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

Optimal. Leaf size=337 \[ \frac{3 \sqrt [4]{a} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 A b-7 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{10 b^{11/4} \sqrt{a+b x^2}}+\frac{3 e^2 \sqrt{e x} \sqrt{a+b x^2} (5 A b-7 a B)}{5 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{3 \sqrt [4]{a} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 A b-7 a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{a+b x^2}}-\frac{e (e x)^{3/2} (5 A b-7 a B)}{5 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}} \]

[Out]

-((5*A*b - 7*a*B)*e*(e*x)^(3/2))/(5*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(7/2))/(5*b*e*Sqrt[a + b*x^2]) + (3*(5*A
*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) - (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5
/2)*(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])/(5*b^(11/4)*Sqrt[a + b*x^2]) + (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5/2)*(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])/(1
0*b^(11/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.252037, antiderivative size = 337, 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 = {459, 288, 329, 305, 220, 1196} \[ \frac{3 e^2 \sqrt{e x} \sqrt{a+b x^2} (5 A b-7 a B)}{5 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{3 \sqrt [4]{a} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 A b-7 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{10 b^{11/4} \sqrt{a+b x^2}}-\frac{3 \sqrt [4]{a} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 A b-7 a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{a+b x^2}}-\frac{e (e x)^{3/2} (5 A b-7 a B)}{5 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*A*b - 7*a*B)*e*(e*x)^(3/2))/(5*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(7/2))/(5*b*e*Sqrt[a + b*x^2]) + (3*(5*A
*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) - (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5
/2)*(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])/(5*b^(11/4)*Sqrt[a + b*x^2]) + (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5/2)*(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])/(1
0*b^(11/4)*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 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{(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}}-\frac{\left (2 \left (-\frac{5 A b}{2}+\frac{7 a B}{2}\right )\right ) \int \frac{(e x)^{5/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{5 b}\\ &=-\frac{(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}}+\frac{\left (3 (5 A b-7 a B) e^2\right ) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{10 b^2}\\ &=-\frac{(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}}+\frac{(3 (5 A b-7 a B) e) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 b^2}\\ &=-\frac{(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}}+\frac{\left (3 \sqrt{a} (5 A b-7 a B) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 b^{5/2}}-\frac{\left (3 \sqrt{a} (5 A b-7 a B) e^2\right ) \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 )}{5 b^{5/2}}\\ &=-\frac{(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{7/2}}{5 b e \sqrt{a+b x^2}}+\frac{3 (5 A b-7 a B) e^2 \sqrt{e x} \sqrt{a+b x^2}}{5 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \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 )}{5 b^{11/4} \sqrt{a+b x^2}}+\frac{3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \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 )}{10 b^{11/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.112121, size = 84, normalized size = 0.25 \[ \frac{2 e (e x)^{3/2} \left (\sqrt{\frac{b x^2}{a}+1} (7 a B-5 A b) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{b x^2}{a}\right )-7 a B+5 A b+b B x^2\right )}{5 b^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 391, normalized size = 1.2 \begin{align*}{\frac{{e}^{2}}{10\,x{b}^{3}}\sqrt{ex} \left ( 30\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) ab-15\,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 ) ab-42\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{2}+21\,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 ){a}^{2}+4\,{b}^{2}B{x}^{4}-10\,A{x}^{2}{b}^{2}+14\,B{x}^{2}ab \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10/x*e^2*(e*x)^(1/2)*(30*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*b-15*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-42*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^2+21*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^
2+4*b^2*B*x^4-10*A*x^2*b^2+14*B*x^2*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{5}{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)^(5/2)*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((B*e^2*x^4 + A*e^2*x^2)*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 [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)**(5/2)*(B*x**2+A)/(b*x**2+a)**(3/2),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{5}{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)^(5/2)*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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