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

Optimal. Leaf size=388 \[ \frac{a^{7/4} e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (49 \sqrt{a} B+25 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{105 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{14 a^2 B e^4 x \sqrt{a+c x^2}}{15 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{14 a^{9/4} B e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c} \]

[Out]

(-10*a*A*e^3*Sqrt[e*x]*Sqrt[a + c*x^2])/(21*c^2) - (14*a*B*e^2*(e*x)^(3/2)*Sqrt[a + c*x^2])/(45*c^2) + (2*A*e*
(e*x)^(5/2)*Sqrt[a + c*x^2])/(7*c) + (2*B*(e*x)^(7/2)*Sqrt[a + c*x^2])/(9*c) + (14*a^2*B*e^4*x*Sqrt[a + c*x^2]
)/(15*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (14*a^(9/4)*B*e^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])/(15*c^(11/4)*Sqrt[e*x]*Sqrt[
a + c*x^2]) + (a^(7/4)*(49*Sqrt[a]*B + 25*A*Sqrt[c])*e^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])/(105*c^(11/4)*Sqrt[e*x]*Sqrt[a + c*x^2
])

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Rubi [A]  time = 0.483376, antiderivative size = 388, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{a^{7/4} e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (49 \sqrt{a} B+25 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{14 a^2 B e^4 x \sqrt{a+c x^2}}{15 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{14 a^{9/4} B e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*a*A*e^3*Sqrt[e*x]*Sqrt[a + c*x^2])/(21*c^2) - (14*a*B*e^2*(e*x)^(3/2)*Sqrt[a + c*x^2])/(45*c^2) + (2*A*e*
(e*x)^(5/2)*Sqrt[a + c*x^2])/(7*c) + (2*B*(e*x)^(7/2)*Sqrt[a + c*x^2])/(9*c) + (14*a^2*B*e^4*x*Sqrt[a + c*x^2]
)/(15*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (14*a^(9/4)*B*e^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])/(15*c^(11/4)*Sqrt[e*x]*Sqrt[
a + c*x^2]) + (a^(7/4)*(49*Sqrt[a]*B + 25*A*Sqrt[c])*e^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])/(105*c^(11/4)*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)^{7/2} (A+B x)}{\sqrt{a+c x^2}} \, dx &=\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{2 \int \frac{(e x)^{5/2} \left (-\frac{7}{2} a B e+\frac{9}{2} A c e x\right )}{\sqrt{a+c x^2}} \, dx}{9 c}\\ &=\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{4 \int \frac{(e x)^{3/2} \left (-\frac{45}{4} a A c e^2-\frac{49}{4} a B c e^2 x\right )}{\sqrt{a+c x^2}} \, dx}{63 c^2}\\ &=-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{8 \int \frac{\sqrt{e x} \left (\frac{147}{8} a^2 B c e^3-\frac{225}{8} a A c^2 e^3 x\right )}{\sqrt{a+c x^2}} \, dx}{315 c^3}\\ &=-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{16 \int \frac{\frac{225}{16} a^2 A c^2 e^4+\frac{441}{16} a^2 B c^2 e^4 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{945 c^4}\\ &=-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{\left (16 \sqrt{x}\right ) \int \frac{\frac{225}{16} a^2 A c^2 e^4+\frac{441}{16} a^2 B c^2 e^4 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{945 c^4 \sqrt{e x}}\\ &=-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{\left (32 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{225}{16} a^2 A c^2 e^4+\frac{441}{16} a^2 B c^2 e^4 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{945 c^4 \sqrt{e x}}\\ &=-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}-\frac{\left (14 a^{5/2} B e^4 \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 )}{15 c^{5/2} \sqrt{e x}}+\frac{\left (2 a^2 \left (49 \sqrt{a} B+25 A \sqrt{c}\right ) e^4 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{105 c^{5/2} \sqrt{e x}}\\ &=-\frac{10 a A e^3 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}-\frac{14 a B e^2 (e x)^{3/2} \sqrt{a+c x^2}}{45 c^2}+\frac{2 A e (e x)^{5/2} \sqrt{a+c x^2}}{7 c}+\frac{2 B (e x)^{7/2} \sqrt{a+c x^2}}{9 c}+\frac{14 a^2 B e^4 x \sqrt{a+c x^2}}{15 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{14 a^{9/4} B e^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 )}{15 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{a^{7/4} \left (49 \sqrt{a} B+25 A \sqrt{c}\right ) e^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 )}{105 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0797752, size = 142, normalized size = 0.37 \[ \frac{2 e^3 \sqrt{e x} \left (75 a^2 A \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )+49 a^2 B x \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )-\left (a+c x^2\right ) \left (a (75 A+49 B x)-5 c x^2 (9 A+7 B x)\right )\right )}{315 c^2 \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*Sqrt[e*x]*(-((a + c*x^2)*(-5*c*x^2*(9*A + 7*B*x) + a*(75*A + 49*B*x))) + 75*a^2*A*Sqrt[1 + (c*x^2)/a]*H
ypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/a)] + 49*a^2*B*x*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4
, -((c*x^2)/a)]))/(315*c^2*Sqrt[a + c*x^2])

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Maple [A]  time = 0.024, size = 344, normalized size = 0.9 \begin{align*}{\frac{{e}^{3}}{315\,x{c}^{3}}\sqrt{ex} \left ( 70\,B{c}^{3}{x}^{6}+75\,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}{a}^{2}+90\,A{c}^{3}{x}^{5}+294\,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}^{3}-147\,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}^{3}-28\,aB{c}^{2}{x}^{4}-60\,aA{c}^{2}{x}^{3}-98\,{a}^{2}Bc{x}^{2}-150\,{a}^{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)^(7/2)*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

1/315*e^3/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)/c^3*(70*B*c^3*x^6+75*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)*a^2+90*A*c^3*x^5+294*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^3-147*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^3-28*a*B*c^2*x^
4-60*a*A*c^2*x^3-98*a^2*B*c*x^2-150*a^2*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{7}{2}}}{\sqrt{c x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x)^(7/2)/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^{3} x^{4} + A e^{3} x^{3}\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)^(7/2)*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*e^3*x^4 + A*e^3*x^3)*sqrt(e*x)/sqrt(c*x^2 + a), 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((e*x)**(7/2)*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

Timed out

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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{7}{2}}}{\sqrt{c x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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