3.817 \(\int \frac{(e x)^{3/2} (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=185 \[ \frac{e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 a^{5/4} b^{9/4} \sqrt{a+b x^2}}-\frac{e \sqrt{e x} (5 a B+A b)}{6 a b^2 \sqrt{a+b x^2}}+\frac{(e x)^{5/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.114255, antiderivative size = 185, 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, 288, 329, 220} \[ \frac{e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{5/4} b^{9/4} \sqrt{a+b x^2}}-\frac{e \sqrt{e x} (5 a B+A b)}{6 a b^2 \sqrt{a+b x^2}}+\frac{(e x)^{5/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(3/2)*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

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

Mathematica [C]  time = 0.148075, size = 105, normalized size = 0.57 \[ \frac{e \sqrt{e x} \left (-5 a^2 B+\left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (5 a B+A b) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )-a b \left (A+7 B x^2\right )+A b^2 x^2\right )}{6 a b^2 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(3/2)*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

(e*Sqrt[e*x]*(-5*a^2*B + A*b^2*x^2 - a*b*(A + 7*B*x^2) + (A*b + 5*a*B)*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hyperge
ometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(6*a*b^2*(a + b*x^2)^(3/2))

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Maple [B]  time = 0.018, size = 429, normalized size = 2.3 \begin{align*}{\frac{e}{12\,ax{b}^{3}} \left ( A\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}{b}^{2}+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}{x}^{2}ab+A\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 ) ab+5\,B\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 ){a}^{2}+2\,A{x}^{3}{b}^{3}-14\,B{x}^{3}a{b}^{2}-2\,Axa{b}^{2}-10\,Bx{a}^{2}b \right ) \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((e*x)^(3/2)*(B*x^2+A)/(b*x^2+a)^(5/2),x)

[Out]

1/12*(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+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)*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+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)*EllipticF(
((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a*b+5*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+2*A*x^3*b^3-14*B*x^3*a*b^2-2*A*x*a*b^2-10*B*x*a^2*b)*e/x*(e*x)^(1/2)/a/b^3
/(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 )} \left (e x\right )^{\frac{3}{2}}}{{\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((e*x)^(3/2)*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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