Integrand size = 17, antiderivative size = 74 \[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\frac {x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac {3 \arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}}+\frac {3 \text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}} \]
1/8*x*(x^4+1)^(3/4)/(x^4+2)+3/32*arctan(1/2*x*2^(3/4)/(x^4+1)^(1/4))*2^(1/ 4)+3/32*arctanh(1/2*x*2^(3/4)/(x^4+1)^(1/4))*2^(1/4)
Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\frac {x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac {3 \arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}}+\frac {3 \text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}} \]
(x*(1 + x^4)^(3/4))/(8*(2 + x^4)) + (3*ArcTan[x/(2^(1/4)*(1 + x^4)^(1/4))] )/(16*2^(3/4)) + (3*ArcTanh[x/(2^(1/4)*(1 + x^4)^(1/4))])/(16*2^(3/4))
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {903, 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.
| \(\displaystyle \int \frac {\left (x^4+1\right )^{3/4}}{\left (x^4+2\right )^2} \, dx\) |
| \(\Big \downarrow \) 903 |
| \(\displaystyle \frac {3}{8} \int \frac {1}{\sqrt [4]{x^4+1} \left (x^4+2\right )}dx+\frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \frac {3}{8} \int \frac {1}{2-\frac {x^4}{x^4+1}}d\frac {x}{\sqrt [4]{x^4+1}}+\frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {3}{8} \left (\frac {\int \frac {1}{\sqrt {2}-\frac {x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}+\frac {\int \frac {1}{\frac {x^2}{\sqrt {x^4+1}}+\sqrt {2}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}\right )+\frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {3}{8} \left (\frac {\int \frac {1}{\sqrt {2}-\frac {x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}+\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}}\right )+\frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{8} \left (\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}}\right )+\frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}\) |
(x*(1 + x^4)^(3/4))/(8*(2 + x^4)) + (3*(ArcTan[x/(2^(1/4)*(1 + x^4)^(1/4)) ]/(2*2^(3/4)) + ArcTanh[x/(2^(1/4)*(1 + x^4)^(1/4))]/(2*2^(3/4))))/8
3.4.17.3.1 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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*n*(p + 1))), x] - Simp[ c*(q/(a*(p + 1))) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]
Time = 2.65 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18
| method | result | size |
| pseudoelliptic | \(\frac {-3 \left (x^{4}+2\right ) \left (2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )-\ln \left (\frac {-x 2^{\frac {3}{4}}-2 \left (x^{4}+1\right )^{\frac {1}{4}}}{x 2^{\frac {3}{4}}-2 \left (x^{4}+1\right )^{\frac {1}{4}}}\right )\right ) 2^{\frac {1}{4}}+8 \left (x^{4}+1\right )^{\frac {3}{4}} x}{64 x^{4}+128}\) | \(87\) |
| trager | \(\frac {\left (x^{4}+1\right )^{\frac {3}{4}} x}{8 x^{4}+16}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{4}+2}\right )}{64}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )}{x^{4}+2}\right )}{64}\) | \(229\) |
| risch | \(\frac {\left (x^{4}+1\right )^{\frac {3}{4}} x}{8 x^{4}+16}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{4}+2}\right )}{64}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )}{x^{4}+2}\right )}{64}\) | \(229\) |
(-3*(x^4+2)*(2*arctan(1/x*2^(1/4)*(x^4+1)^(1/4))-ln((-x*2^(3/4)-2*(x^4+1)^ (1/4))/(x*2^(3/4)-2*(x^4+1)^(1/4))))*2^(1/4)+8*(x^4+1)^(3/4)*x)/(64*x^4+12 8)
Result contains complex when optimal does not.
Time = 3.41 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.23 \[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\frac {3 \cdot 8^{\frac {3}{4}} {\left (x^{4} + 2\right )} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - 3 \cdot 8^{\frac {3}{4}} {\left (-i \, x^{4} - 2 i\right )} \log \left (-\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 i \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 i \, x^{4} + 2 i\right )} - 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - 3 \cdot 8^{\frac {3}{4}} {\left (i \, x^{4} + 2 i\right )} \log \left (-\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 i \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (-3 i \, x^{4} - 2 i\right )} - 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - 3 \cdot 8^{\frac {3}{4}} {\left (x^{4} + 2\right )} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) + 64 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{512 \, {\left (x^{4} + 2\right )}} \]
1/512*(3*8^(3/4)*(x^4 + 2)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 8*8^(1/4)* sqrt(x^4 + 1)*x^2 + 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^(3/4)*x)/(x^4 + 2)) - 3*8^(3/4)*(-I*x^4 - 2*I)*log(-(8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 8*I*8^(1 /4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(3*I*x^4 + 2*I) - 16*(x^4 + 1)^(3/4)*x)/(x ^4 + 2)) - 3*8^(3/4)*(I*x^4 + 2*I)*log(-(8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 8 *I*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(-3*I*x^4 - 2*I) - 16*(x^4 + 1)^(3/ 4)*x)/(x^4 + 2)) - 3*8^(3/4)*(x^4 + 2)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 8*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^(3/4)*x )/(x^4 + 2)) + 64*(x^4 + 1)^(3/4)*x)/(x^4 + 2)
\[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\int \frac {\left (x^{4} + 1\right )^{\frac {3}{4}}}{\left (x^{4} + 2\right )^{2}}\, dx \]
\[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\int { \frac {{\left (x^{4} + 1\right )}^{\frac {3}{4}}}{{\left (x^{4} + 2\right )}^{2}} \,d x } \]
\[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\int { \frac {{\left (x^{4} + 1\right )}^{\frac {3}{4}}}{{\left (x^{4} + 2\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx=\int \frac {{\left (x^4+1\right )}^{3/4}}{{\left (x^4+2\right )}^2} \,d x \]