Integrand size = 26, antiderivative size = 139 \[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\frac {2 B \sqrt {e x} \sqrt {a+b x^2}}{3 b e}+\frac {(3 A b-a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{a} b^{5/4} \sqrt {e} \sqrt {a+b x^2}} \] Output:
2/3*B*(e*x)^(1/2)*(b*x^2+a)^(1/2)/b/e+1/3*(3*A*b-B*a)*(a^(1/2)+b^(1/2)*x)* ((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*( e*x)^(1/2)/a^(1/4)/e^(1/2)),1/2*2^(1/2))/a^(1/4)/b^(5/4)/e^(1/2)/(b*x^2+a) ^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.57 \[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\frac {2 x \left (B \left (a+b x^2\right )+(3 A b-a B) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 b \sqrt {e x} \sqrt {a+b x^2}} \] Input:
Integrate[(A + B*x^2)/(Sqrt[e*x]*Sqrt[a + b*x^2]),x]
Output:
(2*x*(B*(a + b*x^2) + (3*A*b - a*B)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[ 1/4, 1/2, 5/4, -((b*x^2)/a)]))/(3*b*Sqrt[e*x]*Sqrt[a + b*x^2])
Time = 0.25 (sec) , antiderivative size = 152, 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 = {363, 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.
\(\displaystyle \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {(3 A b-a B) \int \frac {1}{\sqrt {e x} \sqrt {b x^2+a}}dx}{3 b}+\frac {2 B \sqrt {e x} \sqrt {a+b x^2}}{3 b e}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 (3 A b-a B) \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{3 b e}+\frac {2 B \sqrt {e x} \sqrt {a+b x^2}}{3 b e}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {(3 A b-a B) \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{a} b^{5/4} e^{3/2} \sqrt {a+b x^2}}+\frac {2 B \sqrt {e x} \sqrt {a+b x^2}}{3 b e}\) |
Input:
Int[(A + B*x^2)/(Sqrt[e*x]*Sqrt[a + b*x^2]),x]
Output:
(2*B*Sqrt[e*x]*Sqrt[a + b*x^2])/(3*b*e) + ((3*A*b - a*B)*(Sqrt[a]*e + Sqrt [b]*e*x)*Sqrt[(a*e^2 + b*e^2*x^2)/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*EllipticF[2 *ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(3*a^(1/4)*b^(5/4)*e ^(3/2)*Sqrt[a + b*x^2])
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[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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 = 0.64 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.21
method | result | size |
risch | \(\frac {2 B x \sqrt {b \,x^{2}+a}}{3 b \sqrt {e x}}+\frac {\left (3 A b -B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{3 b^{2} \sqrt {b e \,x^{3}+a e x}\, \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(168\) |
elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (\frac {2 B \sqrt {b e \,x^{3}+a e x}}{3 b e}+\frac {\left (A -\frac {B a}{3 b}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(169\) |
default | \(\frac {3 A \sqrt {-a b}\, \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b -B \sqrt {-a b}\, \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a +2 B \,b^{2} x^{3}+2 B a b x}{3 \sqrt {b \,x^{2}+a}\, \sqrt {e x}\, b^{2}}\) | \(214\) |
Input:
int((B*x^2+A)/(e*x)^(1/2)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*B/b*x*(b*x^2+a)^(1/2)/(e*x)^(1/2)+1/3*(3*A*b-B*a)/b^2*(-a*b)^(1/2)*((x +1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^(1/2))*b/(-a*b) ^(1/2))^(1/2)*(-b/(-a*b)^(1/2)*x)^(1/2)/(b*e*x^3+a*e*x)^(1/2)*EllipticF((( x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*((b*x^2+a)*e*x)^(1/ 2)/(e*x)^(1/2)/(b*x^2+a)^(1/2)
Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.36 \[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\frac {2 \, {\left (\sqrt {b x^{2} + a} \sqrt {e x} B b - {\left (B a - 3 \, A b\right )} \sqrt {b e} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )}}{3 \, b^{2} e} \] Input:
integrate((B*x^2+A)/(e*x)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
2/3*(sqrt(b*x^2 + a)*sqrt(e*x)*B*b - (B*a - 3*A*b)*sqrt(b*e)*weierstrassPI nverse(-4*a/b, 0, x))/(b^2*e)
Result contains complex when optimal does not.
Time = 1.68 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\frac {A \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {B x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((B*x**2+A)/(e*x)**(1/2)/(b*x**2+a)**(1/2),x)
Output:
A*sqrt(x)*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**2*exp_polar(I*pi)/a)/( 2*sqrt(a)*sqrt(e)*gamma(5/4)) + B*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9 /4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*sqrt(e)*gamma(9/4))
\[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \sqrt {e x}} \,d x } \] Input:
integrate((B*x^2+A)/(e*x)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*sqrt(e*x)), x)
\[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \sqrt {e x}} \,d x } \] Input:
integrate((B*x^2+A)/(e*x)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*sqrt(e*x)), x)
Timed out. \[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\int \frac {B\,x^2+A}{\sqrt {e\,x}\,\sqrt {b\,x^2+a}} \,d x \] Input:
int((A + B*x^2)/((e*x)^(1/2)*(a + b*x^2)^(1/2)),x)
Output:
int((A + B*x^2)/((e*x)^(1/2)*(a + b*x^2)^(1/2)), x)
\[ \int \frac {A+B x^2}{\sqrt {e x} \sqrt {a+b x^2}} \, dx=\frac {2 \sqrt {e}\, \left (\sqrt {x}\, \sqrt {b \,x^{2}+a}+\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{3}+a x}d x \right ) a \right )}{3 e} \] Input:
int((B*x^2+A)/(e*x)^(1/2)/(b*x^2+a)^(1/2),x)
Output:
(2*sqrt(e)*(sqrt(x)*sqrt(a + b*x**2) + int((sqrt(x)*sqrt(a + b*x**2))/(a*x + b*x**3),x)*a))/(3*e)