Integrand size = 21, antiderivative size = 167 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{\sqrt {d} \sqrt {-b c+a d} x}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{\sqrt {d} \sqrt {-b c+a d} x} \]
a^(1/4)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2 ),I)*(-b*x^2/a)^(1/2)/x/d^(1/2)/(a*d-b*c)^(1/2)-a^(1/4)*EllipticPi((b*x^2+ a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/x/d^( 1/2)/(a*d-b*c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 7.77 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=-\frac {6 a c x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\sqrt [4]{a+b x^2} \left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )} \]
(-6*a*c*x*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/((a + b* x^2)^(1/4)*(c + d*x^2)*(-6*a*c*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -( (d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 1/4, 2, 5/2, -((b*x^2)/a), -((d*x^2 )/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))
Time = 0.28 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {310, 993, 1542}
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.
\(\displaystyle \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 310 |
\(\displaystyle \frac {2 \sqrt {-\frac {b x^2}{a}} \int \frac {\sqrt {b x^2+a}}{\sqrt {1-\frac {b x^2+a}{a}} \left (b c-a d+d \left (b x^2+a\right )\right )}d\sqrt [4]{b x^2+a}}{x}\) |
\(\Big \downarrow \) 993 |
\(\displaystyle \frac {2 \sqrt {-\frac {b x^2}{a}} \left (\frac {\int \frac {1}{\left (\sqrt {a d-b c}+\sqrt {d} \sqrt {b x^2+a}\right ) \sqrt {1-\frac {b x^2+a}{a}}}d\sqrt [4]{b x^2+a}}{2 \sqrt {d}}-\frac {\int \frac {1}{\left (\sqrt {a d-b c}-\sqrt {d} \sqrt {b x^2+a}\right ) \sqrt {1-\frac {b x^2+a}{a}}}d\sqrt [4]{b x^2+a}}{2 \sqrt {d}}\right )}{x}\) |
\(\Big \downarrow \) 1542 |
\(\displaystyle \frac {2 \sqrt {-\frac {b x^2}{a}} \left (\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 \sqrt {d} \sqrt {a d-b c}}-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 \sqrt {d} \sqrt {a d-b c}}\right )}{x}\) |
(2*Sqrt[-((b*x^2)/a)]*((a^(1/4)*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(2*Sqrt[d]*Sqrt[-(b*c) + a*d]) - (a^(1/4)*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[ (a + b*x^2)^(1/4)/a^(1/4)], -1])/(2*Sqrt[d]*Sqrt[-(b*c) + a*d])))/x
3.4.25.3.1 Defintions of rubi rules used
Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim p[2*(Sqrt[(-b)*(x^2/a)]/x) Subst[Int[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d* x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* b) Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b) Int[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ {q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x ], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (d \,x^{2}+c \right )}d x\]
Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\int \frac {1}{\sqrt [4]{a + b x^{2}} \left (c + d x^{2}\right )}\, dx \]
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}} \,d x } \]
\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{1/4}\,\left (d\,x^2+c\right )} \,d x \]