\(\int \frac {(c+d x) (g+h x)}{(a+b x) (e+f x)^{3/2}} \, dx\) [154]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x) (g+h x)}{(a+b x) (e+f x)^{3/2}} \, dx=\frac {2 (b c f (-f g+e h)+a d f h (e+f x)+b d e (-2 e h+f (g-h x)))}{b f^2 (-b e+a f) \sqrt {e+f x}}-\frac {2 (b c-a d) (b g-a h) \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )}{b^{3/2} (-b e+a f)^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (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.

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

\(\Big \downarrow \) 163

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*(b*c*f^2*g + 2*b*d*e^2*h - e*f*(b*d*g + b*c*h + a*d*h) + d*f*(b*e - a*f 
)*h*x))/(b*f^2*(b*e - a*f)*Sqrt[e + f*x]) - (2*(b*c - a*d)*(b*g - a*h)*Arc 
Tanh[(Sqrt[b]*Sqrt[e + f*x])/Sqrt[b*e - a*f]])/(b^(3/2)*(b*e - a*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 = 141, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {2 d h \sqrt {f x +e}}{b}-\frac {2 \left (-c e f h +c g \,f^{2}+d \,e^{2} h -d e f g \right )}{\left (a f -b e \right ) \sqrt {f x +e}}-\frac {2 f^{2} \left (a^{2} d h -a b c h -a b d g +b^{2} c g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{\left (a f -b e \right ) b \sqrt {\left (a f -b e \right ) b}}}{f^{2}}\)	\(141\)
default	\(\frac {\frac {2 d h \sqrt {f x +e}}{b}-\frac {2 \left (-c e f h +c g \,f^{2}+d \,e^{2} h -d e f g \right )}{\left (a f -b e \right ) \sqrt {f x +e}}-\frac {2 f^{2} \left (a^{2} d h -a b c h -a b d g +b^{2} c g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{\left (a f -b e \right ) b \sqrt {\left (a f -b e \right ) b}}}{f^{2}}\)	\(141\)
pseudoelliptic	\(\frac {-2 \sqrt {f x +e}\, f^{2} \left (a h -b g \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )+2 \sqrt {\left (a f -b e \right ) b}\, \left (\left (-c g \,f^{2}+e \left (\left (-h x +g \right ) d +c h \right ) f -2 d \,e^{2} h \right ) b +a d f h \left (f x +e \right )\right )}{f^{2} b \left (a f -b e \right ) \sqrt {f x +e}\, \sqrt {\left (a f -b e \right ) b}}\)	\(145\)
risch	\(\frac {2 d h \sqrt {f x +e}}{b \,f^{2}}-\frac {2 \left (-\frac {\left (c e f h -c g \,f^{2}-d \,e^{2} h +d e f g \right ) b}{\left (a f -b e \right ) \sqrt {f x +e}}+\frac {f^{2} \left (a^{2} d h -a b c h -a b d g +b^{2} c g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{\left (a f -b e \right ) \sqrt {\left (a f -b e \right ) b}}\right )}{b \,f^{2}}\)	\(146\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (109) = 218\).

Time = 0.10 (sec) , antiderivative size = 656, normalized size of antiderivative = 5.33 \[ \int \frac {(c+d x) (g+h x)}{(a+b x) (e+f x)^{3/2}} \, dx=\left [-\frac {{\left ({\left (b^{2} c - a b d\right )} e f^{2} g - {\left (a b c - a^{2} d\right )} e f^{2} h + {\left ({\left (b^{2} c - a b d\right )} f^{3} g - {\left (a b c - a^{2} d\right )} f^{3} h\right )} x\right )} \sqrt {b^{2} e - a b f} \log \left (\frac {b f x + 2 \, b e - a f + 2 \, \sqrt {b^{2} e - a b f} \sqrt {f x + e}}{b x + a}\right ) - 2 \, {\left ({\left (b^{3} d e^{2} f - 2 \, a b^{2} d e f^{2} + a^{2} b d f^{3}\right )} h x - {\left (b^{3} d e^{2} f + a b^{2} c f^{3} - {\left (b^{3} c + a b^{2} d\right )} e f^{2}\right )} g + {\left (2 \, b^{3} d e^{3} - {\left (b^{3} c + 3 \, a b^{2} d\right )} e^{2} f + {\left (a b^{2} c + a^{2} b d\right )} e f^{2}\right )} h\right )} \sqrt {f x + e}}{b^{4} e^{3} f^{2} - 2 \, a b^{3} e^{2} f^{3} + a^{2} b^{2} e f^{4} + {\left (b^{4} e^{2} f^{3} - 2 \, a b^{3} e f^{4} + a^{2} b^{2} f^{5}\right )} x}, \frac {2 \, {\left ({\left ({\left (b^{2} c - a b d\right )} e f^{2} g - {\left (a b c - a^{2} d\right )} e f^{2} h + {\left ({\left (b^{2} c - a b d\right )} f^{3} g - {\left (a b c - a^{2} d\right )} f^{3} h\right )} x\right )} \sqrt {-b^{2} e + a b f} \arctan \left (\frac {\sqrt {-b^{2} e + a b f} \sqrt {f x + e}}{b f x + b e}\right ) + {\left ({\left (b^{3} d e^{2} f - 2 \, a b^{2} d e f^{2} + a^{2} b d f^{3}\right )} h x - {\left (b^{3} d e^{2} f + a b^{2} c f^{3} - {\left (b^{3} c + a b^{2} d\right )} e f^{2}\right )} g + {\left (2 \, b^{3} d e^{3} - {\left (b^{3} c + 3 \, a b^{2} d\right )} e^{2} f + {\left (a b^{2} c + a^{2} b d\right )} e f^{2}\right )} h\right )} \sqrt {f x + e}\right )}}{b^{4} e^{3} f^{2} - 2 \, a b^{3} e^{2} f^{3} + a^{2} b^{2} e f^{4} + {\left (b^{4} e^{2} f^{3} - 2 \, a b^{3} e f^{4} + a^{2} b^{2} f^{5}\right )} x}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 8.19 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.42 \[ \int \frac {(c+d x) (g+h x)}{(a+b x) (e+f x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {\left (c f - d e\right ) \left (e h - f g\right )}{f \sqrt {e + f x} \left (a f - b e\right )} + \frac {d h \sqrt {e + f x}}{b f} - \frac {f \left (a d - b c\right ) \left (a h - b g\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {a f - b e}{b}}} \right )}}{b^{2} \sqrt {\frac {a f - b e}{b}} \left (a f - b e\right )}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {d h x^{2}}{2 b} + \frac {x \left (- a d h + b c h + b d g\right )}{b^{2}} + \frac {\left (a d - b c\right ) \left (a h - b g\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{2}}}{e^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [F(-2)]

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

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.48 \[ \int \frac {(c+d x) (g+h x)}{(a+b x) (e+f x)^{3/2}} \, dx=\frac {2\,d\,h\,\sqrt {e+f\,x}}{b\,f^2}-\frac {2\,\left (b\,c\,f^2\,g+b\,d\,e^2\,h-b\,c\,e\,f\,h-b\,d\,e\,f\,g\right )}{b\,f^2\,\sqrt {e+f\,x}\,\left (a\,f-b\,e\right )}+\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {e+f\,x}\,\left (b^2\,e-a\,b\,f\right )\,\left (a\,d-b\,c\right )\,\left (a\,h-b\,g\right )}{\sqrt {b}\,{\left (a\,f-b\,e\right )}^{3/2}\,\left (2\,b^2\,c\,g+2\,a^2\,d\,h-2\,a\,b\,c\,h-2\,a\,b\,d\,g\right )}\right )\,\left (a\,d-b\,c\right )\,\left (a\,h-b\,g\right )}{b^{3/2}\,{\left (a\,f-b\,e\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.03 \[ \int \frac {(c+d x) (g+h x)}{(a+b x) (e+f x)^{3/2}} \, dx=\frac {-2 \sqrt {b}\, \sqrt {f x +e}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a^{2} d \,f^{2} h +2 \sqrt {b}\, \sqrt {f x +e}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a b c \,f^{2} h +2 \sqrt {b}\, \sqrt {f x +e}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a b d \,f^{2} g -2 \sqrt {b}\, \sqrt {f x +e}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) b^{2} c \,f^{2} g +2 a^{2} b d e \,f^{2} h +2 a^{2} b d \,f^{3} h x +2 a \,b^{2} c e \,f^{2} h -2 a \,b^{2} c \,f^{3} g -6 a \,b^{2} d \,e^{2} f h +2 a \,b^{2} d e \,f^{2} g -4 a \,b^{2} d e \,f^{2} h x -2 b^{3} c \,e^{2} f h +2 b^{3} c e \,f^{2} g +4 b^{3} d \,e^{3} h -2 b^{3} d \,e^{2} f g +2 b^{3} d \,e^{2} f h x}{\sqrt {f x +e}\, b^{2} f^{2} \left (a^{2} f^{2}-2 a b e f +b^{2} e^{2}\right )} \] Input:

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

Output:

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