a^(1/4)*b*x^(3/2)*(b + a*x^2)^(3/4)*ArcTan[(a^(1 /4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x^2 + a*x^4)^(3/4) + (3*b^2*x^(3/2)*(b + a*x^2)^(3/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(32*a^(3/4)*( b*x^2 + a*x^4)^(3/4)) - (a^(1/4)*b*x^(3/2)*(b + a*x^2)^(3/4)*ArcTanh[(a^(1 /4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x^2 + a*x^4)^(3/4) - (3*b^2*x^(3/2)*(b + a*x^2)^(3/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(32*a^(3/4)* (b*x^2 + a*x^4)^(3/4))
3.19.1.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_S ymbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ [{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) && !Integer Q[p] && NeQ[n, j]
Time = 0.36 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(-\frac {\frac {b x \left (\frac {3 b}{32}+a \right ) \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )}{2}+b x \left (\frac {3 b}{32}+a \right ) \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )-\frac {\left (b \left (x^{2}+32\right ) a^{\frac {3}{4}}+4 a^{\frac {7}{4}} x^{4}\right ) \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{16}}{a^{\frac {3}{4}} x}\) | \(129\) |
-1/a^(3/4)*(1/2*b*x*(3/32*b+a)*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)))+b*x*(3/32*b+a)*arctan(1/a^(1/4)*(x^2*(a*x^2+ b))^(1/4)/x)-1/16*(b*(x^2+32)*a^(3/4)+4*a^(7/4)*x^4)*(x^2*(a*x^2+b))^(1/4) )/x
Timed out. \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} - b\right )}{x^{2}}\, dx \]
\[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}}{x^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (102) = 204\).
Time = 0.29 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.27 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {\frac {8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{4}} b^{3} + 3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a b^{3}\right )} x^{4}}{b^{2}} + 256 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b^{2} + \frac {2 \, \sqrt {2} {\left (32 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} {\left (32 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (32 \, a b^{2} + 3 \, b^{3}\right )} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{\left (-a\right )^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (32 \, a b^{2} + 3 \, b^{3}\right )} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{128 \, b} \]
1/128*(8*((a + b/x^2)^(5/4)*b^3 + 3*(a + b/x^2)^(1/4)*a*b^3)*x^4/b^2 + 256 *(a + b/x^2)^(1/4)*b^2 + 2*sqrt(2)*(32*a*b^2 + 3*b^3)*arctan(1/2*sqrt(2)*( sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 2*sqrt( 2)*(32*a*b^2 + 3*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x ^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + sqrt(2)*(32*a*b^2 + 3*b^3)*log(sqrt(2) *(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/(-a)^(3/4) - s qrt(2)*(32*a*b^2 + 3*b^3)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt (-a) + sqrt(a + b/x^2))/(-a)^(3/4))/b
Timed out. \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=-\int \frac {\left (b-a\,x^4\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}}{x^2} \,d x \]