\(\int \frac {(A+B x) \sqrt {d+e x}}{(b x+c x^2)^2} \, dx\) [81]

Optimal result

Integrand size = 26, antiderivative size = 171 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\left (b x+c x^2\right )^2} \, dx=\frac {(b B-2 A c) \sqrt {d+e x}}{b^2 (b+c x)}-\frac {A \sqrt {d+e x}}{b x (b+c x)}-\frac {(2 b B d-4 A c d+A b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3 \sqrt {d}}+\frac {\left (2 b B c d-4 A c^2 d-b^2 B e+3 A b c e\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 \sqrt {c} \sqrt {c d-b e}} \] Output:

(-2*A*c+B*b)*(e*x+d)^(1/2)/b^2/(c*x+b)-A*(e*x+d)^(1/2)/b/x/(c*x+b)-(A*b*e- 
4*A*c*d+2*B*b*d)*arctanh((e*x+d)^(1/2)/d^(1/2))/b^3/d^(1/2)+(3*A*b*c*e-4*A 
*c^2*d-B*b^2*e+2*B*b*c*d)*arctanh(c^(1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/ 
b^3/c^(1/2)/(-b*e+c*d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\left (b x+c x^2\right )^2} \, dx=\frac {\frac {b \sqrt {d+e x} (b B x-A (b+2 c x))}{x (b+c x)}+\frac {\left (4 A c^2 d+b^2 B e-b c (2 B d+3 A e)\right ) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{\sqrt {c} \sqrt {-c d+b e}}-\frac {(2 b B d-4 A c d+A b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{b^3} \] Input:

Integrate[((A + B*x)*Sqrt[d + e*x])/(b*x + c*x^2)^2,x]
 

Output:

((b*Sqrt[d + e*x]*(b*B*x - A*(b + 2*c*x)))/(x*(b + c*x)) + ((4*A*c^2*d + b 
^2*B*e - b*c*(2*B*d + 3*A*e))*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + 
 b*e]])/(Sqrt[c]*Sqrt[-(c*d) + b*e]) - ((2*b*B*d - 4*A*c*d + A*b*e)*ArcTan 
h[Sqrt[d + e*x]/Sqrt[d]])/Sqrt[d])/b^3
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1234, 27, 25, 1197, 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.

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

\(\Big \downarrow \) 1234

\(\displaystyle -\frac {\int \frac {4 A c d-b (2 B d+A e)-(b B-2 A c) e x}{2 \sqrt {d+e x} \left (c x^2+b x\right )}dx}{b^2}-\frac {\sqrt {d+e x} (A b-x (b B-2 A c))}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int -\frac {2 b B d-4 A c d+A b e+(b B-2 A c) e x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{2 b^2}-\frac {\sqrt {d+e x} (A b-x (b B-2 A c))}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {2 b B d-4 A c d+A b e+(b B-2 A c) e x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{2 b^2}-\frac {\sqrt {d+e x} (A b-x (b B-2 A c))}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

\(\displaystyle \frac {e \left (\frac {c (A b e-4 A c d+2 b B d) \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}-\frac {\left (3 A b c e-4 A c^2 d+b^2 (-B) e+2 b B c d\right ) \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}\right )}{b^2}-\frac {\sqrt {d+e x} (A b-x (b B-2 A c))}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {e \left (\frac {\left (3 A b c e-4 A c^2 d+b^2 (-B) e+2 b B c d\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c} e \sqrt {c d-b e}}-\frac {\text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) (A b e-4 A c d+2 b B d)}{b \sqrt {d} e}\right )}{b^2}-\frac {\sqrt {d+e x} (A b-x (b B-2 A c))}{b^2 \left (b x+c x^2\right )}\)

Input:

Int[((A + B*x)*Sqrt[d + e*x])/(b*x + c*x^2)^2,x]
 

Output:

-(((A*b - (b*B - 2*A*c)*x)*Sqrt[d + e*x])/(b^2*(b*x + c*x^2))) + (e*(-(((2 
*b*B*d - 4*A*c*d + A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d]*e)) + 
 ((2*b*B*c*d - 4*A*c^2*d - b^2*B*e + 3*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + 
e*x])/Sqrt[c*d - b*e]])/(b*Sqrt[c]*e*Sqrt[c*d - b*e])))/b^2
 

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 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 1234

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*( 
(f*b - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 
 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g 
*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)* 
(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1 
] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 = 1.25 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(-\frac {-4 \left (A \,c^{2} d -\frac {3 b \left (A e +\frac {2 B d}{3}\right ) c}{4}+\frac {b^{2} B e}{4}\right ) \sqrt {d}\, x \left (c x +b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )+\sqrt {c \left (b e -c d \right )}\, \left (x \left (-4 A c d +b \left (A e +2 B d \right )\right ) \left (c x +b \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )+\left (2 A c x +b \left (-B x +A \right )\right ) \sqrt {d}\, b \sqrt {e x +d}\right )}{\sqrt {d}\, \sqrt {c \left (b e -c d \right )}\, x \,b^{3} \left (c x +b \right )}\)	\(166\)
derivativedivides	\(2 e^{2} \left (-\frac {\frac {\left (\frac {1}{2} A c e b -\frac {1}{2} b^{2} B e \right ) \sqrt {e x +d}}{\left (e x +d \right ) c +b e -c d}+\frac {\left (3 A c e b -4 A \,c^{2} d -b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \sqrt {c \left (b e -c d \right )}}}{b^{3} e^{2}}+\frac {-\frac {A b \sqrt {e x +d}}{2 x}-\frac {\left (A b e -4 A c d +2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{b^{3} e^{2}}\right )\)	\(172\)
default	\(2 e^{2} \left (-\frac {\frac {\left (\frac {1}{2} A c e b -\frac {1}{2} b^{2} B e \right ) \sqrt {e x +d}}{\left (e x +d \right ) c +b e -c d}+\frac {\left (3 A c e b -4 A \,c^{2} d -b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \sqrt {c \left (b e -c d \right )}}}{b^{3} e^{2}}+\frac {-\frac {A b \sqrt {e x +d}}{2 x}-\frac {\left (A b e -4 A c d +2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{b^{3} e^{2}}\right )\)	\(172\)
risch	\(-\frac {A \sqrt {e x +d}}{b^{2} x}-\frac {e \left (-\frac {\left (-A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e \sqrt {d}}+\frac {\frac {2 \left (\frac {1}{2} A c e b -\frac {1}{2} b^{2} B e \right ) \sqrt {e x +d}}{\left (e x +d \right ) c +b e -c d}+\frac {\left (3 A c e b -4 A \,c^{2} d -b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\sqrt {c \left (b e -c d \right )}}}{b e}\right )}{b^{2}}\)	\(175\)

Input:

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

Output:

-(-4*(A*c^2*d-3/4*b*(A*e+2/3*B*d)*c+1/4*b^2*B*e)*d^(1/2)*x*(c*x+b)*arctan( 
c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2))+(c*(b*e-c*d))^(1/2)*(x*(-4*A*c*d+b*(A 
*e+2*B*d))*(c*x+b)*arctanh((e*x+d)^(1/2)/d^(1/2))+(2*A*c*x+b*(-B*x+A))*d^( 
1/2)*b*(e*x+d)^(1/2)))/d^(1/2)/(c*(b*e-c*d))^(1/2)/x/b^3/(c*x+b)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (151) = 302\).

Time = 0.31 (sec) , antiderivative size = 1568, normalized size of antiderivative = 9.17 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\left (b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(sqrt(c^2*d - b*c*e)*((2*(B*b*c^2 - 2*A*c^3)*d^2 - (B*b^2*c - 3*A*b*c 
^2)*d*e)*x^2 + (2*(B*b^2*c - 2*A*b*c^2)*d^2 - (B*b^3 - 3*A*b^2*c)*d*e)*x)* 
log((c*e*x + 2*c*d - b*e + 2*sqrt(c^2*d - b*c*e)*sqrt(e*x + d))/(c*x + b)) 
 - ((A*b^2*c^2*e^2 - 2*(B*b*c^3 - 2*A*c^4)*d^2 + (2*B*b^2*c^2 - 5*A*b*c^3) 
*d*e)*x^2 + (A*b^3*c*e^2 - 2*(B*b^2*c^2 - 2*A*b*c^3)*d^2 + (2*B*b^3*c - 5* 
A*b^2*c^2)*d*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 
2*(A*b^2*c^2*d^2 - A*b^3*c*d*e - ((B*b^2*c^2 - 2*A*b*c^3)*d^2 - (B*b^3*c - 
 2*A*b^2*c^2)*d*e)*x)*sqrt(e*x + d))/((b^3*c^3*d^2 - b^4*c^2*d*e)*x^2 + (b 
^4*c^2*d^2 - b^5*c*d*e)*x), -1/2*(2*sqrt(-c^2*d + b*c*e)*((2*(B*b*c^2 - 2* 
A*c^3)*d^2 - (B*b^2*c - 3*A*b*c^2)*d*e)*x^2 + (2*(B*b^2*c - 2*A*b*c^2)*d^2 
 - (B*b^3 - 3*A*b^2*c)*d*e)*x)*arctan(sqrt(-c^2*d + b*c*e)*sqrt(e*x + d)/( 
c*e*x + c*d)) + ((A*b^2*c^2*e^2 - 2*(B*b*c^3 - 2*A*c^4)*d^2 + (2*B*b^2*c^2 
 - 5*A*b*c^3)*d*e)*x^2 + (A*b^3*c*e^2 - 2*(B*b^2*c^2 - 2*A*b*c^3)*d^2 + (2 
*B*b^3*c - 5*A*b^2*c^2)*d*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) 
 + 2*d)/x) + 2*(A*b^2*c^2*d^2 - A*b^3*c*d*e - ((B*b^2*c^2 - 2*A*b*c^3)*d^2 
 - (B*b^3*c - 2*A*b^2*c^2)*d*e)*x)*sqrt(e*x + d))/((b^3*c^3*d^2 - b^4*c^2* 
d*e)*x^2 + (b^4*c^2*d^2 - b^5*c*d*e)*x), -1/2*(2*((A*b^2*c^2*e^2 - 2*(B*b* 
c^3 - 2*A*c^4)*d^2 + (2*B*b^2*c^2 - 5*A*b*c^3)*d*e)*x^2 + (A*b^3*c*e^2 - 2 
*(B*b^2*c^2 - 2*A*b*c^3)*d^2 + (2*B*b^3*c - 5*A*b^2*c^2)*d*e)*x)*sqrt(-d)* 
arctan(sqrt(-d)/sqrt(e*x + d)) - sqrt(c^2*d - b*c*e)*((2*(B*b*c^2 - 2*A...
 

Sympy [F(-1)]

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

integrate((B*x+A)*(e*x+d)**(1/2)/(c*x**2+b*x)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)*(e*x+d)^(1/2)/(c*x^2+b*x)^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(b*e-c*d>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\left (b x+c x^2\right )^2} \, dx=-\frac {{\left (2 \, B b c d - 4 \, A c^{2} d - B b^{2} e + 3 \, A b c e\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b^{3}} + \frac {{\left (2 \, B b d - 4 \, A c d + A b e\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b^{3} \sqrt {-d}} + \frac {{\left (e x + d\right )}^{\frac {3}{2}} B b e - 2 \, {\left (e x + d\right )}^{\frac {3}{2}} A c e - \sqrt {e x + d} B b d e + 2 \, \sqrt {e x + d} A c d e - \sqrt {e x + d} A b e^{2}}{{\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e\right )} b^{2}} \] Input:

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

Output:

-(2*B*b*c*d - 4*A*c^2*d - B*b^2*e + 3*A*b*c*e)*arctan(sqrt(e*x + d)*c/sqrt 
(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b^3) + (2*B*b*d - 4*A*c*d + A*b*e) 
*arctan(sqrt(e*x + d)/sqrt(-d))/(b^3*sqrt(-d)) + ((e*x + d)^(3/2)*B*b*e - 
2*(e*x + d)^(3/2)*A*c*e - sqrt(e*x + d)*B*b*d*e + 2*sqrt(e*x + d)*A*c*d*e 
- sqrt(e*x + d)*A*b*e^2)/(((e*x + d)^2*c - 2*(e*x + d)*c*d + c*d^2 + (e*x 
+ d)*b*e - b*d*e)*b^2)
 

Mupad [B] (verification not implemented)

Time = 11.70 (sec) , antiderivative size = 2558, normalized size of antiderivative = 14.96 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\left (b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

int(((A + B*x)*(d + e*x)^(1/2))/(b*x + c*x^2)^2,x)
 

Output:

(atan((((-c*(b*e - c*d))^(1/2)*((2*(d + e*x)^(1/2)*(10*A^2*b^2*c^3*e^4 + 3 
2*A^2*c^5*d^2*e^2 + B^2*b^4*c*e^4 + 8*B^2*b^2*c^3*d^2*e^2 - 6*A*B*b^3*c^2* 
e^4 - 32*A^2*b*c^4*d*e^3 - 4*B^2*b^3*c^2*d*e^3 - 32*A*B*b*c^4*d^2*e^2 + 24 
*A*B*b^2*c^3*d*e^3))/b^4 + (((4*A*b^7*c^2*e^4 - 8*A*b^6*c^3*d*e^3 + 4*B*b^ 
7*c^2*d*e^3)/b^6 + ((4*b^7*c^2*e^3 - 8*b^6*c^3*d*e^2)*(-c*(b*e - c*d))^(1/ 
2)*(d + e*x)^(1/2)*(4*A*c^2*d + B*b^2*e - 3*A*b*c*e - 2*B*b*c*d))/(b^4*(b^ 
3*c^2*d - b^4*c*e)))*(-c*(b*e - c*d))^(1/2)*(4*A*c^2*d + B*b^2*e - 3*A*b*c 
*e - 2*B*b*c*d))/(2*(b^3*c^2*d - b^4*c*e)))*(4*A*c^2*d + B*b^2*e - 3*A*b*c 
*e - 2*B*b*c*d)*1i)/(2*(b^3*c^2*d - b^4*c*e)) + ((-c*(b*e - c*d))^(1/2)*(( 
2*(d + e*x)^(1/2)*(10*A^2*b^2*c^3*e^4 + 32*A^2*c^5*d^2*e^2 + B^2*b^4*c*e^4 
 + 8*B^2*b^2*c^3*d^2*e^2 - 6*A*B*b^3*c^2*e^4 - 32*A^2*b*c^4*d*e^3 - 4*B^2* 
b^3*c^2*d*e^3 - 32*A*B*b*c^4*d^2*e^2 + 24*A*B*b^2*c^3*d*e^3))/b^4 - (((4*A 
*b^7*c^2*e^4 - 8*A*b^6*c^3*d*e^3 + 4*B*b^7*c^2*d*e^3)/b^6 - ((4*b^7*c^2*e^ 
3 - 8*b^6*c^3*d*e^2)*(-c*(b*e - c*d))^(1/2)*(d + e*x)^(1/2)*(4*A*c^2*d + B 
*b^2*e - 3*A*b*c*e - 2*B*b*c*d))/(b^4*(b^3*c^2*d - b^4*c*e)))*(-c*(b*e - c 
*d))^(1/2)*(4*A*c^2*d + B*b^2*e - 3*A*b*c*e - 2*B*b*c*d))/(2*(b^3*c^2*d - 
b^4*c*e)))*(4*A*c^2*d + B*b^2*e - 3*A*b*c*e - 2*B*b*c*d)*1i)/(2*(b^3*c^2*d 
 - b^4*c*e)))/((2*(6*A^3*b^2*c^3*e^5 + 32*A^3*c^5*d^2*e^3 - 4*B^3*b^3*c^2* 
d^2*e^3 + A*B^2*b^4*c*e^5 - 32*A^3*b*c^4*d*e^4 + 2*B^3*b^4*c*d*e^4 - 5*A^2 
*B*b^3*c^2*e^5 + 24*A*B^2*b^2*c^3*d^2*e^3 - 16*A*B^2*b^3*c^2*d*e^4 - 48...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1009, normalized size of antiderivative = 5.90 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\left (b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(e*x+d)^(1/2)/(c*x^2+b*x)^2,x)
 

Output:

( - 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c 
*d)))*a*b**2*c*d*e*x + 8*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(s 
qrt(c)*sqrt(b*e - c*d)))*a*b*c**2*d**2*x - 6*sqrt(c)*sqrt(b*e - c*d)*atan( 
(sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*b*c**2*d*e*x**2 + 8*sqrt(c) 
*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*c**3* 
d**2*x**2 + 2*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt 
(b*e - c*d)))*b**4*d*e*x - 4*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c 
)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c*d**2*x + 2*sqrt(c)*sqrt(b*e - c*d)*ata 
n((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c*d*e*x**2 - 4*sqrt(c) 
*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c* 
*2*d**2*x**2 - 2*sqrt(d + e*x)*a*b**3*c*d*e + 2*sqrt(d + e*x)*a*b**2*c**2* 
d**2 - 4*sqrt(d + e*x)*a*b**2*c**2*d*e*x + 4*sqrt(d + e*x)*a*b*c**3*d**2*x 
 + 2*sqrt(d + e*x)*b**4*c*d*e*x - 2*sqrt(d + e*x)*b**3*c**2*d**2*x + sqrt( 
d)*log(sqrt(d + e*x) - sqrt(d))*a*b**3*c*e**2*x - 5*sqrt(d)*log(sqrt(d + e 
*x) - sqrt(d))*a*b**2*c**2*d*e*x + sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a* 
b**2*c**2*e**2*x**2 + 4*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*b*c**3*d**2 
*x - 5*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*b*c**3*d*e*x**2 + 4*sqrt(d)* 
log(sqrt(d + e*x) - sqrt(d))*a*c**4*d**2*x**2 + 2*sqrt(d)*log(sqrt(d + e*x 
) - sqrt(d))*b**4*c*d*e*x - 2*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b**3*c* 
*2*d**2*x + 2*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b**3*c**2*d*e*x**2 -...