\(\int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx\) [491]

Optimal result

Integrand size = 24, antiderivative size = 138 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\frac {2 (b c-a d)^2 \sqrt {e+f x}}{d^3}-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{3 d^2 f^2}+\frac {2 b^2 (e+f x)^{5/2}}{5 d f^2}-\frac {2 (b c-a d)^2 \sqrt {d e-c f} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}} \] Output:

2*(-a*d+b*c)^2*(f*x+e)^(1/2)/d^3-2/3*b*(-2*a*d*f+b*c*f+b*d*e)*(f*x+e)^(3/2 
)/d^2/f^2+2/5*b^2*(f*x+e)^(5/2)/d/f^2-2*(-a*d+b*c)^2*(-c*f+d*e)^(1/2)*arct 
anh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\frac {2 \sqrt {e+f x} \left (15 a^2 d^2 f^2+10 a b d f (-3 c f+d (e+f x))+b^2 \left (15 c^2 f^2-5 c d f (e+f x)+d^2 \left (-2 e^2+e f x+3 f^2 x^2\right )\right )\right )}{15 d^3 f^2}-\frac {2 (b c-a d)^2 \sqrt {-d e+c f} \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2}} \] Input:

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

Output:

(2*Sqrt[e + f*x]*(15*a^2*d^2*f^2 + 10*a*b*d*f*(-3*c*f + d*(e + f*x)) + b^2 
*(15*c^2*f^2 - 5*c*d*f*(e + f*x) + d^2*(-2*e^2 + e*f*x + 3*f^2*x^2))))/(15 
*d^3*f^2) - (2*(b*c - a*d)^2*Sqrt[-(d*e) + c*f]*ArcTan[(Sqrt[d]*Sqrt[e + f 
*x])/Sqrt[-(d*e) + c*f]])/d^(7/2)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {99, 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.

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 (b c-a d)^2 \sqrt {d e-c f} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}}+\frac {2 \sqrt {e+f x} (b c-a d)^2}{d^3}-\frac {2 b (e+f x)^{3/2} (-2 a d f+b c f+b d e)}{3 d^2 f^2}+\frac {2 b^2 (e+f x)^{5/2}}{5 d f^2}\)

Input:

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

Output:

(2*(b*c - a*d)^2*Sqrt[e + f*x])/d^3 - (2*b*(b*d*e + b*c*f - 2*a*d*f)*(e + 
f*x)^(3/2))/(3*d^2*f^2) + (2*b^2*(e + f*x)^(5/2))/(5*d*f^2) - (2*(b*c - a* 
d)^2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^( 
7/2)
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\sqrt {\left (c f -d e \right ) d}\, \left (\left (-\frac {2 \left (f x +e \right ) \left (-\frac {3 f x}{2}+e \right ) b^{2}}{15}+\frac {2 a f \left (f x +e \right ) b}{3}+a^{2} f^{2}\right ) d^{2}-2 c b f \left (\frac {\left (f x +e \right ) b}{6}+a f \right ) d +b^{2} c^{2} f^{2}\right ) \sqrt {f x +e}+f^{2} \left (a d -b c \right )^{2} \left (c f -d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )\right )}{\sqrt {\left (c f -d e \right ) d}\, f^{2} d^{3}}\)	\(154\)
risch	\(\frac {2 \left (3 b^{2} d^{2} f^{2} x^{2}+10 a b \,d^{2} f^{2} x -5 b^{2} c d \,f^{2} x +b^{2} d^{2} e f x +15 a^{2} d^{2} f^{2}-30 a b c d \,f^{2}+10 a b \,d^{2} e f +15 b^{2} c^{2} f^{2}-5 b^{2} c d e f -2 b^{2} d^{2} e^{2}\right ) \sqrt {f x +e}}{15 f^{2} d^{3}}-\frac {2 \left (a^{2} c \,d^{2} f -a^{2} e \,d^{3}-2 a b \,c^{2} d f +2 a b c \,d^{2} e +c^{3} b^{2} f -b^{2} c^{2} d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}\)	\(218\)
derivativedivides	\(\frac {\frac {2 \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+\frac {2 a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+a^{2} d^{2} f^{2} \sqrt {f x +e}-2 a b c d \,f^{2} \sqrt {f x +e}+b^{2} c^{2} f^{2} \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f^{2} \left (a^{2} c \,d^{2} f -a^{2} e \,d^{3}-2 a b \,c^{2} d f +2 a b c \,d^{2} e +c^{3} b^{2} f -b^{2} c^{2} d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{2}}\)	\(220\)
default	\(\frac {\frac {2 \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+\frac {2 a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+a^{2} d^{2} f^{2} \sqrt {f x +e}-2 a b c d \,f^{2} \sqrt {f x +e}+b^{2} c^{2} f^{2} \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f^{2} \left (a^{2} c \,d^{2} f -a^{2} e \,d^{3}-2 a b \,c^{2} d f +2 a b c \,d^{2} e +c^{3} b^{2} f -b^{2} c^{2} d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{2}}\)	\(220\)

Input:

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

Output:

-2*(-((c*f-d*e)*d)^(1/2)*((-2/15*(f*x+e)*(-3/2*f*x+e)*b^2+2/3*a*f*(f*x+e)* 
b+a^2*f^2)*d^2-2*c*b*f*(1/6*(f*x+e)*b+a*f)*d+b^2*c^2*f^2)*(f*x+e)^(1/2)+f^ 
2*(a*d-b*c)^2*(c*f-d*e)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))/((c*f 
-d*e)*d)^(1/2)/f^2/d^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.91 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) + 2 \, {\left (3 \, b^{2} d^{2} f^{2} x^{2} - 2 \, b^{2} d^{2} e^{2} - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} e f + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} + {\left (b^{2} d^{2} e f - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} f^{2}\right )} x\right )} \sqrt {f x + e}}{15 \, d^{3} f^{2}}, -\frac {2 \, {\left (15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) - {\left (3 \, b^{2} d^{2} f^{2} x^{2} - 2 \, b^{2} d^{2} e^{2} - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} e f + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} + {\left (b^{2} d^{2} e f - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} f^{2}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, d^{3} f^{2}}\right ] \] Input:

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

Output:

[1/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*sqrt((d*e - c*f)/d)*log((d*f 
*x + 2*d*e - c*f - 2*sqrt(f*x + e)*d*sqrt((d*e - c*f)/d))/(d*x + c)) + 2*( 
3*b^2*d^2*f^2*x^2 - 2*b^2*d^2*e^2 - 5*(b^2*c*d - 2*a*b*d^2)*e*f + 15*(b^2* 
c^2 - 2*a*b*c*d + a^2*d^2)*f^2 + (b^2*d^2*e*f - 5*(b^2*c*d - 2*a*b*d^2)*f^ 
2)*x)*sqrt(f*x + e))/(d^3*f^2), -2/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)* 
f^2*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e 
 - c*f)) - (3*b^2*d^2*f^2*x^2 - 2*b^2*d^2*e^2 - 5*(b^2*c*d - 2*a*b*d^2)*e* 
f + 15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2 + (b^2*d^2*e*f - 5*(b^2*c*d - 2 
*a*b*d^2)*f^2)*x)*sqrt(f*x + e))/(d^3*f^2)]
 

Sympy [A] (verification not implemented)

Time = 3.49 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (e + f x\right )^{\frac {5}{2}}}{5 d f} + \frac {\left (e + f x\right )^{\frac {3}{2}} \cdot \left (2 a b d f - b^{2} c f - b^{2} d e\right )}{3 d^{2} f} + \frac {\sqrt {e + f x} \left (a^{2} d^{2} f - 2 a b c d f + b^{2} c^{2} f\right )}{d^{3}} - \frac {f \left (a d - b c\right )^{2} \left (c f - d e\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{4} \sqrt {\frac {c f - d e}{d}}}\right )}{f} & \text {for}\: f \neq 0 \\\sqrt {e} \left (\frac {b^{2} x^{2}}{2 d} + \frac {x \left (2 a b d - b^{2} c\right )}{d^{2}} + \frac {\left (a d - b c\right )^{2} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(b**2*(e + f*x)**(5/2)/(5*d*f) + (e + f*x)**(3/2)*(2*a*b*d*f 
- b**2*c*f - b**2*d*e)/(3*d**2*f) + sqrt(e + f*x)*(a**2*d**2*f - 2*a*b*c*d 
*f + b**2*c**2*f)/d**3 - f*(a*d - b*c)**2*(c*f - d*e)*atan(sqrt(e + f*x)/s 
qrt((c*f - d*e)/d))/(d**4*sqrt((c*f - d*e)/d)))/f, Ne(f, 0)), (sqrt(e)*(b* 
*2*x**2/(2*d) + x*(2*a*b*d - b**2*c)/d**2 + (a*d - b*c)**2*Piecewise((x/c, 
 Eq(d, 0)), (log(c + d*x)/d, True))/d**2), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\frac {2 \, {\left (b^{2} c^{2} d e - 2 \, a b c d^{2} e + a^{2} d^{3} e - b^{2} c^{3} f + 2 \, a b c^{2} d f - a^{2} c d^{2} f\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} d^{3}} + \frac {2 \, {\left (3 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{2} d^{4} f^{8} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} d^{4} e f^{8} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} c d^{3} f^{9} + 10 \, {\left (f x + e\right )}^{\frac {3}{2}} a b d^{4} f^{9} + 15 \, \sqrt {f x + e} b^{2} c^{2} d^{2} f^{10} - 30 \, \sqrt {f x + e} a b c d^{3} f^{10} + 15 \, \sqrt {f x + e} a^{2} d^{4} f^{10}\right )}}{15 \, d^{5} f^{10}} \] Input:

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

Output:

2*(b^2*c^2*d*e - 2*a*b*c*d^2*e + a^2*d^3*e - b^2*c^3*f + 2*a*b*c^2*d*f - a 
^2*c*d^2*f)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/(sqrt(-d^2*e + c* 
d*f)*d^3) + 2/15*(3*(f*x + e)^(5/2)*b^2*d^4*f^8 - 5*(f*x + e)^(3/2)*b^2*d^ 
4*e*f^8 - 5*(f*x + e)^(3/2)*b^2*c*d^3*f^9 + 10*(f*x + e)^(3/2)*a*b*d^4*f^9 
 + 15*sqrt(f*x + e)*b^2*c^2*d^2*f^10 - 30*sqrt(f*x + e)*a*b*c*d^3*f^10 + 1 
5*sqrt(f*x + e)*a^2*d^4*f^10)/(d^5*f^10)
 

Mupad [B] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.04 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\sqrt {e+f\,x}\,\left (\frac {2\,{\left (a\,f-b\,e\right )}^2}{d\,f^2}+\frac {\left (\frac {4\,b^2\,e-4\,a\,b\,f}{d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{d^2\,f^4}\right )\,\left (c\,f^3-d\,e\,f^2\right )}{d\,f^2}\right )-{\left (e+f\,x\right )}^{3/2}\,\left (\frac {4\,b^2\,e-4\,a\,b\,f}{3\,d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{3\,d^2\,f^4}\right )+\frac {2\,b^2\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\sqrt {d\,e-c\,f}\,1{}\mathrm {i}}{-f\,a^2\,c\,d^2+e\,a^2\,d^3+2\,f\,a\,b\,c^2\,d-2\,e\,a\,b\,c\,d^2-f\,b^2\,c^3+e\,b^2\,c^2\,d}\right )\,{\left (a\,d-b\,c\right )}^2\,\sqrt {d\,e-c\,f}\,2{}\mathrm {i}}{d^{7/2}} \] Input:

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

Output:

(e + f*x)^(1/2)*((2*(a*f - b*e)^2)/(d*f^2) + (((4*b^2*e - 4*a*b*f)/(d*f^2) 
 + (2*b^2*(c*f^3 - d*e*f^2))/(d^2*f^4))*(c*f^3 - d*e*f^2))/(d*f^2)) - (e + 
 f*x)^(3/2)*((4*b^2*e - 4*a*b*f)/(3*d*f^2) + (2*b^2*(c*f^3 - d*e*f^2))/(3* 
d^2*f^4)) + (2*b^2*(e + f*x)^(5/2))/(5*d*f^2) + (atan((d^(1/2)*(e + f*x)^( 
1/2)*(a*d - b*c)^2*(d*e - c*f)^(1/2)*1i)/(a^2*d^3*e - b^2*c^3*f - a^2*c*d^ 
2*f + b^2*c^2*d*e - 2*a*b*c*d^2*e + 2*a*b*c^2*d*f))*(a*d - b*c)^2*(d*e - c 
*f)^(1/2)*2i)/d^(7/2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.30 \[ \int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx=\frac {-2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) a^{2} d^{2} f^{2}+4 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) a b c d \,f^{2}-2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) b^{2} c^{2} f^{2}+2 \sqrt {f x +e}\, a^{2} d^{3} f^{2}-4 \sqrt {f x +e}\, a b c \,d^{2} f^{2}+\frac {4 \sqrt {f x +e}\, a b \,d^{3} e f}{3}+\frac {4 \sqrt {f x +e}\, a b \,d^{3} f^{2} x}{3}+2 \sqrt {f x +e}\, b^{2} c^{2} d \,f^{2}-\frac {2 \sqrt {f x +e}\, b^{2} c \,d^{2} e f}{3}-\frac {2 \sqrt {f x +e}\, b^{2} c \,d^{2} f^{2} x}{3}-\frac {4 \sqrt {f x +e}\, b^{2} d^{3} e^{2}}{15}+\frac {2 \sqrt {f x +e}\, b^{2} d^{3} e f x}{15}+\frac {2 \sqrt {f x +e}\, b^{2} d^{3} f^{2} x^{2}}{5}}{d^{4} f^{2}} \] Input:

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

Output:

(2*( - 15*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f 
 - d*e)))*a**2*d**2*f**2 + 30*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)* 
d)/(sqrt(d)*sqrt(c*f - d*e)))*a*b*c*d*f**2 - 15*sqrt(d)*sqrt(c*f - d*e)*at 
an((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b**2*c**2*f**2 + 15*sqrt(e 
 + f*x)*a**2*d**3*f**2 - 30*sqrt(e + f*x)*a*b*c*d**2*f**2 + 10*sqrt(e + f* 
x)*a*b*d**3*e*f + 10*sqrt(e + f*x)*a*b*d**3*f**2*x + 15*sqrt(e + f*x)*b**2 
*c**2*d*f**2 - 5*sqrt(e + f*x)*b**2*c*d**2*e*f - 5*sqrt(e + f*x)*b**2*c*d* 
*2*f**2*x - 2*sqrt(e + f*x)*b**2*d**3*e**2 + sqrt(e + f*x)*b**2*d**3*e*f*x 
 + 3*sqrt(e + f*x)*b**2*d**3*f**2*x**2))/(15*d**4*f**2)