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

Optimal. Leaf size=377 \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{4 a^{11/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{2 a^{11/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{(e x)^{3/2} (7 A b-a B)}{2 a^3 e^3 \sqrt{a+b x^2}}-\frac{(e x)^{3/2} (7 A b-a B)}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}+\frac{\sqrt{e x} \sqrt{a+b x^2} (7 A b-a B)}{2 a^3 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A}{a e \sqrt{e x} \left (a+b x^2\right )^{3/2}} \]

[Out]

(-2*A)/(a*e*Sqrt[e*x]*(a + b*x^2)^(3/2)) - ((7*A*b - a*B)*(e*x)^(3/2))/(3*a^2*e^3*(a + b*x^2)^(3/2)) - ((7*A*b
- a*B)*(e*x)^(3/2))/(2*a^3*e^3*Sqrt[a + b*x^2]) + ((7*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(2*a^3*Sqrt[b]*e^
2*(Sqrt[a] + Sqrt[b]*x)) - ((7*A*b - a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*Elli
pticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(2*a^(11/4)*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + ((
7*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])/(4*a^(11/4)*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.292428, antiderivative size = 377, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {453, 290, 329, 305, 220, 1196} \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{4 a^{11/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{2 a^{11/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{(e x)^{3/2} (7 A b-a B)}{2 a^3 e^3 \sqrt{a+b x^2}}-\frac{(e x)^{3/2} (7 A b-a B)}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}+\frac{\sqrt{e x} \sqrt{a+b x^2} (7 A b-a B)}{2 a^3 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A}{a e \sqrt{e x} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.062209, size = 86, normalized size = 0.23 \[ \frac{x \left (2 x^2 \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (a B-7 A b) \, _2F_1\left (\frac{3}{4},\frac{5}{2};\frac{7}{4};-\frac{b x^2}{a}\right )-6 a^2 A\right )}{3 a^3 (e x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.025, size = 771, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(42*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))*x^2*a*b^2-21*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))*x^2*a*b^2-6*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))*x^2*a^2*b+3*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))*x^2*
a^2*b+42*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-21*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-6*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+3*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^3-42*A*x^4*b^3+6
*B*x^4*a*b^2-70*A*x^2*a*b^2+10*B*x^2*a^2*b-24*A*a^2*b)/b/a^3/e/(e*x)^(1/2)/(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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/((b*x^2 + a)^(5/2)*(e*x)^(3/2)), 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^{2} x^{8} + 3 \, a b^{2} e^{2} x^{6} + 3 \, a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(3/2)/(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^2*x^8 + 3*a*b^2*e^2*x^6 + 3*a^2*b*e^2*x^4 + a^3*e^2*x^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((B*x**2+A)/(e*x)**(3/2)/(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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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