3.464 \(\int \frac{A+B x}{(e x)^{5/2} \sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=327 \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B-A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{3 a^{5/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}+\frac{2 B \sqrt{c} x \sqrt{a+c x^2}}{a e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

[Out]

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

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Rubi [A]  time = 0.317513, antiderivative size = 327, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {835, 842, 840, 1198, 220, 1196} \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B-A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 a^{5/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}+\frac{2 B \sqrt{c} x \sqrt{a+c x^2}}{a e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[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}{(e x)^{5/2} \sqrt{a+c x^2}} \, dx &=-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 \int \frac{-\frac{3}{2} a B e+\frac{1}{2} A c e x}{(e x)^{3/2} \sqrt{a+c x^2}} \, dx}{3 a e^2}\\ &=-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}+\frac{4 \int \frac{-\frac{1}{4} a A c e^2+\frac{3}{4} a B c e^2 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{3 a^2 e^4}\\ &=-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}+\frac{\left (4 \sqrt{x}\right ) \int \frac{-\frac{1}{4} a A c e^2+\frac{3}{4} a B c e^2 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{3 a^2 e^4 \sqrt{e x}}\\ &=-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}+\frac{\left (8 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-\frac{1}{4} a A c e^2+\frac{3}{4} a B c e^2 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 a^2 e^4 \sqrt{e x}}\\ &=-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}-\frac{\left (2 B \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 )}{\sqrt{a} e^2 \sqrt{e x}}+\frac{\left (2 \left (3 \sqrt{a} B-A \sqrt{c}\right ) \sqrt{c} \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 a e^2 \sqrt{e x}}\\ &=-\frac{2 A \sqrt{a+c x^2}}{3 a e (e x)^{3/2}}-\frac{2 B \sqrt{a+c x^2}}{a e^2 \sqrt{e x}}+\frac{2 B \sqrt{c} x \sqrt{a+c x^2}}{a e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{2 B \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 )}{a^{3/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{\left (3 \sqrt{a} B-A \sqrt{c}\right ) \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}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 a^{5/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0333972, size = 82, normalized size = 0.25 \[ -\frac{2 x \sqrt{\frac{c x^2}{a}+1} \left (A \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};-\frac{c x^2}{a}\right )+3 B x \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c x^2}{a}\right )\right )}{3 (e x)^{5/2} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 306, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,ax{e}^{2}} \left ( A\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}x+3\,B\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 ) xa-6\,B\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 ) xa+6\,Bc{x}^{3}+2\,Ac{x}^{2}+6\,aBx+2\,aA \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)^(5/2)/(c*x^2+a)^(1/2),x)

[Out]

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

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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} \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+A)/(e*x)^(5/2)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/(sqrt(c*x^2 + a)*(e*x)^(5/2)), 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}}{c e^{3} x^{5} + a e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(5/2)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c*e^3*x^5 + a*e^3*x^3), x)

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Sympy [C]  time = 37.4219, size = 100, normalized size = 0.31 \begin{align*} \frac{A \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} e^{\frac{5}{2}} x^{\frac{3}{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{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} e^{\frac{5}{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+A)/(e*x)**(5/2)/(c*x**2+a)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{\sqrt{c x^{2} + a} \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+A)/(e*x)^(5/2)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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