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

Optimal. Leaf size=341 \[ \frac {4 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (5 \sqrt {a} B+21 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{35 \sqrt [4]{c} e \sqrt {e x} \sqrt {a+c x^2}}-\frac {24 a^{5/4} A \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 e \sqrt {e x} \sqrt {a+c x^2}}+\frac {4 \sqrt {e x} \sqrt {a+c x^2} (5 a B+21 A c x)}{35 e^2}-\frac {2 \left (a+c x^2\right )^{3/2} (7 A-B x)}{7 e \sqrt {e x}}+\frac {24 a A \sqrt {c} x \sqrt {a+c x^2}}{5 e \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

[Out]

-2/7*(-B*x+7*A)*(c*x^2+a)^(3/2)/e/(e*x)^(1/2)+24/5*a*A*x*c^(1/2)*(c*x^2+a)^(1/2)/e/(a^(1/2)+x*c^(1/2))/(e*x)^(
1/2)+4/35*(21*A*c*x+5*B*a)*(e*x)^(1/2)*(c*x^2+a)^(1/2)/e^2-24/5*a^(5/4)*A*c^(1/4)*(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)))*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)/e/(e*x)^(1/2)/(c*x^2+a)^(1/2)+
4/35*a^(5/4)*(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))*(5*B*a^(1/2)+21*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^(1/4)/e/(e*x)^(1/2)/(c*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {4 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (5 \sqrt {a} B+21 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{35 \sqrt [4]{c} e \sqrt {e x} \sqrt {a+c x^2}}-\frac {24 a^{5/4} A \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 e \sqrt {e x} \sqrt {a+c x^2}}+\frac {4 \sqrt {e x} \sqrt {a+c x^2} (5 a B+21 A c x)}{35 e^2}-\frac {2 \left (a+c x^2\right )^{3/2} (7 A-B x)}{7 e \sqrt {e x}}+\frac {24 a A \sqrt {c} x \sqrt {a+c x^2}}{5 e \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a*A*Sqrt[c]*x*Sqrt[a + c*x^2])/(5*e*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (4*Sqrt[e*x]*(5*a*B + 21*A*c*x)*Sqr
t[a + c*x^2])/(35*e^2) - (2*(7*A - B*x)*(a + c*x^2)^(3/2))/(7*e*Sqrt[e*x]) - (24*a^(5/4)*A*c^(1/4)*Sqrt[x]*(Sq
rt[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])/(5*e*Sqrt[e*x]*Sqrt[a + c*x^2]) + (4*a^(5/4)*(5*Sqrt[a]*B + 21*A*Sqrt[c])*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sq
rt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(35*c^(1/4)*e*Sqr
t[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 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 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 )^{3/2}}{(e x)^{3/2}} \, dx &=-\frac {2 (7 A-B x) \left (a+c x^2\right )^{3/2}}{7 e \sqrt {e x}}-\frac {6 \int \frac {(-a B e-7 A c e x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx}{7 e^2}\\ &=\frac {4 \sqrt {e x} (5 a B+21 A c x) \sqrt {a+c x^2}}{35 e^2}-\frac {2 (7 A-B x) \left (a+c x^2\right )^{3/2}}{7 e \sqrt {e x}}-\frac {8 \int \frac {-\frac {5}{2} a^2 B c e^3-\frac {21}{2} a A c^2 e^3 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{35 c e^4}\\ &=\frac {4 \sqrt {e x} (5 a B+21 A c x) \sqrt {a+c x^2}}{35 e^2}-\frac {2 (7 A-B x) \left (a+c x^2\right )^{3/2}}{7 e \sqrt {e x}}-\frac {\left (8 \sqrt {x}\right ) \int \frac {-\frac {5}{2} a^2 B c e^3-\frac {21}{2} a A c^2 e^3 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{35 c e^4 \sqrt {e x}}\\ &=\frac {4 \sqrt {e x} (5 a B+21 A c x) \sqrt {a+c x^2}}{35 e^2}-\frac {2 (7 A-B x) \left (a+c x^2\right )^{3/2}}{7 e \sqrt {e x}}-\frac {\left (16 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {5}{2} a^2 B c e^3-\frac {21}{2} a A c^2 e^3 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{35 c e^4 \sqrt {e x}}\\ &=\frac {4 \sqrt {e x} (5 a B+21 A c x) \sqrt {a+c x^2}}{35 e^2}-\frac {2 (7 A-B x) \left (a+c x^2\right )^{3/2}}{7 e \sqrt {e x}}+\frac {\left (8 a^{3/2} \left (5 \sqrt {a} B+21 A \sqrt {c}\right ) \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{35 e \sqrt {e x}}-\frac {\left (24 a^{3/2} A \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 )}{5 e \sqrt {e x}}\\ &=\frac {24 a A \sqrt {c} x \sqrt {a+c x^2}}{5 e \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {4 \sqrt {e x} (5 a B+21 A c x) \sqrt {a+c x^2}}{35 e^2}-\frac {2 (7 A-B x) \left (a+c x^2\right )^{3/2}}{7 e \sqrt {e x}}-\frac {24 a^{5/4} A \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 )}{5 e \sqrt {e x} \sqrt {a+c x^2}}+\frac {4 a^{5/4} \left (5 \sqrt {a} B+21 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 )}{35 \sqrt [4]{c} e \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 81, normalized size = 0.24 \[ \frac {2 a x \sqrt {a+c x^2} \left (B x \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{a}\right )-A \, _2F_1\left (-\frac {3}{2},-\frac {1}{4};\frac {3}{4};-\frac {c x^2}{a}\right )\right )}{(e x)^{3/2} \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c x^{3} + A c x^{2} + B a x + A a\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{e^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 340, normalized size = 1.00 \[ -\frac {2 \left (-5 B \,c^{3} x^{5}-7 A \,c^{3} x^{4}-20 B a \,c^{2} x^{3}+28 A a \,c^{2} x^{2}-84 \sqrt {2}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, A \,a^{2} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+42 \sqrt {2}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, A \,a^{2} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-15 B \,a^{2} c x +35 A \,a^{2} c -10 \sqrt {-a c}\, \sqrt {2}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, B \,a^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right )}{35 \sqrt {c \,x^{2}+a}\, \sqrt {e x}\, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(42*A*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))
*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*a^2*c-84*A*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))*((c*x+(-a*c)^(1/2))/(-a*c)^(
1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*a^2*c-10*B*(-a*c)^(1/2)*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))*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c
*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*a^2-5*B*c^3*x^5-7*A*c^3*x^4-20*B*a*c^2*x^3+28*A*a*c^2*x^2-15*B*a^2*c*x+35
*A*a^2*c)/(c*x^2+a)^(1/2)/c/e/(e*x)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/(e*x)^(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 \[ \int \frac {{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right )}{{\left (e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 8.62, size = 202, normalized size = 0.59 \[ \frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {A \sqrt {a} c x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {B a^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {B \sqrt {a} c x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________