3.616 \(\int \frac {\sqrt {c x}}{\sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=236 \[ \frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^2}}-\frac {2 \sqrt [4]{a} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^2}}+\frac {2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right )} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {329, 305, 220, 1196} \[ \frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^2}}-\frac {2 \sqrt [4]{a} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^2}}+\frac {2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/Sqrt[a + b*x^2],x]

[Out]

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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

Rule 329

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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {\sqrt {c x}}{\sqrt {a+b x^2}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{c}\\ &=\frac {\left (2 \sqrt {a}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{\sqrt {b}}-\frac {\left (2 \sqrt {a}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a} c}}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{\sqrt {b}}\\ &=\frac {2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 \sqrt [4]{a} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^2}}+\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 56, normalized size = 0.24 \[ \frac {2 x \sqrt {c x} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )}{3 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[c*x]*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^2)/a)])/(3*Sqrt[a + b*x^2])

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x}}{\sqrt {b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x)/sqrt(b*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x}}{\sqrt {b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 132, normalized size = 0.56 \[ \frac {\sqrt {c x}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (2 \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-\EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) a}{\sqrt {b \,x^{2}+a}\, b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x)^(1/2)/(b*x^2+a)^(1/2)*a/b*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1
/2))^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*(2*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))-Ellip
ticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x}}{\sqrt {b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c\,x}}{\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(a + b*x^2)^(1/2),x)

[Out]

int((c*x)^(1/2)/(a + b*x^2)^(1/2), x)

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sympy [C]  time = 0.93, size = 44, normalized size = 0.19 \[ \frac {\sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c)*x**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(7/4))

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