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

Optimal. Leaf size=397 \[ \frac{2 a^{9/4} e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (25 \sqrt{a} B-77 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{1155 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 a^2 A e^3 x \sqrt{a+c x^2}}{15 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{4 a^{9/4} A e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 a e^2 \sqrt{e x} \sqrt{a+c x^2} (25 a B-77 A c x)}{1155 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c} \]

[Out]

(-4*a^2*A*e^3*x*Sqrt[a + c*x^2])/(15*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (2*a*e^2*Sqrt[e*x]*(25*a*B - 7
7*A*c*x)*Sqrt[a + c*x^2])/(1155*c^2) - (10*a*B*e^2*Sqrt[e*x]*(a + c*x^2)^(3/2))/(77*c^2) + (2*A*e*(e*x)^(3/2)*
(a + c*x^2)^(3/2))/(9*c) + (2*B*(e*x)^(5/2)*(a + c*x^2)^(3/2))/(11*c) + (4*a^(9/4)*A*e^3*Sqrt[x]*(Sqrt[a] + Sq
rt[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^
(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(9/4)*(25*Sqrt[a]*B - 77*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])/(1155*c^(9/4)*S
qrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.48878, antiderivative size = 397, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{2 a^{9/4} e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (25 \sqrt{a} B-77 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{1155 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 a^2 A e^3 x \sqrt{a+c x^2}}{15 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{4 a^{9/4} A e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 a e^2 \sqrt{e x} \sqrt{a+c x^2} (25 a B-77 A c x)}{1155 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*A*e^3*x*Sqrt[a + c*x^2])/(15*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (2*a*e^2*Sqrt[e*x]*(25*a*B - 7
7*A*c*x)*Sqrt[a + c*x^2])/(1155*c^2) - (10*a*B*e^2*Sqrt[e*x]*(a + c*x^2)^(3/2))/(77*c^2) + (2*A*e*(e*x)^(3/2)*
(a + c*x^2)^(3/2))/(9*c) + (2*B*(e*x)^(5/2)*(a + c*x^2)^(3/2))/(11*c) + (4*a^(9/4)*A*e^3*Sqrt[x]*(Sqrt[a] + Sq
rt[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^
(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(9/4)*(25*Sqrt[a]*B - 77*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])/(1155*c^(9/4)*S
qrt[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) \sqrt{a+c x^2} \, dx &=\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{2 \int (e x)^{3/2} \left (-\frac{5}{2} a B e+\frac{11}{2} A c e x\right ) \sqrt{a+c x^2} \, dx}{11 c}\\ &=\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{4 \int \sqrt{e x} \left (-\frac{33}{4} a A c e^2-\frac{45}{4} a B c e^2 x\right ) \sqrt{a+c x^2} \, dx}{99 c^2}\\ &=-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{8 \int \frac{\left (\frac{45}{8} a^2 B c e^3-\frac{231}{8} a A c^2 e^3 x\right ) \sqrt{a+c x^2}}{\sqrt{e x}} \, dx}{693 c^3}\\ &=\frac{2 a e^2 \sqrt{e x} (25 a B-77 A c x) \sqrt{a+c x^2}}{1155 c^2}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{32 \int \frac{\frac{225}{16} a^3 B c^2 e^5-\frac{693}{16} a^2 A c^3 e^5 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{10395 c^4 e^2}\\ &=\frac{2 a e^2 \sqrt{e x} (25 a B-77 A c x) \sqrt{a+c x^2}}{1155 c^2}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{\left (32 \sqrt{x}\right ) \int \frac{\frac{225}{16} a^3 B c^2 e^5-\frac{693}{16} a^2 A c^3 e^5 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{10395 c^4 e^2 \sqrt{e x}}\\ &=\frac{2 a e^2 \sqrt{e x} (25 a B-77 A c x) \sqrt{a+c x^2}}{1155 c^2}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{\left (64 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{225}{16} a^3 B c^2 e^5-\frac{693}{16} a^2 A c^3 e^5 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{10395 c^4 e^2 \sqrt{e x}}\\ &=\frac{2 a e^2 \sqrt{e x} (25 a B-77 A c x) \sqrt{a+c x^2}}{1155 c^2}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{\left (4 a^{5/2} \left (25 \sqrt{a} B-77 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 )}{1155 c^2 \sqrt{e x}}+\frac{\left (4 a^{5/2} 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 )}{15 c^{3/2} \sqrt{e x}}\\ &=-\frac{4 a^2 A e^3 x \sqrt{a+c x^2}}{15 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 a e^2 \sqrt{e x} (25 a B-77 A c x) \sqrt{a+c x^2}}{1155 c^2}-\frac{10 a B e^2 \sqrt{e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac{2 A e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{9 c}+\frac{2 B (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac{4 a^{9/4} 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 )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 a^{9/4} \left (25 \sqrt{a} B-77 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 )}{1155 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.119318, size = 133, normalized size = 0.34 \[ \frac{2 e^2 \sqrt{e x} \sqrt{a+c x^2} \left (45 a^2 B \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{a}\right )-\left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} (45 a B-7 c x (11 A+9 B x))-77 a A c x \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )\right )}{693 c^2 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]*(-((a + c*x^2)*Sqrt[1 + (c*x^2)/a]*(45*a*B - 7*c*x*(11*A + 9*B*x))) + 45*a^2*
B*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/a)] - 77*a*A*c*x*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c*x^2)/a)]
))/(693*c^2*Sqrt[1 + (c*x^2)/a])

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Maple [A]  time = 0.023, size = 360, normalized size = 0.9 \begin{align*}{\frac{2\,{e}^{2}}{3465\,x{c}^{3}}\sqrt{ex} \left ( 315\,B{c}^{4}{x}^{7}+385\,A{c}^{4}{x}^{6}+231\,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}^{3}c-462\,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}^{3}c+75\,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}^{3}+405\,aB{c}^{3}{x}^{5}+539\,aA{c}^{3}{x}^{4}-60\,{a}^{2}B{c}^{2}{x}^{3}+154\,{a}^{2}A{c}^{2}{x}^{2}-150\,{a}^{3}Bcx \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)^(1/2),x)

[Out]

2/3465*e^2/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)*(315*B*c^4*x^7+385*A*c^4*x^6+231*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^3*c-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)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/
2))*a^3*c+75*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^3+405*a*B*c^
3*x^5+539*a*A*c^3*x^4-60*a^2*B*c^2*x^3+154*a^2*A*c^2*x^2-150*a^3*B*c*x)/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + a}{\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)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)*(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 e^{2} x^{3} + 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 129.284, size = 97, normalized size = 0.24 \begin{align*} \frac{A \sqrt{a} e^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{11}{4}\right )} + \frac{B \sqrt{a} e^{\frac{5}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{13}{4}\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)**(1/2),x)

[Out]

A*sqrt(a)*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(11/4)) +
B*sqrt(a)*e**(5/2)*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(13/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + a}{\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)^(1/2),x, algorithm="giac")

[Out]

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