3.818 \(\int \frac{\sqrt{e x} (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=344 \[ -\frac{\sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{4 a^{7/4} b^{7/4} \sqrt{a+b x^2}}-\frac{\sqrt{e x} \sqrt{a+b x^2} (a B+A b)}{2 a^2 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{\sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{2 a^{7/4} b^{7/4} \sqrt{a+b x^2}}+\frac{(e x)^{3/2} (a B+A b)}{2 a^2 b e \sqrt{a+b x^2}}+\frac{(e x)^{3/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

[Out]

((A*b - a*B)*(e*x)^(3/2))/(3*a*b*e*(a + b*x^2)^(3/2)) + ((A*b + a*B)*(e*x)^(3/2))/(2*a^2*b*e*Sqrt[a + b*x^2])
- ((A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(2*a^2*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)) + ((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])/(2*a^(7/4)*b^(7/4)*Sqrt[a + b*x^2]) - ((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])/(4*a^(7/4)*b^(7/4)*S
qrt[a + b*x^2])

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Rubi [A]  time = 0.258791, antiderivative size = 344, 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 = {457, 290, 329, 305, 220, 1196} \[ -\frac{\sqrt{e x} \sqrt{a+b x^2} (a B+A b)}{2 a^2 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{\sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{4 a^{7/4} b^{7/4} \sqrt{a+b x^2}}+\frac{\sqrt{e} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{2 a^{7/4} b^{7/4} \sqrt{a+b x^2}}+\frac{(e x)^{3/2} (a B+A b)}{2 a^2 b e \sqrt{a+b x^2}}+\frac{(e x)^{3/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[e*x]*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

((A*b - a*B)*(e*x)^(3/2))/(3*a*b*e*(a + b*x^2)^(3/2)) + ((A*b + a*B)*(e*x)^(3/2))/(2*a^2*b*e*Sqrt[a + b*x^2])
- ((A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(2*a^2*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)) + ((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])/(2*a^(7/4)*b^(7/4)*Sqrt[a + b*x^2]) - ((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])/(4*a^(7/4)*b^(7/4)*S
qrt[a + b*x^2])

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 \frac{\sqrt{e x} \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac{(A b-a B) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(A b+a B) \int \frac{\sqrt{e x}}{\left (a+b x^2\right )^{3/2}} \, dx}{2 a b}\\ &=\frac{(A b-a B) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(A b+a B) (e x)^{3/2}}{2 a^2 b e \sqrt{a+b x^2}}-\frac{(A b+a B) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{4 a^2 b}\\ &=\frac{(A b-a B) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(A b+a B) (e x)^{3/2}}{2 a^2 b e \sqrt{a+b x^2}}-\frac{(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 )}{2 a^2 b e}\\ &=\frac{(A b-a B) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(A b+a B) (e x)^{3/2}}{2 a^2 b e \sqrt{a+b x^2}}-\frac{(A b+a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{2 a^{3/2} b^{3/2}}+\frac{(A b+a B) \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 )}{2 a^{3/2} b^{3/2}}\\ &=\frac{(A b-a B) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(A b+a B) (e x)^{3/2}}{2 a^2 b e \sqrt{a+b x^2}}-\frac{(A b+a B) \sqrt{e x} \sqrt{a+b x^2}}{2 a^2 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{(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 )}{2 a^{7/4} b^{7/4} \sqrt{a+b x^2}}-\frac{(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 )}{4 a^{7/4} b^{7/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.104803, size = 84, normalized size = 0.24 \[ \frac{2 x \sqrt{e x} \left (\left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (a B+A b) \, _2F_1\left (\frac{3}{4},\frac{5}{2};\frac{7}{4};-\frac{b x^2}{a}\right )-a^2 B\right )}{3 a^2 b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[e*x]*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

(2*x*Sqrt[e*x]*(-(a^2*B) + (A*b + a*B)*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[3/4, 5/2, 7/4, -((b*x
^2)/a)]))/(3*a^2*b*(a + b*x^2)^(3/2))

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Maple [B]  time = 0.021, size = 764, normalized size = 2.2 \begin{align*} \text{result too large to display} \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)^(5/2),x)

[Out]

-1/12*(6*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))*x^2*a*b^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))*x^2*a*b^2+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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/
2))^(1/2),1/2*2^(1/2))*x^2*a^2*b-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))*x^2*a
^2*b+6*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-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^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)*EllipticE(((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^3-6*A*x^4*b^3-6*B*x
^4*a*b^2-10*A*x^2*a*b^2-2*B*x^2*a^2*b)*(e*x)^(1/2)/b^2/a^2/x/(b*x^2+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} \sqrt{e x}}{{\left (b x^{2} + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(e*x)/(b*x^2 + a)^(5/2), 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^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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Sympy [C]  time = 162.735, size = 94, normalized size = 0.27 \begin{align*} \frac{A \sqrt{e} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{5}{2} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B \sqrt{e} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{7}{4}, \frac{5}{2} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{2}} \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)**(5/2),x)

[Out]

A*sqrt(e)*x**(3/2)*gamma(3/4)*hyper((3/4, 5/2), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/2)*gamma(7/4)) + B*
sqrt(e)*x**(7/2)*gamma(7/4)*hyper((7/4, 5/2), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/2)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} \sqrt{e x}}{{\left (b x^{2} + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(e*x)/(b*x^2 + a)^(5/2), x)