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

Optimal. Leaf size=342 \[ -\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-5 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{a+b x^2} (3 A b-5 a B)}{5 a^2 e^3 \sqrt{e x}}-\frac{2 \sqrt{b} \sqrt{e x} \sqrt{a+b x^2} (3 A b-5 a B)}{5 a^2 e^4 \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-5 a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt{a+b x^2}}-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.251416, antiderivative size = 342, 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 = {453, 325, 329, 305, 220, 1196} \[ \frac{2 \sqrt{a+b x^2} (3 A b-5 a B)}{5 a^2 e^3 \sqrt{e x}}-\frac{2 \sqrt{b} \sqrt{e x} \sqrt{a+b x^2} (3 A b-5 a B)}{5 a^2 e^4 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt{a+b x^2}}+\frac{2 \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-5 a B) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt{a+b x^2}}-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[a + b*x^2])/(5*a*e*(e*x)^(5/2)) + (2*(3*A*b - 5*a*B)*Sqrt[a + b*x^2])/(5*a^2*e^3*Sqrt[e*x]) - (2*Sq
rt[b]*(3*A*b - 5*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*a^2*e^4*(Sqrt[a] + Sqrt[b]*x)) + (2*b^(1/4)*(3*A*b - 5*a*B
)*(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])/(5*a^(7/4)*e^(7/2)*Sqrt[a + b*x^2]) - (b^(1/4)*(3*A*b - 5*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])/(5*a^(7
/4)*e^(7/2)*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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)^{7/2} \sqrt{a+b x^2}} \, dx &=-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}}-\frac{(3 A b-5 a B) \int \frac{1}{(e x)^{3/2} \sqrt{a+b x^2}} \, dx}{5 a e^2}\\ &=-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}}+\frac{2 (3 A b-5 a B) \sqrt{a+b x^2}}{5 a^2 e^3 \sqrt{e x}}-\frac{(b (3 A b-5 a B)) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{5 a^2 e^4}\\ &=-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}}+\frac{2 (3 A b-5 a B) \sqrt{a+b x^2}}{5 a^2 e^3 \sqrt{e x}}-\frac{(2 b (3 A b-5 a B)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 a^2 e^5}\\ &=-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}}+\frac{2 (3 A b-5 a B) \sqrt{a+b x^2}}{5 a^2 e^3 \sqrt{e x}}-\frac{\left (2 \sqrt{b} (3 A b-5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 a^{3/2} e^4}+\frac{\left (2 \sqrt{b} (3 A b-5 a B)\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 )}{5 a^{3/2} e^4}\\ &=-\frac{2 A \sqrt{a+b x^2}}{5 a e (e x)^{5/2}}+\frac{2 (3 A b-5 a B) \sqrt{a+b x^2}}{5 a^2 e^3 \sqrt{e x}}-\frac{2 \sqrt{b} (3 A b-5 a B) \sqrt{e x} \sqrt{a+b x^2}}{5 a^2 e^4 \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 \sqrt [4]{b} (3 A b-5 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 )}{5 a^{7/4} e^{7/2} \sqrt{a+b x^2}}-\frac{\sqrt [4]{b} (3 A b-5 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 )}{5 a^{7/4} e^{7/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0464111, size = 82, normalized size = 0.24 \[ -\frac{2 x \left (x^2 \sqrt{\frac{b x^2}{a}+1} (5 a B-3 A b) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{b x^2}{a}\right )+A \left (a+b x^2\right )\right )}{5 a (e x)^{7/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.018, size = 417, normalized size = 1.2 \begin{align*} -{\frac{1}{5\,{x}^{2}{e}^{3}{a}^{2}} \left ( 6\,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 ){x}^{2}ab-3\,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 ){x}^{2}ab-10\,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 ){x}^{2}{a}^{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 ){x}^{2}{a}^{2}-6\,A{b}^{2}{x}^{4}+10\,B{x}^{4}ab-4\,aAb{x}^{2}+10\,B{x}^{2}{a}^{2}+2\,A{a}^{2} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x)

[Out]

-1/5/x^2*(6*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-3*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-10*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+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))*x^2*a^2
-6*A*b^2*x^4+10*B*x^4*a*b-4*a*A*b*x^2+10*B*x^2*a^2+2*A*a^2)/(b*x^2+a)^(1/2)/e^3/(e*x)^(1/2)/a^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{b x^{2} + a} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*(e*x)^(7/2)), 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 e^{4} x^{6} + a e^{4} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b*e^4*x^6 + a*e^4*x^4), x)

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Sympy [C]  time = 136.821, size = 104, normalized size = 0.3 \begin{align*} \frac{A \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} e^{\frac{7}{2}} x^{\frac{5}{2}} \Gamma \left (- \frac{1}{4}\right )} + \frac{B \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} e^{\frac{7}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(e*x)**(7/2)/(b*x**2+a)**(1/2),x)

[Out]

A*gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**(7/2)*x**(5/2)*gamma(-1/4))
+ B*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**(7/2)*sqrt(x)*gamma(3/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{b x^{2} + a} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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