Integrand size = 24, antiderivative size = 148 \[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {2 \sqrt {d x} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{2},\frac {3}{2},\frac {5}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d \sqrt {a+b x^2+c x^4}} \]
2*AppellF1(1/4,3/2,3/2,5/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4 *a*c+b^2)^(1/2)))*(d*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)/a/d/(c*x^4+b*x^2+a)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(366\) vs. \(2(148)=296\).
Time = 11.37 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.47 \[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {x \left (-5 \left (b^2-2 a c+b c x^2\right )-5 \left (b^2-6 a c\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+b c x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{5 a \left (-b^2+4 a c\right ) \sqrt {d x} \sqrt {a+b x^2+c x^4}} \]
(x*(-5*(b^2 - 2*a*c + b*c*x^2) - 5*(b^2 - 6*a*c)*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[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])] + b*c*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[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*a*(-b^2 + 4*a*c)*Sqrt[d*x]*Sqrt[a + b*x^2 + c*x^4])
Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1461, 394}
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.
\(\displaystyle \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1461 |
\(\displaystyle \frac {\sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \int \frac {1}{\sqrt {d x} \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}}dx}{a \sqrt {a+b x^2+c x^4}}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {2 \sqrt {d x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{2},\frac {3}{2},\frac {5}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d \sqrt {a+b x^2+c x^4}}\) |
(2*Sqrt[d*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, 3/2, 3/2, 5/4, (-2*c*x^2)/(b - Sq rt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(a*d*Sqrt[a + b*x^2 + c*x^4])
3.12.3.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + R t[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2])))^F racPart[p])) Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2 *c*(x^2/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]
\[\int \frac {1}{\sqrt {d x}\, \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x}} \,d x } \]
integral(sqrt(c*x^4 + b*x^2 + a)*sqrt(d*x)/(c^2*d*x^9 + 2*b*c*d*x^7 + (b^2 + 2*a*c)*d*x^5 + 2*a*b*d*x^3 + a^2*d*x), x)
\[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {d x} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x}} \,d x } \]
\[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {d\,x}\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \]