\(\int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx\) [155]

Optimal result

Integrand size = 27, antiderivative size = 180 \[ \int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx=\frac {2 (d e-c f) (f g-e h)}{f (b e-a f)^2 \sqrt {e+f x}}-\frac {(b c-a d) (b g-a h) \sqrt {e+f x}}{b (b e-a f)^2 (a+b x)}-\frac {\left (a^2 d f h+b^2 (2 d e g-3 c f g+2 c e h)+a b (d f g-4 d e h+c f h)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{3/2} (b e-a f)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx=\frac {-a^2 d f h (e+f x)-b^2 (2 d e (-f g+e h) x+c f (e g+3 f g x-2 e h x))+a b \left (d \left (3 e f g-2 e^2 h+f^2 g x\right )+c f (-2 f g+3 e h+f h x)\right )}{b f (b e-a f)^2 (a+b x) \sqrt {e+f x}}+\frac {\left (a^2 d f h+b^2 (2 d e g-3 c f g+2 c e h)+a b (d f g-4 d e h+c f h)\right ) \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )}{b^{3/2} (-b e+a f)^{5/2}} \] Input:

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

Output:

(-(a^2*d*f*h*(e + f*x)) - b^2*(2*d*e*(-(f*g) + e*h)*x + c*f*(e*g + 3*f*g*x 
 - 2*e*h*x)) + a*b*(d*(3*e*f*g - 2*e^2*h + f^2*g*x) + c*f*(-2*f*g + 3*e*h 
+ f*h*x)))/(b*f*(b*e - a*f)^2*(a + b*x)*Sqrt[e + f*x]) + ((a^2*d*f*h + b^2 
*(2*d*e*g - 3*c*f*g + 2*c*e*h) + a*b*(d*f*g - 4*d*e*h + c*f*h))*ArcTan[(Sq 
rt[b]*Sqrt[e + f*x])/Sqrt[-(b*e) + a*f]])/(b^(3/2)*(-(b*e) + a*f)^(5/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {161, 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.

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

\(\Big \downarrow \) 161

\(\displaystyle \frac {\left (a^2 d f h+a b (c f h-4 d e h+d f g)+b^2 (2 c e h-3 c f g+2 d e g)\right ) \int \frac {1}{(a+b x) \sqrt {e+f x}}dx}{2 b (b e-a f)^2}-\frac {x \left (a^2 d f^2 h-a b f^2 (c h+d g)+b^2 (c f (3 f g-2 e h)-2 d e (f g-e h))\right )+a^2 d e f h+2 a b \left (-\frac {3}{2} e f (c h+d g)+c f^2 g+d e^2 h\right )+b^2 c e f g}{b f (a+b x) \sqrt {e+f x} (b e-a f)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\left (a^2 d f h+a b (c f h-4 d e h+d f g)+b^2 (2 c e h-3 c f g+2 d e g)\right ) \int \frac {1}{a+\frac {b (e+f x)}{f}-\frac {b e}{f}}d\sqrt {e+f x}}{b f (b e-a f)^2}-\frac {x \left (a^2 d f^2 h-a b f^2 (c h+d g)+b^2 (c f (3 f g-2 e h)-2 d e (f g-e h))\right )+a^2 d e f h+2 a b \left (-\frac {3}{2} e f (c h+d g)+c f^2 g+d e^2 h\right )+b^2 c e f g}{b f (a+b x) \sqrt {e+f x} (b e-a f)^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right ) \left (a^2 d f h+a b (c f h-4 d e h+d f g)+b^2 (2 c e h-3 c f g+2 d e g)\right )}{b^{3/2} (b e-a f)^{5/2}}-\frac {x \left (a^2 d f^2 h-a b f^2 (c h+d g)+b^2 (c f (3 f g-2 e h)-2 d e (f g-e h))\right )+a^2 d e f h+2 a b \left (-\frac {3}{2} e f (c h+d g)+c f^2 g+d e^2 h\right )+b^2 c e f g}{b f (a+b x) \sqrt {e+f x} (b e-a f)^2}\)

Input:

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

Output:

-((b^2*c*e*f*g + a^2*d*e*f*h + 2*a*b*(c*f^2*g + d*e^2*h - (3*e*f*(d*g + c* 
h))/2) + (a^2*d*f^2*h - a*b*f^2*(d*g + c*h) + b^2*(c*f*(3*f*g - 2*e*h) - 2 
*d*e*(f*g - e*h)))*x)/(b*f*(b*e - a*f)^2*(a + b*x)*Sqrt[e + f*x])) - ((a^2 
*d*f*h + b^2*(2*d*e*g - 3*c*f*g + 2*c*e*h) + a*b*(d*f*g - 4*d*e*h + c*f*h) 
)*ArcTanh[(Sqrt[b]*Sqrt[e + f*x])/Sqrt[b*e - a*f]])/(b^(3/2)*(b*e - a*f)^( 
5/2))
 

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 161

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_)) 
*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 
 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m + n + 2)) 
 + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1 
) - c*d*(f*g + e*h)*(m + 1) + d^2*e*g*(m + n + 2)))*x)/(b*d*(b*c - a*d)^2*( 
m + 1)*(n + 1)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Simp[(a^2*d^2*f* 
h*(2 + 3*n + 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*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
 d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b*c - a*d)^2*(m + 1)*( 
n + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h}, x] && LtQ[m, -1] && LtQ[n, -1]
 

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 = 0.45 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {\frac {2 f \left (-\frac {f \left (a^{2} d h -a b c h -a b d g +b^{2} c g \right ) \sqrt {f x +e}}{2 b \left (\left (f x +e \right ) b +a f -b e \right )}+\frac {\left (a^{2} d f h +a b c f h -4 a b d e h +a b d f g +2 b^{2} c e h -3 b^{2} c f g +2 b^{2} d e g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 b \sqrt {\left (a f -b e \right ) b}}\right )}{\left (a f -b e \right )^{2}}-\frac {2 \left (-c e f h +c g \,f^{2}+d \,e^{2} h -d e f g \right )}{\left (a f -b e \right )^{2} \sqrt {f x +e}}}{f}\)	\(209\)
default	\(\frac {\frac {2 f \left (-\frac {f \left (a^{2} d h -a b c h -a b d g +b^{2} c g \right ) \sqrt {f x +e}}{2 b \left (\left (f x +e \right ) b +a f -b e \right )}+\frac {\left (a^{2} d f h +a b c f h -4 a b d e h +a b d f g +2 b^{2} c e h -3 b^{2} c f g +2 b^{2} d e g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 b \sqrt {\left (a f -b e \right ) b}}\right )}{\left (a f -b e \right )^{2}}-\frac {2 \left (-c e f h +c g \,f^{2}+d \,e^{2} h -d e f g \right )}{\left (a f -b e \right )^{2} \sqrt {f x +e}}}{f}\)	\(209\)
pseudoelliptic	\(-\frac {-\left (\left (-3 c f g +2 e \left (c h +d g \right )\right ) b^{2}+a \left (f \left (c h +d g \right )-4 d e h \right ) b +a^{2} d f h \right ) \left (b x +a \right ) \sqrt {f x +e}\, f \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )+\sqrt {\left (a f -b e \right ) b}\, \left (\left (3 x c \,f^{2} g +\left (-2 d g x +c \left (-2 h x +g \right )\right ) e f +2 d \,e^{2} h x \right ) b^{2}-3 a \left (\frac {\left (d g x -2 c \left (-\frac {h x}{2}+g \right )\right ) f^{2}}{3}+e \left (c h +d g \right ) f -\frac {2 d \,e^{2} h}{3}\right ) b +a^{2} d f h \left (f x +e \right )\right )}{\sqrt {f x +e}\, \sqrt {\left (a f -b e \right ) b}\, b f \left (b x +a \right ) \left (a f -b e \right )^{2}}\)	\(234\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (166) = 332\).

Time = 0.14 (sec) , antiderivative size = 1437, normalized size of antiderivative = 7.98 \[ \int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(sqrt(b^2*e - a*b*f)*(((2*b^3*d*e*f^2 - (3*b^3*c - a*b^2*d)*f^3)*g + 
(2*(b^3*c - 2*a*b^2*d)*e*f^2 + (a*b^2*c + a^2*b*d)*f^3)*h)*x^2 + (2*a*b^2* 
d*e^2*f - (3*a*b^2*c - a^2*b*d)*e*f^2)*g + (2*(a*b^2*c - 2*a^2*b*d)*e^2*f 
+ (a^2*b*c + a^3*d)*e*f^2)*h + ((2*b^3*d*e^2*f - 3*(b^3*c - a*b^2*d)*e*f^2 
 - (3*a*b^2*c - a^2*b*d)*f^3)*g + (2*(b^3*c - 2*a*b^2*d)*e^2*f + 3*(a*b^2* 
c - a^2*b*d)*e*f^2 + (a^2*b*c + a^3*d)*f^3)*h)*x)*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*((2*a^2*b^2*c*f^3 - 
 (b^4*c - 3*a*b^3*d)*e^2*f - (a*b^3*c + 3*a^2*b^2*d)*e*f^2)*g - (2*a*b^3*d 
*e^3 - (3*a*b^3*c + a^2*b^2*d)*e^2*f + (3*a^2*b^2*c - a^3*b*d)*e*f^2)*h + 
((2*b^4*d*e^2*f - (3*b^4*c + a*b^3*d)*e*f^2 + (3*a*b^3*c - a^2*b^2*d)*f^3) 
*g - (2*b^4*d*e^3 - 2*(b^4*c + a*b^3*d)*e^2*f + (a*b^3*c + a^2*b^2*d)*e*f^ 
2 + (a^2*b^2*c - a^3*b*d)*f^3)*h)*x)*sqrt(f*x + e))/(a*b^5*e^4*f - 3*a^2*b 
^4*e^3*f^2 + 3*a^3*b^3*e^2*f^3 - a^4*b^2*e*f^4 + (b^6*e^3*f^2 - 3*a*b^5*e^ 
2*f^3 + 3*a^2*b^4*e*f^4 - a^3*b^3*f^5)*x^2 + (b^6*e^4*f - 2*a*b^5*e^3*f^2 
+ 2*a^3*b^3*e*f^4 - a^4*b^2*f^5)*x), (sqrt(-b^2*e + a*b*f)*(((2*b^3*d*e*f^ 
2 - (3*b^3*c - a*b^2*d)*f^3)*g + (2*(b^3*c - 2*a*b^2*d)*e*f^2 + (a*b^2*c + 
 a^2*b*d)*f^3)*h)*x^2 + (2*a*b^2*d*e^2*f - (3*a*b^2*c - a^2*b*d)*e*f^2)*g 
+ (2*(a*b^2*c - 2*a^2*b*d)*e^2*f + (a^2*b*c + a^3*d)*e*f^2)*h + ((2*b^3*d* 
e^2*f - 3*(b^3*c - a*b^2*d)*e*f^2 - (3*a*b^2*c - a^2*b*d)*f^3)*g + (2*(b^3 
*c - 2*a*b^2*d)*e^2*f + 3*(a*b^2*c - a^2*b*d)*e*f^2 + (a^2*b*c + a^3*d)...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((d*x+c)*(h*x+g)/(b*x+a)^2/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (166) = 332\).

Time = 0.14 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.00 \[ \int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx=\frac {{\left (2 \, b^{2} d e g - 3 \, b^{2} c f g + a b d f g + 2 \, b^{2} c e h - 4 \, a b d e h + a b c f h + a^{2} d f h\right )} \arctan \left (\frac {\sqrt {f x + e} b}{\sqrt {-b^{2} e + a b f}}\right )}{{\left (b^{3} e^{2} - 2 \, a b^{2} e f + a^{2} b f^{2}\right )} \sqrt {-b^{2} e + a b f}} + \frac {2 \, {\left (f x + e\right )} b^{2} d e f g - 2 \, b^{2} d e^{2} f g - 3 \, {\left (f x + e\right )} b^{2} c f^{2} g + {\left (f x + e\right )} a b d f^{2} g + 2 \, b^{2} c e f^{2} g + 2 \, a b d e f^{2} g - 2 \, a b c f^{3} g - 2 \, {\left (f x + e\right )} b^{2} d e^{2} h + 2 \, b^{2} d e^{3} h + 2 \, {\left (f x + e\right )} b^{2} c e f h - 2 \, b^{2} c e^{2} f h - 2 \, a b d e^{2} f h + {\left (f x + e\right )} a b c f^{2} h - {\left (f x + e\right )} a^{2} d f^{2} h + 2 \, a b c e f^{2} h}{{\left (b^{3} e^{2} f - 2 \, a b^{2} e f^{2} + a^{2} b f^{3}\right )} {\left ({\left (f x + e\right )}^{\frac {3}{2}} b - \sqrt {f x + e} b e + \sqrt {f x + e} a f\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.49 \[ \int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {e+f\,x}\,\left (a^2\,b\,f^2-2\,a\,b^2\,e\,f+b^3\,e^2\right )}{\sqrt {b}\,{\left (a\,f-b\,e\right )}^{5/2}}\right )\,\left (2\,b^2\,c\,e\,h-3\,b^2\,c\,f\,g+2\,b^2\,d\,e\,g+a^2\,d\,f\,h+a\,b\,c\,f\,h-4\,a\,b\,d\,e\,h+a\,b\,d\,f\,g\right )}{b^{3/2}\,{\left (a\,f-b\,e\right )}^{5/2}}-\frac {\frac {2\,\left (c\,f^2\,g+d\,e^2\,h-c\,e\,f\,h-d\,e\,f\,g\right )}{a\,f-b\,e}-\frac {\left (e+f\,x\right )\,\left (a\,b\,c\,f^2\,h-a^2\,d\,f^2\,h-2\,b^2\,d\,e^2\,h-3\,b^2\,c\,f^2\,g+a\,b\,d\,f^2\,g+2\,b^2\,c\,e\,f\,h+2\,b^2\,d\,e\,f\,g\right )}{b\,{\left (a\,f-b\,e\right )}^2}}{\sqrt {e+f\,x}\,\left (a\,f^2-b\,e\,f\right )+b\,f\,{\left (e+f\,x\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 1089, normalized size of antiderivative = 6.05 \[ \int \frac {(c+d x) (g+h x)}{(a+b x)^2 (e+f x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(b)*sqrt(e + f*x)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqr 
t(a*f - b*e)))*a**3*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**2*b*c*f**2*h - 4*sqrt(b)*s 
qrt(e + f*x)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b* 
e)))*a**2*b*d*e*f*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**2*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)))*a**2*b 
*d*f**2*h*x + 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*b**2*c*e*f*h - 3*sqrt(b)*sqrt(e + f*x)*sqr 
t(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c*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)))*a*b**2*c*f**2*h*x + 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*b**2*d*e*f*g - 
4*sqrt(b)*sqrt(e + f*x)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sq 
rt(a*f - b*e)))*a*b**2*d*e*f*h*x + sqrt(b)*sqrt(e + f*x)*sqrt(a*f - b*e)*a 
tan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*d*f**2*g*x + 2*sqr 
t(b)*sqrt(e + f*x)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a* 
f - b*e)))*b**3*c*e*f*h*x - 3*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**3*c*f**2*g*x + 2*sqrt(b)*sq 
rt(e + f*x)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - ...