Integrand size = 30, antiderivative size = 104 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\frac {x^3}{6 b^2 \left (b+a x^4\right )^{3/4}}+\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{3/4} b^2}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{3/4} b^2} \]
-1/6*x^3/b^2/(a*x^4+b)^(3/4)+1/8*arctan(2^(1/4)*a^(1/4)*x/(a*x^4+b)^(1/4)) *2^(1/4)/a^(3/4)/b^2-1/8*arctanh(2^(1/4)*a^(1/4)*x/(a*x^4+b)^(1/4))*2^(1/4 )/a^(3/4)/b^2
Time = 0.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\frac {\frac {4 x^3}{\left (b+a x^4\right )^{3/4}}-\frac {3 \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{a^{3/4}}+\frac {3 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{a^{3/4}}}{24 b^2} \]
-1/24*((4*x^3)/(b + a*x^4)^(3/4) - (3*2^(1/4)*ArcTan[(2^(1/4)*a^(1/4)*x)/( b + a*x^4)^(1/4)])/a^(3/4) + (3*2^(1/4)*ArcTanh[(2^(1/4)*a^(1/4)*x)/(b + a *x^4)^(1/4)])/a^(3/4))/b^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.92 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1388, 1013, 25, 1012}
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.
\(\displaystyle \int \frac {x^2}{\left (a x^4+b\right )^{3/4} \left (a^2 x^8-b^2\right )} \, dx\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \int \frac {x^2}{\left (a x^4-b\right ) \left (a x^4+b\right )^{7/4}}dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\left (\frac {a x^4}{b}+1\right )^{3/4} \int -\frac {x^2}{\left (b-a x^4\right ) \left (\frac {a x^4}{b}+1\right )^{7/4}}dx}{b \left (a x^4+b\right )^{3/4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\left (\frac {a x^4}{b}+1\right )^{3/4} \int \frac {x^2}{\left (b-a x^4\right ) \left (\frac {a x^4}{b}+1\right )^{7/4}}dx}{b \left (a x^4+b\right )^{3/4}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {4 x^3 \operatorname {Gamma}\left (\frac {7}{4}\right ) \left (11 \left (-4 a^2 x^8-3 a b x^4+7 b^2\right ) \operatorname {Hypergeometric2F1}\left (1,1,\frac {11}{4},-\frac {2 a x^4}{b-a x^4}\right )-32 a x^4 \left (a x^4+b\right ) \operatorname {Hypergeometric2F1}\left (2,2,\frac {15}{4},-\frac {2 a x^4}{b-a x^4}\right )\right )}{693 b^2 \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (b-a x^4\right )^2 \left (a x^4+b\right )^{3/4}}\) |
(-4*x^3*Gamma[7/4]*(11*(7*b^2 - 3*a*b*x^4 - 4*a^2*x^8)*Hypergeometric2F1[1 , 1, 11/4, (-2*a*x^4)/(b - a*x^4)] - 32*a*x^4*(b + a*x^4)*Hypergeometric2F 1[2, 2, 15/4, (-2*a*x^4)/(b - a*x^4)]))/(693*b^2*(b - a*x^4)^2*(b + a*x^4) ^(3/4)*Gamma[3/4])
3.15.99.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Time = 0.47 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19
method | result | size |
pseudoelliptic | \(-\frac {a \,x^{3}+\frac {3 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right ) 2^{\frac {1}{4}} a^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {3}{4}}}{4}+\frac {3 \ln \left (\frac {-x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} a^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {3}{4}}}{8}}{6 \left (a \,x^{4}+b \right )^{\frac {3}{4}} a \,b^{2}}\) | \(124\) |
-1/6/(a*x^4+b)^(3/4)*(a*x^3+3/4*arctan(1/2*2^(3/4)/a^(1/4)/x*(a*x^4+b)^(1/ 4))*2^(1/4)*a^(1/4)*(a*x^4+b)^(3/4)+3/8*ln((-x*2^(1/4)*a^(1/4)-(a*x^4+b)^( 1/4))/(x*2^(1/4)*a^(1/4)-(a*x^4+b)^(1/4)))*2^(1/4)*a^(1/4)*(a*x^4+b)^(3/4) )/a/b^2
Timed out. \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\int \frac {x^{2}}{\left (a x^{4} - b\right ) \left (a x^{4} + b\right )^{\frac {7}{4}}}\, dx \]
\[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\int { \frac {x^{2}}{{\left (a^{2} x^{8} - b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\int { \frac {x^{2}}{{\left (a^{2} x^{8} - b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\int \frac {x^2}{\left (b^2-a^2\,x^8\right )\,{\left (a\,x^4+b\right )}^{3/4}} \,d x \]