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

Optimal. Leaf size=187 \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+5 A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{\sqrt{e x} (a B+5 A b)}{6 a^2 b e \sqrt{a+b x^2}}+\frac{\sqrt{e x} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

[Out]

((A*b - a*B)*Sqrt[e*x])/(3*a*b*e*(a + b*x^2)^(3/2)) + ((5*A*b + a*B)*Sqrt[e*x])/(6*a^2*b*e*Sqrt[a + b*x^2]) +
((5*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)*Sqr
t[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(12*a^(9/4)*b^(5/4)*Sqrt[e]*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.111794, antiderivative size = 187, 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 = {457, 290, 329, 220} \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+5 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{\sqrt{e x} (a B+5 A b)}{6 a^2 b e \sqrt{a+b x^2}}+\frac{\sqrt{e x} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*Sqrt[e*x])/(3*a*b*e*(a + b*x^2)^(3/2)) + ((5*A*b + a*B)*Sqrt[e*x])/(6*a^2*b*e*Sqrt[a + b*x^2]) +
((5*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)*Sqr
t[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(12*a^(9/4)*b^(5/4)*Sqrt[e]*Sqrt[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 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} \left (a+b x^2\right )^{5/2}} \, dx &=\frac{(A b-a B) \sqrt{e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{\left (\frac{5 A b}{2}+\frac{a B}{2}\right ) \int \frac{1}{\sqrt{e x} \left (a+b x^2\right )^{3/2}} \, dx}{3 a b}\\ &=\frac{(A b-a B) \sqrt{e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(5 A b+a B) \sqrt{e x}}{6 a^2 b e \sqrt{a+b x^2}}+\frac{(5 A b+a B) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{12 a^2 b}\\ &=\frac{(A b-a B) \sqrt{e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(5 A b+a B) \sqrt{e x}}{6 a^2 b e \sqrt{a+b x^2}}+\frac{(5 A b+a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{6 a^2 b e}\\ &=\frac{(A b-a B) \sqrt{e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac{(5 A b+a B) \sqrt{e x}}{6 a^2 b e \sqrt{a+b x^2}}+\frac{(5 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 )}{12 a^{9/4} b^{5/4} \sqrt{e} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0823972, size = 108, normalized size = 0.58 \[ \frac{-a^2 B x+x \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (a B+5 A b) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+a b x \left (7 A+B x^2\right )+5 A b^2 x^3}{6 a^2 b \sqrt{e x} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*B*x) + 5*A*b^2*x^3 + a*b*x*(7*A + B*x^2) + (5*A*b + a*B)*x*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hypergeometr
ic2F1[1/4, 1/2, 5/4, -((b*x^2)/a)])/(6*a^2*b*Sqrt[e*x]*(a + b*x^2)^(3/2))

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Maple [B]  time = 0.023, size = 425, normalized size = 2.3 \begin{align*}{\frac{1}{12\,{b}^{2}{a}^{2}} \left ( 5\,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}{x}^{2}{b}^{2}+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}{x}^{2}ab+5\,A\sqrt{-ab}\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 ) ab+B\sqrt{-ab}\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 ){a}^{2}+10\,A{x}^{3}{b}^{3}+2\,B{x}^{3}a{b}^{2}+14\,Axa{b}^{2}-2\,Bx{a}^{2}b \right ){\frac{1}{\sqrt{ex}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x)

[Out]

1/12*(5*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)*x^2*b^2+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
pticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x^2*a*b+5*A*(-a*b)^(1/2)*((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)*Elliptic
F(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a*b+B*(-a*b)^(1/2)*((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+10*A*x^3*b^3+2*B*x^3*a*b^2+14*A*x*a*b^2-2*B*x*a^2*b)/(e*x)^(1/2)/a^2/b^2/(
b*x^2+a)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/((b*x^2 + a)^(5/2)*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 x^{2} + A\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{b^{3} e x^{7} + 3 \, a b^{2} e x^{5} + 3 \, a^{2} b e x^{3} + a^{3} e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b^3*e*x^7 + 3*a*b^2*e*x^5 + 3*a^2*b*e*x^3 + a^3*e*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(b*x**2+a)**(5/2)/(e*x)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="giac")

[Out]

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