3.815 \(\int \frac{(e x)^{7/2} (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=208 \[ \frac{5 e^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (A b-3 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 \sqrt [4]{a} b^{13/4} \sqrt{a+b x^2}}-\frac{5 e^3 \sqrt{e x} (A b-3 a B)}{6 b^3 \sqrt{a+b x^2}}-\frac{e (e x)^{5/2} (A b-3 a B)}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}} \]

[Out]

-((A*b - 3*a*B)*e*(e*x)^(5/2))/(3*b^2*(a + b*x^2)^(3/2)) + (2*B*(e*x)^(9/2))/(3*b*e*(a + b*x^2)^(3/2)) - (5*(A
*b - 3*a*B)*e^3*Sqrt[e*x])/(6*b^3*Sqrt[a + b*x^2]) + (5*(A*b - 3*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])/(12*a^(1/4)*b
^(13/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.132842, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {459, 288, 329, 220} \[ -\frac{5 e^3 \sqrt{e x} (A b-3 a B)}{6 b^3 \sqrt{a+b x^2}}+\frac{5 e^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (A b-3 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 \sqrt [4]{a} b^{13/4} \sqrt{a+b x^2}}-\frac{e (e x)^{5/2} (A b-3 a B)}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

-((A*b - 3*a*B)*e*(e*x)^(5/2))/(3*b^2*(a + b*x^2)^(3/2)) + (2*B*(e*x)^(9/2))/(3*b*e*(a + b*x^2)^(3/2)) - (5*(A
*b - 3*a*B)*e^3*Sqrt[e*x])/(6*b^3*Sqrt[a + b*x^2]) + (5*(A*b - 3*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])/(12*a^(1/4)*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 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 )^{5/2}} \, dx &=\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}}-\frac{\left (2 \left (-\frac{3 A b}{2}+\frac{9 a B}{2}\right )\right ) \int \frac{(e x)^{7/2}}{\left (a+b x^2\right )^{5/2}} \, dx}{3 b}\\ &=-\frac{(A b-3 a B) e (e x)^{5/2}}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}}+\frac{\left (5 (A b-3 a B) e^2\right ) \int \frac{(e x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{6 b^2}\\ &=-\frac{(A b-3 a B) e (e x)^{5/2}}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}}-\frac{5 (A b-3 a B) e^3 \sqrt{e x}}{6 b^3 \sqrt{a+b x^2}}+\frac{\left (5 (A b-3 a B) e^4\right ) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{12 b^3}\\ &=-\frac{(A b-3 a B) e (e x)^{5/2}}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}}-\frac{5 (A b-3 a B) e^3 \sqrt{e x}}{6 b^3 \sqrt{a+b x^2}}+\frac{\left (5 (A b-3 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 )}{6 b^3}\\ &=-\frac{(A b-3 a B) e (e x)^{5/2}}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac{2 B (e x)^{9/2}}{3 b e \left (a+b x^2\right )^{3/2}}-\frac{5 (A b-3 a B) e^3 \sqrt{e x}}{6 b^3 \sqrt{a+b x^2}}+\frac{5 (A b-3 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 )}{12 \sqrt [4]{a} b^{13/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.173096, size = 116, normalized size = 0.56 \[ \frac{e^3 \sqrt{e x} \left (15 a^2 B+5 \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (A b-3 a B) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+a \left (21 b B x^2-5 A b\right )+b^2 x^2 \left (4 B x^2-7 A\right )\right )}{6 b^3 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

(e^3*Sqrt[e*x]*(15*a^2*B + b^2*x^2*(-7*A + 4*B*x^2) + a*(-5*A*b + 21*b*B*x^2) + 5*(A*b - 3*a*B)*(a + b*x^2)*Sq
rt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(6*b^3*(a + b*x^2)^(3/2))

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Maple [B]  time = 0.034, size = 439, normalized size = 2.1 \begin{align*}{\frac{{e}^{3}}{12\,x{b}^{4}} \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}-15\,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+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-15\,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}+8\,B{x}^{5}{b}^{3}-14\,A{x}^{3}{b}^{3}+42\,B{x}^{3}a{b}^{2}-10\,Axa{b}^{2}+30\,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)^(7/2)*(B*x^2+A)/(b*x^2+a)^(5/2),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^2-15*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)*E
llipticF(((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)*Ellip
ticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a*b-15*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+8*B*x^5*b^3-14*A*x^3*b^3+42*B*x^3*a*b^2-10*A*x*a*b^2+30*B*x*a^2*b)*e
^3/x*(e*x)^(1/2)/b^4/(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{7}{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)^(7/2)*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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