Integrand size = 30, antiderivative size = 456 \[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\frac {\left (-1+\sqrt [4]{-1}\right ) \arctan \left (\frac {(-1)^{7/8} \sqrt {2+\sqrt {2}} \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}{(-1)^{3/4} \sqrt [4]{a^2-2 b} x^2+\sqrt {b+a x^4}}\right )}{8 \left (a^2-2 b\right )^{5/8}}+\frac {i \left (i \sqrt {2}+\sqrt {2 \left (3+2 \sqrt {2}\right )}\right ) \arctan \left (\frac {(-1)^{7/8} \left (-2+\sqrt {2}\right ) \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}{(-1)^{3/4} \sqrt {2-\sqrt {2}} \sqrt [4]{a^2-2 b} x^2+\sqrt {2-\sqrt {2}} \sqrt {b+a x^4}}\right )}{16 \left (a^2-2 b\right )^{5/8}}+\frac {\left (\sqrt {2}-i \sqrt {2 \left (3+2 \sqrt {2}\right )}\right ) \text {arctanh}\left (\frac {(-1)^{7/8} \sqrt [4]{a^2-2 b} x^2-\sqrt [8]{-1} \sqrt {b+a x^4}}{\sqrt {2-\sqrt {2}} \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}\right )}{16 \left (a^2-2 b\right )^{5/8}}+\frac {\left (-1+\sqrt [4]{-1}\right ) \text {arctanh}\left (\frac {(-1)^{7/8} \sqrt [4]{a^2-2 b} x^2-\sqrt [8]{-1} \sqrt {b+a x^4}}{\sqrt {2+\sqrt {2}} \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}\right )}{8 \left (a^2-2 b\right )^{5/8}} \]
1/8*(-1+(-1)^(1/4))*arctan((-1)^(7/8)*(2+2^(1/2))^(1/2)*(a^2-2*b)^(1/8)*x* (a*x^4+b)^(1/4)/((-1)^(3/4)*(a^2-2*b)^(1/4)*x^2+(a*x^4+b)^(1/2)))/(a^2-2*b )^(5/8)+1/16*I*(I*2^(1/2)+2+2^(1/2))*arctan((-1)^(7/8)*(-2+2^(1/2))*(a^2-2 *b)^(1/8)*x*(a*x^4+b)^(1/4)/((-1)^(3/4)*(2-2^(1/2))^(1/2)*(a^2-2*b)^(1/4)* x^2+(2-2^(1/2))^(1/2)*(a*x^4+b)^(1/2)))/(a^2-2*b)^(5/8)+1/16*(2^(1/2)-I*(2 +2^(1/2)))*arctanh(((-1)^(7/8)*(a^2-2*b)^(1/4)*x^2-(-1)^(1/8)*(a*x^4+b)^(1 /2))/(2-2^(1/2))^(1/2)/(a^2-2*b)^(1/8)/x/(a*x^4+b)^(1/4))/(a^2-2*b)^(5/8)+ 1/8*(-1+(-1)^(1/4))*arctanh(((-1)^(7/8)*(a^2-2*b)^(1/4)*x^2-(-1)^(1/8)*(a* x^4+b)^(1/2))/(2+2^(1/2))^(1/2)/(a^2-2*b)^(1/8)/x/(a*x^4+b)^(1/4))/(a^2-2* b)^(5/8)
Time = 10.51 (sec) , antiderivative size = 418, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\frac {-\frac {\sqrt [4]{a-\sqrt {a^2-2 b}} \arctan \left (\frac {\sqrt [4]{a^2-a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a^2-a \sqrt {a^2-2 b}-2 b}}+\frac {\sqrt [4]{a+\sqrt {a^2-2 b}} \arctan \left (\frac {\sqrt [4]{a^2+a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a+\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a^2+a \sqrt {a^2-2 b}-2 b}}-\frac {\sqrt [4]{a-\sqrt {a^2-2 b}} \text {arctanh}\left (\frac {\sqrt [4]{a^2-a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a^2-a \sqrt {a^2-2 b}-2 b}}+\frac {\sqrt [4]{a+\sqrt {a^2-2 b}} \text {arctanh}\left (\frac {\sqrt [4]{a^2+a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a+\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a^2+a \sqrt {a^2-2 b}-2 b}}}{4 \sqrt {a^2-2 b}} \]
(-(((a - Sqrt[a^2 - 2*b])^(1/4)*ArcTan[((a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1 /4)*x)/((a - Sqrt[a^2 - 2*b])^(1/4)*(b + a*x^4)^(1/4))])/(a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)) + ((a + Sqrt[a^2 - 2*b])^(1/4)*ArcTan[((a^2 + a*Sqrt [a^2 - 2*b] - 2*b)^(1/4)*x)/((a + Sqrt[a^2 - 2*b])^(1/4)*(b + a*x^4)^(1/4) )])/(a^2 + a*Sqrt[a^2 - 2*b] - 2*b)^(1/4) - ((a - Sqrt[a^2 - 2*b])^(1/4)*A rcTanh[((a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)*x)/((a - Sqrt[a^2 - 2*b])^(1 /4)*(b + a*x^4)^(1/4))])/(a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1/4) + ((a + Sqr t[a^2 - 2*b])^(1/4)*ArcTanh[((a^2 + a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)*x)/((a + Sqrt[a^2 - 2*b])^(1/4)*(b + a*x^4)^(1/4))])/(a^2 + a*Sqrt[a^2 - 2*b] - 2 *b)^(1/4))/(4*Sqrt[a^2 - 2*b])
Time = 0.88 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1852, 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.
| \(\displaystyle \int \frac {x^4}{\sqrt [4]{a x^4+b} \left (2 a x^4+b+2 x^8\right )} \, dx\) |
| \(\Big \downarrow \) 1852 |
| \(\displaystyle \int \left (\frac {1-\frac {a}{\sqrt {a^2-2 b}}}{\left (-2 \sqrt {a^2-2 b}+2 a+4 x^4\right ) \sqrt [4]{a x^4+b}}+\frac {\frac {a}{\sqrt {a^2-2 b}}+1}{\left (2 \sqrt {a^2-2 b}+2 a+4 x^4\right ) \sqrt [4]{a x^4+b}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [4]{a-\sqrt {a^2-2 b}} \arctan \left (\frac {x \sqrt [4]{-a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2-2 b} \sqrt [4]{-a \sqrt {a^2-2 b}+a^2-2 b}}+\frac {\sqrt [4]{\sqrt {a^2-2 b}+a} \arctan \left (\frac {x \sqrt [4]{a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2-2 b} \sqrt [4]{a \sqrt {a^2-2 b}+a^2-2 b}}-\frac {\sqrt [4]{a-\sqrt {a^2-2 b}} \text {arctanh}\left (\frac {x \sqrt [4]{-a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2-2 b} \sqrt [4]{-a \sqrt {a^2-2 b}+a^2-2 b}}+\frac {\sqrt [4]{\sqrt {a^2-2 b}+a} \text {arctanh}\left (\frac {x \sqrt [4]{a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2-2 b} \sqrt [4]{a \sqrt {a^2-2 b}+a^2-2 b}}\) |
-1/4*((a - Sqrt[a^2 - 2*b])^(1/4)*ArcTan[((a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^ (1/4)*x)/((a - Sqrt[a^2 - 2*b])^(1/4)*(b + a*x^4)^(1/4))])/(Sqrt[a^2 - 2*b ]*(a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)) + ((a + Sqrt[a^2 - 2*b])^(1/4)*Ar cTan[((a^2 + a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)*x)/((a + Sqrt[a^2 - 2*b])^(1/4 )*(b + a*x^4)^(1/4))])/(4*Sqrt[a^2 - 2*b]*(a^2 + a*Sqrt[a^2 - 2*b] - 2*b)^ (1/4)) - ((a - Sqrt[a^2 - 2*b])^(1/4)*ArcTanh[((a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)*x)/((a - Sqrt[a^2 - 2*b])^(1/4)*(b + a*x^4)^(1/4))])/(4*Sqrt[a^ 2 - 2*b]*(a^2 - a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)) + ((a + Sqrt[a^2 - 2*b])^( 1/4)*ArcTanh[((a^2 + a*Sqrt[a^2 - 2*b] - 2*b)^(1/4)*x)/((a + Sqrt[a^2 - 2* b])^(1/4)*(b + a*x^4)^(1/4))])/(4*Sqrt[a^2 - 2*b]*(a^2 + a*Sqrt[a^2 - 2*b] - 2*b)^(1/4))
3.31.46.3.1 Defintions of rubi rules used
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q, ( f*x)^m/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, f, q, n}, x ] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && !IntegerQ[q] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.09
| method | result | size |
| pseudoelliptic | \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a^{2}+2 b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{5}}\right )}{8}\) | \(42\) |
Timed out. \[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\int \frac {x^{4}}{\sqrt [4]{a x^{4} + b} \left (2 a x^{4} + b + 2 x^{8}\right )}\, dx \]
\[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\int { \frac {x^{4}}{{\left (2 \, x^{8} + 2 \, a x^{4} + b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\int { \frac {x^{4}}{{\left (2 \, x^{8} + 2 \, a x^{4} + b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\sqrt [4]{b+a x^4} \left (b+2 a x^4+2 x^8\right )} \, dx=\int \frac {x^4}{{\left (a\,x^4+b\right )}^{1/4}\,\left (2\,x^8+2\,a\,x^4+b\right )} \,d x \]