Integrand size = 23, antiderivative size = 83 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b} \]
1/4*arctan(1/2*a^(1/4)*x*2^(3/4)/(a*x^4+b)^(1/4))*2^(1/4)/a^(1/4)/b+1/4*ar ctanh(1/2*a^(1/4)*x*2^(3/4)/(a*x^4+b)^(1/4))*2^(1/4)/a^(1/4)/b
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b} \]
(ArcTan[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))] + ArcTanh[(a^(1/4)*x)/(2^ (1/4)*(b + a*x^4)^(1/4))])/(2*2^(3/4)*a^(1/4)*b)
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {902, 756, 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.
| \(\displaystyle \int \frac {1}{\sqrt [4]{a x^4+b} \left (a x^4+2 b\right )} \, dx\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \int \frac {1}{2 b-\frac {a b x^4}{a x^4+b}}d\frac {x}{\sqrt [4]{a x^4+b}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {2}-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {2} b}+\frac {\int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}+\sqrt {2}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {2} b}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {2}-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {2} b}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}\) |
ArcTan[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))]/(2*2^(3/4)*a^(1/4)*b) + Ar cTanh[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))]/(2*2^(3/4)*a^(1/4)*b)
3.12.15.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b , c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Time = 3.79 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96
| method | result | size |
| pseudoelliptic | \(\frac {2^{\frac {1}{4}} \left (-2 \arctan \left (\frac {2^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {2^{\frac {3}{4}} a^{\frac {1}{4}} x +2 \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{-2^{\frac {3}{4}} a^{\frac {1}{4}} x +2 \left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right )\right )}{8 a^{\frac {1}{4}} b}\) | \(80\) |
1/8*2^(1/4)*(-2*arctan(1/a^(1/4)/x*2^(1/4)*(a*x^4+b)^(1/4))+ln((2^(3/4)*a^ (1/4)*x+2*(a*x^4+b)^(1/4))/(-2^(3/4)*a^(1/4)*x+2*(a*x^4+b)^(1/4))))/a^(1/4 )/b
Result contains complex when optimal does not.
Time = 105.62 (sec) , antiderivative size = 509, normalized size of antiderivative = 6.13 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\frac {1}{8} \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {8 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} a b^{3} x^{2} \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {1}{a b^{4}}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 \, a b x^{4} + 2 \, b^{2}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) - \frac {1}{8} \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {8 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} a b^{3} x^{2} \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} - 2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {1}{a b^{4}}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 \, a b x^{4} + 2 \, b^{2}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) + \frac {1}{8} i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {8 i \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} a b^{3} x^{2} \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {1}{a b^{4}}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (-3 i \, a b x^{4} - 2 i \, b^{2}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) - \frac {1}{8} i \, \left (\frac {1}{8}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {-8 i \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} a b^{3} x^{2} \left (\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {1}{a b^{4}}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 i \, a b x^{4} + 2 i \, b^{2}\right )} \left (\frac {1}{a b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) \]
1/8*(1/8)^(1/4)*(1/(a*b^4))^(1/4)*log(1/2*(8*(1/8)^(3/4)*sqrt(a*x^4 + b)*a *b^3*x^2*(1/(a*b^4))^(3/4) + 2*sqrt(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt( 1/(a*b^4)) + 2*(a*x^4 + b)^(3/4)*x + (1/8)^(1/4)*(3*a*b*x^4 + 2*b^2)*(1/(a *b^4))^(1/4))/(a*x^4 + 2*b)) - 1/8*(1/8)^(1/4)*(1/(a*b^4))^(1/4)*log(-1/2* (8*(1/8)^(3/4)*sqrt(a*x^4 + b)*a*b^3*x^2*(1/(a*b^4))^(3/4) - 2*sqrt(1/2)*( a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(1/(a*b^4)) - 2*(a*x^4 + b)^(3/4)*x + (1/8) ^(1/4)*(3*a*b*x^4 + 2*b^2)*(1/(a*b^4))^(1/4))/(a*x^4 + 2*b)) + 1/8*I*(1/8) ^(1/4)*(1/(a*b^4))^(1/4)*log(-1/2*(8*I*(1/8)^(3/4)*sqrt(a*x^4 + b)*a*b^3*x ^2*(1/(a*b^4))^(3/4) + 2*sqrt(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(1/(a*b ^4)) - 2*(a*x^4 + b)^(3/4)*x + (1/8)^(1/4)*(-3*I*a*b*x^4 - 2*I*b^2)*(1/(a* b^4))^(1/4))/(a*x^4 + 2*b)) - 1/8*I*(1/8)^(1/4)*(1/(a*b^4))^(1/4)*log(-1/2 *(-8*I*(1/8)^(3/4)*sqrt(a*x^4 + b)*a*b^3*x^2*(1/(a*b^4))^(3/4) + 2*sqrt(1/ 2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(1/(a*b^4)) - 2*(a*x^4 + b)^(3/4)*x + ( 1/8)^(1/4)*(3*I*a*b*x^4 + 2*I*b^2)*(1/(a*b^4))^(1/4))/(a*x^4 + 2*b))
\[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\int \frac {1}{\sqrt [4]{a x^{4} + b} \left (a x^{4} + 2 b\right )}\, dx \]
\[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\int { \frac {1}{{\left (a x^{4} + 2 \, b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\int { \frac {1}{{\left (a x^{4} + 2 \, b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (2 b+a x^4\right )} \, dx=\int \frac {1}{{\left (a\,x^4+b\right )}^{1/4}\,\left (a\,x^4+2\,b\right )} \,d x \]