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

Optimal. Leaf size=296 \[ \frac{e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B+A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e \sqrt{e x} (A+B x)}{c \sqrt{a+c x^2}}+\frac{3 B e^2 x \sqrt{a+c x^2}}{c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{3 \sqrt [4]{a} B e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{7/4} \sqrt{e x} \sqrt{a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.266385, antiderivative size = 296, 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 = {819, 842, 840, 1198, 220, 1196} \[ \frac{e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B+A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e \sqrt{e x} (A+B x)}{c \sqrt{a+c x^2}}+\frac{3 B e^2 x \sqrt{a+c x^2}}{c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{3 \sqrt [4]{a} B e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{7/4} \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 819

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

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

Mathematica [C]  time = 0.051173, size = 102, normalized size = 0.34 \[ \frac{e \sqrt{e x} \left (A \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )+B x \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )-A-B x\right )}{c \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.029, size = 290, normalized size = 1. \begin{align*}{\frac{e}{2\,x{c}^{2}}\sqrt{ex} \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}-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 ) a+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 ) a-2\,Bc{x}^{2}-2\,Acx \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 52.8504, size = 94, normalized size = 0.32 \begin{align*} \frac{A e^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{9}{4}\right )} + \frac{B e^{\frac{3}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 3/2), (9/4,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(9/4)) + B
*e**(3/2)*x**(7/2)*gamma(7/4)*hyper((3/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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