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

Optimal. Leaf size=174 \[ \frac{e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-5 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{6 \sqrt [4]{a} b^{9/4} \sqrt{a+b x^2}}-\frac{e \sqrt{e x} (3 A b-5 a B)}{3 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{5/2}}{3 b e \sqrt{a+b x^2}} \]

[Out]

-((3*A*b - 5*a*B)*e*Sqrt[e*x])/(3*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(5/2))/(3*b*e*Sqrt[a + b*x^2]) + ((3*A*b -
5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sq
rt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(6*a^(1/4)*b^(9/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.111641, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {459, 288, 329, 220} \[ \frac{e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{6 \sqrt [4]{a} b^{9/4} \sqrt{a+b x^2}}-\frac{e \sqrt{e x} (3 A b-5 a B)}{3 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{5/2}}{3 b e \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A*b - 5*a*B)*e*Sqrt[e*x])/(3*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(5/2))/(3*b*e*Sqrt[a + b*x^2]) + ((3*A*b -
5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sq
rt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(6*a^(1/4)*b^(9/4)*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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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{(e x)^{3/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac{2 B (e x)^{5/2}}{3 b e \sqrt{a+b x^2}}-\frac{\left (2 \left (-\frac{3 A b}{2}+\frac{5 a B}{2}\right )\right ) \int \frac{(e x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac{(3 A b-5 a B) e \sqrt{e x}}{3 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{5/2}}{3 b e \sqrt{a+b x^2}}+\frac{\left ((3 A b-5 a B) e^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{6 b^2}\\ &=-\frac{(3 A b-5 a B) e \sqrt{e x}}{3 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{5/2}}{3 b e \sqrt{a+b x^2}}+\frac{((3 A b-5 a B) e) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{3 b^2}\\ &=-\frac{(3 A b-5 a B) e \sqrt{e x}}{3 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{5/2}}{3 b e \sqrt{a+b x^2}}+\frac{(3 A b-5 a B) e^{3/2} \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 )}{6 \sqrt [4]{a} b^{9/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.104776, size = 85, normalized size = 0.49 \[ \frac{e \sqrt{e x} \left (\sqrt{\frac{b x^2}{a}+1} (3 A b-5 a B) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+5 a B-3 A b+2 b B x^2\right )}{3 b^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 225, normalized size = 1.3 \begin{align*}{\frac{e}{6\,x{b}^{3}}\sqrt{ex} \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-5\,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 ) \sqrt{-ab}a+4\,{b}^{2}B{x}^{3}-6\,Ax{b}^{2}+10\,Bxab \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*e/x*(e*x)^(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-5*
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+4*b^2*B*x^3-6*A*x*b^2+10*
B*x*a*b)/(b*x^2+a)^(1/2)/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B e x^{3} + A e x\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*e*x^3 + A*e*x)*sqrt(b*x^2 + a)*sqrt(e*x)/(b^2*x^4 + 2*a*b*x^2 + a^2), x)

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Sympy [C]  time = 61.2484, size = 94, normalized size = 0.54 \begin{align*} \frac{A e^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{9}{4}\right )} + \frac{B e^{\frac{3}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(9/4)) + B
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*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 (e x\right )^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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