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

Optimal. Leaf size=338 \[ -\frac{a^{5/4} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (9 A b-7 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{15 b^{11/4} \sqrt{a+b x^2}}+\frac{2 a^{5/4} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (9 A b-7 a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{a+b x^2}}-\frac{2 a e^2 \sqrt{e x} \sqrt{a+b x^2} (9 A b-7 a B)}{15 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 e (e x)^{3/2} \sqrt{a+b x^2} (9 A b-7 a B)}{45 b^2}+\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e} \]

[Out]

(2*(9*A*b - 7*a*B)*e*(e*x)^(3/2)*Sqrt[a + b*x^2])/(45*b^2) + (2*B*(e*x)^(7/2)*Sqrt[a + b*x^2])/(9*b*e) - (2*a*
(9*A*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) + (2*a^(5/4)*(9*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])/(15*b^(11/4)*Sqrt[a + b*x^2]) - (a^(5/4)*(9*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]
)/(15*b^(11/4)*Sqrt[a + b*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.257584, antiderivative size = 338, 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, 321, 329, 305, 220, 1196} \[ -\frac{a^{5/4} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (9 A b-7 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{a+b x^2}}+\frac{2 a^{5/4} e^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (9 A b-7 a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{a+b x^2}}-\frac{2 a e^2 \sqrt{e x} \sqrt{a+b x^2} (9 A b-7 a B)}{15 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 e (e x)^{3/2} \sqrt{a+b x^2} (9 A b-7 a B)}{45 b^2}+\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9*A*b - 7*a*B)*e*(e*x)^(3/2)*Sqrt[a + b*x^2])/(45*b^2) + (2*B*(e*x)^(7/2)*Sqrt[a + b*x^2])/(9*b*e) - (2*a*
(9*A*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) + (2*a^(5/4)*(9*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])/(15*b^(11/4)*Sqrt[a + b*x^2]) - (a^(5/4)*(9*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]
)/(15*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 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 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 )}{\sqrt{a+b x^2}} \, dx &=\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e}-\frac{\left (2 \left (-\frac{9 A b}{2}+\frac{7 a B}{2}\right )\right ) \int \frac{(e x)^{5/2}}{\sqrt{a+b x^2}} \, dx}{9 b}\\ &=\frac{2 (9 A b-7 a B) e (e x)^{3/2} \sqrt{a+b x^2}}{45 b^2}+\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e}-\frac{\left (a (9 A b-7 a B) e^2\right ) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{15 b^2}\\ &=\frac{2 (9 A b-7 a B) e (e x)^{3/2} \sqrt{a+b x^2}}{45 b^2}+\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e}-\frac{(2 a (9 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 )}{15 b^2}\\ &=\frac{2 (9 A b-7 a B) e (e x)^{3/2} \sqrt{a+b x^2}}{45 b^2}+\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e}-\frac{\left (2 a^{3/2} (9 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 )}{15 b^{5/2}}+\frac{\left (2 a^{3/2} (9 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 )}{15 b^{5/2}}\\ &=\frac{2 (9 A b-7 a B) e (e x)^{3/2} \sqrt{a+b x^2}}{45 b^2}+\frac{2 B (e x)^{7/2} \sqrt{a+b x^2}}{9 b e}-\frac{2 a (9 A b-7 a B) e^2 \sqrt{e x} \sqrt{a+b x^2}}{15 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 a^{5/4} (9 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 )}{15 b^{11/4} \sqrt{a+b x^2}}-\frac{a^{5/4} (9 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 )}{15 b^{11/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.111846, size = 96, normalized size = 0.28 \[ \frac{2 e (e x)^{3/2} \left (a \sqrt{\frac{b x^2}{a}+1} (7 a B-9 A b) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )-\left (a+b x^2\right ) \left (7 a B-9 A b-5 b B x^2\right )\right )}{45 b^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(e*x)^(3/2)*(-((a + b*x^2)*(-9*A*b + 7*a*B - 5*b*B*x^2)) + a*(-9*A*b + 7*a*B)*Sqrt[1 + (b*x^2)/a]*Hyperge
ometric2F1[1/2, 3/4, 7/4, -((b*x^2)/a)]))/(45*b^2*Sqrt[a + b*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.025, size = 417, normalized size = 1.2 \begin{align*} -{\frac{{e}^{2}}{45\,x{b}^{3}}\sqrt{ex} \left ( -10\,B{x}^{6}{b}^{3}+54\,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 ){a}^{2}b-27\,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 ){a}^{2}b-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}^{3}+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}^{3}-18\,A{x}^{4}{b}^{3}+4\,B{x}^{4}a{b}^{2}-18\,A{x}^{2}a{b}^{2}+14\,B{x}^{2}{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)^(5/2)*(B*x^2+A)/(b*x^2+a)^(1/2),x)

[Out]

-1/45/x*e^2*(e*x)^(1/2)/(b*x^2+a)^(1/2)/b^3*(-10*B*x^6*b^3+54*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^2*b-27*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^2*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^3+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^3-18*A*x^4*b^3+4*B*x^4*a*b^2-18*A*x^2*a*b^2+14*B*x^2*a^2*b)

________________________________________________________________________________________

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}}}{\sqrt{b x^{2} + a}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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{e x}}{\sqrt{b x^{2} + a}}, 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)^(1/2),x, algorithm="fricas")

[Out]

integral((B*e^2*x^4 + A*e^2*x^2)*sqrt(e*x)/sqrt(b*x^2 + a), x)

________________________________________________________________________________________

Sympy [C]  time = 98.2592, size = 94, normalized size = 0.28 \begin{align*} \frac{A e^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} + \frac{B e^{\frac{5}{2}} x^{\frac{11}{2}} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{15}{4}\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)**(1/2),x)

[Out]

A*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(11/4)) +
B*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((1/2, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(15/4))

________________________________________________________________________________________

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}}}{\sqrt{b x^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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