3.807 \(\int \frac{(e x)^{7/2} (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=211 \[ -\frac{5 a^{3/4} e^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-9 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{42 b^{13/4} \sqrt{a+b x^2}}+\frac{5 e^3 \sqrt{e x} \sqrt{a+b x^2} (7 A b-9 a B)}{21 b^3}-\frac{e (e x)^{5/2} (7 A b-9 a B)}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}} \]

[Out]

-((7*A*b - 9*a*B)*e*(e*x)^(5/2))/(7*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(9/2))/(7*b*e*Sqrt[a + b*x^2]) + (5*(7*A
*b - 9*a*B)*e^3*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*b^3) - (5*a^(3/4)*(7*A*b - 9*a*B)*e^(7/2)*(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])/(4
2*b^(13/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.137075, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {459, 288, 321, 329, 220} \[ -\frac{5 a^{3/4} e^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-9 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{42 b^{13/4} \sqrt{a+b x^2}}+\frac{5 e^3 \sqrt{e x} \sqrt{a+b x^2} (7 A b-9 a B)}{21 b^3}-\frac{e (e x)^{5/2} (7 A b-9 a B)}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

-((7*A*b - 9*a*B)*e*(e*x)^(5/2))/(7*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(9/2))/(7*b*e*Sqrt[a + b*x^2]) + (5*(7*A
*b - 9*a*B)*e^3*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*b^3) - (5*a^(3/4)*(7*A*b - 9*a*B)*e^(7/2)*(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])/(4
2*b^(13/4)*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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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)^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}}-\frac{\left (2 \left (-\frac{7 A b}{2}+\frac{9 a B}{2}\right )\right ) \int \frac{(e x)^{7/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{7 b}\\ &=-\frac{(7 A b-9 a B) e (e x)^{5/2}}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}}+\frac{\left (5 (7 A b-9 a B) e^2\right ) \int \frac{(e x)^{3/2}}{\sqrt{a+b x^2}} \, dx}{14 b^2}\\ &=-\frac{(7 A b-9 a B) e (e x)^{5/2}}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}}+\frac{5 (7 A b-9 a B) e^3 \sqrt{e x} \sqrt{a+b x^2}}{21 b^3}-\frac{\left (5 a (7 A b-9 a B) e^4\right ) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{42 b^3}\\ &=-\frac{(7 A b-9 a B) e (e x)^{5/2}}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}}+\frac{5 (7 A b-9 a B) e^3 \sqrt{e x} \sqrt{a+b x^2}}{21 b^3}-\frac{\left (5 a (7 A b-9 a B) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{21 b^3}\\ &=-\frac{(7 A b-9 a B) e (e x)^{5/2}}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}}+\frac{5 (7 A b-9 a B) e^3 \sqrt{e x} \sqrt{a+b x^2}}{21 b^3}-\frac{5 a^{3/4} (7 A b-9 a B) e^{7/2} \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 )}{42 b^{13/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.143431, size = 111, normalized size = 0.53 \[ \frac{e^3 \sqrt{e x} \left (-45 a^2 B+5 a \sqrt{\frac{b x^2}{a}+1} (9 a B-7 A b) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+a b \left (35 A-18 B x^2\right )+2 b^2 x^2 \left (7 A+3 B x^2\right )\right )}{21 b^3 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

(e^3*Sqrt[e*x]*(-45*a^2*B + a*b*(35*A - 18*B*x^2) + 2*b^2*x^2*(7*A + 3*B*x^2) + 5*a*(-7*A*b + 9*a*B)*Sqrt[1 +
(b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(21*b^3*Sqrt[a + b*x^2])

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Maple [A]  time = 0.035, size = 252, normalized size = 1.2 \begin{align*} -{\frac{{e}^{3}}{42\,x{b}^{4}}\sqrt{ex} \left ( 35\,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-45\,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}-12\,B{x}^{5}{b}^{3}-28\,A{x}^{3}{b}^{3}+36\,B{x}^{3}a{b}^{2}-70\,Axa{b}^{2}+90\,Bx{a}^{2}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/42*e^3/x*(e*x)^(1/2)*(35*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-45*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^2-12*B*x^5*b^3-28
*A*x^3*b^3+36*B*x^3*a*b^2-70*A*x*a*b^2+90*B*x*a^2*b)/(b*x^2+a)^(1/2)/b^4

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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{7}{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)^(7/2)*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(e*x)^(7/2)/(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^{3} x^{5} + A e^{3} x^{3}\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)^(7/2)*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral((B*e^3*x^5 + A*e^3*x^3)*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)**(7/2)*(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{7}{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)^(7/2)*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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