3.796 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{\sqrt{e x}} \, dx\)

Optimal. Leaf size=214 \[ \frac{4 a^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (11 A b-a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{77 b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (11 A b-a B)}{77 b e}+\frac{4 a \sqrt{e x} \sqrt{a+b x^2} (11 A b-a B)}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e} \]

[Out]

(4*a*(11*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(77*b*e) + (2*(11*A*b - a*B)*Sqrt[e*x]*(a + b*x^2)^(3/2))/(77*b
*e) + (2*B*Sqrt[e*x]*(a + b*x^2)^(5/2))/(11*b*e) + (4*a^(7/4)*(11*A*b - a*B)*(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])/(77*b^(5/4)*Sq
rt[e]*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.136428, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {459, 279, 329, 220} \[ \frac{4 a^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (11 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{77 b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (11 A b-a B)}{77 b e}+\frac{4 a \sqrt{e x} \sqrt{a+b x^2} (11 A b-a B)}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/Sqrt[e*x],x]

[Out]

(4*a*(11*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(77*b*e) + (2*(11*A*b - a*B)*Sqrt[e*x]*(a + b*x^2)^(3/2))/(77*b
*e) + (2*B*Sqrt[e*x]*(a + b*x^2)^(5/2))/(11*b*e) + (4*a^(7/4)*(11*A*b - a*B)*(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])/(77*b^(5/4)*Sq
rt[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 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 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{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{\sqrt{e x}} \, dx &=\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e}-\frac{\left (2 \left (-\frac{11 A b}{2}+\frac{a B}{2}\right )\right ) \int \frac{\left (a+b x^2\right )^{3/2}}{\sqrt{e x}} \, dx}{11 b}\\ &=\frac{2 (11 A b-a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac{(6 a (11 A b-a B)) \int \frac{\sqrt{a+b x^2}}{\sqrt{e x}} \, dx}{77 b}\\ &=\frac{4 a (11 A b-a B) \sqrt{e x} \sqrt{a+b x^2}}{77 b e}+\frac{2 (11 A b-a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac{\left (4 a^2 (11 A b-a B)\right ) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{77 b}\\ &=\frac{4 a (11 A b-a B) \sqrt{e x} \sqrt{a+b x^2}}{77 b e}+\frac{2 (11 A b-a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac{\left (8 a^2 (11 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 )}{77 b e}\\ &=\frac{4 a (11 A b-a B) \sqrt{e x} \sqrt{a+b x^2}}{77 b e}+\frac{2 (11 A b-a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac{4 a^{7/4} (11 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 )}{77 b^{5/4} \sqrt{e} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0754132, size = 96, normalized size = 0.45 \[ \frac{2 x \sqrt{a+b x^2} \left (a (11 A b-a B) \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )+B \sqrt{\frac{b x^2}{a}+1} \left (a+b x^2\right )^2\right )}{11 b \sqrt{e x} \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/Sqrt[e*x],x]

[Out]

(2*x*Sqrt[a + b*x^2]*(B*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/a] + a*(11*A*b - a*B)*Hypergeometric2F1[-3/2, 1/4, 5/4,
-((b*x^2)/a)]))/(11*b*Sqrt[e*x]*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.013, size = 272, normalized size = 1.3 \begin{align*}{\frac{2}{77\,{b}^{2}} \left ( 7\,B{x}^{7}{b}^{4}+22\,A\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{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-ab}{a}^{2}b+11\,A{x}^{5}{b}^{4}-2\,B\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{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-ab}{a}^{3}+20\,B{x}^{5}a{b}^{3}+44\,A{x}^{3}a{b}^{3}+17\,B{x}^{3}{a}^{2}{b}^{2}+33\,Ax{a}^{2}{b}^{2}+4\,Bx{a}^{3}b \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)^(3/2)*(B*x^2+A)/(e*x)^(1/2),x)

[Out]

2/77/(b*x^2+a)^(1/2)*(7*B*x^7*b^4+22*A*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))*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-a
*b)^(1/2)*a^2*b+11*A*x^5*b^4-2*B*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*El
lipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-a*b)^(1
/2)*a^3+20*B*x^5*a*b^3+44*A*x^3*a*b^3+17*B*x^3*a^2*b^2+33*A*x*a^2*b^2+4*B*x*a^3*b)/b^2/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/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 b x^{4} +{\left (B a + A b\right )} x^{2} + A a\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(1/2),x, algorithm="fricas")

[Out]

integral((B*b*x^4 + (B*a + A*b)*x^2 + A*a)*sqrt(b*x^2 + a)*sqrt(e*x)/(e*x), x)

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Sympy [C]  time = 12.2051, size = 199, normalized size = 0.93 \begin{align*} \frac{A a^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{5}{4}\right )} + \frac{A \sqrt{a} 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{e} \Gamma \left (\frac{9}{4}\right )} + \frac{B a^{\frac{3}{2}} 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{e} \Gamma \left (\frac{9}{4}\right )} + \frac{B \sqrt{a} b x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/(e*x)**(1/2),x)

[Out]

A*a**(3/2)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(5/4)) + A*
sqrt(a)*b*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(9/4)) + B*
a**(3/2)*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(9/4)) + B*s
qrt(a)*b*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(13/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(1/2),x, algorithm="giac")

[Out]

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