\(\int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx\) [384]

Optimal result

Integrand size = 21, antiderivative size = 90 \[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt [4]{c} \sqrt {b x^2+c x^4}} \]

[Out]

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)))*EllipticF(sin(2*arct
an(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^(1/4)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2057, 335, 226} \[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt [4]{c} \sqrt {b x^2+c x^4}} \]

[In]

Int[Sqrt[x]/Sqrt[b*x^2 + c*x^4],x]

[Out]

(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])/(b^(1/4)*c^(1/4)*Sqrt[b*x^2 + c*x^4])

Rule 226

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]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )}{\sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

Integrate[Sqrt[x]/Sqrt[b*x^2 + c*x^4],x]

[Out]

(2*x^(3/2)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)])/Sqrt[x^2*(b + c*x^2)]

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {\sqrt {x}\, \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, c}\)	\(106\)

[In]

int(x^(1/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )}{\sqrt {c}} \]

[In]

integrate(x^(1/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

2*weierstrassPInverse(-4*b/c, 0, x)/sqrt(c)

Sympy [F]

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

[In]

integrate(x**(1/2)/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]

[In]

integrate(x^(1/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)/sqrt(c*x^4 + b*x^2), x)

Giac [F]

\[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]

[In]

integrate(x^(1/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(x)/sqrt(c*x^4 + b*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {\sqrt {x}}{\sqrt {c\,x^4+b\,x^2}} \,d x \]

[In]

int(x^(1/2)/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^(1/2)/(b*x^2 + c*x^4)^(1/2), x)