Integrand size = 31, antiderivative size = 295 \[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=-\frac {2 d \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {3}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {f x} \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{3/2} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f^3 \sqrt {a+b x^2+c x^4}} \]
2/3*e*(f*x)^(3/2)*AppellF1(3/4,1/2,1/2,7/4,-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(-1/4,1/2,1/2,3/4,-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/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2)
Time = 11.54 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.21 \[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\frac {2 x \left (-21 d \left (a+b x^2+c x^4\right )+7 (b d+a e) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+9 c d x^4 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{21 a (f x)^{3/2} \sqrt {a+b x^2+c x^4}} \]
(2*x*(-21*d*(a + b*x^2 + c*x^4) + 7*(b*d + a*e)*x^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])]*AppellF1[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 9*c*d*x^4*Sqr t[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqr t[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a* c])]))/(21*a*(f*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4])
Time = 0.49 (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.
\(\displaystyle \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1674 |
\(\displaystyle \int \left (\frac {d}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}}+\frac {e \sqrt {f x}}{f^2 \sqrt {a+b x^2+c x^4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e (f x)^{3/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f^3 \sqrt {a+b x^2+c x^4}}-\frac {2 d \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (-\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {3}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {f x} \sqrt {a+b x^2+c x^4}}\) |
(-2*d*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, 3/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f*Sqrt[f*x]*Sqrt[a + b*x ^2 + c*x^4]) + (2*e*(f*x)^(3/2)*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[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/ (3*f^3*Sqrt[a + b*x^2 + c*x^4])
3.3.15.3.1 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])
\[\int \frac {e \,x^{2}+d}{\left (f x \right )^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}}d x\]
\[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {c x^{4} + b x^{2} + a} \left (f x\right )^{\frac {3}{2}}} \,d x } \]
integral(sqrt(c*x^4 + b*x^2 + a)*(e*x^2 + d)*sqrt(f*x)/(c*f^2*x^6 + b*f^2* x^4 + a*f^2*x^2), x)
\[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {d + e x^{2}}{\left (f x\right )^{\frac {3}{2}} \sqrt {a + b x^{2} + c x^{4}}}\, dx \]
\[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {c x^{4} + b x^{2} + a} \left (f x\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {c x^{4} + b x^{2} + a} \left (f x\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {e\,x^2+d}{{\left (f\,x\right )}^{3/2}\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]