3.457 \(\int \frac{(A+B x) (a+c x^2)^{5/2}}{(e x)^{11/2}} \, dx\)

Optimal. Leaf size=375 \[ \frac{8 \sqrt [4]{a} c^{7/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{21 e^5 \sqrt{e x} \sqrt{a+c x^2}}-\frac{8 c^2 \sqrt{a+c x^2} (7 A-5 B x)}{21 e^5 \sqrt{e x}}-\frac{4 c \left (a+c x^2\right )^{3/2} (7 A+15 B x)}{63 e^3 (e x)^{5/2}}-\frac{2 \left (a+c x^2\right )^{5/2} (7 A+9 B x)}{63 e (e x)^{9/2}}+\frac{16 A c^{5/2} x \sqrt{a+c x^2}}{3 e^5 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{16 \sqrt [4]{a} A c^{9/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 e^5 \sqrt{e x} \sqrt{a+c x^2}} \]

[Out]

(-8*c^2*(7*A - 5*B*x)*Sqrt[a + c*x^2])/(21*e^5*Sqrt[e*x]) + (16*A*c^(5/2)*x*Sqrt[a + c*x^2])/(3*e^5*Sqrt[e*x]*
(Sqrt[a] + Sqrt[c]*x)) - (4*c*(7*A + 15*B*x)*(a + c*x^2)^(3/2))/(63*e^3*(e*x)^(5/2)) - (2*(7*A + 9*B*x)*(a + c
*x^2)^(5/2))/(63*e*(e*x)^(9/2)) - (16*a^(1/4)*A*c^(9/4)*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])/(3*e^5*Sqrt[e*x]*Sqrt[a + c*x^2]) + (8*
a^(1/4)*(5*Sqrt[a]*B + 7*A*Sqrt[c])*c^(7/4)*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])/(21*e^5*Sqrt[e*x]*Sqrt[a + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.423479, antiderivative size = 375, 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 = {811, 813, 842, 840, 1198, 220, 1196} \[ -\frac{8 c^2 \sqrt{a+c x^2} (7 A-5 B x)}{21 e^5 \sqrt{e x}}+\frac{8 \sqrt [4]{a} c^{7/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{21 e^5 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 c \left (a+c x^2\right )^{3/2} (7 A+15 B x)}{63 e^3 (e x)^{5/2}}-\frac{2 \left (a+c x^2\right )^{5/2} (7 A+9 B x)}{63 e (e x)^{9/2}}+\frac{16 A c^{5/2} x \sqrt{a+c x^2}}{3 e^5 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{16 \sqrt [4]{a} A c^{9/4} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 e^5 \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/(e*x)^(11/2),x]

[Out]

(-8*c^2*(7*A - 5*B*x)*Sqrt[a + c*x^2])/(21*e^5*Sqrt[e*x]) + (16*A*c^(5/2)*x*Sqrt[a + c*x^2])/(3*e^5*Sqrt[e*x]*
(Sqrt[a] + Sqrt[c]*x)) - (4*c*(7*A + 15*B*x)*(a + c*x^2)^(3/2))/(63*e^3*(e*x)^(5/2)) - (2*(7*A + 9*B*x)*(a + c
*x^2)^(5/2))/(63*e*(e*x)^(9/2)) - (16*a^(1/4)*A*c^(9/4)*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])/(3*e^5*Sqrt[e*x]*Sqrt[a + c*x^2]) + (8*
a^(1/4)*(5*Sqrt[a]*B + 7*A*Sqrt[c])*c^(7/4)*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])/(21*e^5*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 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 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)^{11/2}} \, dx &=-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}-\frac{10 \int \frac{\left (-7 a A c e^2-9 a B c e^2 x\right ) \left (a+c x^2\right )^{3/2}}{(e x)^{7/2}} \, dx}{63 a e^4}\\ &=-\frac{4 c (7 A+15 B x) \left (a+c x^2\right )^{3/2}}{63 e^3 (e x)^{5/2}}-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}+\frac{4 \int \frac{\left (21 a^2 A c^2 e^4+45 a^2 B c^2 e^4 x\right ) \sqrt{a+c x^2}}{(e x)^{3/2}} \, dx}{63 a^2 e^8}\\ &=-\frac{8 c^2 (7 A-5 B x) \sqrt{a+c x^2}}{21 e^5 \sqrt{e x}}-\frac{4 c (7 A+15 B x) \left (a+c x^2\right )^{3/2}}{63 e^3 (e x)^{5/2}}-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}-\frac{8 \int \frac{-45 a^3 B c^2 e^5-63 a^2 A c^3 e^5 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{189 a^2 e^{10}}\\ &=-\frac{8 c^2 (7 A-5 B x) \sqrt{a+c x^2}}{21 e^5 \sqrt{e x}}-\frac{4 c (7 A+15 B x) \left (a+c x^2\right )^{3/2}}{63 e^3 (e x)^{5/2}}-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}-\frac{\left (8 \sqrt{x}\right ) \int \frac{-45 a^3 B c^2 e^5-63 a^2 A c^3 e^5 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{189 a^2 e^{10} \sqrt{e x}}\\ &=-\frac{8 c^2 (7 A-5 B x) \sqrt{a+c x^2}}{21 e^5 \sqrt{e x}}-\frac{4 c (7 A+15 B x) \left (a+c x^2\right )^{3/2}}{63 e^3 (e x)^{5/2}}-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}-\frac{\left (16 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-45 a^3 B c^2 e^5-63 a^2 A c^3 e^5 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{189 a^2 e^{10} \sqrt{e x}}\\ &=-\frac{8 c^2 (7 A-5 B x) \sqrt{a+c x^2}}{21 e^5 \sqrt{e x}}-\frac{4 c (7 A+15 B x) \left (a+c x^2\right )^{3/2}}{63 e^3 (e x)^{5/2}}-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}+\frac{\left (16 \sqrt{a} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) c^2 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{21 e^5 \sqrt{e x}}-\frac{\left (16 \sqrt{a} A c^{5/2} \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 )}{3 e^5 \sqrt{e x}}\\ &=-\frac{8 c^2 (7 A-5 B x) \sqrt{a+c x^2}}{21 e^5 \sqrt{e x}}+\frac{16 A c^{5/2} x \sqrt{a+c x^2}}{3 e^5 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{4 c (7 A+15 B x) \left (a+c x^2\right )^{3/2}}{63 e^3 (e x)^{5/2}}-\frac{2 (7 A+9 B x) \left (a+c x^2\right )^{5/2}}{63 e (e x)^{9/2}}-\frac{16 \sqrt [4]{a} A c^{9/4} \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 )}{3 e^5 \sqrt{e x} \sqrt{a+c x^2}}+\frac{8 \sqrt [4]{a} \left (5 \sqrt{a} B+7 A \sqrt{c}\right ) c^{7/4} \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 )}{21 e^5 \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0363533, size = 91, normalized size = 0.24 \[ -\frac{2 a^2 \sqrt{e x} \sqrt{a+c x^2} \left (7 A \, _2F_1\left (-\frac{5}{2},-\frac{9}{4};-\frac{5}{4};-\frac{c x^2}{a}\right )+9 B x \, _2F_1\left (-\frac{5}{2},-\frac{7}{4};-\frac{3}{4};-\frac{c x^2}{a}\right )\right )}{63 e^6 x^5 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/(e*x)^(11/2),x]

[Out]

(-2*a^2*Sqrt[e*x]*Sqrt[a + c*x^2]*(7*A*Hypergeometric2F1[-5/2, -9/4, -5/4, -((c*x^2)/a)] + 9*B*x*Hypergeometri
c2F1[-5/2, -7/4, -3/4, -((c*x^2)/a)]))/(63*e^6*x^5*Sqrt[1 + (c*x^2)/a])

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 366, normalized size = 1. \begin{align*} -{\frac{2}{63\,{x}^{4}{e}^{5}} \left ( 84\,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}^{4}a{c}^{2}-168\,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}^{4}a{c}^{2}-60\,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}^{4}ac-21\,B{c}^{3}{x}^{7}+105\,A{c}^{3}{x}^{6}+27\,aB{c}^{2}{x}^{5}+133\,aA{c}^{2}{x}^{4}+57\,{a}^{2}Bc{x}^{3}+35\,{a}^{2}Ac{x}^{2}+9\,{a}^{3}Bx+7\,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)^(11/2),x)

[Out]

-2/63/x^4*(84*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^4*a*c^2-168*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)*Ellip
ticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*c^2-60*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^4*a*c-21*B*c^3*x^7+105*A*c^3*x^6+27*a*B*c^2*x^5+133*a*A*c^2*x^4+
57*a^2*B*c*x^3+35*a^2*A*c*x^2+9*a^3*B*x+7*A*a^3)/(c*x^2+a)^(1/2)/e^5/(e*x)^(1/2)

________________________________________________________________________________________

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{11}{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)^(11/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^(5/2)*(B*x + A)/(e*x)^(11/2), x)

________________________________________________________________________________________

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^{6} x^{6}}, 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)^(11/2),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^6*
x^6), x)

________________________________________________________________________________________

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)**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{11}{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)^(11/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + a)^(5/2)*(B*x + A)/(e*x)^(11/2), x)