3.442 \(\int (e x)^{5/2} (A+B x) (a+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=438 \[ \frac{4 a^{13/4} e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (65 \sqrt{a} B-231 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15015 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{4 a^2 e^2 \sqrt{e x} \sqrt{a+c x^2} (65 a B-231 A c x)}{15015 c^2}-\frac{8 a^3 A e^3 x \sqrt{a+c x^2}}{65 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{8 a^{13/4} A e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{65 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 a e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2} (13 a B-77 A c x)}{3003 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c} \]

[Out]

(-8*a^3*A*e^3*x*Sqrt[a + c*x^2])/(65*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (4*a^2*e^2*Sqrt[e*x]*(65*a*B -
231*A*c*x)*Sqrt[a + c*x^2])/(15015*c^2) + (2*a*e^2*Sqrt[e*x]*(13*a*B - 77*A*c*x)*(a + c*x^2)^(3/2))/(3003*c^2
) - (2*a*B*e^2*Sqrt[e*x]*(a + c*x^2)^(5/2))/(33*c^2) + (2*A*e*(e*x)^(3/2)*(a + c*x^2)^(5/2))/(13*c) + (2*B*(e*
x)^(5/2)*(a + c*x^2)^(5/2))/(15*c) + (8*a^(13/4)*A*e^3*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])/(65*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) +
(4*a^(13/4)*(65*Sqrt[a]*B - 231*A*Sqrt[c])*e^3*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])/(15015*c^(9/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.577302, antiderivative size = 438, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {833, 815, 842, 840, 1198, 220, 1196} \[ \frac{4 a^2 e^2 \sqrt{e x} \sqrt{a+c x^2} (65 a B-231 A c x)}{15015 c^2}+\frac{4 a^{13/4} e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (65 \sqrt{a} B-231 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15015 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{8 a^3 A e^3 x \sqrt{a+c x^2}}{65 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{8 a^{13/4} A e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{65 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 a e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2} (13 a B-77 A c x)}{3003 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(5/2)*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(-8*a^3*A*e^3*x*Sqrt[a + c*x^2])/(65*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (4*a^2*e^2*Sqrt[e*x]*(65*a*B -
231*A*c*x)*Sqrt[a + c*x^2])/(15015*c^2) + (2*a*e^2*Sqrt[e*x]*(13*a*B - 77*A*c*x)*(a + c*x^2)^(3/2))/(3003*c^2
) - (2*a*B*e^2*Sqrt[e*x]*(a + c*x^2)^(5/2))/(33*c^2) + (2*A*e*(e*x)^(3/2)*(a + c*x^2)^(5/2))/(13*c) + (2*B*(e*
x)^(5/2)*(a + c*x^2)^(5/2))/(15*c) + (8*a^(13/4)*A*e^3*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])/(65*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) +
(4*a^(13/4)*(65*Sqrt[a]*B - 231*A*Sqrt[c])*e^3*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])/(15015*c^(9/4)*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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (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 (e x)^{5/2} (A+B x) \left (a+c x^2\right )^{3/2} \, dx &=\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{2 \int (e x)^{3/2} \left (-\frac{5}{2} a B e+\frac{15}{2} A c e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{15 c}\\ &=\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{4 \int \sqrt{e x} \left (-\frac{45}{4} a A c e^2-\frac{65}{4} a B c e^2 x\right ) \left (a+c x^2\right )^{3/2} \, dx}{195 c^2}\\ &=-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{8 \int \frac{\left (\frac{65}{8} a^2 B c e^3-\frac{495}{8} a A c^2 e^3 x\right ) \left (a+c x^2\right )^{3/2}}{\sqrt{e x}} \, dx}{2145 c^3}\\ &=\frac{2 a e^2 \sqrt{e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{32 \int \frac{\left (\frac{585}{16} a^3 B c^2 e^5-\frac{3465}{16} a^2 A c^3 e^5 x\right ) \sqrt{a+c x^2}}{\sqrt{e x}} \, dx}{45045 c^4 e^2}\\ &=\frac{4 a^2 e^2 \sqrt{e x} (65 a B-231 A c x) \sqrt{a+c x^2}}{15015 c^2}+\frac{2 a e^2 \sqrt{e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{128 \int \frac{\frac{2925}{32} a^4 B c^3 e^7-\frac{10395}{32} a^3 A c^4 e^7 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{675675 c^5 e^4}\\ &=\frac{4 a^2 e^2 \sqrt{e x} (65 a B-231 A c x) \sqrt{a+c x^2}}{15015 c^2}+\frac{2 a e^2 \sqrt{e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{\left (128 \sqrt{x}\right ) \int \frac{\frac{2925}{32} a^4 B c^3 e^7-\frac{10395}{32} a^3 A c^4 e^7 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{675675 c^5 e^4 \sqrt{e x}}\\ &=\frac{4 a^2 e^2 \sqrt{e x} (65 a B-231 A c x) \sqrt{a+c x^2}}{15015 c^2}+\frac{2 a e^2 \sqrt{e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{\left (256 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{2925}{32} a^4 B c^3 e^7-\frac{10395}{32} a^3 A c^4 e^7 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{675675 c^5 e^4 \sqrt{e x}}\\ &=\frac{4 a^2 e^2 \sqrt{e x} (65 a B-231 A c x) \sqrt{a+c x^2}}{15015 c^2}+\frac{2 a e^2 \sqrt{e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{\left (8 a^{7/2} \left (65 \sqrt{a} B-231 A \sqrt{c}\right ) e^3 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{15015 c^2 \sqrt{e x}}+\frac{\left (8 a^{7/2} A e^3 \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 )}{65 c^{3/2} \sqrt{e x}}\\ &=-\frac{8 a^3 A e^3 x \sqrt{a+c x^2}}{65 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{4 a^2 e^2 \sqrt{e x} (65 a B-231 A c x) \sqrt{a+c x^2}}{15015 c^2}+\frac{2 a e^2 \sqrt{e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac{2 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac{8 a^{13/4} A e^3 \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 )}{65 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{4 a^{13/4} \left (65 \sqrt{a} B-231 A \sqrt{c}\right ) e^3 \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 )}{15015 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.148204, size = 137, normalized size = 0.31 \[ \frac{2 e^2 \sqrt{e x} \sqrt{a+c x^2} \left (-165 a^2 A c x \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )+65 a^3 B \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{a}\right )-\left (a+c x^2\right )^2 \sqrt{\frac{c x^2}{a}+1} (65 a B-11 c x (15 A+13 B x))\right )}{2145 c^2 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^(5/2)*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(2*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]*(-((a + c*x^2)^2*Sqrt[1 + (c*x^2)/a]*(65*a*B - 11*c*x*(15*A + 13*B*x))) + 65*
a^3*B*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/a)] - 165*a^2*A*c*x*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x
^2)/a)]))/(2145*c^2*Sqrt[1 + (c*x^2)/a])

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Maple [A]  time = 0.026, size = 384, normalized size = 0.9 \begin{align*}{\frac{2\,{e}^{2}}{15015\,x{c}^{3}}\sqrt{ex} \left ( 1001\,B{x}^{9}{c}^{5}+1155\,A{x}^{8}{c}^{5}+2548\,B{x}^{7}a{c}^{4}+3080\,A{x}^{6}a{c}^{4}+462\,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 ){a}^{4}c-924\,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 ){a}^{4}c+130\,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 ) \sqrt{-ac}{a}^{4}+1703\,B{x}^{5}{a}^{2}{c}^{3}+2233\,A{x}^{4}{a}^{2}{c}^{3}-104\,B{x}^{3}{a}^{3}{c}^{2}+308\,A{x}^{2}{a}^{3}{c}^{2}-260\,Bx{a}^{4}c \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*e^2/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)*(1001*B*x^9*c^5+1155*A*x^8*c^5+2548*B*x^7*a*c^4+3080*A*x^6*a*c^4+462
*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^4*c-924*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))*a^4*c+130*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*c)^(1/2)*a^4+1703*B*x^5*a^2*c^3+2233*A*x^4*a^2*c^3-104*B*x^3*a^3*c^2+308*A*x^2*a^3*c^2-260*B*x*
a^4*c)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(B*x+A)*(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(B*x+A)*(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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