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

Optimal. Leaf size=400 \[ -\frac {4 a^{11/4} e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (77 \sqrt {a} B+65 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5005 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {8 a^{13/4} B e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{65 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {8 a^3 B e^2 x \sqrt {a+c x^2}}{65 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {4 a^2 e \sqrt {e x} \sqrt {a+c x^2} (65 A+77 B x)}{5005 c}-\frac {2 a e \sqrt {e x} \left (a+c x^2\right )^{3/2} (39 A+77 B x)}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c} \]

[Out]

2/13*B*(e*x)^(3/2)*(c*x^2+a)^(5/2)/c-2/3003*a*e*(77*B*x+39*A)*(c*x^2+a)^(3/2)*(e*x)^(1/2)/c+2/11*A*e*(c*x^2+a)
^(5/2)*(e*x)^(1/2)/c-8/65*a^3*B*e^2*x*(c*x^2+a)^(1/2)/c^(3/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)-4/5005*a^2*e*(77
*B*x+65*A)*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c+8/65*a^(13/4)*B*e^2*(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^(7/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)-4/5005*a^(11/4)
*e^2*(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*a
rctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(77*B*a^(1/2)+65*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^(7/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.50, antiderivative size = 400, normalized size of antiderivative = 1.00, 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} \[ -\frac {4 a^{11/4} e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (77 \sqrt {a} B+65 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5005 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {4 a^2 e \sqrt {e x} \sqrt {a+c x^2} (65 A+77 B x)}{5005 c}-\frac {8 a^3 B e^2 x \sqrt {a+c x^2}}{65 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {8 a^{13/4} B e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{65 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 a e \sqrt {e x} \left (a+c x^2\right )^{3/2} (39 A+77 B x)}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*e*Sqrt[e*x]*(65*A + 77*B*x)*Sqrt[a + c*x^2])/(5005*c) - (8*a^3*B*e^2*x*Sqrt[a + c*x^2])/(65*c^(3/2)*Sq
rt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (2*a*e*Sqrt[e*x]*(39*A + 77*B*x)*(a + c*x^2)^(3/2))/(3003*c) + (2*A*e*Sqrt[e*
x]*(a + c*x^2)^(5/2))/(11*c) + (2*B*(e*x)^(3/2)*(a + c*x^2)^(5/2))/(13*c) + (8*a^(13/4)*B*e^2*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^(11/4)*(77*Sqrt[a]*B + 65*A*Sqrt[c])*e^2*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])/(5005*c^(
7/4)*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 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 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 (e x)^{3/2} (A+B x) \left (a+c x^2\right )^{3/2} \, dx &=\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 \int \sqrt {e x} \left (-\frac {3}{2} a B e+\frac {13}{2} A c e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{13 c}\\ &=\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {4 \int \frac {\left (-\frac {13}{4} a A c e^2-\frac {33}{4} a B c e^2 x\right ) \left (a+c x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{143 c^2}\\ &=-\frac {2 a e \sqrt {e x} (39 A+77 B x) \left (a+c x^2\right )^{3/2}}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {16 \int \frac {\left (-\frac {117}{8} a^2 A c^2 e^4-\frac {231}{8} a^2 B c^2 e^4 x\right ) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx}{3003 c^3 e^2}\\ &=-\frac {4 a^2 e \sqrt {e x} (65 A+77 B x) \sqrt {a+c x^2}}{5005 c}-\frac {2 a e \sqrt {e x} (39 A+77 B x) \left (a+c x^2\right )^{3/2}}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {64 \int \frac {-\frac {585}{16} a^3 A c^3 e^6-\frac {693}{16} a^3 B c^3 e^6 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{45045 c^4 e^4}\\ &=-\frac {4 a^2 e \sqrt {e x} (65 A+77 B x) \sqrt {a+c x^2}}{5005 c}-\frac {2 a e \sqrt {e x} (39 A+77 B x) \left (a+c x^2\right )^{3/2}}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {\left (64 \sqrt {x}\right ) \int \frac {-\frac {585}{16} a^3 A c^3 e^6-\frac {693}{16} a^3 B c^3 e^6 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{45045 c^4 e^4 \sqrt {e x}}\\ &=-\frac {4 a^2 e \sqrt {e x} (65 A+77 B x) \sqrt {a+c x^2}}{5005 c}-\frac {2 a e \sqrt {e x} (39 A+77 B x) \left (a+c x^2\right )^{3/2}}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {\left (128 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {585}{16} a^3 A c^3 e^6-\frac {693}{16} a^3 B c^3 e^6 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{45045 c^4 e^4 \sqrt {e x}}\\ &=-\frac {4 a^2 e \sqrt {e x} (65 A+77 B x) \sqrt {a+c x^2}}{5005 c}-\frac {2 a e \sqrt {e x} (39 A+77 B x) \left (a+c x^2\right )^{3/2}}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {\left (8 a^{7/2} B e^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 )}{65 c^{3/2} \sqrt {e x}}-\frac {\left (8 a^3 \left (77 \sqrt {a} B+65 A \sqrt {c}\right ) e^2 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{5005 c^{3/2} \sqrt {e x}}\\ &=-\frac {4 a^2 e \sqrt {e x} (65 A+77 B x) \sqrt {a+c x^2}}{5005 c}-\frac {8 a^3 B e^2 x \sqrt {a+c x^2}}{65 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {2 a e \sqrt {e x} (39 A+77 B x) \left (a+c x^2\right )^{3/2}}{3003 c}+\frac {2 A e \sqrt {e x} \left (a+c x^2\right )^{5/2}}{11 c}+\frac {2 B (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {8 a^{13/4} B e^2 \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^{11/4} \left (77 \sqrt {a} B+65 A \sqrt {c}\right ) e^2 \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 )}{5005 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.12, size = 124, normalized size = 0.31 \[ \frac {2 e \sqrt {e x} \sqrt {a+c x^2} \left (-13 a^2 A \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{a}\right )-11 a^2 B x \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )+\left (a+c x^2\right )^2 \sqrt {\frac {c x^2}{a}+1} (13 A+11 B x)\right )}{143 c \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[e*x]*Sqrt[a + c*x^2]*((13*A + 11*B*x)*(a + c*x^2)^2*Sqrt[1 + (c*x^2)/a] - 13*a^2*A*Hypergeometric2F1
[-3/2, 1/4, 5/4, -((c*x^2)/a)] - 11*a^2*B*x*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x^2)/a)]))/(143*c*Sqrt[1 +
(c*x^2)/a])

________________________________________________________________________________________

fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B c e x^{4} + A c e x^{3} + B a e x^{2} + A a e x\right )} \sqrt {c x^{2} + a} \sqrt {e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(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)^(3/2), x)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 366, normalized size = 0.92 \[ -\frac {2 \sqrt {e x}\, \left (-1155 B \,c^{4} x^{8}-1365 A \,c^{4} x^{7}-3080 B a \,c^{3} x^{6}-3900 A a \,c^{3} x^{5}-2233 B \,a^{2} c^{2} x^{4}-3315 A \,a^{2} c^{2} x^{3}-308 B \,a^{3} c \,x^{2}-780 A \,a^{3} c x +924 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-462 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+390 \sqrt {2}\, \sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{15015 \sqrt {c \,x^{2}+a}\, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15015*e/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)/c^2*(-1155*B*c^4*x^8-1365*A*c^4*x^7-3080*B*a*c^3*x^6+390*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^3-3900*A*a*c^3*x^5+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)*Ell
ipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^4-462*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^4-2233*B*a^2*c^2*x^4-3315*A*a^2*c^2*x^3-308*B*a^3*c*x^2-780*A*a^3*c*x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(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)^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,x\right )}^{3/2}\,{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 22.27, size = 199, normalized size = 0.50 \[ \frac {A a^{\frac {3}{2}} e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {A \sqrt {a} c e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} + \frac {B a^{\frac {3}{2}} e^{\frac {3}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} + \frac {B \sqrt {a} c e^{\frac {3}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________