3.438 \(\int \frac{(A+B x) \sqrt{a+c x^2}}{(e x)^{3/2}} \, dx\)

Optimal. Leaf size=300 \[ \frac{2 \sqrt [4]{a} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{a} B+3 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{c} e \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 \sqrt{a+c x^2} (3 A-B x)}{3 e \sqrt{e x}}+\frac{4 A \sqrt{c} x \sqrt{a+c x^2}}{e \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 \sqrt [4]{a} A \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{e \sqrt{e x} \sqrt{a+c x^2}} \]

[Out]

(-2*(3*A - B*x)*Sqrt[a + c*x^2])/(3*e*Sqrt[e*x]) + (4*A*Sqrt[c]*x*Sqrt[a + c*x^2])/(e*Sqrt[e*x]*(Sqrt[a] + Sqr
t[c]*x)) - (4*a^(1/4)*A*c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*Ellipt
icE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(e*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(1/4)*(Sqrt[a]*B + 3*A*Sqr
t[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt
[x])/a^(1/4)], 1/2])/(3*c^(1/4)*e*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.260199, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {813, 842, 840, 1198, 220, 1196} \[ -\frac{2 \sqrt{a+c x^2} (3 A-B x)}{3 e \sqrt{e x}}+\frac{2 \sqrt [4]{a} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{a} B+3 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{c} e \sqrt{e x} \sqrt{a+c x^2}}+\frac{4 A \sqrt{c} x \sqrt{a+c x^2}}{e \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 \sqrt [4]{a} A \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{e \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*A - B*x)*Sqrt[a + c*x^2])/(3*e*Sqrt[e*x]) + (4*A*Sqrt[c]*x*Sqrt[a + c*x^2])/(e*Sqrt[e*x]*(Sqrt[a] + Sqr
t[c]*x)) - (4*a^(1/4)*A*c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*Ellipt
icE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(e*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(1/4)*(Sqrt[a]*B + 3*A*Sqr
t[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt
[x])/a^(1/4)], 1/2])/(3*c^(1/4)*e*Sqrt[e*x]*Sqrt[a + c*x^2])

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 842

Int[((f_) + (g_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[x]/Sqrt[e*x], Int[
(f + g*x)/(Sqrt[x]*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, e, f, g}, x]

Rule 840

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f + g*x^2)/Sqrt[
a + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, c, f, g}, x]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/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) \sqrt{a+c x^2}}{(e x)^{3/2}} \, dx &=-\frac{2 (3 A-B x) \sqrt{a+c x^2}}{3 e \sqrt{e x}}-\frac{2 \int \frac{-a B e-3 A c e x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{3 e^2}\\ &=-\frac{2 (3 A-B x) \sqrt{a+c x^2}}{3 e \sqrt{e x}}-\frac{\left (2 \sqrt{x}\right ) \int \frac{-a B e-3 A c e x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{3 e^2 \sqrt{e x}}\\ &=-\frac{2 (3 A-B x) \sqrt{a+c x^2}}{3 e \sqrt{e x}}-\frac{\left (4 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-a B e-3 A c e x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 e^2 \sqrt{e x}}\\ &=-\frac{2 (3 A-B x) \sqrt{a+c x^2}}{3 e \sqrt{e x}}+\frac{\left (4 \sqrt{a} \left (\sqrt{a} B+3 A \sqrt{c}\right ) \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 e \sqrt{e x}}-\frac{\left (4 \sqrt{a} A \sqrt{c} \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{e \sqrt{e x}}\\ &=-\frac{2 (3 A-B x) \sqrt{a+c x^2}}{3 e \sqrt{e x}}+\frac{4 A \sqrt{c} x \sqrt{a+c x^2}}{e \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 \sqrt [4]{a} A \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{e \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 \sqrt [4]{a} \left (\sqrt{a} B+3 A \sqrt{c}\right ) \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{c} e \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0312356, size = 80, normalized size = 0.27 \[ \frac{2 x \sqrt{a+c x^2} \left (B x \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{a}\right )-A \, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-\frac{c x^2}{a}\right )\right )}{(e x)^{3/2} \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 310, normalized size = 1. \begin{align*} -{\frac{2}{3\,ce} \left ( 3\,A\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) ac-6\,A\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) ac-B\sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{2}\sqrt{{ \left ( -cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{-{cx{\frac{1}{\sqrt{-ac}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ac}a-B{c}^{2}{x}^{3}+3\,A{c}^{2}{x}^{2}-aBcx+3\,aAc \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(3/2),x)

[Out]

-2/3*(3*A*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c
)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a*c-6*A*((c*x+(-a*c)^(1/2))/(-a*
c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticE(((c*x+(-a
*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a*c-B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c
)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2
^(1/2))*(-a*c)^(1/2)*a-B*c^2*x^3+3*A*c^2*x^2-a*B*c*x+3*a*A*c)/(c*x^2+a)^(1/2)/e/(e*x)^(1/2)/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{\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+A)*(c*x^2+a)^(1/2)/(e*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/(e*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 4.369, size = 100, normalized size = 0.33 \begin{align*} \frac{A \sqrt{a} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{B \sqrt{a} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(1/2)/(e*x)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{\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+A)*(c*x^2+a)^(1/2)/(e*x)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/(e*x)^(3/2), x)