3.803 \(\int \frac{A+B x^2}{\sqrt{e x} \sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=139 \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{a} b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{2 B \sqrt{e x} \sqrt{a+b x^2}}{3 b e} \]

[Out]

(2*B*Sqrt[e*x]*Sqrt[a + b*x^2])/(3*b*e) + ((3*A*b - a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqr
t[b]*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)*Sqrt[e]*Sqrt[a
+ b*x^2])

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Rubi [A]  time = 0.0811609, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {459, 329, 220} \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{a} b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{2 B \sqrt{e x} \sqrt{a+b x^2}}{3 b e} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(Sqrt[e*x]*Sqrt[a + b*x^2]),x]

[Out]

(2*B*Sqrt[e*x]*Sqrt[a + b*x^2])/(3*b*e) + ((3*A*b - a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqr
t[b]*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)*Sqrt[e]*Sqrt[a
+ b*x^2])

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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 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]

Rubi steps

\begin{align*} \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{\left (2 \left (-\frac{3 A b}{2}+\frac{a B}{2}\right )\right ) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{3 b}\\ &=\frac{2 B \sqrt{e x} \sqrt{a+b x^2}}{3 b e}+\frac{(2 (3 A b-a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{3 b e}\\ &=\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}} F\left (2 \tan ^{-1}\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}}\\ \end{align*}

Mathematica [C]  time = 0.0554849, size = 79, normalized size = 0.57 \[ \frac{2 x \left (\sqrt{\frac{b x^2}{a}+1} (3 A b-a B) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+B \left (a+b x^2\right )\right )}{3 b \sqrt{e x} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(Sqrt[e*x]*Sqrt[a + b*x^2]),x]

[Out]

(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])

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Maple [A]  time = 0.016, size = 214, normalized size = 1.5 \begin{align*}{\frac{1}{3\,{b}^{2}} \left ( 3\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}b-B\sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{ \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}a+2\,{b}^{2}B{x}^{3}+2\,Bxab \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(e*x)^(1/2)/(b*x^2+a)^(1/2),x)

[Out]

1/3/(b*x^2+a)^(1/2)*(3*A*((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)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*b-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)*(-x*b/(-a*b)^(1/2))^(
1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*a+2*b^2*B*x^3+2*B*x*a*b)/(e*x
)^(1/2)/b^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{b x^{2} + a} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*sqrt(e*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{b e x^{3} + a e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b*e*x^3 + a*e*x), x)

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Sympy [C]  time = 2.37705, size = 94, normalized size = 0.68 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(e*x)**(1/2)/(b*x**2+a)**(1/2),x)

[Out]

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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{b x^{2} + a} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*sqrt(e*x)), x)