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

Optimal. Leaf size=253 \[ \frac{\sqrt [4]{a} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\frac{A \sqrt{c}}{\sqrt{a}}+B\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 \sqrt [4]{a} B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 B x \sqrt{a+c x^2}}{\sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

[Out]

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

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Rubi [A]  time = 0.180883, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {842, 840, 1198, 220, 1196} \[ \frac{\sqrt [4]{a} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\frac{A \sqrt{c}}{\sqrt{a}}+B\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 \sqrt [4]{a} B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 B x \sqrt{a+c x^2}}{\sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{e x} \sqrt{a+c x^2}} \, dx &=\frac{\sqrt{x} \int \frac{A+B x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{\sqrt{e x}}\\ &=\frac{\left (2 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{A+B x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{\sqrt{e x}}\\ &=\frac{\left (2 \left (A+\frac{\sqrt{a} B}{\sqrt{c}}\right ) \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{\sqrt{e x}}-\frac{\left (2 \sqrt{a} B \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 )}{\sqrt{c} \sqrt{e x}}\\ &=\frac{2 B x \sqrt{a+c x^2}}{\sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 \sqrt [4]{a} B \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 )}{c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{\left (\sqrt{a} B+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 )}{\sqrt [4]{a} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0277151, size = 82, normalized size = 0.32 \[ \frac{2 x \sqrt{\frac{c x^2}{a}+1} \left (3 A \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )+B x \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )\right )}{3 \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 175, normalized size = 0.7 \begin{align*}{\frac{\sqrt{2}}{{c}^{2}}\sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{{ \left ( -cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{-{cx{\frac{1}{\sqrt{-ac}}}}}\sqrt{-ac} \left ( A{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}},{\frac{\sqrt{2}}{2}} \right ) c+B{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ac}-2\,B{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) \sqrt{-ac} \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)/(e*x)^(1/2)/(c*x^2+a)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)/(sqrt(c*x^2 + a)*sqrt(e*x)), 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}}{c e x^{3} + a e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 4.77581, size = 94, normalized size = 0.37 \begin{align*} \frac{A \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \sqrt{e} \Gamma \left (\frac{5}{4}\right )} + \frac{B x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \sqrt{e} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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