Integrand size = 26, antiderivative size = 87 \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]
-1/4*arctan(2^(1/4)*a^(1/4)*x/(a*x^4-b)^(1/4))*2^(1/4)/a^(3/4)/b+1/4*arcta nh(2^(1/4)*a^(1/4)*x/(a*x^4-b)^(1/4))*2^(1/4)/a^(3/4)/b
Time = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\frac {-\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]
(-ArcTan[(2^(1/4)*a^(1/4)*x)/(-b + a*x^4)^(1/4)] + ArcTanh[(2^(1/4)*a^(1/4 )*x)/(-b + a*x^4)^(1/4)])/(2*2^(3/4)*a^(3/4)*b)
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {996, 27, 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.
\(\displaystyle \int \frac {x^2}{\left (a x^4-b\right )^{3/4} \left (a x^4+b\right )} \, dx\) |
\(\Big \downarrow \) 996 |
\(\displaystyle \int \frac {x^2}{b \sqrt {a x^4-b} \left (1-\frac {2 a x^4}{a x^4-b}\right )}d\frac {x}{\sqrt [4]{a x^4-b}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x^2}{\sqrt {a x^4-b} \left (1-\frac {2 a x^4}{a x^4-b}\right )}d\frac {x}{\sqrt [4]{a x^4-b}}}{b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {\frac {\int \frac {1}{1-\frac {\sqrt {2} \sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {2} \sqrt {a}}-\frac {\int \frac {1}{\frac {\sqrt {2} \sqrt {a} x^2}{\sqrt {a x^4-b}}+1}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {2} \sqrt {a}}}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\int \frac {1}{1-\frac {\sqrt {2} \sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {2} \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4}}}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4}}}{b}\) |
(-1/2*ArcTan[(2^(1/4)*a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2^(3/4)*a^(3/4)) + A rcTanh[(2^(1/4)*a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*2^(3/4)*a^(3/4)))/b
3.12.84.3.1 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[((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[(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_)*((c_) + (d_.)*(x_)^(n_))^(q_.) , x_Symbol] :> With[{k = Denominator[p]}, Simp[k*(a^(p + (m + 1)/n)/n) Su bst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p + q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && RationalQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]
Time = 3.91 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-2^{\frac {1}{4}} a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2^{\frac {1}{4}} a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right )\right )}{8 a^{\frac {3}{4}} b}\) | \(87\) |
1/8*2^(1/4)/a^(3/4)*(ln((-2^(1/4)*a^(1/4)*x-(a*x^4-b)^(1/4))/(2^(1/4)*a^(1 /4)*x-(a*x^4-b)^(1/4)))+2*arctan(1/2*2^(3/4)/a^(1/4)/x*(a*x^4-b)^(1/4)))/b
Timed out. \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int \frac {x^{2}}{\left (a x^{4} - b\right )^{\frac {3}{4}} \left (a x^{4} + b\right )}\, dx \]
\[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int \frac {x^2}{\left (a\,x^4+b\right )\,{\left (a\,x^4-b\right )}^{3/4}} \,d x \]