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

Optimal. Leaf size=378 \[ \frac{8 a^{7/4} \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (7 \sqrt{a} B+5 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{21 e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{16 a^2 B \sqrt{c} x \sqrt{a+c x^2}}{3 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{16 a^{9/4} B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{20 \left (a+c x^2\right )^{3/2} (7 a B-3 A c x)}{63 e^2 \sqrt{e x}}+\frac{8 a c \sqrt{e x} \sqrt{a+c x^2} (5 A+7 B x)}{21 e^3}-\frac{2 \left (a+c x^2\right )^{5/2} (3 A-B x)}{9 e (e x)^{3/2}} \]

[Out]

(8*a*c*Sqrt[e*x]*(5*A + 7*B*x)*Sqrt[a + c*x^2])/(21*e^3) + (16*a^2*B*Sqrt[c]*x*Sqrt[a + c*x^2])/(3*e^2*Sqrt[e*
x]*(Sqrt[a] + Sqrt[c]*x)) - (20*(7*a*B - 3*A*c*x)*(a + c*x^2)^(3/2))/(63*e^2*Sqrt[e*x]) - (2*(3*A - B*x)*(a +
c*x^2)^(5/2))/(9*e*(e*x)^(3/2)) - (16*a^(9/4)*B*c^(1/4)*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^2*Sqrt[e*x]*Sqrt[a + c*x^2]) + (8*
a^(7/4)*(7*Sqrt[a]*B + 5*A*Sqrt[c])*c^(1/4)*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^2*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.384894, antiderivative size = 378, 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, 815, 842, 840, 1198, 220, 1196} \[ \frac{8 a^{7/4} \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (7 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{21 e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{16 a^2 B \sqrt{c} x \sqrt{a+c x^2}}{3 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{16 a^{9/4} B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{20 \left (a+c x^2\right )^{3/2} (7 a B-3 A c x)}{63 e^2 \sqrt{e x}}+\frac{8 a c \sqrt{e x} \sqrt{a+c x^2} (5 A+7 B x)}{21 e^3}-\frac{2 \left (a+c x^2\right )^{5/2} (3 A-B x)}{9 e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*c*Sqrt[e*x]*(5*A + 7*B*x)*Sqrt[a + c*x^2])/(21*e^3) + (16*a^2*B*Sqrt[c]*x*Sqrt[a + c*x^2])/(3*e^2*Sqrt[e*
x]*(Sqrt[a] + Sqrt[c]*x)) - (20*(7*a*B - 3*A*c*x)*(a + c*x^2)^(3/2))/(63*e^2*Sqrt[e*x]) - (2*(3*A - B*x)*(a +
c*x^2)^(5/2))/(9*e*(e*x)^(3/2)) - (16*a^(9/4)*B*c^(1/4)*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^2*Sqrt[e*x]*Sqrt[a + c*x^2]) + (8*
a^(7/4)*(7*Sqrt[a]*B + 5*A*Sqrt[c])*c^(1/4)*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^2*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 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 \frac{(A+B x) \left (a+c x^2\right )^{5/2}}{(e x)^{5/2}} \, dx &=-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}-\frac{10 \int \frac{(-3 a B e-9 A c e x) \left (a+c x^2\right )^{3/2}}{(e x)^{3/2}} \, dx}{27 e^2}\\ &=-\frac{20 (7 a B-3 A c x) \left (a+c x^2\right )^{3/2}}{63 e^2 \sqrt{e x}}-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}+\frac{20 \int \frac{\left (9 a A c e^2+21 a B c e^2 x\right ) \sqrt{a+c x^2}}{\sqrt{e x}} \, dx}{63 e^4}\\ &=\frac{8 a c \sqrt{e x} (5 A+7 B x) \sqrt{a+c x^2}}{21 e^3}-\frac{20 (7 a B-3 A c x) \left (a+c x^2\right )^{3/2}}{63 e^2 \sqrt{e x}}-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}+\frac{16 \int \frac{\frac{45}{2} a^2 A c^2 e^4+\frac{63}{2} a^2 B c^2 e^4 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{189 c e^6}\\ &=\frac{8 a c \sqrt{e x} (5 A+7 B x) \sqrt{a+c x^2}}{21 e^3}-\frac{20 (7 a B-3 A c x) \left (a+c x^2\right )^{3/2}}{63 e^2 \sqrt{e x}}-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}+\frac{\left (16 \sqrt{x}\right ) \int \frac{\frac{45}{2} a^2 A c^2 e^4+\frac{63}{2} a^2 B c^2 e^4 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{189 c e^6 \sqrt{e x}}\\ &=\frac{8 a c \sqrt{e x} (5 A+7 B x) \sqrt{a+c x^2}}{21 e^3}-\frac{20 (7 a B-3 A c x) \left (a+c x^2\right )^{3/2}}{63 e^2 \sqrt{e x}}-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}+\frac{\left (32 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{45}{2} a^2 A c^2 e^4+\frac{63}{2} a^2 B c^2 e^4 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{189 c e^6 \sqrt{e x}}\\ &=\frac{8 a c \sqrt{e x} (5 A+7 B x) \sqrt{a+c x^2}}{21 e^3}-\frac{20 (7 a B-3 A c x) \left (a+c x^2\right )^{3/2}}{63 e^2 \sqrt{e x}}-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}-\frac{\left (16 a^{5/2} B \sqrt{c} \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^2 \sqrt{e x}}+\frac{\left (16 a^2 \left (7 \sqrt{a} B+5 A \sqrt{c}\right ) \sqrt{c} \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{21 e^2 \sqrt{e x}}\\ &=\frac{8 a c \sqrt{e x} (5 A+7 B x) \sqrt{a+c x^2}}{21 e^3}+\frac{16 a^2 B \sqrt{c} x \sqrt{a+c x^2}}{3 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{20 (7 a B-3 A c x) \left (a+c x^2\right )^{3/2}}{63 e^2 \sqrt{e x}}-\frac{2 (3 A-B x) \left (a+c x^2\right )^{5/2}}{9 e (e x)^{3/2}}-\frac{16 a^{9/4} B \sqrt [4]{c} \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^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{8 a^{7/4} \left (7 \sqrt{a} B+5 A \sqrt{c}\right ) \sqrt [4]{c} \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^2 \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0327021, size = 85, normalized size = 0.22 \[ -\frac{2 a^2 x \sqrt{a+c x^2} \left (A \, _2F_1\left (-\frac{5}{2},-\frac{3}{4};\frac{1}{4};-\frac{c x^2}{a}\right )+3 B x \, _2F_1\left (-\frac{5}{2},-\frac{1}{4};\frac{3}{4};-\frac{c x^2}{a}\right )\right )}{3 (e x)^{5/2} \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 359, normalized size = 1. \begin{align*}{\frac{2}{63\,x{e}^{2}} \left ( 7\,B{c}^{3}{x}^{7}+60\,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{a}^{2}+9\,A{c}^{3}{x}^{6}+168\,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{a}^{3}-84\,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{a}^{3}+35\,aB{c}^{2}{x}^{5}+57\,aA{c}^{2}{x}^{4}-35\,{a}^{2}Bc{x}^{3}+27\,{a}^{2}Ac{x}^{2}-63\,{a}^{3}Bx-21\,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)^(5/2),x)

[Out]

2/63/x*(7*B*c^3*x^7+60*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*a^
2+9*A*c^3*x^6+168*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*a^3-84*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)*Ellipt
icF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a^3+35*a*B*c^2*x^5+57*a*A*c^2*x^4-35*a^2*B*c*x^3+27
*a^2*A*c*x^2-63*a^3*B*x-21*A*a^3)/(c*x^2+a)^(1/2)/e^2/(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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^(5/2)*(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 (\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^{3} x^{3}}, 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)^(5/2),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^3*
x^3), x)

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Sympy [C]  time = 67.0077, size = 308, normalized size = 0.81 \begin{align*} \frac{A a^{\frac{5}{2}} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{A a^{\frac{3}{2}} c \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{A \sqrt{a} c^{2} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} + \frac{B a^{\frac{5}{2}} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{B a^{\frac{3}{2}} c x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{e^{\frac{5}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B \sqrt{a} c^{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 e^{\frac{5}{2}} \Gamma \left (\frac{11}{4}\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)**(5/2),x)

[Out]

A*a**(5/2)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*x**(3/2)*gamma(1/4))
+ A*a**(3/2)*c*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**2*exp_polar(I*pi)/a)/(e**(5/2)*gamma(5/4)) +
A*sqrt(a)*c**2*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*gamma(9/4
)) + B*a**(5/2)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*sqrt(x)*gamma(3/
4)) + B*a**(3/2)*c*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**2*exp_polar(I*pi)/a)/(e**(5/2)*gamma(7/
4)) + B*sqrt(a)*c**2*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*gam
ma(11/4))

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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