3.814 \(\int \frac {A+B x^2}{(e x)^{7/2} (a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=379 \[ -\frac {3 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (7 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{10 a^{11/4} e^{7/2} \sqrt {a+b x^2}}+\frac {3 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (7 A b-5 a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{11/4} e^{7/2} \sqrt {a+b x^2}}-\frac {3 \sqrt {b} \sqrt {e x} \sqrt {a+b x^2} (7 A b-5 a B)}{5 a^3 e^4 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {3 \sqrt {a+b x^2} (7 A b-5 a B)}{5 a^3 e^3 \sqrt {e x}}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}} \]

[Out]

-2/5*A/a/e/(e*x)^(5/2)/(b*x^2+a)^(1/2)+1/5*(-7*A*b+5*B*a)/a^2/e^3/(e*x)^(1/2)/(b*x^2+a)^(1/2)+3/5*(7*A*b-5*B*a
)*(b*x^2+a)^(1/2)/a^3/e^3/(e*x)^(1/2)-3/5*(7*A*b-5*B*a)*b^(1/2)*(e*x)^(1/2)*(b*x^2+a)^(1/2)/a^3/e^4/(a^(1/2)+x
*b^(1/2))+3/5*b^(1/4)*(7*A*b-5*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)/a^(11/4)/e^(7/2)/(b*x^2+a)^(1/2)-3/10*b^(1/4)*(7
*A*b-5*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)))*EllipticF(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)/a^(11/4)/e^(7/2)/(b*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 379, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {453, 290, 325, 329, 305, 220, 1196} \[ \frac {3 \sqrt {a+b x^2} (7 A b-5 a B)}{5 a^3 e^3 \sqrt {e x}}-\frac {3 \sqrt {b} \sqrt {e x} \sqrt {a+b x^2} (7 A b-5 a B)}{5 a^3 e^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}-\frac {3 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (7 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{10 a^{11/4} e^{7/2} \sqrt {a+b x^2}}+\frac {3 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (7 A b-5 a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{11/4} e^{7/2} \sqrt {a+b x^2}}-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/2)),x]

[Out]

(-2*A)/(5*a*e*(e*x)^(5/2)*Sqrt[a + b*x^2]) - (7*A*b - 5*a*B)/(5*a^2*e^3*Sqrt[e*x]*Sqrt[a + b*x^2]) + (3*(7*A*b
- 5*a*B)*Sqrt[a + b*x^2])/(5*a^3*e^3*Sqrt[e*x]) - (3*Sqrt[b]*(7*A*b - 5*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*a^
3*e^4*(Sqrt[a] + Sqrt[b]*x)) + (3*b^(1/4)*(7*A*b - 5*a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sq
rt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(5*a^(11/4)*e^(7/2)*Sqrt[a + b*x^
2]) - (3*b^(1/4)*(7*A*b - 5*a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*A
rcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(10*a^(11/4)*e^(7/2)*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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 325

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

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 {A+B x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/2}} \, dx &=-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}}-\frac {(7 A b-5 a B) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{3/2}} \, dx}{5 a e^2}\\ &=-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}-\frac {(3 (7 A b-5 a B)) \int \frac {1}{(e x)^{3/2} \sqrt {a+b x^2}} \, dx}{10 a^2 e^2}\\ &=-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}+\frac {3 (7 A b-5 a B) \sqrt {a+b x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {(3 b (7 A b-5 a B)) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{10 a^3 e^4}\\ &=-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}+\frac {3 (7 A b-5 a B) \sqrt {a+b x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {(3 b (7 A b-5 a B)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 a^3 e^5}\\ &=-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}+\frac {3 (7 A b-5 a B) \sqrt {a+b x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {\left (3 \sqrt {b} (7 A b-5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 a^{5/2} e^4}+\frac {\left (3 \sqrt {b} (7 A b-5 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 )}{5 a^{5/2} e^4}\\ &=-\frac {2 A}{5 a e (e x)^{5/2} \sqrt {a+b x^2}}-\frac {7 A b-5 a B}{5 a^2 e^3 \sqrt {e x} \sqrt {a+b x^2}}+\frac {3 (7 A b-5 a B) \sqrt {a+b x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {3 \sqrt {b} (7 A b-5 a B) \sqrt {e x} \sqrt {a+b x^2}}{5 a^3 e^4 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {3 \sqrt [4]{b} (7 A b-5 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 )}{5 a^{11/4} e^{7/2} \sqrt {a+b x^2}}-\frac {3 \sqrt [4]{b} (7 A b-5 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 )}{10 a^{11/4} e^{7/2} \sqrt {a+b x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 78, normalized size = 0.21 \[ \frac {x \left (2 x^2 \sqrt {\frac {b x^2}{a}+1} (7 A b-5 a B) \, _2F_1\left (-\frac {1}{4},\frac {3}{2};\frac {3}{4};-\frac {b x^2}{a}\right )-2 a A\right )}{5 a^2 (e x)^{7/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/2)),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{b^{2} e^{4} x^{8} + 2 \, a b e^{4} x^{6} + a^{2} e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b^2*e^4*x^8 + 2*a*b*e^4*x^6 + a^2*e^4*x^4), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/((b*x^2 + a)^(3/2)*(e*x)^(7/2)), x)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 417, normalized size = 1.10 \[ -\frac {-42 A \,b^{2} x^{4}+30 B a b \,x^{4}+42 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-21 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-30 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+15 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-28 A a b \,x^{2}+20 B \,a^{2} x^{2}+4 A \,a^{2}}{10 \sqrt {b \,x^{2}+a}\, \sqrt {e x}\, a^{3} e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(3/2),x)

[Out]

-1/10/x^2*(42*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))*x^2*a*b-21*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)*Elli
pticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-30*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))*x^2*a^2+15*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))*x^2*a^2-42*A*b^2*x^4+30*B*a*b*x^4-28*A*a*b*x^2+20*B*a^2*x^2+4*A*a^2)/(b*x^2+a)^(1/2)/e^3/(e*x)^(1/2)/a^3

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/((b*x^2 + a)^(3/2)*(e*x)^(7/2)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {B\,x^2+A}{{\left (e\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/2)),x)

[Out]

int((A + B*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/2)), x)

________________________________________________________________________________________

sympy [C]  time = 132.17, size = 104, normalized size = 0.27 \[ \frac {A \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(e*x)**(7/2)/(b*x**2+a)**(3/2),x)

[Out]

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

________________________________________________________________________________________