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

Optimal. Leaf size=337 \[ \frac{2 a^{5/4} \sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}-\frac{4 a^{5/4} \sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}+\frac{4 a \sqrt{e x} \sqrt{a+b x^2} (3 A b-a B)}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 (e x)^{3/2} \sqrt{a+b x^2} (3 A b-a B)}{15 b e}+\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e} \]

[Out]

(2*(3*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*b*e) + (4*a*(3*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*b^(
3/2)*(Sqrt[a] + Sqrt[b]*x)) + (2*B*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*b*e) - (4*a^(5/4)*(3*A*b - a*B)*Sqrt[e]*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)
*Sqrt[e])], 1/2])/(15*b^(7/4)*Sqrt[a + b*x^2]) + (2*a^(5/4)*(3*A*b - a*B)*Sqrt[e]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(
a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(15*b^(7/
4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.269853, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {459, 279, 329, 305, 220, 1196} \[ \frac{2 a^{5/4} \sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}-\frac{4 a^{5/4} \sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}+\frac{4 a \sqrt{e x} \sqrt{a+b x^2} (3 A b-a B)}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 (e x)^{3/2} \sqrt{a+b x^2} (3 A b-a B)}{15 b e}+\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*b*e) + (4*a*(3*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*b^(
3/2)*(Sqrt[a] + Sqrt[b]*x)) + (2*B*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*b*e) - (4*a^(5/4)*(3*A*b - a*B)*Sqrt[e]*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)
*Sqrt[e])], 1/2])/(15*b^(7/4)*Sqrt[a + b*x^2]) + (2*a^(5/4)*(3*A*b - a*B)*Sqrt[e]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(
a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(15*b^(7/
4)*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 279

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

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 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 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 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 \sqrt{e x} \sqrt{a+b x^2} \left (A+B x^2\right ) \, dx &=\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}-\frac{\left (2 \left (-\frac{9 A b}{2}+\frac{3 a B}{2}\right )\right ) \int \sqrt{e x} \sqrt{a+b x^2} \, dx}{9 b}\\ &=\frac{2 (3 A b-a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 b e}+\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}+\frac{(2 a (3 A b-a B)) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{15 b}\\ &=\frac{2 (3 A b-a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 b e}+\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}+\frac{(4 a (3 A b-a B)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 b e}\\ &=\frac{2 (3 A b-a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 b e}+\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}+\frac{\left (4 a^{3/2} (3 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 b^{3/2}}-\frac{\left (4 a^{3/2} (3 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a} e}}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 b^{3/2}}\\ &=\frac{2 (3 A b-a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 b e}+\frac{4 a (3 A b-a B) \sqrt{e x} \sqrt{a+b x^2}}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}-\frac{4 a^{5/4} (3 A b-a B) \sqrt{e} \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{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}+\frac{2 a^{5/4} (3 A b-a B) \sqrt{e} \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 )}{15 b^{7/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0873176, size = 93, normalized size = 0.28 \[ \frac{2 x \sqrt{e x} \sqrt{a+b x^2} \left ((3 A b-a B) \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )+B \sqrt{\frac{b x^2}{a}+1} \left (a+b x^2\right )\right )}{9 b \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[e*x]*Sqrt[a + b*x^2]*(B*(a + b*x^2)*Sqrt[1 + (b*x^2)/a] + (3*A*b - a*B)*Hypergeometric2F1[-1/2, 3/4,
7/4, -((b*x^2)/a)]))/(9*b*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.029, size = 414, normalized size = 1.2 \begin{align*}{\frac{2}{45\,{b}^{2}x}\sqrt{ex} \left ( 5\,B{x}^{6}{b}^{3}+18\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{2}b-9\,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 ){a}^{2}b-6\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}+3\,B\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 ){a}^{3}+9\,A{x}^{4}{b}^{3}+7\,B{x}^{4}a{b}^{2}+9\,A{x}^{2}a{b}^{2}+2\,B{x}^{2}{a}^{2}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \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]

2/45*(e*x)^(1/2)/(b*x^2+a)^(1/2)/b^2*(5*B*x^6*b^3+18*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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),
1/2*2^(1/2))*a^2*b-9*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^2*b-6*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)*Elli
pticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^3+3*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^3+9*A*x^4*b^3+7*B*x^4*a*b^2+9*A*x^2*a*b^2+2*B*x^2*a^2*b)/x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B x^{2} + A\right )} \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 ({\left (B x^{2} + A\right )} \sqrt{b x^{2} + a} \sqrt{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), x)

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Sympy [C]  time = 3.38528, size = 95, normalized size = 0.28 \begin{align*} \frac{A \sqrt{a} \left (e x\right )^{\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 e \Gamma \left (\frac{7}{4}\right )} + \frac{B \sqrt{a} \left (e x\right )^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{3} \Gamma \left (\frac{11}{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(a)*(e*x)**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e*gamma(7/4)) + B*sq
rt(a)*(e*x)**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**3*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B x^{2} + A\right )} \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)