3.440 \(\int \frac{(A+B x) \sqrt{a+c x^2}}{(e x)^{7/2}} \, dx\)

Optimal. Leaf size=338 \[ \frac{2 c^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 \sqrt{a} B+3 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 a^{3/4} e^3 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 A c^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} e^3 \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 \sqrt{a+c x^2} (3 A+5 B x)}{15 e (e x)^{5/2}}+\frac{4 A c^{3/2} x \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}} \]

[Out]

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

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Rubi [A]  time = 0.349788, antiderivative size = 338, 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 = {811, 835, 842, 840, 1198, 220, 1196} \[ \frac{2 c^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 \sqrt{a} B+3 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} e^3 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 A c^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} e^3 \sqrt{e x} \sqrt{a+c x^2}}-\frac{2 \sqrt{a+c x^2} (3 A+5 B x)}{15 e (e x)^{5/2}}+\frac{4 A c^{3/2} x \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + c*x^2])/(e*x)^(7/2),x]

[Out]

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

Rule 811

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*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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) \sqrt{a+c x^2}}{(e x)^{7/2}} \, dx &=-\frac{2 (3 A+5 B x) \sqrt{a+c x^2}}{15 e (e x)^{5/2}}-\frac{2 \int \frac{-3 a A c e^2-5 a B c e^2 x}{(e x)^{3/2} \sqrt{a+c x^2}} \, dx}{15 a e^4}\\ &=-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}}-\frac{2 (3 A+5 B x) \sqrt{a+c x^2}}{15 e (e x)^{5/2}}+\frac{4 \int \frac{\frac{5}{2} a^2 B c e^3+\frac{3}{2} a A c^2 e^3 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{15 a^2 e^6}\\ &=-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}}-\frac{2 (3 A+5 B x) \sqrt{a+c x^2}}{15 e (e x)^{5/2}}+\frac{\left (4 \sqrt{x}\right ) \int \frac{\frac{5}{2} a^2 B c e^3+\frac{3}{2} a A c^2 e^3 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{15 a^2 e^6 \sqrt{e x}}\\ &=-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}}-\frac{2 (3 A+5 B x) \sqrt{a+c x^2}}{15 e (e x)^{5/2}}+\frac{\left (8 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{5}{2} a^2 B c e^3+\frac{3}{2} a A c^2 e^3 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15 a^2 e^6 \sqrt{e x}}\\ &=-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}}-\frac{2 (3 A+5 B x) \sqrt{a+c x^2}}{15 e (e x)^{5/2}}+\frac{\left (4 \left (5 \sqrt{a} B+3 A \sqrt{c}\right ) c \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{a} e^3 \sqrt{e x}}-\frac{\left (4 A c^{3/2} \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 \sqrt{a} e^3 \sqrt{e x}}\\ &=-\frac{4 A c \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x}}-\frac{2 (3 A+5 B x) \sqrt{a+c x^2}}{15 e (e x)^{5/2}}+\frac{4 A c^{3/2} x \sqrt{a+c x^2}}{5 a e^3 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 A c^{5/4} \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 a^{3/4} e^3 \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 \left (5 \sqrt{a} B+3 A \sqrt{c}\right ) c^{3/4} \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 a^{3/4} e^3 \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0281791, size = 83, normalized size = 0.25 \[ -\frac{2 x \sqrt{a+c x^2} \left (3 A \, _2F_1\left (-\frac{5}{4},-\frac{1}{2};-\frac{1}{4};-\frac{c x^2}{a}\right )+5 B x \, _2F_1\left (-\frac{3}{4},-\frac{1}{2};\frac{1}{4};-\frac{c x^2}{a}\right )\right )}{15 (e x)^{7/2} \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + c*x^2])/(e*x)^(7/2),x]

[Out]

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

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Maple [A]  time = 0.031, size = 331, normalized size = 1. \begin{align*}{\frac{2}{15\,{x}^{2}{e}^{3}a} \left ( 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 ){x}^{2}ac-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 ){x}^{2}ac+5\,B\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 ){x}^{2}a-6\,A{c}^{2}{x}^{4}-5\,aBc{x}^{3}-9\,aAc{x}^{2}-5\,{a}^{2}Bx-3\,A{a}^{2} \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)*(c*x^2+a)^(1/2)/(e*x)^(7/2),x)

[Out]

2/15/x^2*(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))*x^2*a*c-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))*x^2*a*c+5*B*(-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))*x^2*a-6*A*c^2*x^4-5*a*B*c*x^3-9*a*A*c*x^2-5*a^2*B*x-3*A*a^2)/(c*x^2+a)^(1/2
)/e^3/(e*x)^(1/2)/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{\left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/(e*x)^(7/2), 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}}{e^{4} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 111.234, size = 107, normalized size = 0.32 \begin{align*} \frac{A \sqrt{a} \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{7}{2}} x^{\frac{5}{2}} \Gamma \left (- \frac{1}{4}\right )} + \frac{B \sqrt{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 e^{\frac{7}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(1/2)/(e*x)**(7/2),x)

[Out]

A*sqrt(a)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*x**(5/2)*gamma(-1/4))
+ B*sqrt(a)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*x**(3/2)*gamma(1/4)
)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{\left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(7/2),x, algorithm="giac")

[Out]

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