\(\int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx\) [114]

Optimal result

Integrand size = 24, antiderivative size = 121 \[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {\sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a x+b x^3+c x^5}} \]

[Out]

1/2*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)
*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*x^(1/2)*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1
/2))^2)^(1/2)/a^(1/4)/c^(1/4)/(c*x^5+b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 121, 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 = {1928, 1117} \[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {\sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a x+b x^3+c x^5}} \]

[In]

Int[Sqrt[x]/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

(Sqrt[x]*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/
4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(1/4)*c^(1/4)*Sqrt[a*x + b*x^3 + c*x^5])

Rule 1117

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1928

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[x^(q/2)*(Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))]/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)]), Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{\sqrt {a x+b x^3+c x^5}} \\ & = \frac {\sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a x+b x^3+c x^5}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.08 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.60 \[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=-\frac {i \sqrt {x} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {x \left (a+b x^2+c x^4\right )}} \]

[In]

Integrate[Sqrt[x]/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

((-I)*Sqrt[x]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2
- 4*a*c])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[
b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[x*(a + b*x^2 + c*x^4)])

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \sqrt {-\frac {2 \left (\sqrt {-4 a c +b^{2}}\, x^{2}-b \,x^{2}-2 a \right )}{a}}\, \sqrt {\frac {\sqrt {-4 a c +b^{2}}\, x^{2}+b \,x^{2}+2 a}{a}}\, F\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {2}\, \sqrt {\frac {b \sqrt {-4 a c +b^{2}}-2 a c +b^{2}}{a c}}}{2}\right )}{2 \sqrt {x}\, \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}\)	\(177\)

[In]

int(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/x^(1/2)*(x*(c*x^4+b*x^2+a))^(1/2)/(c*x^4+b*x^2+a)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*((-4*a*c+b^2)^(1/2
)*x^2-b*x^2-2*a)/a)^(1/2)*(1/a*((-4*a*c+b^2)^(1/2)*x^2+b*x^2+2*a))^(1/2)*EllipticF(1/2*x*2^(1/2)*((-b+(-4*a*c+
b^2)^(1/2))/a)^(1/2),1/2*2^(1/2)*((b*(-4*a*c+b^2)^(1/2)-2*a*c+b^2)/a/c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b\right )} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c})}{2 \, a \sqrt {c}} \]

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(1/2)*(c*sqrt((b^2 - 4*a*c)/c^2) + b)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)*elliptic_f(arcsin(sqrt(1
/2)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)/x), 1/2*(b*c*sqrt((b^2 - 4*a*c)/c^2) + b^2 - 2*a*c)/(a*c))/(a*sqrt
(c))

Sympy [F]

\[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {\sqrt {x}}{\sqrt {x \left (a + b x^{2} + c x^{4}\right )}}\, dx \]

[In]

integrate(x**(1/2)/(c*x**5+b*x**3+a*x)**(1/2),x)

[Out]

Integral(sqrt(x)/sqrt(x*(a + b*x**2 + c*x**4)), x)

Maxima [F]

\[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \]

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)/sqrt(c*x^5 + b*x^3 + a*x), x)

Giac [F]

\[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \]

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(x)/sqrt(c*x^5 + b*x^3 + a*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {\sqrt {x}}{\sqrt {c\,x^5+b\,x^3+a\,x}} \,d x \]

[In]

int(x^(1/2)/(a*x + b*x^3 + c*x^5)^(1/2),x)

[Out]

int(x^(1/2)/(a*x + b*x^3 + c*x^5)^(1/2), x)