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

Optimal. Leaf size=335 \[ -\frac{\sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B-5 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{12 a^{9/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}+\frac{B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{B x \sqrt{a+c x^2}}{2 a^2 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}} \]

[Out]

(Sqrt[e*x]*(A + B*x))/(3*a*e*(a + c*x^2)^(3/2)) + (Sqrt[e*x]*(5*A + 3*B*x))/(6*a^2*e*Sqrt[a + c*x^2]) - (B*x*S
qrt[a + c*x^2])/(2*a^2*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (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])/(2*a^(7/4)*c^(3/4)*Sqrt[e*
x]*Sqrt[a + c*x^2]) - ((3*Sqrt[a]*B - 5*A*Sqrt[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + S
qrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(12*a^(9/4)*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2
])

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Rubi [A]  time = 0.358285, antiderivative size = 335, 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 = {823, 842, 840, 1198, 220, 1196} \[ -\frac{\sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (3 \sqrt{a} B-5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{12 a^{9/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}+\frac{B \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{B x \sqrt{a+c x^2}}{2 a^2 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*(A + B*x))/(3*a*e*(a + c*x^2)^(3/2)) + (Sqrt[e*x]*(5*A + 3*B*x))/(6*a^2*e*Sqrt[a + c*x^2]) - (B*x*S
qrt[a + c*x^2])/(2*a^2*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (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])/(2*a^(7/4)*c^(3/4)*Sqrt[e*
x]*Sqrt[a + c*x^2]) - ((3*Sqrt[a]*B - 5*A*Sqrt[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + S
qrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(12*a^(9/4)*c^(3/4)*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 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} \left (a+c x^2\right )^{5/2}} \, dx &=\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}}-\frac{\int \frac{-\frac{5}{2} a A c e^2-\frac{3}{2} a B c e^2 x}{\sqrt{e x} \left (a+c x^2\right )^{3/2}} \, dx}{3 a^2 c e^2}\\ &=\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}+\frac{\int \frac{\frac{5}{4} a^2 A c^2 e^4-\frac{3}{4} a^2 B c^2 e^4 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{3 a^4 c^2 e^4}\\ &=\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}+\frac{\sqrt{x} \int \frac{\frac{5}{4} a^2 A c^2 e^4-\frac{3}{4} a^2 B c^2 e^4 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{3 a^4 c^2 e^4 \sqrt{e x}}\\ &=\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}+\frac{\left (2 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{5}{4} a^2 A c^2 e^4-\frac{3}{4} a^2 B c^2 e^4 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 a^4 c^2 e^4 \sqrt{e x}}\\ &=\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}+\frac{\left (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 )}{2 a^{3/2} \sqrt{c} \sqrt{e x}}-\frac{\left (\left (3 \sqrt{a} B-5 A \sqrt{c}\right ) \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{6 a^2 \sqrt{c} \sqrt{e x}}\\ &=\frac{\sqrt{e x} (A+B x)}{3 a e \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (5 A+3 B x)}{6 a^2 e \sqrt{a+c x^2}}-\frac{B x \sqrt{a+c x^2}}{2 a^2 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{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 )}{2 a^{7/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{\left (3 \sqrt{a} B-5 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 )}{12 a^{9/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.113803, size = 140, normalized size = 0.42 \[ \frac{x \left (7 a A-B x \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )+5 a B x+5 A c x^2+3 B c x^3\right )+5 A x \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )}{6 a^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*A*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/a)] + x*(7*a*A + 5*a*B*x + 5
*A*c*x^2 + 3*B*c*x^3 - B*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)]))/(6
*a^2*Sqrt[e*x]*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.036, size = 581, normalized size = 1.7 \begin{align*}{\frac{1}{12\,{a}^{2}c} \left ( 5\,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 ) \sqrt{-ac}{x}^{2}c+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 ){x}^{2}ac-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 ){x}^{2}ac+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+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}^{2}-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}+6\,B{c}^{2}{x}^{4}+10\,A{c}^{2}{x}^{3}+10\,aBc{x}^{2}+14\,aAcx \right ){\frac{1}{\sqrt{ex}}} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \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)^(5/2),x)

[Out]

1/12*(5*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^2*c+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)*Elli
pticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*c-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^2*a*c+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+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^2-6*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+6*B*c^2*x^4+10*A*c^2*x^3+10*a*B*c*x^2+14*a*A*c*x)/c/(e
*x)^(1/2)/a^2/(c*x^2+a)^(3/2)

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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{5}{2}} \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)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + a)^(5/2)*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^{3} e x^{7} + 3 \, a c^{2} e x^{5} + 3 \, a^{2} c e x^{3} + a^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c^3*e*x^7 + 3*a*c^2*e*x^5 + 3*a^2*c*e*x^3 + a^3*e*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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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{5}{2}} \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)^(5/2),x, algorithm="giac")

[Out]

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