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

Optimal. Leaf size=369 \[ \frac {8 a^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (77 \sqrt {a} B+195 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3003 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {16 a^{13/4} B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{39 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {16 a^3 B x \sqrt {a+c x^2}}{39 \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {8 a^2 \sqrt {e x} \sqrt {a+c x^2} (195 A+77 B x)}{3003 e}+\frac {20 a \sqrt {e x} \left (a+c x^2\right )^{3/2} (117 A+77 B x)}{9009 e}+\frac {2 \sqrt {e x} \left (a+c x^2\right )^{5/2} (13 A+11 B x)}{143 e} \]

[Out]

20/9009*a*(77*B*x+117*A)*(c*x^2+a)^(3/2)*(e*x)^(1/2)/e+2/143*(11*B*x+13*A)*(c*x^2+a)^(5/2)*(e*x)^(1/2)/e+16/39
*a^3*B*x*(c*x^2+a)^(1/2)/c^(1/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+8/3003*a^2*(77*B*x+195*A)*(e*x)^(1/2)*(c*x^2+
a)^(1/2)/e-16/39*a^(13/4)*B*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1
/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^
(1/2)+x*c^(1/2))^2)^(1/2)/c^(3/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)+8/3003*a^(11/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^
(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2
^(1/2))*(77*B*a^(1/2)+195*A*c^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/c^(3/
4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {815, 842, 840, 1198, 220, 1196} \[ \frac {8 a^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (77 \sqrt {a} B+195 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3003 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {8 a^2 \sqrt {e x} \sqrt {a+c x^2} (195 A+77 B x)}{3003 e}-\frac {16 a^{13/4} B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{39 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {16 a^3 B x \sqrt {a+c x^2}}{39 \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {20 a \sqrt {e x} \left (a+c x^2\right )^{3/2} (117 A+77 B x)}{9009 e}+\frac {2 \sqrt {e x} \left (a+c x^2\right )^{5/2} (13 A+11 B x)}{143 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Sqrt[e*x]*(195*A + 77*B*x)*Sqrt[a + c*x^2])/(3003*e) + (16*a^3*B*x*Sqrt[a + c*x^2])/(39*Sqrt[c]*Sqrt[e*
x]*(Sqrt[a] + Sqrt[c]*x)) + (20*a*Sqrt[e*x]*(117*A + 77*B*x)*(a + c*x^2)^(3/2))/(9009*e) + (2*Sqrt[e*x]*(13*A
+ 11*B*x)*(a + c*x^2)^(5/2))/(143*e) - (16*a^(13/4)*B*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])/(39*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) +
(8*a^(11/4)*(77*Sqrt[a]*B + 195*A*Sqrt[c])*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])/(3003*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{\sqrt {e x}} \, dx &=\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}+\frac {20 \int \frac {\left (\frac {13}{2} a A c e^2+\frac {11}{2} a B c e^2 x\right ) \left (a+c x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{143 c e^2}\\ &=\frac {20 a \sqrt {e x} (117 A+77 B x) \left (a+c x^2\right )^{3/2}}{9009 e}+\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}+\frac {80 \int \frac {\left (\frac {117}{4} a^2 A c^2 e^4+\frac {77}{4} a^2 B c^2 e^4 x\right ) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx}{3003 c^2 e^4}\\ &=\frac {8 a^2 \sqrt {e x} (195 A+77 B x) \sqrt {a+c x^2}}{3003 e}+\frac {20 a \sqrt {e x} (117 A+77 B x) \left (a+c x^2\right )^{3/2}}{9009 e}+\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}+\frac {64 \int \frac {\frac {585}{8} a^3 A c^3 e^6+\frac {231}{8} a^3 B c^3 e^6 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{9009 c^3 e^6}\\ &=\frac {8 a^2 \sqrt {e x} (195 A+77 B x) \sqrt {a+c x^2}}{3003 e}+\frac {20 a \sqrt {e x} (117 A+77 B x) \left (a+c x^2\right )^{3/2}}{9009 e}+\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}+\frac {\left (64 \sqrt {x}\right ) \int \frac {\frac {585}{8} a^3 A c^3 e^6+\frac {231}{8} a^3 B c^3 e^6 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{9009 c^3 e^6 \sqrt {e x}}\\ &=\frac {8 a^2 \sqrt {e x} (195 A+77 B x) \sqrt {a+c x^2}}{3003 e}+\frac {20 a \sqrt {e x} (117 A+77 B x) \left (a+c x^2\right )^{3/2}}{9009 e}+\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}+\frac {\left (128 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {585}{8} a^3 A c^3 e^6+\frac {231}{8} a^3 B c^3 e^6 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{9009 c^3 e^6 \sqrt {e x}}\\ &=\frac {8 a^2 \sqrt {e x} (195 A+77 B x) \sqrt {a+c x^2}}{3003 e}+\frac {20 a \sqrt {e x} (117 A+77 B x) \left (a+c x^2\right )^{3/2}}{9009 e}+\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}-\frac {\left (16 a^{7/2} B \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 )}{39 \sqrt {c} \sqrt {e x}}+\frac {\left (16 a^3 \left (77 \sqrt {a} B+195 A \sqrt {c}\right ) \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{3003 \sqrt {c} \sqrt {e x}}\\ &=\frac {8 a^2 \sqrt {e x} (195 A+77 B x) \sqrt {a+c x^2}}{3003 e}+\frac {16 a^3 B x \sqrt {a+c x^2}}{39 \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {20 a \sqrt {e x} (117 A+77 B x) \left (a+c x^2\right )^{3/2}}{9009 e}+\frac {2 \sqrt {e x} (13 A+11 B x) \left (a+c x^2\right )^{5/2}}{143 e}-\frac {16 a^{13/4} B \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 )}{39 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {8 a^{11/4} \left (77 \sqrt {a} B+195 A \sqrt {c}\right ) \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 )}{3003 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 85, normalized size = 0.23 \[ \frac {2 a^2 x \sqrt {a+c x^2} \left (3 A \, _2F_1\left (-\frac {5}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{a}\right )+B x \, _2F_1\left (-\frac {5}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )\right )}{3 \sqrt {e x} \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c^{2} x^{5} + A c^{2} x^{4} + 2 \, B a c x^{3} + 2 \, A a c x^{2} + B a^{2} x + A a^{2}\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(1/2),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*x)
, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{\sqrt {e x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 362, normalized size = 0.98 \[ \frac {\frac {2 B \,c^{4} x^{8}}{13}+\frac {2 A \,c^{4} x^{7}}{11}+\frac {74 B a \,c^{3} x^{6}}{117}+\frac {62 A a \,c^{3} x^{5}}{77}+\frac {118 B \,a^{2} c^{2} x^{4}}{117}+\frac {122 A \,a^{2} c^{2} x^{3}}{77}+\frac {62 B \,a^{3} c \,x^{2}}{117}+\frac {74 A \,a^{3} c x}{77}+\frac {16 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{39}-\frac {8 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{39}+\frac {40 \sqrt {2}\, \sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{77}}{\sqrt {c \,x^{2}+a}\, \sqrt {e x}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(1/2),x)

[Out]

2/9009/(c*x^2+a)^(1/2)/c*(693*B*c^4*x^8+819*A*c^4*x^7+2849*B*a*c^3*x^6+2340*A*2^(1/2)*(-a*c)^(1/2)*((c*x+(-a*c
)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*EllipticF(((
c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^3+3627*A*a*c^3*x^5+1848*B*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a
*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*EllipticE(((c*x+(-a*c)^(
1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^4-924*B*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)
^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*
2^(1/2))*a^4+4543*B*a^2*c^2*x^4+7137*A*a^2*c^2*x^3+2387*B*a^3*c*x^2+4329*A*a^3*c*x)/(e*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{\sqrt {e x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right )}{\sqrt {e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(5/2)*(A + B*x))/(e*x)^(1/2),x)

[Out]

int(((a + c*x^2)^(5/2)*(A + B*x))/(e*x)^(1/2), x)

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sympy [C]  time = 18.20, size = 301, normalized size = 0.82 \[ \frac {A a^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {A a^{\frac {3}{2}} c x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{\sqrt {e} \Gamma \left (\frac {9}{4}\right )} + \frac {A \sqrt {a} c^{2} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {13}{4}\right )} + \frac {B a^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {7}{4}\right )} + \frac {B a^{\frac {3}{2}} c x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{\sqrt {e} \Gamma \left (\frac {11}{4}\right )} + \frac {B \sqrt {a} c^{2} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/(e*x)**(1/2),x)

[Out]

A*a**(5/2)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(5/4)) + A*
a**(3/2)*c*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/a)/(sqrt(e)*gamma(9/4)) + A*s
qrt(a)*c**2*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(13/4))
+ B*a**(5/2)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(7/4)) +
B*a**(3/2)*c*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(sqrt(e)*gamma(11/4))
+ B*sqrt(a)*c**2*x**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma
(15/4))

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