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

Optimal. Leaf size=326 \[ -\frac{a^{3/4} e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{6 a^{5/4} B e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}-\frac{6 a B e^2 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c} \]

[Out]

(2*A*e*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c) + (2*B*(e*x)^(3/2)*Sqrt[a + c*x^2])/(5*c) - (6*a*B*e^2*x*Sqrt[a + c*x^
2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (6*a^(5/4)*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])/(5*c^(7/4)*Sqrt[e*x]*Sqrt[a
+ c*x^2]) - (a^(3/4)*(9*Sqrt[a]*B + 5*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])/(15*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.335053, antiderivative size = 326, 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 = {833, 842, 840, 1198, 220, 1196} \[ -\frac{a^{3/4} e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{6 a^{5/4} B e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}-\frac{6 a B e^2 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*e*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c) + (2*B*(e*x)^(3/2)*Sqrt[a + c*x^2])/(5*c) - (6*a*B*e^2*x*Sqrt[a + c*x^
2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (6*a^(5/4)*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])/(5*c^(7/4)*Sqrt[e*x]*Sqrt[a
+ c*x^2]) - (a^(3/4)*(9*Sqrt[a]*B + 5*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])/(15*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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)}{\sqrt{a+c x^2}} \, dx &=\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c}+\frac{2 \int \frac{\sqrt{e x} \left (-\frac{3}{2} a B e+\frac{5}{2} A c e x\right )}{\sqrt{a+c x^2}} \, dx}{5 c}\\ &=\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c}+\frac{4 \int \frac{-\frac{5}{4} a A c e^2-\frac{9}{4} a B c e^2 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{15 c^2}\\ &=\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c}+\frac{\left (4 \sqrt{x}\right ) \int \frac{-\frac{5}{4} a A c e^2-\frac{9}{4} a B c e^2 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{15 c^2 \sqrt{e x}}\\ &=\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c}+\frac{\left (8 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-\frac{5}{4} a A c e^2-\frac{9}{4} a B c e^2 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15 c^2 \sqrt{e x}}\\ &=\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c}+\frac{\left (6 a^{3/2} 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 )}{5 c^{3/2} \sqrt{e x}}-\frac{\left (2 a \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) e^2 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15 c^{3/2} \sqrt{e x}}\\ &=\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c}-\frac{6 a B e^2 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{6 a^{5/4} 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 )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{a^{3/4} \left (9 \sqrt{a} B+5 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 )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0531168, size = 118, normalized size = 0.36 \[ \frac{2 e \sqrt{e x} \left (\left (a+c x^2\right ) (5 A+3 B x)-5 a A \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )-3 a B x \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )\right )}{15 c \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[e*x]*((5*A + 3*B*x)*(a + c*x^2) - 5*a*A*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^
2)/a)] - 3*a*B*x*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)]))/(15*c*Sqrt[a + c*x^2])

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Maple [A]  time = 0.014, size = 316, normalized size = 1. \begin{align*} -{\frac{e}{15\,x{c}^{2}}\sqrt{ex} \left ( 5\,A\sqrt{-ac}\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+18\,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}-9\,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}^{2}-6\,B{c}^{2}{x}^{4}-10\,A{c}^{2}{x}^{3}-6\,aBc{x}^{2}-10\,aAcx \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)^(1/2),x)

[Out]

-1/15*e/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)/c^2*(5*A*(-a*c)^(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)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(
1/2),1/2*2^(1/2))*a+18*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^2-9*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)*Elli
pticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2-6*B*c^2*x^4-10*A*c^2*x^3-6*a*B*c*x^2-10*a*A*c*x
)

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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}}}{\sqrt{c x^{2} + a}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C]  time = 16.0488, size = 94, normalized size = 0.29 \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{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \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{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \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)**(1/2),x)

[Out]

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + B*
e**(3/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*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}}}{\sqrt{c x^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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