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

Optimal. Leaf size=327 \[ \frac{\sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{a} B+3 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 a^{7/4} \sqrt [4]{c} e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{3 A \sqrt{c} x \sqrt{a+c x^2}}{a^2 e \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{3 A \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{7/4} e \sqrt{e x} \sqrt{a+c x^2}}+\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.335623, antiderivative size = 327, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {823, 835, 842, 840, 1198, 220, 1196} \[ \frac{\sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{a} B+3 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} \sqrt [4]{c} e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{3 A \sqrt{c} x \sqrt{a+c x^2}}{a^2 e \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{3 A \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{7/4} e \sqrt{e x} \sqrt{a+c x^2}}+\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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)^{3/2} \left (a+c x^2\right )^{3/2}} \, dx &=\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}}-\frac{\int \frac{-\frac{3}{2} a A c e^2-\frac{1}{2} a B c e^2 x}{(e x)^{3/2} \sqrt{a+c x^2}} \, dx}{a^2 c e^2}\\ &=\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{2 \int \frac{\frac{1}{4} a^2 B c e^3+\frac{3}{4} a A c^2 e^3 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{a^3 c e^4}\\ &=\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{\left (2 \sqrt{x}\right ) \int \frac{\frac{1}{4} a^2 B c e^3+\frac{3}{4} a A c^2 e^3 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{a^3 c e^4 \sqrt{e x}}\\ &=\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{\left (4 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{1}{4} a^2 B c e^3+\frac{3}{4} a A c^2 e^3 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{a^3 c e^4 \sqrt{e x}}\\ &=\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{\left (\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 )}{a^{3/2} e \sqrt{e x}}-\frac{\left (3 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 )}{a^{3/2} e \sqrt{e x}}\\ &=\frac{A+B x}{a e \sqrt{e x} \sqrt{a+c x^2}}-\frac{3 A \sqrt{a+c x^2}}{a^2 e \sqrt{e x}}+\frac{3 A \sqrt{c} x \sqrt{a+c x^2}}{a^2 e \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{3 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 )}{a^{7/4} e \sqrt{e x} \sqrt{a+c x^2}}+\frac{\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 )}{2 a^{7/4} \sqrt [4]{c} e \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0487739, size = 100, normalized size = 0.31 \[ \frac{x \left (-3 A \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c x^2}{a}\right )+B x \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )+A+B x\right )}{a (e x)^{3/2} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 304, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,ce{a}^{2}} \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+6\,A{c}^{2}{x}^{2}-2\,aBcx+4\,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)/(e*x)^(3/2)/(c*x^2+a)^(3/2),x)

[Out]

-1/2*(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+6*A*c^2*x^2-2*a*B*c*x+4*a*A*c)/(c*x^2+a)^(1/2)/c/e/(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 + A}{{\left (c x^{2} + a\right )}^{\frac{3}{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+A)/(e*x)^(3/2)/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + a}{\left (B x + A\right )} \sqrt{e x}}{c^{2} e^{2} x^{6} + 2 \, a c e^{2} x^{4} + a^{2} e^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 65.6325, size = 97, normalized size = 0.3 \begin{align*} \frac{A \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{3}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} e^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{B \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{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)/(e*x)**(3/2)/(c*x**2+a)**(3/2),x)

[Out]

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

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

[Out]

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