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

Optimal. Leaf size=367 \[ \frac {4 a^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+9 A b) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt {a+b x^2}}-\frac {8 a^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+9 A b) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt {a+b x^2}}+\frac {2 (e x)^{3/2} \left (a+b x^2\right )^{3/2} (a B+9 A b)}{9 a e^3}+\frac {4 (e x)^{3/2} \sqrt {a+b x^2} (a B+9 A b)}{15 e^3}+\frac {8 a \sqrt {e x} \sqrt {a+b x^2} (a B+9 A b)}{15 \sqrt {b} e^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}} \]

[Out]

2/9*(9*A*b+B*a)*(e*x)^(3/2)*(b*x^2+a)^(3/2)/a/e^3-2*A*(b*x^2+a)^(5/2)/a/e/(e*x)^(1/2)+4/15*(9*A*b+B*a)*(e*x)^(
3/2)*(b*x^2+a)^(1/2)/e^3+8/15*a*(9*A*b+B*a)*(e*x)^(1/2)*(b*x^2+a)^(1/2)/e^2/b^(1/2)/(a^(1/2)+x*b^(1/2))-8/15*a
^(5/4)*(9*A*b+B*a)*(cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/
2)/a^(1/4)/e^(1/2)))*EllipticE(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1
/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(3/4)/e^(3/2)/(b*x^2+a)^(1/2)+4/15*a^(5/4)*(9*A*b+B*a)*(cos(2*a
rctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))*Ellipt
icF(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*
b^(1/2))^2)^(1/2)/b^(3/4)/e^(3/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {453, 279, 329, 305, 220, 1196} \[ \frac {4 a^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+9 A b) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt {a+b x^2}}-\frac {8 a^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+9 A b) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt {a+b x^2}}+\frac {2 (e x)^{3/2} \left (a+b x^2\right )^{3/2} (a B+9 A b)}{9 a e^3}+\frac {4 (e x)^{3/2} \sqrt {a+b x^2} (a B+9 A b)}{15 e^3}+\frac {8 a \sqrt {e x} \sqrt {a+b x^2} (a B+9 A b)}{15 \sqrt {b} e^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(9*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*e^3) + (8*a*(9*A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*Sqr
t[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) + (2*(9*A*b + a*B)*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*a*e^3) - (2*A*(a + b*x^2)
^(5/2))/(a*e*Sqrt[e*x]) - (8*a^(5/4)*(9*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x
)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (
4*a^(5/4)*(9*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^
(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{(e x)^{3/2}} \, dx &=-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}}+\frac {(9 A b+a B) \int \sqrt {e x} \left (a+b x^2\right )^{3/2} \, dx}{a e^2}\\ &=\frac {2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}}+\frac {(2 (9 A b+a B)) \int \sqrt {e x} \sqrt {a+b x^2} \, dx}{3 e^2}\\ &=\frac {4 (9 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 e^3}+\frac {2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}}+\frac {(4 a (9 A b+a B)) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{15 e^2}\\ &=\frac {4 (9 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 e^3}+\frac {2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}}+\frac {(8 a (9 A b+a B)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 e^3}\\ &=\frac {4 (9 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 e^3}+\frac {2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}}+\frac {\left (8 a^{3/2} (9 A b+a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 \sqrt {b} e^2}-\frac {\left (8 a^{3/2} (9 A b+a B)\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a} e}}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 \sqrt {b} e^2}\\ &=\frac {4 (9 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 e^3}+\frac {8 a (9 A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{15 \sqrt {b} e^2 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt {e x}}-\frac {8 a^{5/4} (9 A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt {a+b x^2}}+\frac {4 a^{5/4} (9 A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 84, normalized size = 0.23 \[ \frac {2 x \sqrt {a+b x^2} \left (\frac {x^2 (a B+9 A b) \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )}{\sqrt {\frac {b x^2}{a}+1}}-\frac {3 A \left (a+b x^2\right )^2}{a}\right )}{3 (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[a + b*x^2]*((-3*A*(a + b*x^2)^2)/a + ((9*A*b + a*B)*x^2*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^2)/
a)])/Sqrt[1 + (b*x^2)/a]))/(3*(e*x)^(3/2))

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b x^{4} + {\left (B a + A b\right )} x^{2} + A a\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{e^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b*x^4 + (B*a + A*b)*x^2 + A*a)*sqrt(b*x^2 + a)*sqrt(e*x)/(e^2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 421, normalized size = 1.15 \[ \frac {\frac {2 B \,b^{3} x^{6}}{9}+\frac {2 A \,b^{3} x^{4}}{5}+\frac {32 B a \,b^{2} x^{4}}{45}-\frac {8 A a \,b^{2} x^{2}}{5}+\frac {22 B \,a^{2} b \,x^{2}}{45}+\frac {24 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A \,a^{2} b \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {12 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A \,a^{2} b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {8 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{3} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{15}-\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{3} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{15}-2 A \,a^{2} b}{\sqrt {b \,x^{2}+a}\, \sqrt {e x}\, b e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*B*b^3*x^6+108*A*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/
2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2*b-54*A*((b*x
+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2
)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2*b+12*B*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^
(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticE(((b*x+(-a*b)^(1/2
))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^3-6*B*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2
))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/
2))*a^3+9*A*b^3*x^4+16*B*a*b^2*x^4-36*A*a*b^2*x^2+11*B*a^2*b*x^2-45*A*a^2*b)/(b*x^2+a)^(1/2)/b/e/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.49, size = 202, normalized size = 0.55 \[ \frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {A \sqrt {a} b x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {B a^{\frac {3}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {B \sqrt {a} b x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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