Integrand size = 21, antiderivative size = 162 \[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=-\frac {d x \left (a+b x^4\right )^{3/4}}{4 c (b c-a d) \left (c+d x^4\right )}+\frac {(4 b c-3 a d) \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} (b c-a d)^{5/4}}+\frac {(4 b c-3 a d) \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} (b c-a d)^{5/4}} \]
-1/4*d*x*(b*x^4+a)^(3/4)/c/(-a*d+b*c)/(d*x^4+c)+1/8*(-3*a*d+4*b*c)*arctan( (-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(7/4)/(-a*d+b*c)^(5/4)+1/8*( -3*a*d+4*b*c)*arctanh((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(7/4)/ (-a*d+b*c)^(5/4)
Result contains complex when optimal does not.
Time = 2.40 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \left (-\frac {(2-2 i) c^{3/4} d x \left (a+b x^4\right )^{3/4}}{(b c-a d) \left (c+d x^4\right )}+\frac {(4 b c-3 a d) \arctan \left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}-\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{(b c-a d)^{5/4}}+\frac {(4 b c-3 a d) \text {arctanh}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}+\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{(b c-a d)^{5/4}}\right )}{c^{7/4}} \]
((1/16 + I/16)*(((-2 + 2*I)*c^(3/4)*d*x*(a + b*x^4)^(3/4))/((b*c - a*d)*(c + d*x^4)) + ((4*b*c - 3*a*d)*ArcTan[(((1 - I)*(b*c - a*d)^(1/4)*x^2)/(c^( 1/4)*(a + b*x^4)^(1/4)) - ((1 + I)*c^(1/4)*(a + b*x^4)^(1/4))/(b*c - a*d)^ (1/4))/(2*x)])/(b*c - a*d)^(5/4) + ((4*b*c - 3*a*d)*ArcTanh[(((1 - I)*(b*c - a*d)^(1/4)*x^2)/(c^(1/4)*(a + b*x^4)^(1/4)) + ((1 + I)*c^(1/4)*(a + b*x ^4)^(1/4))/(b*c - a*d)^(1/4))/(2*x)])/(b*c - a*d)^(5/4)))/c^(7/4)
Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {907, 902, 756, 218, 221}
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.
| \(\displaystyle \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 907 |
| \(\displaystyle \frac {(4 b c-3 a d) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{4 c (b c-a d)}-\frac {d x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \frac {(4 b c-3 a d) \int \frac {1}{c-\frac {(b c-a d) x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 c (b c-a d)}-\frac {d x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {(4 b c-3 a d) \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{4 c (b c-a d)}-\frac {d x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(4 b c-3 a d) \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{4 c (b c-a d)}-\frac {d x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(4 b c-3 a d) \left (\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{4 c (b c-a d)}-\frac {d x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right ) (b c-a d)}\) |
-1/4*(d*x*(a + b*x^4)^(3/4))/(c*(b*c - a*d)*(c + d*x^4)) + ((4*b*c - 3*a*d )*(ArcTan[((b*c - a*d)^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/4))]/(2*c^(3/4)*(b *c - a*d)^(1/4)) + ArcTanh[((b*c - a*d)^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/4 ))]/(2*c^(3/4)*(b*c - a*d)^(1/4))))/(4*c*(b*c - a*d))
3.3.8.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Simp[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q} , x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] || ! LtQ[q, -1]) && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(342\) vs. \(2(134)=268\).
Time = 4.40 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.12
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {3 \left (a d -\frac {4 b c}{3}\right ) \sqrt {2}\, \left (d \,x^{4}+c \right ) \ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )}{32}+\frac {3 \left (a d -\frac {4 b c}{3}\right ) \sqrt {2}\, \left (d \,x^{4}+c \right ) \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x -\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )}{16}-\frac {3 \left (a d -\frac {4 b c}{3}\right ) \sqrt {2}\, \left (d \,x^{4}+c \right ) \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x +\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )}{16}+\frac {d \left (b \,x^{4}+a \right )^{\frac {3}{4}} x c \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}}}{4}}{c^{2} \left (a d -b c \right ) \left (d \,x^{4}+c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}}}\) | \(343\) |
3/16*(-1/2*(a*d-4/3*b*c)*2^(1/2)*(d*x^4+c)*ln((-((a*d-b*c)/c)^(1/4)*(b*x^4 +a)^(1/4)*2^(1/2)*x+((a*d-b*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2))/(((a*d-b*c)/c )^(1/4)*(b*x^4+a)^(1/4)*2^(1/2)*x+((a*d-b*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2)) )+(a*d-4/3*b*c)*2^(1/2)*(d*x^4+c)*arctan((((a*d-b*c)/c)^(1/4)*x-2^(1/2)*(b *x^4+a)^(1/4))/((a*d-b*c)/c)^(1/4)/x)-(a*d-4/3*b*c)*2^(1/2)*(d*x^4+c)*arct an((((a*d-b*c)/c)^(1/4)*x+2^(1/2)*(b*x^4+a)^(1/4))/((a*d-b*c)/c)^(1/4)/x)+ 4/3*d*(b*x^4+a)^(3/4)*x*c*((a*d-b*c)/c)^(1/4))/((a*d-b*c)/c)^(1/4)/c^2/(a* d-b*c)/(d*x^4+c)
Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\int \frac {1}{\sqrt [4]{a + b x^{4}} \left (c + d x^{4}\right )^{2}}\, dx \]
\[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (d x^{4} + c\right )}^{2}} \,d x } \]
\[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (d x^{4} + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{1/4}\,{\left (d\,x^4+c\right )}^2} \,d x \]