Integrand size = 26, antiderivative size = 138 \[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=-\frac {2 A \sqrt {a+b x^2}}{3 a e (e x)^{3/2}}-\frac {(A b-3 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{3 a^{5/4} \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}} \]
-2/3*A*(b*x^2+a)^(1/2)/a/e/(e*x)^(3/2)-1/3*(A*b-3*B*a)*(cos(2*arctan(b^(1/ 4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2) /a^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1 /2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^( 1/2)/a^(5/4)/b^(1/4)/e^(5/2)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.59 \[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=-\frac {2 x \left (A \left (a+b x^2\right )+(A b-3 a B) x^2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 a (e x)^{5/2} \sqrt {a+b x^2}} \]
(-2*x*(A*(a + b*x^2) + (A*b - 3*a*B)*x^2*Sqrt[1 + (b*x^2)/a]*Hypergeometri c2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(3*a*(e*x)^(5/2)*Sqrt[a + b*x^2])
Time = 0.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {359, 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.
\(\displaystyle \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(A b-3 a B) \int \frac {1}{\sqrt {e x} \sqrt {b x^2+a}}dx}{3 a e^2}-\frac {2 A \sqrt {a+b x^2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {2 (A b-3 a B) \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{3 a e^3}-\frac {2 A \sqrt {a+b x^2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {(A b-3 a B) \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{3 a^{5/4} \sqrt [4]{b} e^{7/2} \sqrt {a+b x^2}}-\frac {2 A \sqrt {a+b x^2}}{3 a e (e x)^{3/2}}\) |
(-2*A*Sqrt[a + b*x^2])/(3*a*e*(e*x)^(3/2)) - ((A*b - 3*a*B)*(Sqrt[a]*e + S qrt[b]*e*x)*Sqrt[(a*e^2 + b*e^2*x^2)/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*Elliptic F[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(3*a^(5/4)*b^(1/4 )*e^(7/2)*Sqrt[a + b*x^2])
3.9.5.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]
Time = 3.05 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.29
method | result | size |
risch | \(-\frac {2 A \sqrt {b \,x^{2}+a}}{3 a x \,e^{2} \sqrt {e x}}-\frac {\left (A b -3 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{3 a b \sqrt {b e \,x^{3}+a e x}\, e^{2} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(178\) |
elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {2 A \sqrt {b e \,x^{3}+a e x}}{3 e^{3} a \,x^{2}}+\frac {\left (\frac {B}{e^{2}}-\frac {b A}{3 a \,e^{2}}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(179\) |
default | \(-\frac {A \sqrt {2}\, \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b x -3 B \sqrt {2}\, \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a x +2 A \,b^{2} x^{2}+2 a b A}{3 \sqrt {b \,x^{2}+a}\, x b a \,e^{2} \sqrt {e x}}\) | \(223\) |
-2/3/a*A*(b*x^2+a)^(1/2)/x/e^2/(e*x)^(1/2)-1/3*(A*b-3*B*a)/a*(-a*b)^(1/2)/ b*((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^ (1/2)*b)^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b*e*x^3+a*e*x)^(1/2)*EllipticF(( (x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))/e^2*((b*x^2+a)*e*x)^ (1/2)/(e*x)^(1/2)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.43 \[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=\frac {2 \, {\left ({\left (3 \, B a - A b\right )} \sqrt {b e} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - \sqrt {b x^{2} + a} \sqrt {e x} A b\right )}}{3 \, a b e^{3} x^{2}} \]
2/3*((3*B*a - A*b)*sqrt(b*e)*x^2*weierstrassPInverse(-4*a/b, 0, x) - sqrt( b*x^2 + a)*sqrt(e*x)*A*b)/(a*b*e^3*x^2)
Result contains complex when optimal does not.
Time = 4.82 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=\frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \]
A*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt (a)*e**(5/2)*x**(3/2)*gamma(1/4)) + B*sqrt(x)*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**(5/2)*gamma(5/4))
\[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \left (e x\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \left (e x\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B x^2}{(e x)^{5/2} \sqrt {a+b x^2}} \, dx=\int \frac {B\,x^2+A}{{\left (e\,x\right )}^{5/2}\,\sqrt {b\,x^2+a}} \,d x \]