Integrand size = 26, antiderivative size = 338 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {2 (9 A b-7 a B) e (e x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {2 a (9 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^2}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 a^{5/4} (9 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}}-\frac {a^{5/4} (9 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}} \]
2/45*(9*A*b-7*B*a)*e*(e*x)^(3/2)*(b*x^2+a)^(1/2)/b^2+2/9*B*(e*x)^(7/2)*(b* x^2+a)^(1/2)/b/e-2/15*a*(9*A*b-7*B*a)*e^2*(e*x)^(1/2)*(b*x^2+a)^(1/2)/b^(5 /2)/(a^(1/2)+x*b^(1/2))+2/15*a^(5/4)*(9*A*b-7*B*a)*e^(5/2)*(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^(11/4)/(b*x^2+a)^(1/2)-1/15*a^(5/4)*(9*A*b-7*B*a)*e^(5/2)*(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)/b^(11/4)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.28 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {2 e (e x)^{3/2} \left (-\left (\left (a+b x^2\right ) \left (-9 A b+7 a B-5 b B x^2\right )\right )+a (-9 A b+7 a B) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{45 b^2 \sqrt {a+b x^2}} \]
(2*e*(e*x)^(3/2)*(-((a + b*x^2)*(-9*A*b + 7*a*B - 5*b*B*x^2)) + a*(-9*A*b + 7*a*B)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^2)/a) ]))/(45*b^2*Sqrt[a + b*x^2])
Time = 0.41 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {363, 262, 266, 834, 27, 761, 1510}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(9 A b-7 a B) \int \frac {(e x)^{5/2}}{\sqrt {b x^2+a}}dx}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(9 A b-7 a B) \left (\frac {2 e (e x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {3 a e^2 \int \frac {\sqrt {e x}}{\sqrt {b x^2+a}}dx}{5 b}\right )}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {(9 A b-7 a B) \left (\frac {2 e (e x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a e \int \frac {e x}{\sqrt {b x^2+a}}d\sqrt {e x}}{5 b}\right )}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {(9 A b-7 a B) \left (\frac {2 e (e x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a e \left (\frac {\sqrt {a} e \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}-\frac {\sqrt {a} e \int \frac {\sqrt {a} e-\sqrt {b} e x}{\sqrt {a} e \sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(9 A b-7 a B) \left (\frac {2 e (e x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a e \left (\frac {\sqrt {a} e \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a} e-\sqrt {b} e x}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(9 A b-7 a B) \left (\frac {2 e (e x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a e \left (\frac {\sqrt [4]{a} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a} e-\sqrt {b} e x}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {(9 A b-7 a B) \left (\frac {2 e (e x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a e \left (\frac {\sqrt [4]{a} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {e^2 \sqrt {e x} \sqrt {a+b x^2}}{\sqrt {a} e+\sqrt {b} e x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}\) |
(2*B*(e*x)^(7/2)*Sqrt[a + b*x^2])/(9*b*e) + ((9*A*b - 7*a*B)*((2*e*(e*x)^( 3/2)*Sqrt[a + b*x^2])/(5*b) - (6*a*e*(-((-((e^2*Sqrt[e*x]*Sqrt[a + b*x^2]) /(Sqrt[a]*e + Sqrt[b]*e*x)) + (a^(1/4)*Sqrt[e]*(Sqrt[a]*e + Sqrt[b]*e*x)*S qrt[(a*e^2 + b*e^2*x^2)/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*EllipticE[2*ArcTan[(b ^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqr t[b]) + (a^(1/4)*Sqrt[e]*(Sqrt[a]*e + Sqrt[b]*e*x)*Sqrt[(a*e^2 + b*e^2*x^2 )/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^( 1/4)*Sqrt[e])], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^2])))/(5*b)))/(9*b)
3.8.100.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
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)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 3.05 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.72
| method | result | size |
| risch | \(\frac {2 x^{2} \left (5 b B \,x^{2}+9 A b -7 B a \right ) \sqrt {b \,x^{2}+a}\, e^{3}}{45 b^{2} \sqrt {e x}}-\frac {a \left (9 A b -7 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) e^{3} \sqrt {\left (b \,x^{2}+a \right ) e x}}{15 b^{3} \sqrt {b e \,x^{3}+a e x}\, \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(242\) |
| elliptic | \(\frac {\sqrt {e x}\, \sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (\frac {2 B \,e^{2} x^{3} \sqrt {b e \,x^{3}+a e x}}{9 b}+\frac {2 \left (A \,e^{3}-\frac {7 B \,e^{3} a}{9 b}\right ) x \sqrt {b e \,x^{3}+a e x}}{5 b e}-\frac {3 \left (A \,e^{3}-\frac {7 B \,e^{3} a}{9 b}\right ) a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b^{2} \sqrt {b e \,x^{3}+a e x}}\right )}{e x \sqrt {b \,x^{2}+a}}\) | \(275\) |
| default | \(-\frac {e^{2} \sqrt {e x}\, \left (-10 b^{3} B \,x^{6}+54 A \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} b -27 A \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} b -42 B \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{3}+21 B \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{3}-18 A \,b^{3} x^{4}+4 B a \,b^{2} x^{4}-18 a A \,b^{2} x^{2}+14 B \,a^{2} b \,x^{2}\right )}{45 x \sqrt {b \,x^{2}+a}\, b^{3}}\) | \(417\) |
2/45*x^2*(5*B*b*x^2+9*A*b-7*B*a)*(b*x^2+a)^(1/2)/b^2*e^3/(e*x)^(1/2)-1/15* a*(9*A*b-7*B*a)/b^3*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2) *(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b *e*x^3+a*e*x)^(1/2)*(-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b )^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*EllipticF(((x+(-a*b)^(1/2)/b) /(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2)))*e^3*((b*x^2+a)*e*x)^(1/2)/(e*x)^(1/2) /(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.27 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {2 \, {\left (3 \, {\left (7 \, B a^{2} - 9 \, A a b\right )} \sqrt {b e} e^{2} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) - {\left (5 \, B b^{2} e^{2} x^{3} - {\left (7 \, B a b - 9 \, A b^{2}\right )} e^{2} x\right )} \sqrt {b x^{2} + a} \sqrt {e x}\right )}}{45 \, b^{3}} \]
-2/45*(3*(7*B*a^2 - 9*A*a*b)*sqrt(b*e)*e^2*weierstrassZeta(-4*a/b, 0, weie rstrassPInverse(-4*a/b, 0, x)) - (5*B*b^2*e^2*x^3 - (7*B*a*b - 9*A*b^2)*e^ 2*x)*sqrt(b*x^2 + a)*sqrt(e*x))/b^3
Result contains complex when optimal does not.
Time = 13.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.28 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {A e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} + \frac {B e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \]
A*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**2*exp_polar (I*pi)/a)/(2*sqrt(a)*gamma(11/4)) + B*e**(5/2)*x**(11/2)*gamma(11/4)*hyper ((1/2, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(15/4))
\[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {5}{2}}}{\sqrt {b x^{2} + a}} \,d x } \]
\[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {5}{2}}}{\sqrt {b x^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^{5/2}}{\sqrt {b\,x^2+a}} \,d x \]