Integrand size = 31, antiderivative size = 295 \[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\frac {2 d \sqrt {f x} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{5/2} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 f^3 \sqrt {a+b x^2+c x^4}} \]
2/5*e*(f*x)^(5/2)*AppellF1(5/4,1/2,1/2,9/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)) ,-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2) *(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/f^3/(c*x^4+b*x^2+a)^(1/2)+2*d*Ap pellF1(1/4,1/2,1/2,5/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c +b^2)^(1/2)))*(f*x)^(1/2)*(1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c* x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/f/(c*x^4+b*x^2+a)^(1/2)
Time = 11.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.82 \[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\frac {2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \left (5 d x \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+e x^3 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{5 \sqrt {f x} \sqrt {a+b x^2+c x^4}} \]
(2*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*(5*d*x*AppellF1[1 /4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqr t[b^2 - 4*a*c])] + e*x^3*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b + Sqrt [b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(5*Sqrt[f*x]*Sqrt[a + b*x^2 + c*x^4])
Time = 0.47 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1674, 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.
\(\displaystyle \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1674 |
\(\displaystyle \int \left (\frac {d}{\sqrt {f x} \sqrt {a+b x^2+c x^4}}+\frac {e (f x)^{3/2}}{f^2 \sqrt {a+b x^2+c x^4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 d \sqrt {f x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{5/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 f^3 \sqrt {a+b x^2+c x^4}}\) |
(2*d*Sqrt[f*x]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x ^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f*Sqrt[a + b*x^2 + c*x^4]) + (2*e*(f*x)^(5/2)*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]* Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[5/4, 1/2, 1/2, 9/4, ( -2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(5 *f^3*Sqrt[a + b*x^2 + c*x^4])
3.3.14.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
\[\int \frac {e \,x^{2}+d}{\sqrt {f x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}d x\]
\[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {c x^{4} + b x^{2} + a} \sqrt {f x}} \,d x } \]
\[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {d + e x^{2}}{\sqrt {f x} \sqrt {a + b x^{2} + c x^{4}}}\, dx \]
\[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {c x^{4} + b x^{2} + a} \sqrt {f x}} \,d x } \]
\[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {c x^{4} + b x^{2} + a} \sqrt {f x}} \,d x } \]
Timed out. \[ \int \frac {d+e x^2}{\sqrt {f x} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {e\,x^2+d}{\sqrt {f\,x}\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]