\(\int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx\) [90]

Optimal result

Integrand size = 22, antiderivative size = 146 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\frac {2 (d f g+d e h-2 c f h) \sqrt {e+f x}}{d^3}-\frac {(d e-c f) (d g-c h) \sqrt {e+f x}}{d^3 (c+d x)}+\frac {2 h (e+f x)^{3/2}}{3 d^2}-\frac {\sqrt {d e-c f} (3 d f g+2 d e h-5 c f h) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}} \] Output:

2*(-2*c*f*h+d*e*h+d*f*g)*(f*x+e)^(1/2)/d^3-(-c*f+d*e)*(-c*h+d*g)*(f*x+e)^( 
1/2)/d^3/(d*x+c)+2/3*h*(f*x+e)^(3/2)/d^2-(-c*f+d*e)^(1/2)*(-5*c*f*h+2*d*e* 
h+3*d*f*g)*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\frac {\sqrt {e+f x} \left (-15 c^2 f h+c d (9 f g+11 e h-10 f h x)+d^2 (2 f x (3 g+h x)+e (-3 g+8 h x))\right )}{3 d^3 (c+d x)}-\frac {\sqrt {-d e+c f} (3 d f g+2 d e h-5 c f h) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2}} \] Input:

Integrate[((e + f*x)^(3/2)*(g + h*x))/(c + d*x)^2,x]
 

Output:

(Sqrt[e + f*x]*(-15*c^2*f*h + c*d*(9*f*g + 11*e*h - 10*f*h*x) + d^2*(2*f*x 
*(3*g + h*x) + e*(-3*g + 8*h*x))))/(3*d^3*(c + d*x)) - (Sqrt[-(d*e) + c*f] 
*(3*d*f*g + 2*d*e*h - 5*c*f*h)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) 
+ c*f]])/d^(7/2)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {87, 60, 60, 73, 221}

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.

\(\displaystyle \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(-5 c f h+2 d e h+3 d f g) \int \frac {(e+f x)^{3/2}}{c+d x}dx}{2 d (d e-c f)}-\frac {(e+f x)^{5/2} (d g-c h)}{d (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(-5 c f h+2 d e h+3 d f g) \left (\frac {(d e-c f) \int \frac {\sqrt {e+f x}}{c+d x}dx}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{2 d (d e-c f)}-\frac {(e+f x)^{5/2} (d g-c h)}{d (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(-5 c f h+2 d e h+3 d f g) \left (\frac {(d e-c f) \left (\frac {(d e-c f) \int \frac {1}{(c+d x) \sqrt {e+f x}}dx}{d}+\frac {2 \sqrt {e+f x}}{d}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{2 d (d e-c f)}-\frac {(e+f x)^{5/2} (d g-c h)}{d (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(-5 c f h+2 d e h+3 d f g) \left (\frac {(d e-c f) \left (\frac {2 (d e-c f) \int \frac {1}{c+\frac {d (e+f x)}{f}-\frac {d e}{f}}d\sqrt {e+f x}}{d f}+\frac {2 \sqrt {e+f x}}{d}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{2 d (d e-c f)}-\frac {(e+f x)^{5/2} (d g-c h)}{d (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 221

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

Input:

Int[((e + f*x)^(3/2)*(g + h*x))/(c + d*x)^2,x]
 

Output:

-(((d*g - c*h)*(e + f*x)^(5/2))/(d*(d*e - c*f)*(c + d*x))) + ((3*d*f*g + 2 
*d*e*h - 5*c*f*h)*((2*(e + f*x)^(3/2))/(3*d) + ((d*e - c*f)*((2*Sqrt[e + f 
*x])/d - (2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f 
]])/d^(3/2)))/d))/(2*d*(d*e - c*f))
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {2 \left (-f d h x +6 c f h -4 d e h -3 d f g \right ) \sqrt {f x +e}}{3 d^{3}}+\frac {\left (2 c f -2 d e \right ) \left (\frac {\left (-\frac {1}{2} c f h +\frac {1}{2} d f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) d +c f -d e}+\frac {\left (5 c f h -2 d e h -3 d f g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \sqrt {\left (c f -d e \right ) d}}\right )}{d^{3}}\)	\(138\)
pseudoelliptic	\(-\frac {5 \left (-\left (c f -d e \right ) \left (\frac {\left (-2 e h -3 f g \right ) d}{5}+c f h \right ) \left (x d +c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\left (\frac {\left (-2 x \left (\frac {h x}{3}+g \right ) f +e \left (-\frac {8 h x}{3}+g \right )\right ) d^{2}}{5}-\frac {11 c \left (\frac {\left (-10 h x +9 g \right ) f}{11}+e h \right ) d}{15}+c^{2} f h \right ) \sqrt {f x +e}\, \sqrt {\left (c f -d e \right ) d}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{3} \left (x d +c \right )}\)	\(150\)
derivativedivides	\(-\frac {2 \left (-\frac {h \left (f x +e \right )^{\frac {3}{2}} d}{3}+2 c f h \sqrt {f x +e}-d e h \sqrt {f x +e}-d f g \sqrt {f x +e}\right )}{d^{3}}+\frac {\frac {2 \left (-\frac {1}{2} c^{2} f^{2} h +\frac {1}{2} c d e h f +\frac {1}{2} c d \,f^{2} g -\frac {1}{2} d^{2} e g f \right ) \sqrt {f x +e}}{\left (f x +e \right ) d +c f -d e}+\frac {\left (5 c^{2} f^{2} h -7 c d e h f -3 c d \,f^{2} g +2 d^{2} e^{2} h +3 d^{2} e g f \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}}{d^{3}}\)	\(197\)
default	\(-\frac {2 \left (-\frac {h \left (f x +e \right )^{\frac {3}{2}} d}{3}+2 c f h \sqrt {f x +e}-d e h \sqrt {f x +e}-d f g \sqrt {f x +e}\right )}{d^{3}}+\frac {\frac {2 \left (-\frac {1}{2} c^{2} f^{2} h +\frac {1}{2} c d e h f +\frac {1}{2} c d \,f^{2} g -\frac {1}{2} d^{2} e g f \right ) \sqrt {f x +e}}{\left (f x +e \right ) d +c f -d e}+\frac {\left (5 c^{2} f^{2} h -7 c d e h f -3 c d \,f^{2} g +2 d^{2} e^{2} h +3 d^{2} e g f \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}}{d^{3}}\)	\(197\)

Input:

int((f*x+e)^(3/2)*(h*x+g)/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/3*(-d*f*h*x+6*c*f*h-4*d*e*h-3*d*f*g)*(f*x+e)^(1/2)/d^3+1/d^3*(2*c*f-2*d 
*e)*((-1/2*c*f*h+1/2*d*f*g)*(f*x+e)^(1/2)/((f*x+e)*d+c*f-d*e)+1/2*(5*c*f*h 
-2*d*e*h-3*d*f*g)/((c*f-d*e)*d)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d) 
^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.63 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\left [-\frac {3 \, {\left (3 \, c d f g + {\left (2 \, c d e - 5 \, c^{2} f\right )} h + {\left (3 \, d^{2} f g + {\left (2 \, d^{2} e - 5 \, c d f\right )} h\right )} x\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f + 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) - 2 \, {\left (2 \, d^{2} f h x^{2} - 3 \, {\left (d^{2} e - 3 \, c d f\right )} g + {\left (11 \, c d e - 15 \, c^{2} f\right )} h + 2 \, {\left (3 \, d^{2} f g + {\left (4 \, d^{2} e - 5 \, c d f\right )} h\right )} x\right )} \sqrt {f x + e}}{6 \, {\left (d^{4} x + c d^{3}\right )}}, -\frac {3 \, {\left (3 \, c d f g + {\left (2 \, c d e - 5 \, c^{2} f\right )} h + {\left (3 \, d^{2} f g + {\left (2 \, d^{2} e - 5 \, c d f\right )} h\right )} x\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) - {\left (2 \, d^{2} f h x^{2} - 3 \, {\left (d^{2} e - 3 \, c d f\right )} g + {\left (11 \, c d e - 15 \, c^{2} f\right )} h + 2 \, {\left (3 \, d^{2} f g + {\left (4 \, d^{2} e - 5 \, c d f\right )} h\right )} x\right )} \sqrt {f x + e}}{3 \, {\left (d^{4} x + c d^{3}\right )}}\right ] \] Input:

integrate((f*x+e)^(3/2)*(h*x+g)/(d*x+c)^2,x, algorithm="fricas")
 

Output:

[-1/6*(3*(3*c*d*f*g + (2*c*d*e - 5*c^2*f)*h + (3*d^2*f*g + (2*d^2*e - 5*c* 
d*f)*h)*x)*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*(2*d^2*f*h*x^2 - 3*(d^2*e - 3*c*d*f) 
*g + (11*c*d*e - 15*c^2*f)*h + 2*(3*d^2*f*g + (4*d^2*e - 5*c*d*f)*h)*x)*sq 
rt(f*x + e))/(d^4*x + c*d^3), -1/3*(3*(3*c*d*f*g + (2*c*d*e - 5*c^2*f)*h + 
 (3*d^2*f*g + (2*d^2*e - 5*c*d*f)*h)*x)*sqrt(-(d*e - c*f)/d)*arctan(-sqrt( 
f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e - c*f)) - (2*d^2*f*h*x^2 - 3*(d^2*e - 
 3*c*d*f)*g + (11*c*d*e - 15*c^2*f)*h + 2*(3*d^2*f*g + (4*d^2*e - 5*c*d*f) 
*h)*x)*sqrt(f*x + e))/(d^4*x + c*d^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\text {Timed out} \] Input:

integrate((f*x+e)**(3/2)*(h*x+g)/(d*x+c)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x+e)^(3/2)*(h*x+g)/(d*x+c)^2,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 = 225, normalized size of antiderivative = 1.54 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\frac {{\left (3 \, d^{2} e f g - 3 \, c d f^{2} g + 2 \, d^{2} e^{2} h - 7 \, c d e f h + 5 \, c^{2} f^{2} h\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} d^{3}} - \frac {\sqrt {f x + e} d^{2} e f g - \sqrt {f x + e} c d f^{2} g - \sqrt {f x + e} c d e f h + \sqrt {f x + e} c^{2} f^{2} h}{{\left ({\left (f x + e\right )} d - d e + c f\right )} d^{3}} + \frac {2 \, {\left (3 \, \sqrt {f x + e} d^{4} f g + {\left (f x + e\right )}^{\frac {3}{2}} d^{4} h + 3 \, \sqrt {f x + e} d^{4} e h - 6 \, \sqrt {f x + e} c d^{3} f h\right )}}{3 \, d^{6}} \] Input:

integrate((f*x+e)^(3/2)*(h*x+g)/(d*x+c)^2,x, algorithm="giac")
 

Output:

(3*d^2*e*f*g - 3*c*d*f^2*g + 2*d^2*e^2*h - 7*c*d*e*f*h + 5*c^2*f^2*h)*arct 
an(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/(sqrt(-d^2*e + c*d*f)*d^3) - (sqr 
t(f*x + e)*d^2*e*f*g - sqrt(f*x + e)*c*d*f^2*g - sqrt(f*x + e)*c*d*e*f*h + 
 sqrt(f*x + e)*c^2*f^2*h)/(((f*x + e)*d - d*e + c*f)*d^3) + 2/3*(3*sqrt(f* 
x + e)*d^4*f*g + (f*x + e)^(3/2)*d^4*h + 3*sqrt(f*x + e)*d^4*e*h - 6*sqrt( 
f*x + e)*c*d^3*f*h)/d^6
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.20 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\frac {2\,h\,{\left (e+f\,x\right )}^{3/2}}{3\,d^2}-\frac {\sqrt {e+f\,x}\,\left (h\,c^2\,f^2-g\,c\,d\,f^2-e\,h\,c\,d\,f+e\,g\,d^2\,f\right )}{d^4\,\left (e+f\,x\right )-d^4\,e+c\,d^3\,f}-\sqrt {e+f\,x}\,\left (\frac {2\,e\,h-2\,f\,g}{d^2}-\frac {2\,h\,\left (2\,d^2\,e-2\,c\,d\,f\right )}{d^4}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,1{}\mathrm {i}}{\sqrt {d\,e-c\,f}}\right )\,\sqrt {d\,e-c\,f}\,\left (2\,d\,e\,h-5\,c\,f\,h+3\,d\,f\,g\right )\,1{}\mathrm {i}}{d^{7/2}} \] Input:

int(((e + f*x)^(3/2)*(g + h*x))/(c + d*x)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.60 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{(c+d x)^2} \, dx=\frac {15 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c^{2} f h -6 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c d e h -9 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c d f g +15 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c d f h x -6 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) d^{2} e h x -9 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) d^{2} f g x -15 \sqrt {f x +e}\, c^{2} d f h +11 \sqrt {f x +e}\, c \,d^{2} e h +9 \sqrt {f x +e}\, c \,d^{2} f g -10 \sqrt {f x +e}\, c \,d^{2} f h x -3 \sqrt {f x +e}\, d^{3} e g +8 \sqrt {f x +e}\, d^{3} e h x +6 \sqrt {f x +e}\, d^{3} f g x +2 \sqrt {f x +e}\, d^{3} f h \,x^{2}}{3 d^{4} \left (d x +c \right )} \] Input:

int((f*x+e)^(3/2)*(h*x+g)/(d*x+c)^2,x)
 

Output:

(15*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e 
)))*c**2*f*h - 6*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*s 
qrt(c*f - d*e)))*c*d*e*h - 9*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d 
)/(sqrt(d)*sqrt(c*f - d*e)))*c*d*f*g + 15*sqrt(d)*sqrt(c*f - d*e)*atan((sq 
rt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*c*d*f*h*x - 6*sqrt(d)*sqrt(c*f - 
 d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*d**2*e*h*x - 9*sqr 
t(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*d** 
2*f*g*x - 15*sqrt(e + f*x)*c**2*d*f*h + 11*sqrt(e + f*x)*c*d**2*e*h + 9*sq 
rt(e + f*x)*c*d**2*f*g - 10*sqrt(e + f*x)*c*d**2*f*h*x - 3*sqrt(e + f*x)*d 
**3*e*g + 8*sqrt(e + f*x)*d**3*e*h*x + 6*sqrt(e + f*x)*d**3*f*g*x + 2*sqrt 
(e + f*x)*d**3*f*h*x**2)/(3*d**4*(c + d*x))