3.456 \(\int \frac{(A+B x) (a+c x^2)^{5/2}}{(e x)^{9/2}} \, dx\)

Optimal. Leaf size=377 \[ \frac{8 a^{3/4} c^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (63 \sqrt{a} B+25 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{105 e^4 \sqrt{e x} \sqrt{a+c x^2}}-\frac{48 a^{5/4} B c^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 e^4 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 \left (a+c x^2\right )^{3/2} (21 a B+25 A c x)}{105 e^2 (e x)^{5/2}}-\frac{8 c \sqrt{a+c x^2} (63 a B-25 A c x)}{105 e^4 \sqrt{e x}}-\frac{2 \left (a+c x^2\right )^{5/2} (5 A-7 B x)}{35 e (e x)^{7/2}}+\frac{48 a B c^{3/2} x \sqrt{a+c x^2}}{5 e^4 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

[Out]

(48*a*B*c^(3/2)*x*Sqrt[a + c*x^2])/(5*e^4*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (8*c*(63*a*B - 25*A*c*x)*Sqrt[a +
c*x^2])/(105*e^4*Sqrt[e*x]) - (4*(21*a*B + 25*A*c*x)*(a + c*x^2)^(3/2))/(105*e^2*(e*x)^(5/2)) - (2*(5*A - 7*B
*x)*(a + c*x^2)^(5/2))/(35*e*(e*x)^(7/2)) - (48*a^(5/4)*B*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*e^4*Sqrt[e*x]*Sqrt[a + c*x
^2]) + (8*a^(3/4)*(63*Sqrt[a]*B + 25*A*Sqrt[c])*c^(5/4)*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*e^4*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.413166, antiderivative size = 377, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {813, 811, 842, 840, 1198, 220, 1196} \[ \frac{8 a^{3/4} c^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (63 \sqrt{a} B+25 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 e^4 \sqrt{e x} \sqrt{a+c x^2}}-\frac{48 a^{5/4} B c^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 e^4 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 \left (a+c x^2\right )^{3/2} (21 a B+25 A c x)}{105 e^2 (e x)^{5/2}}-\frac{8 c \sqrt{a+c x^2} (63 a B-25 A c x)}{105 e^4 \sqrt{e x}}-\frac{2 \left (a+c x^2\right )^{5/2} (5 A-7 B x)}{35 e (e x)^{7/2}}+\frac{48 a B c^{3/2} x \sqrt{a+c x^2}}{5 e^4 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/(e*x)^(9/2),x]

[Out]

(48*a*B*c^(3/2)*x*Sqrt[a + c*x^2])/(5*e^4*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (8*c*(63*a*B - 25*A*c*x)*Sqrt[a +
c*x^2])/(105*e^4*Sqrt[e*x]) - (4*(21*a*B + 25*A*c*x)*(a + c*x^2)^(3/2))/(105*e^2*(e*x)^(5/2)) - (2*(5*A - 7*B
*x)*(a + c*x^2)^(5/2))/(35*e*(e*x)^(7/2)) - (48*a^(5/4)*B*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*e^4*Sqrt[e*x]*Sqrt[a + c*x
^2]) + (8*a^(3/4)*(63*Sqrt[a]*B + 25*A*Sqrt[c])*c^(5/4)*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*e^4*Sqrt[e*x]*Sqrt[a + c*x^2])

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 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) \left (a+c x^2\right )^{5/2}}{(e x)^{9/2}} \, dx &=-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}-\frac{2 \int \frac{(-7 a B e-5 A c e x) \left (a+c x^2\right )^{3/2}}{(e x)^{7/2}} \, dx}{7 e^2}\\ &=-\frac{4 (21 a B+25 A c x) \left (a+c x^2\right )^{3/2}}{105 e^2 (e x)^{5/2}}-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}+\frac{4 \int \frac{\left (21 a^2 B c e^3+25 a A c^2 e^3 x\right ) \sqrt{a+c x^2}}{(e x)^{3/2}} \, dx}{35 a e^6}\\ &=-\frac{8 c (63 a B-25 A c x) \sqrt{a+c x^2}}{105 e^4 \sqrt{e x}}-\frac{4 (21 a B+25 A c x) \left (a+c x^2\right )^{3/2}}{105 e^2 (e x)^{5/2}}-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}-\frac{8 \int \frac{-25 a^2 A c^2 e^4-63 a^2 B c^2 e^4 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{105 a e^8}\\ &=-\frac{8 c (63 a B-25 A c x) \sqrt{a+c x^2}}{105 e^4 \sqrt{e x}}-\frac{4 (21 a B+25 A c x) \left (a+c x^2\right )^{3/2}}{105 e^2 (e x)^{5/2}}-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}-\frac{\left (8 \sqrt{x}\right ) \int \frac{-25 a^2 A c^2 e^4-63 a^2 B c^2 e^4 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{105 a e^8 \sqrt{e x}}\\ &=-\frac{8 c (63 a B-25 A c x) \sqrt{a+c x^2}}{105 e^4 \sqrt{e x}}-\frac{4 (21 a B+25 A c x) \left (a+c x^2\right )^{3/2}}{105 e^2 (e x)^{5/2}}-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}-\frac{\left (16 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-25 a^2 A c^2 e^4-63 a^2 B c^2 e^4 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{105 a e^8 \sqrt{e x}}\\ &=-\frac{8 c (63 a B-25 A c x) \sqrt{a+c x^2}}{105 e^4 \sqrt{e x}}-\frac{4 (21 a B+25 A c x) \left (a+c x^2\right )^{3/2}}{105 e^2 (e x)^{5/2}}-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}-\frac{\left (48 a^{3/2} B 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 e^4 \sqrt{e x}}+\frac{\left (16 a \left (63 \sqrt{a} B+25 A \sqrt{c}\right ) c^{3/2} \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{105 e^4 \sqrt{e x}}\\ &=\frac{48 a B c^{3/2} x \sqrt{a+c x^2}}{5 e^4 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{8 c (63 a B-25 A c x) \sqrt{a+c x^2}}{105 e^4 \sqrt{e x}}-\frac{4 (21 a B+25 A c x) \left (a+c x^2\right )^{3/2}}{105 e^2 (e x)^{5/2}}-\frac{2 (5 A-7 B x) \left (a+c x^2\right )^{5/2}}{35 e (e x)^{7/2}}-\frac{48 a^{5/4} B 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 e^4 \sqrt{e x} \sqrt{a+c x^2}}+\frac{8 a^{3/4} \left (63 \sqrt{a} B+25 A \sqrt{c}\right ) 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}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 e^4 \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0372933, size = 91, normalized size = 0.24 \[ -\frac{2 a^2 \sqrt{e x} \sqrt{a+c x^2} \left (5 A \, _2F_1\left (-\frac{5}{2},-\frac{7}{4};-\frac{3}{4};-\frac{c x^2}{a}\right )+7 B x \, _2F_1\left (-\frac{5}{2},-\frac{5}{4};-\frac{1}{4};-\frac{c x^2}{a}\right )\right )}{35 e^5 x^4 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/(e*x)^(9/2),x]

[Out]

(-2*a^2*Sqrt[e*x]*Sqrt[a + c*x^2]*(5*A*Hypergeometric2F1[-5/2, -7/4, -3/4, -((c*x^2)/a)] + 7*B*x*Hypergeometri
c2F1[-5/2, -5/4, -1/4, -((c*x^2)/a)]))/(35*e^5*x^4*Sqrt[1 + (c*x^2)/a])

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Maple [A]  time = 0.022, size = 366, normalized size = 1. \begin{align*}{\frac{2}{105\,{x}^{3}{e}^{4}} \left ( 100\,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}^{3}ac+504\,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}^{3}{a}^{2}c-252\,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}^{3}{a}^{2}c+21\,B{c}^{3}{x}^{7}+35\,A{c}^{3}{x}^{6}-231\,aB{c}^{2}{x}^{5}-45\,aA{c}^{2}{x}^{4}-273\,{a}^{2}Bc{x}^{3}-95\,{a}^{2}Ac{x}^{2}-21\,{a}^{3}Bx-15\,A{a}^{3} \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)^(5/2)/(e*x)^(9/2),x)

[Out]

2/105/x^3*(100*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^3*a*c+504*
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^3*a^2*c-252*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))*x^3*a^2*c+21*B*c^3*x^7+35*A*c^3*x^6-231*a*B*c^2*x^5-45*a*A*c^2*x^4
-273*a^2*B*c*x^3-95*a^2*A*c*x^2-21*a^3*B*x-15*A*a^3)/(c*x^2+a)^(1/2)/e^4/(e*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + a\right )}^{\frac{5}{2}}{\left (B x + A\right )}}{\left (e x\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^(5/2)*(B*x + A)/(e*x)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B c^{2} x^{5} + A c^{2} x^{4} + 2 \, B a c x^{3} + 2 \, A a c x^{2} + B a^{2} x + A a^{2}\right )} \sqrt{c x^{2} + a} \sqrt{e x}}{e^{5} x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(9/2),x, algorithm="fricas")

[Out]

integral((B*c^2*x^5 + A*c^2*x^4 + 2*B*a*c*x^3 + 2*A*a*c*x^2 + B*a^2*x + A*a^2)*sqrt(c*x^2 + a)*sqrt(e*x)/(e^5*
x^5), 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)*(c*x**2+a)**(5/2)/(e*x)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + a\right )}^{\frac{5}{2}}{\left (B x + A\right )}}{\left (e x\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(9/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + a)^(5/2)*(B*x + A)/(e*x)^(9/2), x)