Integrand size = 28, antiderivative size = 130 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 B \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {(b B-3 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} c^{5/4} \sqrt {b x^2+c x^4}} \]
2/3*B*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)-1/3*(-3*A*c+B*b)*x*(cos(2*arctan(c^(1/ 4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*Ellip ticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2 ))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/b^(1/4)/c^(5/4)/(c*x^4+b*x^2)^( 1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (B \left (b+c x^2\right )+(-b B+3 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{3 c \sqrt {x^2 \left (b+c x^2\right )}} \]
(2*x^(3/2)*(B*(b + c*x^2) + (-(b*B) + 3*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeom etric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(3*c*Sqrt[x^2*(b + c*x^2)])
Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1945, 1431, 266, 761}
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.
| \(\displaystyle \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
| \(\Big \downarrow \) 1945 |
| \(\displaystyle \frac {2 B \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {(b B-3 A c) \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {2 B \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {x \sqrt {b+c x^2} (b B-3 A c) \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 B \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 x \sqrt {b+c x^2} (b B-3 A c) \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2 B \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (b B-3 A c) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} c^{5/4} \sqrt {b x^2+c x^4}}\) |
(2*B*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - ((b*B - 3*A*c)*x*(Sqrt[b] + Sqrt [c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/ 4)*Sqrt[x])/b^(1/4)], 1/2])/(3*b^(1/4)*c^(5/4)*Sqrt[b*x^2 + c*x^4])
3.3.50.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 2.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \(\frac {2 B \,x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{3 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (3 A c -B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c^{2} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(175\) |
| default | \(\frac {\sqrt {x}\, \left (3 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c -B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b +2 B \,c^{2} x^{3}+2 B b c x \right )}{3 \sqrt {x^{4} c +b \,x^{2}}\, c^{2}}\) | \(216\) |
2/3*B/c*x^(3/2)*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)+1/3*(3*A*c-B*b)/c^2*(-b*c) ^(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2) )*c/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)/(c*x^3+b*x)^(1/2)*Ellipt icF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^(1/2)/(x^2* (c*x^2+b))^(1/2)*(x*(c*x^2+b))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left ({\left (B b - 3 \, A c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} B c \sqrt {x}\right )}}{3 \, c^{2} x} \]
-2/3*((B*b - 3*A*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x ^4 + b*x^2)*B*c*sqrt(x))/(c^2*x)
\[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {\sqrt {x} \left (A + B x^{2}\right )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]
\[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \sqrt {x}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
\[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \sqrt {x}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {\sqrt {x}\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \]