\(\int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx\) [15]

Optimal result

Integrand size = 30, antiderivative size = 233 \[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}{3 e}-\frac {f^2 \left (4 a-\frac {b^2 f^2}{e^2}\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{4 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{4 \sqrt {2} e^{5/2} \sqrt {2 d e-b f^2}} \] Output:

1/3*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(3/2)/e-f^2*(4*a-b^2*f^2/e^2)*(d+e 
*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/(4*b*f^2+8*e*(e*x+f*(a+x*(b*f^2+e^2* 
x)/f^2)^(1/2)))-1/8*f^2*(-b^2*f^2+4*a*e^2)*arctanh(2^(1/2)*e^(1/2)*(d+e*x+ 
f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^(1/2))*2^(1/2)/e^(5/2)/( 
-b*f^2+2*d*e)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.97 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.40 \[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\frac {\sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}} \left (3 b^2 f^4+4 b e f^2 \left (d+3 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+4 e^2 \left (-a f^2+2 (d+2 e x) \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )\right )}{12 e^2 \left (b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}+\frac {a f^2 \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\sqrt {2} \sqrt {e} \sqrt {-2 d e+b f^2}}-\frac {b^2 f^4 \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{4 e^{5/2} \sqrt {-4 d e+2 b f^2}} \] Input:

Integrate[Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]],x]
 

Output:

(Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]]*(3*b^2*f^4 + 4*b*e*f^2*(d 
 + 3*e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 4*e^2*(-(a*f^2) + 2*(d + 2*e 
*x)*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))))/(12*e^2*(b*f^2 + 2*e*(e*x + 
 f*Sqrt[a + x*(b + (e^2*x)/f^2)]))) + (a*f^2*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[ 
d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2*d*e + b*f^2]])/(Sqrt[2 
]*Sqrt[e]*Sqrt[-2*d*e + b*f^2]) - (b^2*f^4*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[d 
+ e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2*d*e + b*f^2]])/(4*e^(5/2 
)*Sqrt[-4*d*e + 2*b*f^2])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2541, 1192, 1580, 25, 1467, 2009}

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.

\(\displaystyle \int \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x} \, dx\)

\(\Big \downarrow \) 2541

\(\displaystyle 2 \int \frac {\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}} \left (e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\)

\(\Big \downarrow \) 1192

\(\displaystyle 4 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) \left (e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}\)

\(\Big \downarrow \) 1580

\(\displaystyle 4 \left (\frac {\int -\frac {8 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 e^3-4 \left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) e^2+f^2 \left (4 a e^2-b^2 f^2\right ) e}{-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{16 e^3}+\frac {f^2 \left (4 a-\frac {b^2 f^2}{e^2}\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{16 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 4 \left (\frac {f^2 \left (4 a-\frac {b^2 f^2}{e^2}\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{16 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\int \frac {8 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 e^3-4 \left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) e^2+f^2 \left (4 a e^2-b^2 f^2\right ) e}{-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{16 e^3}\right )\)

\(\Big \downarrow \) 1467

\(\displaystyle 4 \left (\frac {f^2 \left (4 a-\frac {b^2 f^2}{e^2}\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{16 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\int \left (\frac {e f^2 \left (4 a e^2-b^2 f^2\right )}{-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}-4 e^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{16 e^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \left (\frac {f^2 \left (4 a-\frac {b^2 f^2}{e^2}\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{16 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\frac {\sqrt {e} f^2 \left (4 a e^2-b^2 f^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{\sqrt {2} \sqrt {2 d e-b f^2}}-\frac {4}{3} e^2 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{16 e^3}\right )\)

Input:

Int[Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]],x]
 

Output:

4*((f^2*(4*a - (b^2*f^2)/e^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^ 
2]])/(16*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])) 
) - ((-4*e^2*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2))/3 + (Sqrt[ 
e]*f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[ 
a + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e - b*f^2]])/(Sqrt[2]*Sqrt[2*d*e - b*f 
^2]))/(16*e^3))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1192

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^( 
2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + 
 c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 1580

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) 
^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
 + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* 
(q + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* 
e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b 
*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e 
}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2541

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2   Subst[Int[(g + h*x^n)^p*((d 
^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x 
)^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e 
, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
 

Maple [F]

\[\int \sqrt {d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}d x\]

Input:

int((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x)
 

Output:

int((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 692, normalized size of antiderivative = 2.97 \[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\left [-\frac {3 \, {\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt {-2 \, b e f^{2} + 4 \, d e^{2}} \log \left (-b^{2} f^{4} + 4 \, {\left (b d e - a e^{2}\right )} f^{2} - 4 \, {\left (b e^{2} f^{2} - 2 \, d e^{3}\right )} x - 2 \, {\left (2 \, \sqrt {-2 \, b e f^{2} + 4 \, d e^{2}} e f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt {-2 \, b e f^{2} + 4 \, d e^{2}} {\left (b f^{2} + 2 \, e^{2} x\right )}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} + 4 \, {\left (b e f^{3} - 2 \, d e^{2} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) - 4 \, {\left (3 \, b^{2} e f^{4} - 2 \, b d e^{2} f^{2} - 8 \, d^{2} e^{3} + 10 \, {\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x - 2 \, {\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{48 \, {\left (b e^{3} f^{2} - 2 \, d e^{4}\right )}}, \frac {3 \, {\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt {2 \, b e f^{2} - 4 \, d e^{2}} \arctan \left (\frac {\sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} {\left (\sqrt {2 \, b e f^{2} - 4 \, d e^{2}} f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt {2 \, b e f^{2} - 4 \, d e^{2}} {\left (e x + d\right )}\right )}}{2 \, {\left (a e f^{2} - d^{2} e + {\left (b e f^{2} - 2 \, d e^{2}\right )} x\right )}}\right ) + 2 \, {\left (3 \, b^{2} e f^{4} - 2 \, b d e^{2} f^{2} - 8 \, d^{2} e^{3} + 10 \, {\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x - 2 \, {\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{24 \, {\left (b e^{3} f^{2} - 2 \, d e^{4}\right )}}\right ] \] Input:

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="fricas")
 

Output:

[-1/48*(3*(b^2*f^4 - 4*a*e^2*f^2)*sqrt(-2*b*e*f^2 + 4*d*e^2)*log(-b^2*f^4 
+ 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x - 2*(2*sqrt(-2*b*e*f^2 
 + 4*d*e^2)*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(-2*b*e*f^2 + 
4*d*e^2)*(b*f^2 + 2*e^2*x))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/ 
f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) 
 - 4*(3*b^2*e*f^4 - 2*b*d*e^2*f^2 - 8*d^2*e^3 + 10*(b*e^3*f^2 - 2*d*e^4)*x 
 - 2*(b*e^2*f^3 - 2*d*e^3*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e 
*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/(b*e^3*f^2 - 2*d*e^4), 
1/24*(3*(b^2*f^4 - 4*a*e^2*f^2)*sqrt(2*b*e*f^2 - 4*d*e^2)*arctan(1/2*sqrt( 
e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*(sqrt(2*b*e*f^2 - 4*d*e 
^2)*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(2*b*e*f^2 - 4*d*e^2)*(e 
*x + d))/(a*e*f^2 - d^2*e + (b*e*f^2 - 2*d*e^2)*x)) + 2*(3*b^2*e*f^4 - 2*b 
*d*e^2*f^2 - 8*d^2*e^3 + 10*(b*e^3*f^2 - 2*d*e^4)*x - 2*(b*e^2*f^3 - 2*d*e 
^3*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + 
e^2*x^2 + a*f^2)/f^2) + d))/(b*e^3*f^2 - 2*d*e^4)]
 

Sympy [F]

\[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\int \sqrt {d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}}\, dx \] Input:

integrate((d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**(1/2),x)
 

Output:

Integral(sqrt(d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2)), x)
 

Maxima [F]

\[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\int { \sqrt {e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d} \,d x } \] Input:

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d), x)
 

Giac [F]

\[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\int { \sqrt {e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d} \,d x } \] Input:

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\int \sqrt {d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}} \,d x \] Input:

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(1/2),x)
 

Output:

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}} \, dx=\int \sqrt {\sqrt {b \,f^{2} x +e^{2} x^{2}+a \,f^{2}}+d +e x}d x \] Input:

int((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(sqrt(sqrt(a*f**2 + b*f**2*x + e**2*x**2) + d + e*x),x)