Integrand size = 24, antiderivative size = 297 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\frac {2 \sqrt {e x} (5 A+3 B x) \sqrt {a+c x^2}}{15 e}+\frac {4 a B x \sqrt {a+c x^2}}{5 \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {4 a^{5/4} B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 a^{3/4} \left (3 \sqrt {a} B+5 A \sqrt {c}\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {e x} \sqrt {a+c x^2}} \]
4/5*a*B*x*(c*x^2+a)^(1/2)/c^(1/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+2/15*(3* B*x+5*A)*(e*x)^(1/2)*(c*x^2+a)^(1/2)/e-4/5*a^(5/4)*B*(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)))*Ellipti cE(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^(3/4)/(e*x)^(1/2)/(c*x^ 2+a)^(1/2)+2/15*a^(3/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/c os(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))*(3*B*a^(1/2)+5*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^(3/4)/(e*x)^(1/2)/(c*x^2+a )^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.28 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\frac {2 x \sqrt {a+c x^2} \left (3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{a}\right )+B x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{a}\right )\right )}{3 \sqrt {e x} \sqrt {1+\frac {c x^2}{a}}} \]
(2*x*Sqrt[a + c*x^2]*(3*A*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/a)] + B*x*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c*x^2)/a)]))/(3*Sqrt[e*x]*Sqrt[ 1 + (c*x^2)/a])
Time = 0.39 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {548, 27, 556, 555, 1512, 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.
| \(\displaystyle \int \frac {\sqrt {a+c x^2} (A+B x)}{\sqrt {e x}} \, dx\) |
| \(\Big \downarrow \) 548 |
| \(\displaystyle \frac {4}{15} a \int \frac {5 A+3 B x}{2 \sqrt {e x} \sqrt {c x^2+a}}dx+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{15} a \int \frac {5 A+3 B x}{\sqrt {e x} \sqrt {c x^2+a}}dx+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 556 |
| \(\displaystyle \frac {2 a \sqrt {x} \int \frac {5 A+3 B x}{\sqrt {x} \sqrt {c x^2+a}}dx}{15 \sqrt {e x}}+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 555 |
| \(\displaystyle \frac {4 a \sqrt {x} \int \frac {5 A+3 B x}{\sqrt {c x^2+a}}d\sqrt {x}}{15 \sqrt {e x}}+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {4 a \sqrt {x} \left (\left (\frac {3 \sqrt {a} B}{\sqrt {c}}+5 A\right ) \int \frac {1}{\sqrt {c x^2+a}}d\sqrt {x}-\frac {3 \sqrt {a} B \int \frac {\sqrt {a}-\sqrt {c} x}{\sqrt {a} \sqrt {c x^2+a}}d\sqrt {x}}{\sqrt {c}}\right )}{15 \sqrt {e x}}+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 a \sqrt {x} \left (\left (\frac {3 \sqrt {a} B}{\sqrt {c}}+5 A\right ) \int \frac {1}{\sqrt {c x^2+a}}d\sqrt {x}-\frac {3 B \int \frac {\sqrt {a}-\sqrt {c} x}{\sqrt {c x^2+a}}d\sqrt {x}}{\sqrt {c}}\right )}{15 \sqrt {e x}}+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {4 a \sqrt {x} \left (\frac {\left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (\frac {3 \sqrt {a} B}{\sqrt {c}}+5 A\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^2}}-\frac {3 B \int \frac {\sqrt {a}-\sqrt {c} x}{\sqrt {c x^2+a}}d\sqrt {x}}{\sqrt {c}}\right )}{15 \sqrt {e x}}+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {4 a \sqrt {x} \left (\frac {\left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (\frac {3 \sqrt {a} B}{\sqrt {c}}+5 A\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^2}}-\frac {3 B \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {a+c x^2}}-\frac {\sqrt {x} \sqrt {a+c x^2}}{\sqrt {a}+\sqrt {c} x}\right )}{\sqrt {c}}\right )}{15 \sqrt {e x}}+\frac {2 \sqrt {e x} \sqrt {a+c x^2} (5 A+3 B x)}{15 e}\) |
(2*Sqrt[e*x]*(5*A + 3*B*x)*Sqrt[a + c*x^2])/(15*e) + (4*a*Sqrt[x]*((-3*B*( -((Sqrt[x]*Sqrt[a + c*x^2])/(Sqrt[a] + Sqrt[c]*x)) + (a^(1/4)*(Sqrt[a] + S qrt[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])/(c^(1/4)*Sqrt[a + c*x^2])))/Sqrt[c] + ((5*A + (3*Sqrt[a]*B)/Sqrt[c])*(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])/(2*a^ (1/4)*c^(1/4)*Sqrt[a + c*x^2])))/(15*Sqrt[e*x])
3.5.37.3.1 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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[(e*x)^(m + 1)*(c*(m + 2*p + 2) + d*(m + 2*p + 1)*x)*((a + b*x^ 2)^p/(e*(m + 2*p + 1)*(m + 2*p + 2))), x] + Simp[2*a*(p/((m + 2*p + 1)*(m + 2*p + 2))) Int[(e*x)^m*(a + b*x^2)^(p - 1)*(c*(m + 2*p + 2) + d*(m + 2*p + 1)*x), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && GtQ[p, 0] && (IntegerQ[ p] || !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Simp[2 Subst[Int[(f + g*x^2)/Sqrt[a + c*x^4], x], x, Sqrt[x]], x] /; Free Q[{a, c, f, g}, x]
Int[((c_) + (d_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symb ol] :> Simp[Sqrt[x]/Sqrt[e*x] Int[(c + d*x)/(Sqrt[x]*Sqrt[a + b*x^2]), x] , x] /; FreeQ[{a, b, c, d, e}, x]
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[((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]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[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]
Time = 0.39 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(\frac {\frac {2 A \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a c}\, a}{3}+\frac {4 B \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right ) a^{2}}{5}-\frac {2 B \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right ) a^{2}}{5}+\frac {2 B \,c^{2} x^{4}}{5}+\frac {2 A \,c^{2} x^{3}}{3}+\frac {2 a B c \,x^{2}}{5}+\frac {2 a A c x}{3}}{\sqrt {c \,x^{2}+a}\, c \sqrt {e x}}\) | \(312\) |
| risch | \(\frac {2 \left (3 B x +5 A \right ) x \sqrt {c \,x^{2}+a}}{15 \sqrt {e x}}+\frac {2 a \left (\frac {5 A \sqrt {-a c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c e \,x^{3}+a e x}}+\frac {3 B \sqrt {-a c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, \left (-\frac {2 \sqrt {-a c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-a c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{c \sqrt {c e \,x^{3}+a e x}}\right ) \sqrt {\left (c \,x^{2}+a \right ) e x}}{15 \sqrt {e x}\, \sqrt {c \,x^{2}+a}}\) | \(329\) |
| elliptic | \(\frac {\sqrt {\left (c \,x^{2}+a \right ) e x}\, \left (\frac {2 B x \sqrt {c e \,x^{3}+a e x}}{5 e}+\frac {2 A \sqrt {c e \,x^{3}+a e x}}{3 e}+\frac {2 a A \sqrt {-a c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{3 c \sqrt {c e \,x^{3}+a e x}}+\frac {2 B a \sqrt {-a c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, \left (-\frac {2 \sqrt {-a c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-a c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{5 c \sqrt {c e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {c \,x^{2}+a}}\) | \(342\) |
2/15/(c*x^2+a)^(1/2)/c*(5*A*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2 )*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x/(-a*c)^(1/2)*c)^(1/2)*Ellip ticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c)^(1/2)*a+6 *B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(- a*c)^(1/2))^(1/2)*(-x/(-a*c)^(1/2)*c)^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/ (-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2-3*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2)) ^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x/(-a*c)^(1/2)*c )^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2 +3*B*c^2*x^4+5*A*c^2*x^3+3*a*B*c*x^2+5*a*A*c*x)/(e*x)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.26 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\frac {2 \, {\left (10 \, \sqrt {c e} A a {\rm weierstrassPInverse}\left (-\frac {4 \, a}{c}, 0, x\right ) - 6 \, \sqrt {c e} B a {\rm weierstrassZeta}\left (-\frac {4 \, a}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{c}, 0, x\right )\right ) + {\left (3 \, B c x + 5 \, A c\right )} \sqrt {c x^{2} + a} \sqrt {e x}\right )}}{15 \, c e} \]
2/15*(10*sqrt(c*e)*A*a*weierstrassPInverse(-4*a/c, 0, x) - 6*sqrt(c*e)*B*a *weierstrassZeta(-4*a/c, 0, weierstrassPInverse(-4*a/c, 0, x)) + (3*B*c*x + 5*A*c)*sqrt(c*x^2 + a)*sqrt(e*x))/(c*e)
Result contains complex when optimal does not.
Time = 2.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.33 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\frac {A \sqrt {a} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {B \sqrt {a} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {7}{4}\right )} \]
A*sqrt(a)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**2*exp_polar(I *pi)/a)/(2*sqrt(e)*gamma(5/4)) + B*sqrt(a)*x**(3/2)*gamma(3/4)*hyper((-1/2 , 3/4), (7/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(e)*gamma(7/4))
\[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\int { \frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\sqrt {e x}} \,d x } \]
\[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\int { \frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\sqrt {e x}} \,d x } \]
Timed out. \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx=\int \frac {\sqrt {c\,x^2+a}\,\left (A+B\,x\right )}{\sqrt {e\,x}} \,d x \]