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

Optimal. Leaf size=213 \[ -\frac{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{12 a^{13/4} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{e x} (3 A b-a B)}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{\sqrt{e x} (3 A b-a B)}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.141613, antiderivative size = 213, 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 = {453, 290, 329, 220} \[ -\frac{5 \sqrt{e x} (3 A b-a B)}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{\sqrt{e x} (3 A b-a B)}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{13/4} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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{A+B x^2}{(e x)^{5/2} \left (a+b x^2\right )^{5/2}} \, dx &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}}-\frac{(3 A b-a B) \int \frac{1}{\sqrt{e x} \left (a+b x^2\right )^{5/2}} \, dx}{a e^2}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}}-\frac{(3 A b-a B) \sqrt{e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{(5 (3 A b-a B)) \int \frac{1}{\sqrt{e x} \left (a+b x^2\right )^{3/2}} \, dx}{6 a^2 e^2}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}}-\frac{(3 A b-a B) \sqrt{e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{5 (3 A b-a B) \sqrt{e x}}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{(5 (3 A b-a B)) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{12 a^3 e^2}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}}-\frac{(3 A b-a B) \sqrt{e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{5 (3 A b-a B) \sqrt{e x}}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{(5 (3 A b-a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{6 a^3 e^3}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}}-\frac{(3 A b-a B) \sqrt{e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{5 (3 A b-a B) \sqrt{e x}}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{5 (3 A b-a B) \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^{13/4} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0765028, size = 120, normalized size = 0.56 \[ \frac{x \left (a^2 \left (7 B x^2-4 A\right )+5 x^2 \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (a B-3 A b) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+a \left (5 b B x^4-21 A b x^2\right )-15 A b^2 x^4\right )}{6 a^3 (e x)^{5/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.025, size = 446, normalized size = 2.1 \begin{align*} -{\frac{1}{12\,x{e}^{2}{a}^{3}b} \left ( 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 ) \sqrt{-ab}{x}^{3}{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}^{3}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 ) \sqrt{-ab}xab-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{a}^{2}+30\,A{x}^{4}{b}^{3}-10\,B{x}^{4}a{b}^{2}+42\,A{x}^{2}a{b}^{2}-14\,B{x}^{2}{a}^{2}b+8\,A{a}^{2}b \right ){\frac{1}{\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((B*x^2+A)/(e*x)^(5/2)/(b*x^2+a)^(5/2),x)

[Out]

-1/12*(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)^(1/2)*x^3*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)*
EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x^3*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)^(1/2)*x*a*b-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)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b
)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x*a^2+30*A*x^4*b^3-10*B*x^4*a*b^2+42*A*x^2*a*b^2-14*B*x^2*a^2*b+8*A*a
^2*b)/x/e^2/(e*x)^(1/2)/a^3/b/(b*x^2+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(5/2)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{b^{3} e^{3} x^{9} + 3 \, a b^{2} e^{3} x^{7} + 3 \, a^{2} b e^{3} x^{5} + a^{3} e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b^3*e^3*x^9 + 3*a*b^2*e^3*x^7 + 3*a^2*b*e^3*x^5 + a^3*e^3*x^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((B*x**2+A)/(e*x)**(5/2)/(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{B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(5/2)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

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