\(\int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx\) [170]

Optimal result

Integrand size = 27, antiderivative size = 123 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=-\frac {2 (b e-a f) (f g-e h)}{f^2 (d e-c f) \sqrt {e+f x}}+\frac {2 b h \sqrt {e+f x}}{d f^2}+\frac {2 (b c-a d) (d g-c h) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (d e-c f)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=\frac {2 (a d f (-f g+e h)+b c f h (e+f x)+b d e (-2 e h+f (g-h x)))}{d f^2 (-d e+c f) \sqrt {e+f x}}-\frac {2 (-b c+a d) (d g-c h) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{3/2} (-d e+c f)^{3/2}} \] Input:

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

Output:

(2*(a*d*f*(-(f*g) + e*h) + b*c*f*h*(e + f*x) + b*d*e*(-2*e*h + f*(g - h*x) 
)))/(d*f^2*(-(d*e) + c*f)*Sqrt[e + f*x]) - (2*(-(b*c) + a*d)*(d*g - c*h)*A 
rcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(d^(3/2)*(-(d*e) + c*f) 
^(3/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {163, 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.

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

\(\Big \downarrow \) 163

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*(a*d*f^2*g + 2*b*d*e^2*h - e*f*(b*d*g + b*c*h + a*d*h) + b*f*(d*e - c*f 
)*h*x))/(d*f^2*(d*e - c*f)*Sqrt[e + f*x]) + (2*(b*c - a*d)*(d*g - c*h)*Arc 
Tanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(3/2)*(d*e - c*f)^(3/2))
 

Defintions of rubi rules used

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 163

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n 
+ 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* 
(m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + 
d*x)^(n + 1), x] - Simp[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f 
*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c* 
d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2* 
d*(b*c - a*d)*(m + 1)*(m + n + 3))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 
1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
 

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 = 0.40 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {2 h b \sqrt {f x +e}}{d}-\frac {2 \left (-a f h e +g \,f^{2} a +b \,e^{2} h -b f g e \right )}{\left (c f -d e \right ) \sqrt {f x +e}}+\frac {2 f^{2} \left (a c d h -a \,d^{2} g -b \,c^{2} h +b c d g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) d \sqrt {\left (c f -d e \right ) d}}}{f^{2}}\)	\(140\)
default	\(\frac {\frac {2 h b \sqrt {f x +e}}{d}-\frac {2 \left (-a f h e +g \,f^{2} a +b \,e^{2} h -b f g e \right )}{\left (c f -d e \right ) \sqrt {f x +e}}+\frac {2 f^{2} \left (a c d h -a \,d^{2} g -b \,c^{2} h +b c d g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) d \sqrt {\left (c f -d e \right ) d}}}{f^{2}}\)	\(140\)
pseudoelliptic	\(\frac {2 \sqrt {f x +e}\, f^{2} \left (c h -d g \right ) \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+2 \left (\left (-g \,f^{2} a +\left (\left (-h x +g \right ) b +a h \right ) e f -2 b \,e^{2} h \right ) d +b c f h \left (f x +e \right )\right ) \sqrt {\left (c f -d e \right ) d}}{f^{2} d \left (c f -d e \right ) \sqrt {f x +e}\, \sqrt {\left (c f -d e \right ) d}}\)	\(144\)
risch	\(\frac {2 b h \sqrt {f x +e}}{d \,f^{2}}+\frac {\frac {2 \left (a f h e -g \,f^{2} a -b \,e^{2} h +b f g e \right ) d}{\left (c f -d e \right ) \sqrt {f x +e}}+\frac {2 f^{2} \left (a c d h -a \,d^{2} g -b \,c^{2} h +b c d g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}}{f^{2} d}\)	\(145\)

Input:

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

Output:

2/f^2*(h*b/d*(f*x+e)^(1/2)-(-a*e*f*h+a*f^2*g+b*e^2*h-b*e*f*g)/(c*f-d*e)/(f 
*x+e)^(1/2)+f^2*(a*c*d*h-a*d^2*g-b*c^2*h+b*c*d*g)/(c*f-d*e)/d/((c*f-d*e)*d 
)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (109) = 218\).

Time = 0.09 (sec) , antiderivative size = 657, normalized size of antiderivative = 5.34 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=\left [-\frac {{\left ({\left (b c d - a d^{2}\right )} e f^{2} g - {\left (b c^{2} - a c d\right )} e f^{2} h + {\left ({\left (b c d - a d^{2}\right )} f^{3} g - {\left (b c^{2} - a c d\right )} f^{3} h\right )} x\right )} \sqrt {d^{2} e - c d f} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {d^{2} e - c d f} \sqrt {f x + e}}{d x + c}\right ) - 2 \, {\left ({\left (b d^{3} e^{2} f - 2 \, b c d^{2} e f^{2} + b c^{2} d f^{3}\right )} h x - {\left (b d^{3} e^{2} f + a c d^{2} f^{3} - {\left (b c d^{2} + a d^{3}\right )} e f^{2}\right )} g + {\left (2 \, b d^{3} e^{3} - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{2} f + {\left (b c^{2} d + a c d^{2}\right )} e f^{2}\right )} h\right )} \sqrt {f x + e}}{d^{4} e^{3} f^{2} - 2 \, c d^{3} e^{2} f^{3} + c^{2} d^{2} e f^{4} + {\left (d^{4} e^{2} f^{3} - 2 \, c d^{3} e f^{4} + c^{2} d^{2} f^{5}\right )} x}, -\frac {2 \, {\left ({\left ({\left (b c d - a d^{2}\right )} e f^{2} g - {\left (b c^{2} - a c d\right )} e f^{2} h + {\left ({\left (b c d - a d^{2}\right )} f^{3} g - {\left (b c^{2} - a c d\right )} f^{3} h\right )} x\right )} \sqrt {-d^{2} e + c d f} \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) - {\left ({\left (b d^{3} e^{2} f - 2 \, b c d^{2} e f^{2} + b c^{2} d f^{3}\right )} h x - {\left (b d^{3} e^{2} f + a c d^{2} f^{3} - {\left (b c d^{2} + a d^{3}\right )} e f^{2}\right )} g + {\left (2 \, b d^{3} e^{3} - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{2} f + {\left (b c^{2} d + a c d^{2}\right )} e f^{2}\right )} h\right )} \sqrt {f x + e}\right )}}{d^{4} e^{3} f^{2} - 2 \, c d^{3} e^{2} f^{3} + c^{2} d^{2} e f^{4} + {\left (d^{4} e^{2} f^{3} - 2 \, c d^{3} e f^{4} + c^{2} d^{2} f^{5}\right )} x}\right ] \] Input:

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

Output:

[-(((b*c*d - a*d^2)*e*f^2*g - (b*c^2 - a*c*d)*e*f^2*h + ((b*c*d - a*d^2)*f 
^3*g - (b*c^2 - a*c*d)*f^3*h)*x)*sqrt(d^2*e - c*d*f)*log((d*f*x + 2*d*e - 
c*f - 2*sqrt(d^2*e - c*d*f)*sqrt(f*x + e))/(d*x + c)) - 2*((b*d^3*e^2*f - 
2*b*c*d^2*e*f^2 + b*c^2*d*f^3)*h*x - (b*d^3*e^2*f + a*c*d^2*f^3 - (b*c*d^2 
 + a*d^3)*e*f^2)*g + (2*b*d^3*e^3 - (3*b*c*d^2 + a*d^3)*e^2*f + (b*c^2*d + 
 a*c*d^2)*e*f^2)*h)*sqrt(f*x + e))/(d^4*e^3*f^2 - 2*c*d^3*e^2*f^3 + c^2*d^ 
2*e*f^4 + (d^4*e^2*f^3 - 2*c*d^3*e*f^4 + c^2*d^2*f^5)*x), -2*(((b*c*d - a* 
d^2)*e*f^2*g - (b*c^2 - a*c*d)*e*f^2*h + ((b*c*d - a*d^2)*f^3*g - (b*c^2 - 
 a*c*d)*f^3*h)*x)*sqrt(-d^2*e + c*d*f)*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f* 
x + e)/(d*f*x + d*e)) - ((b*d^3*e^2*f - 2*b*c*d^2*e*f^2 + b*c^2*d*f^3)*h*x 
 - (b*d^3*e^2*f + a*c*d^2*f^3 - (b*c*d^2 + a*d^3)*e*f^2)*g + (2*b*d^3*e^3 
- (3*b*c*d^2 + a*d^3)*e^2*f + (b*c^2*d + a*c*d^2)*e*f^2)*h)*sqrt(f*x + e)) 
/(d^4*e^3*f^2 - 2*c*d^3*e^2*f^3 + c^2*d^2*e*f^4 + (d^4*e^2*f^3 - 2*c*d^3*e 
*f^4 + c^2*d^2*f^5)*x)]
 

Sympy [A] (verification not implemented)

Time = 8.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b h \sqrt {e + f x}}{d f} + \frac {\left (a f - b e\right ) \left (e h - f g\right )}{f \sqrt {e + f x} \left (c f - d e\right )} + \frac {f \left (a d - b c\right ) \left (c h - d g\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{2} \sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b h x^{2}}{2 d} + \frac {x \left (a d h - b c h + b d g\right )}{d^{2}} - \frac {\left (a d - b c\right ) \left (c h - d g\right ) \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}}}{e^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(b*h*sqrt(e + f*x)/(d*f) + (a*f - b*e)*(e*h - f*g)/(f*sqrt(e 
+ f*x)*(c*f - d*e)) + f*(a*d - b*c)*(c*h - d*g)*atan(sqrt(e + f*x)/sqrt((c 
*f - d*e)/d))/(d**2*sqrt((c*f - d*e)/d)*(c*f - d*e)))/f, Ne(f, 0)), ((b*h* 
x**2/(2*d) + x*(a*d*h - b*c*h + b*d*g)/d**2 - (a*d - b*c)*(c*h - d*g)*Piec 
ewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True))/d**2)/e**(3/2), True))
 

Maxima [F(-2)]

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

integrate((b*x+a)*(h*x+g)/(d*x+c)/(f*x+e)^(3/2),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 = 143, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=-\frac {2 \, {\left (b c d g - a d^{2} g - b c^{2} h + a c d h\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{2} e - c d f\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (b e f g - a f^{2} g - b e^{2} h + a e f h\right )}}{{\left (d e f^{2} - c f^{3}\right )} \sqrt {f x + e}} + \frac {2 \, \sqrt {f x + e} b h}{d f^{2}} \] Input:

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

Output:

-2*(b*c*d*g - a*d^2*g - b*c^2*h + a*c*d*h)*arctan(sqrt(f*x + e)*d/sqrt(-d^ 
2*e + c*d*f))/((d^2*e - c*d*f)*sqrt(-d^2*e + c*d*f)) - 2*(b*e*f*g - a*f^2* 
g - b*e^2*h + a*e*f*h)/((d*e*f^2 - c*f^3)*sqrt(f*x + e)) + 2*sqrt(f*x + e) 
*b*h/(d*f^2)
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.48 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=\frac {2\,b\,h\,\sqrt {e+f\,x}}{d\,f^2}-\frac {2\,\left (a\,d\,f^2\,g+b\,d\,e^2\,h-a\,d\,e\,f\,h-b\,d\,e\,f\,g\right )}{d\,f^2\,\sqrt {e+f\,x}\,\left (c\,f-d\,e\right )}+\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {e+f\,x}\,\left (d^2\,e-c\,d\,f\right )\,\left (a\,d-b\,c\right )\,\left (c\,h-d\,g\right )}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{3/2}\,\left (2\,a\,d^2\,g+2\,b\,c^2\,h-2\,a\,c\,d\,h-2\,b\,c\,d\,g\right )}\right )\,\left (a\,d-b\,c\right )\,\left (c\,h-d\,g\right )}{d^{3/2}\,{\left (c\,f-d\,e\right )}^{3/2}} \] Input:

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

Output:

(2*b*h*(e + f*x)^(1/2))/(d*f^2) - (2*(a*d*f^2*g + b*d*e^2*h - a*d*e*f*h - 
b*d*e*f*g))/(d*f^2*(e + f*x)^(1/2)*(c*f - d*e)) + (2*atan((2*(e + f*x)^(1/ 
2)*(d^2*e - c*d*f)*(a*d - b*c)*(c*h - d*g))/(d^(1/2)*(c*f - d*e)^(3/2)*(2* 
a*d^2*g + 2*b*c^2*h - 2*a*c*d*h - 2*b*c*d*g)))*(a*d - b*c)*(c*h - d*g))/(d 
^(3/2)*(c*f - d*e)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.03 \[ \int \frac {(a+b x) (g+h x)}{(c+d x) (e+f x)^{3/2}} \, dx=\frac {2 \sqrt {d}\, \sqrt {f x +e}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) a c d \,f^{2} h -2 \sqrt {d}\, \sqrt {f x +e}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) a \,d^{2} f^{2} g -2 \sqrt {d}\, \sqrt {f x +e}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) b \,c^{2} f^{2} h +2 \sqrt {d}\, \sqrt {f x +e}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) b c d \,f^{2} g +2 a c \,d^{2} e \,f^{2} h -2 a c \,d^{2} f^{3} g -2 a \,d^{3} e^{2} f h +2 a \,d^{3} e \,f^{2} g +2 b \,c^{2} d e \,f^{2} h +2 b \,c^{2} d \,f^{3} h x -6 b c \,d^{2} e^{2} f h +2 b c \,d^{2} e \,f^{2} g -4 b c \,d^{2} e \,f^{2} h x +4 b \,d^{3} e^{3} h -2 b \,d^{3} e^{2} f g +2 b \,d^{3} e^{2} f h x}{\sqrt {f x +e}\, d^{2} f^{2} \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right )} \] Input:

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

Output:

(2*(sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)* 
sqrt(c*f - d*e)))*a*c*d*f**2*h - sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*ata 
n((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*d**2*f**2*g - sqrt(d)*sqr 
t(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e) 
))*b*c**2*f**2*h + sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f* 
x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c*d*f**2*g + a*c*d**2*e*f**2*h - a*c*d* 
*2*f**3*g - a*d**3*e**2*f*h + a*d**3*e*f**2*g + b*c**2*d*e*f**2*h + b*c**2 
*d*f**3*h*x - 3*b*c*d**2*e**2*f*h + b*c*d**2*e*f**2*g - 2*b*c*d**2*e*f**2* 
h*x + 2*b*d**3*e**3*h - b*d**3*e**2*f*g + b*d**3*e**2*f*h*x))/(sqrt(e + f* 
x)*d**2*f**2*(c**2*f**2 - 2*c*d*e*f + d**2*e**2))