\(\int \frac {1}{\sqrt {e+f x} (a c+(b c+a d) x+b d x^2)^2} \, dx\) [399]

Optimal result

Integrand size = 31, antiderivative size = 256 \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {\sqrt {e+f x} \left (a b d^2 e-a^2 d^2 f+b^2 c (d e-c f)+b d (2 b d e-(b c+a d) f) x\right )}{(b c-a d)^2 (b e-a f) (d e-c f) \left (a c+(b c+a d) x+b d x^2\right )}+\frac {b^{3/2} (4 b d e+b c f-5 a d f) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{(b c-a d)^3 (b e-a f)^{3/2}}-\frac {d^{3/2} (4 b d e-5 b c f+a d f) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(b c-a d)^3 (d e-c f)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {\sqrt {e+f x} \left (-a^2 d^2 f+a b d^2 (e-f x)+b^2 \left (-c^2 f+2 d^2 e x+c d (e-f x)\right )\right )}{(b c-a d)^2 (b e-a f) (-d e+c f) (a+b x) (c+d x)}+\frac {b^{3/2} (4 b d e+b c f-5 a d f) \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )}{(b c-a d)^3 (-b e+a f)^{3/2}}+\frac {d^{3/2} (4 b d e-5 b c f+a d f) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-b c+a d)^3 (-d e+c f)^{3/2}} \] Input:

Integrate[1/(Sqrt[e + f*x]*(a*c + (b*c + a*d)*x + b*d*x^2)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {1165, 27, 1197, 25, 27, 1480, 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.

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

\(\Big \downarrow \) 1165

\(\displaystyle -\frac {\int \frac {4 b^2 d^2 e^2-3 b d (b c+a d) f e-(b c+a d)^2 f^2+6 a b c d f^2+b d f (2 b d e-(b c+a d) f) x}{2 \sqrt {e+f x} \left (b d x^2+(b c+a d) x+a c\right )}dx}{(b c-a d)^2 (b e-a f) (d e-c f)}-\frac {\sqrt {e+f x} \left (-a^2 d^2 f+b d x (2 b d e-f (a d+b c))+a b d^2 e+b^2 c (d e-c f)\right )}{(b c-a d)^2 (b e-a f) (d e-c f) \left (x (a d+b c)+a c+b d x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {4 b^2 d^2 e^2-3 b d (b c+a d) f e-(b c+a d)^2 f^2+6 a b c d f^2+b d f (2 b d e-(b c+a d) f) x}{\sqrt {e+f x} \left (b d x^2+(b c+a d) x+a c\right )}dx}{2 (b c-a d)^2 (b e-a f) (d e-c f)}-\frac {\sqrt {e+f x} \left (-a^2 d^2 f+b d x (2 b d e-f (a d+b c))+a b d^2 e+b^2 c (d e-c f)\right )}{(b c-a d)^2 (b e-a f) (d e-c f) \left (x (a d+b c)+a c+b d x^2\right )}\)

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

\(\displaystyle \frac {f \left (\frac {b d^2 (b e-a f) (a d f-5 b c f+4 b d e) \int \frac {1}{b d (e+f x)-b (d e-c f)}d\sqrt {e+f x}}{f (b c-a d)}-\frac {b^2 d (d e-c f) (-5 a d f+b c f+4 b d e) \int \frac {1}{b d (e+f x)-d (b e-a f)}d\sqrt {e+f x}}{f (b c-a d)}\right )}{(b c-a d)^2 (b e-a f) (d e-c f)}-\frac {\sqrt {e+f x} \left (-a^2 d^2 f+b d x (2 b d e-f (a d+b c))+a b d^2 e+b^2 c (d e-c f)\right )}{(b c-a d)^2 (b e-a f) (d e-c f) \left (x (a d+b c)+a c+b d x^2\right )}\)

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[e + f*x]*(a*c + (b*c + a*d)*x + b*d*x^2)^2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(2 f^{3} \left (\frac {d^{2} \left (\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 \left (c f -d e \right ) \left (d \left (f x +e \right )+c f -d e \right )}+\frac {\left (a d f -5 b c f +4 b d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}\right )}{f^{3} \left (a d -b c \right )^{3}}+\frac {b^{2} \left (\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 \left (a f -b e \right ) \left (b \left (f x +e \right )+a f -b e \right )}+\frac {\left (5 a d f -b c f -4 b d e \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 \left (a f -b e \right ) \sqrt {\left (a f -b e \right ) b}}\right )}{f^{3} \left (a d -b c \right )^{3}}\right )\)	\(256\)
default	\(2 f^{3} \left (\frac {d^{2} \left (\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 \left (c f -d e \right ) \left (d \left (f x +e \right )+c f -d e \right )}+\frac {\left (a d f -5 b c f +4 b d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}\right )}{f^{3} \left (a d -b c \right )^{3}}+\frac {b^{2} \left (\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 \left (a f -b e \right ) \left (b \left (f x +e \right )+a f -b e \right )}+\frac {\left (5 a d f -b c f -4 b d e \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 \left (a f -b e \right ) \sqrt {\left (a f -b e \right ) b}}\right )}{f^{3} \left (a d -b c \right )^{3}}\right )\)	\(256\)
pseudoelliptic	\(\frac {5 \left (\frac {\left (-c f -4 d e \right ) b}{5}+a d f \right ) b^{2} \left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}\, \left (d x +c \right ) \left (b x +a \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )+\sqrt {\left (a f -b e \right ) b}\, \left (d^{2} \left (a f -b e \right ) \left (d x +c \right ) \left (\left (-5 c f +4 d e \right ) b +a d f \right ) \left (b x +a \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\sqrt {f x +e}\, \left (a d -b c \right ) \sqrt {\left (c f -d e \right ) d}\, \left (\left (-2 d^{2} e x -c \left (-f x +e \right ) d +c^{2} f \right ) b^{2}-a \,d^{2} \left (-f x +e \right ) b +a^{2} d^{2} f \right )\right )}{\sqrt {\left (c f -d e \right ) d}\, \sqrt {\left (a f -b e \right ) b}\, \left (a f -b e \right ) \left (b x +a \right ) \left (a d -b c \right )^{3} \left (c f -d e \right ) \left (d x +c \right )}\)	\(296\)

Input:

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

Output:

2*f^3*(d^2/f^3/(a*d-b*c)^3*(1/2*f*(a*d-b*c)/(c*f-d*e)*(f*x+e)^(1/2)/(d*(f* 
x+e)+c*f-d*e)+1/2*(a*d*f-5*b*c*f+4*b*d*e)/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*ar 
ctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))+b^2/f^3/(a*d-b*c)^3*(1/2*f*(a*d 
-b*c)/(a*f-b*e)*(f*x+e)^(1/2)/(b*(f*x+e)+a*f-b*e)+1/2*(5*a*d*f-b*c*f-4*b*d 
*e)/(a*f-b*e)/((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. 1096 vs. \(2 (234) = 468\).

Time = 4.91 (sec) , antiderivative size = 4502, normalized size of antiderivative = 17.59 \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(f*x+e)^(1/2)/(a*c+(a*d+b*c)*x+b*d*x^2)^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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (234) = 468\).

Time = 0.41 (sec) , antiderivative size = 619, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {{\left (4 \, b^{3} d e + b^{3} c f - 5 \, a b^{2} d f\right )} \arctan \left (\frac {\sqrt {f x + e} b}{\sqrt {-b^{2} e + a b f}}\right )}{{\left (b^{4} c^{3} e - 3 \, a b^{3} c^{2} d e + 3 \, a^{2} b^{2} c d^{2} e - a^{3} b d^{3} e - a b^{3} c^{3} f + 3 \, a^{2} b^{2} c^{2} d f - 3 \, a^{3} b c d^{2} f + a^{4} d^{3} f\right )} \sqrt {-b^{2} e + a b f}} + \frac {{\left (4 \, b d^{3} e - 5 \, b c d^{2} f + a d^{3} f\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (b^{3} c^{3} d e - 3 \, a b^{2} c^{2} d^{2} e + 3 \, a^{2} b c d^{3} e - a^{3} d^{4} e - b^{3} c^{4} f + 3 \, a b^{2} c^{3} d f - 3 \, a^{2} b c^{2} d^{2} f + a^{3} c d^{3} f\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} d^{2} e f - 2 \, \sqrt {f x + e} b^{2} d^{2} e^{2} f - {\left (f x + e\right )}^{\frac {3}{2}} b^{2} c d f^{2} - {\left (f x + e\right )}^{\frac {3}{2}} a b d^{2} f^{2} + 2 \, \sqrt {f x + e} b^{2} c d e f^{2} + 2 \, \sqrt {f x + e} a b d^{2} e f^{2} - \sqrt {f x + e} b^{2} c^{2} f^{3} - \sqrt {f x + e} a^{2} d^{2} f^{3}}{{\left (b^{3} c^{2} d e^{2} - 2 \, a b^{2} c d^{2} e^{2} + a^{2} b d^{3} e^{2} - b^{3} c^{3} e f + a b^{2} c^{2} d e f + a^{2} b c d^{2} e f - a^{3} d^{3} e f + a b^{2} c^{3} f^{2} - 2 \, a^{2} b c^{2} d f^{2} + a^{3} c d^{2} f^{2}\right )} {\left ({\left (f x + e\right )}^{2} b d - 2 \, {\left (f x + e\right )} b d e + b d e^{2} + {\left (f x + e\right )} b c f + {\left (f x + e\right )} a d f - b c e f - a d e f + a c f^{2}\right )}} \] Input:

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

Output:

-(4*b^3*d*e + b^3*c*f - 5*a*b^2*d*f)*arctan(sqrt(f*x + e)*b/sqrt(-b^2*e + 
a*b*f))/((b^4*c^3*e - 3*a*b^3*c^2*d*e + 3*a^2*b^2*c*d^2*e - a^3*b*d^3*e - 
a*b^3*c^3*f + 3*a^2*b^2*c^2*d*f - 3*a^3*b*c*d^2*f + a^4*d^3*f)*sqrt(-b^2*e 
 + a*b*f)) + (4*b*d^3*e - 5*b*c*d^2*f + a*d^3*f)*arctan(sqrt(f*x + e)*d/sq 
rt(-d^2*e + c*d*f))/((b^3*c^3*d*e - 3*a*b^2*c^2*d^2*e + 3*a^2*b*c*d^3*e - 
a^3*d^4*e - b^3*c^4*f + 3*a*b^2*c^3*d*f - 3*a^2*b*c^2*d^2*f + a^3*c*d^3*f) 
*sqrt(-d^2*e + c*d*f)) - (2*(f*x + e)^(3/2)*b^2*d^2*e*f - 2*sqrt(f*x + e)* 
b^2*d^2*e^2*f - (f*x + e)^(3/2)*b^2*c*d*f^2 - (f*x + e)^(3/2)*a*b*d^2*f^2 
+ 2*sqrt(f*x + e)*b^2*c*d*e*f^2 + 2*sqrt(f*x + e)*a*b*d^2*e*f^2 - sqrt(f*x 
 + e)*b^2*c^2*f^3 - sqrt(f*x + e)*a^2*d^2*f^3)/((b^3*c^2*d*e^2 - 2*a*b^2*c 
*d^2*e^2 + a^2*b*d^3*e^2 - b^3*c^3*e*f + a*b^2*c^2*d*e*f + a^2*b*c*d^2*e*f 
 - a^3*d^3*e*f + a*b^2*c^3*f^2 - 2*a^2*b*c^2*d*f^2 + a^3*c*d^2*f^2)*((f*x 
+ e)^2*b*d - 2*(f*x + e)*b*d*e + b*d*e^2 + (f*x + e)*b*c*f + (f*x + e)*a*d 
*f - b*c*e*f - a*d*e*f + a*c*f^2))
 

Mupad [B] (verification not implemented)

Time = 10.36 (sec) , antiderivative size = 36222, normalized size of antiderivative = 141.49 \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(((e + f*x)^(1/2)*(a^2*d^2*f^3 + b^2*c^2*f^3 + 2*b^2*d^2*e^2*f - 2*a*b*d^2 
*e*f^2 - 2*b^2*c*d*e*f^2))/((a^2*d^2 + b^2*c^2 - 2*a*b*c*d)*(a*c*f^2 + b*d 
*e^2 - a*d*e*f - b*c*e*f)) + (b*d*(e + f*x)^(3/2)*(a*d*f^2 + b*c*f^2 - 2*b 
*d*e*f))/((a^2*d^2 + b^2*c^2 - 2*a*b*c*d)*(a*c*f^2 + b*d*e^2 - a*d*e*f - b 
*c*e*f)))/((e + f*x)*(a*d*f + b*c*f - 2*b*d*e) + b*d*(e + f*x)^2 + a*c*f^2 
 + b*d*e^2 - a*d*e*f - b*c*e*f) - (atan(((((2*(e + f*x)^(1/2)*(a^4*b^3*d^7 
*f^6 + b^7*c^4*d^3*f^6 + 32*b^7*d^7*e^4*f^2 + 50*a^2*b^5*c^2*d^5*f^6 + 26* 
a^2*b^5*d^7*e^2*f^4 + 26*b^7*c^2*d^5*e^2*f^4 - 10*a*b^6*c^3*d^4*f^6 - 10*a 
^3*b^4*c*d^6*f^6 - 64*a*b^6*d^7*e^3*f^3 + 6*a^3*b^4*d^7*e*f^5 - 64*b^7*c*d 
^6*e^3*f^3 + 6*b^7*c^3*d^4*e*f^5 + 140*a*b^6*c*d^6*e^2*f^4 - 70*a*b^6*c^2* 
d^5*e*f^5 - 70*a^2*b^5*c*d^6*e*f^5))/(2*a*b^5*c^6*e*f^3 - a^4*b^2*d^6*e^4 
- a^6*c^2*d^4*f^4 - b^6*c^4*d^2*e^4 - a^6*d^6*e^2*f^2 - b^6*c^6*e^2*f^2 - 
6*a^2*b^4*c^2*d^4*e^4 - 6*a^4*b^2*c^4*d^2*f^4 - a^2*b^4*c^6*f^4 + 2*a^5*b* 
d^6*e^3*f + 2*a^6*c*d^5*e*f^3 + 2*b^6*c^5*d*e^3*f + 4*a*b^5*c^3*d^3*e^4 + 
4*a^3*b^3*c*d^5*e^4 + 4*a^3*b^3*c^5*d*f^4 + 4*a^5*b*c^3*d^3*f^4 - 6*a*b^5* 
c^4*d^2*e^3*f - 6*a^2*b^4*c^5*d*e*f^3 - 6*a^4*b^2*c*d^5*e^3*f - 6*a^5*b*c^ 
2*d^4*e*f^3 + 4*a^2*b^4*c^3*d^3*e^3*f + 4*a^3*b^3*c^2*d^4*e^3*f + 4*a^3*b^ 
3*c^4*d^2*e*f^3 + 4*a^4*b^2*c^3*d^3*e*f^3 + 9*a^2*b^4*c^4*d^2*e^2*f^2 - 16 
*a^3*b^3*c^3*d^3*e^2*f^2 + 9*a^4*b^2*c^2*d^4*e^2*f^2) + ((-b^3*(a*f - b*e) 
^3)^(1/2)*((40*a^2*b^9*c^8*d^3*f^7 - 160*a^3*b^8*c^7*d^4*f^7 + 344*a^4*...
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 4120, normalized size of antiderivative = 16.09 \[ \int \frac {1}{\sqrt {e+f x} \left (a c+(b c+a d) x+b d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(5*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e) 
))*a**2*b*c**3*d*f**3 - 10*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/ 
(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*c**2*d**2*e*f**2 + 5*sqrt(b)*sqrt(a*f - 
b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*c**2*d**2*f* 
*3*x + 5*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f 
- b*e)))*a**2*b*c*d**3*e**2*f - 10*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + 
f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*c*d**3*e*f**2*x + 5*sqrt(b)*sqrt 
(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*d**4* 
e**2*f*x - sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a* 
f - b*e)))*a*b**2*c**4*f**3 - 2*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x 
)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c**3*d*e*f**2 + 4*sqrt(b)*sqrt(a*f 
- b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c**3*d*f** 
3*x + 7*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - 
 b*e)))*a*b**2*c**2*d**2*e**2*f - 12*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e 
+ f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c**2*d**2*e*f**2*x + 5*sqrt(b) 
*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2* 
c**2*d**2*f**3*x**2 - 4*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sq 
rt(b)*sqrt(a*f - b*e)))*a*b**2*c*d**3*e**3 + 12*sqrt(b)*sqrt(a*f - b*e)*at 
an((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c*d**3*e**2*f*x - 1 
0*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*...