\(\int \frac {A+B x}{\sqrt {d+e x} (2 A B d-A^2 e-B^2 e x^2)} \, dx\) [139]

Optimal result

Integrand size = 38, antiderivative size = 76 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {B} \sqrt {2 B d-A e} \sqrt {d+e x}}{2 B d-A e+B e x}\right )}{\sqrt {B} e \sqrt {2 B d-A e}} \] Output:

2^(1/2)*arctanh(2^(1/2)*B^(1/2)*(-A*e+2*B*d)^(1/2)*(e*x+d)^(1/2)/(B*e*x-A* 
e+2*B*d))/B^(1/2)/e/(-A*e+2*B*d)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.93 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.32 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=\frac {-\frac {\left (-i \sqrt {A} \sqrt {e}+\sqrt {-2 B d+A e}\right ) \arctan \left (\frac {\sqrt {B} \sqrt {d+e x}}{\sqrt {-B d-i \sqrt {A} \sqrt {e} \sqrt {-2 B d+A e}}}\right )}{\sqrt {-B d-i \sqrt {A} \sqrt {e} \sqrt {-2 B d+A e}}}-\frac {\left (i \sqrt {A} \sqrt {e}+\sqrt {-2 B d+A e}\right ) \arctan \left (\frac {\sqrt {B} \sqrt {d+e x}}{\sqrt {-B d+i \sqrt {A} \sqrt {e} \sqrt {-2 B d+A e}}}\right )}{\sqrt {-B d+i \sqrt {A} \sqrt {e} \sqrt {-2 B d+A e}}}}{\sqrt {B} e \sqrt {-2 B d+A e}} \] Input:

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

Output:

(-((((-I)*Sqrt[A]*Sqrt[e] + Sqrt[-2*B*d + A*e])*ArcTan[(Sqrt[B]*Sqrt[d + e 
*x])/Sqrt[-(B*d) - I*Sqrt[A]*Sqrt[e]*Sqrt[-2*B*d + A*e]]])/Sqrt[-(B*d) - I 
*Sqrt[A]*Sqrt[e]*Sqrt[-2*B*d + A*e]]) - ((I*Sqrt[A]*Sqrt[e] + Sqrt[-2*B*d 
+ A*e])*ArcTan[(Sqrt[B]*Sqrt[d + e*x])/Sqrt[-(B*d) + I*Sqrt[A]*Sqrt[e]*Sqr 
t[-2*B*d + A*e]]])/Sqrt[-(B*d) + I*Sqrt[A]*Sqrt[e]*Sqrt[-2*B*d + A*e]])/(S 
qrt[B]*e*Sqrt[-2*B*d + A*e])
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(162\) vs. \(2(76)=152\).

Time = 0.41 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {654, 1478, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 654

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

\(\Big \downarrow \) 1478

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle 2 \left (\frac {\log \left (\sqrt {2} \sqrt {B} \sqrt {d+e x} \sqrt {2 B d-A e}-A e+B (d+e x)+B d\right )}{2 \sqrt {2} \sqrt {B} e \sqrt {2 B d-A e}}-\frac {\log \left (-\sqrt {2} \sqrt {B} \sqrt {d+e x} \sqrt {2 B d-A e}-A e+B (d+e x)+B d\right )}{2 \sqrt {2} \sqrt {B} e \sqrt {2 B d-A e}}\right )\)

Input:

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

Output:

2*(-1/2*Log[B*d - A*e - Sqrt[2]*Sqrt[B]*Sqrt[2*B*d - A*e]*Sqrt[d + e*x] + 
B*(d + e*x)]/(Sqrt[2]*Sqrt[B]*e*Sqrt[2*B*d - A*e]) + Log[B*d - A*e + Sqrt[ 
2]*Sqrt[B]*Sqrt[2*B*d - A*e]*Sqrt[d + e*x] + B*(d + e*x)]/(2*Sqrt[2]*Sqrt[ 
B]*e*Sqrt[2*B*d - A*e]))
 

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 654

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

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1478

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(196\) vs. \(2(62)=124\).

Time = 1.88 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.59

method	result	size
pseudoelliptic	\(-\frac {-\frac {\left (A e B +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\right ) \operatorname {arctanh}\left (\frac {B \sqrt {e x +d}}{\sqrt {B^{2} d +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{\sqrt {B^{2} d +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}+\frac {\left (A e B -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\right ) \operatorname {arctanh}\left (\frac {B \sqrt {e x +d}}{\sqrt {B^{2} d -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{\sqrt {B^{2} d -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}}{e \sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}\)	\(197\)
derivativedivides	\(-\frac {2 B^{2} \left (-\frac {\left (A e B +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\right ) \operatorname {arctanh}\left (\frac {B \sqrt {e x +d}}{\sqrt {B^{2} d +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{2 B^{2} \sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\, \sqrt {B^{2} d +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}-\frac {\left (-A e B +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\right ) \operatorname {arctanh}\left (\frac {B \sqrt {e x +d}}{\sqrt {B^{2} d -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{2 B^{2} \sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\, \sqrt {B^{2} d -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{e}\)	\(223\)
default	\(-\frac {2 B^{2} \left (-\frac {\left (A e B +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\right ) \operatorname {arctanh}\left (\frac {B \sqrt {e x +d}}{\sqrt {B^{2} d +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{2 B^{2} \sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\, \sqrt {B^{2} d +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}-\frac {\left (-A e B +\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\right ) \operatorname {arctanh}\left (\frac {B \sqrt {e x +d}}{\sqrt {B^{2} d -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{2 B^{2} \sqrt {-e \,B^{2} A \left (A e -2 B d \right )}\, \sqrt {B^{2} d -\sqrt {-e \,B^{2} A \left (A e -2 B d \right )}}}\right )}{e}\)	\(223\)

Input:

int((B*x+A)/(e*x+d)^(1/2)/(-B^2*e*x^2-A^2*e+2*A*B*d),x,method=_RETURNVERBO 
SE)
 

Output:

-1/e/(-e*B^2*A*(A*e-2*B*d))^(1/2)*(-(A*e*B+(-e*B^2*A*(A*e-2*B*d))^(1/2))/( 
B^2*d+(-e*B^2*A*(A*e-2*B*d))^(1/2))^(1/2)*arctanh(B*(e*x+d)^(1/2)/(B^2*d+( 
-e*B^2*A*(A*e-2*B*d))^(1/2))^(1/2))+(A*e*B-(-e*B^2*A*(A*e-2*B*d))^(1/2))*a 
rctanh(B*(e*x+d)^(1/2)/(B^2*d-(-e*B^2*A*(A*e-2*B*d))^(1/2))^(1/2))/(B^2*d- 
(-e*B^2*A*(A*e-2*B*d))^(1/2))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.11 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=\left [\frac {\sqrt {2} \log \left (\frac {B^{2} e^{2} x^{2} + 8 \, B^{2} d^{2} - 6 \, A B d e + A^{2} e^{2} + 4 \, {\left (2 \, B^{2} d e - A B e^{2}\right )} x + \frac {2 \, \sqrt {2} {\left (4 \, B^{3} d^{2} - 4 \, A B^{2} d e + A^{2} B e^{2} + {\left (2 \, B^{3} d e - A B^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{\sqrt {2 \, B^{2} d - A B e}}}{B^{2} e x^{2} - 2 \, A B d + A^{2} e}\right )}{2 \, \sqrt {2 \, B^{2} d - A B e} e}, -\frac {\sqrt {2} \sqrt {-\frac {1}{2 \, B^{2} d - A B e}} \arctan \left (\frac {\sqrt {2} {\left (B e x + 2 \, B d - A e\right )} \sqrt {-\frac {1}{2 \, B^{2} d - A B e}}}{2 \, \sqrt {e x + d}}\right )}{e}\right ] \] Input:

integrate((B*x+A)/(e*x+d)^(1/2)/(-B^2*e*x^2-A^2*e+2*A*B*d),x, algorithm="f 
ricas")
 

Output:

[1/2*sqrt(2)*log((B^2*e^2*x^2 + 8*B^2*d^2 - 6*A*B*d*e + A^2*e^2 + 4*(2*B^2 
*d*e - A*B*e^2)*x + 2*sqrt(2)*(4*B^3*d^2 - 4*A*B^2*d*e + A^2*B*e^2 + (2*B^ 
3*d*e - A*B^2*e^2)*x)*sqrt(e*x + d)/sqrt(2*B^2*d - A*B*e))/(B^2*e*x^2 - 2* 
A*B*d + A^2*e))/(sqrt(2*B^2*d - A*B*e)*e), -sqrt(2)*sqrt(-1/(2*B^2*d - A*B 
*e))*arctan(1/2*sqrt(2)*(B*e*x + 2*B*d - A*e)*sqrt(-1/(2*B^2*d - A*B*e))/s 
qrt(e*x + d))/e]
 

Sympy [F]

\[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=- \int \frac {A}{A^{2} e \sqrt {d + e x} - 2 A B d \sqrt {d + e x} + B^{2} e x^{2} \sqrt {d + e x}}\, dx - \int \frac {B x}{A^{2} e \sqrt {d + e x} - 2 A B d \sqrt {d + e x} + B^{2} e x^{2} \sqrt {d + e x}}\, dx \] Input:

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

Output:

-Integral(A/(A**2*e*sqrt(d + e*x) - 2*A*B*d*sqrt(d + e*x) + B**2*e*x**2*sq 
rt(d + e*x)), x) - Integral(B*x/(A**2*e*sqrt(d + e*x) - 2*A*B*d*sqrt(d + e 
*x) + B**2*e*x**2*sqrt(d + e*x)), x)
 

Maxima [F]

\[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=\int { -\frac {B x + A}{{\left (B^{2} e x^{2} - 2 \, A B d + A^{2} e\right )} \sqrt {e x + d}} \,d x } \] Input:

integrate((B*x+A)/(e*x+d)^(1/2)/(-B^2*e*x^2-A^2*e+2*A*B*d),x, algorithm="m 
axima")
 

Output:

-integrate((B*x + A)/((B^2*e*x^2 - 2*A*B*d + A^2*e)*sqrt(e*x + d)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1495 vs. \(2 (62) = 124\).

Time = 0.39 (sec) , antiderivative size = 1495, normalized size of antiderivative = 19.67 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)/(e*x+d)^(1/2)/(-B^2*e*x^2-A^2*e+2*A*B*d),x, algorithm="g 
iac")
 

Output:

(2*A*B^5*d^2*e^3 - A^2*B^4*d*e^4 - (4*A*B^3*d^2*e - 4*A^2*B^2*d*e^2 + A^3* 
B*e^3 + (2*A*B^2*d*e - A^2*B*e^2)*sqrt(2*A*B*d*e - A^2*e^2))*B^2*e^2*sgn(e 
) - (4*A*B^3*d^2*e - 4*A^2*B^2*d*e^2 + A^3*B*e^3 + (4*B^3*d^2 - 4*A*B^2*d* 
e + A^2*B*e^2)*sqrt(2*A*B*d*e - A^2*e^2))*B^2*e^2 - (2*A*B^4*d^2*e^2 - 3*A 
^2*B^3*d*e^3 + A^3*B^2*e^4 - (2*B^4*d^2*e - 3*A*B^3*d*e^2 + A^2*B^2*e^3)*s 
qrt(2*A*B*d*e - A^2*e^2))*abs(B)*abs(e)*sgn(e) - (4*B^5*d^3*e - 8*A*B^4*d^ 
2*e^2 + 5*A^2*B^3*d*e^3 - A^3*B^2*e^4 + (2*B^4*d^2*e - 3*A*B^3*d*e^2 + A^2 
*B^2*e^3)*sqrt(2*A*B*d*e - A^2*e^2))*abs(B)*abs(e) + (2*A*B^5*d^2*e^3 - A^ 
2*B^4*d*e^4 - sqrt(2*A*B*d*e - A^2*e^2)*A*B^4*d*e^3)*sgn(e) - (2*B^5*d^2*e 
^2 - A*B^4*d*e^3)*sqrt(2*A*B*d*e - A^2*e^2))*arctan(sqrt(e*x + d)/sqrt(-(B 
^2*d*e + sqrt(B^4*d^2*e^2 - (B^2*d^2*e - 2*A*B*d*e^2 + A^2*e^3)*B^2*e))/(B 
^2*e)))/((2*sqrt(2)*sqrt(-2*B^2*d + A*B*e)*B^5*d^3*e^2 - 2*sqrt(2)*sqrt(-A 
*B*e)*B^5*d^3*e^2 - 5*sqrt(2)*sqrt(-2*B^2*d + A*B*e)*A*B^4*d^2*e^3 + 5*sqr 
t(2)*sqrt(-A*B*e)*A*B^4*d^2*e^3 + 4*sqrt(2)*sqrt(-2*B^2*d + A*B*e)*A^2*B^3 
*d*e^4 - 4*sqrt(2)*sqrt(-A*B*e)*A^2*B^3*d*e^4 - sqrt(2)*sqrt(-2*B^2*d + A* 
B*e)*A^3*B^2*e^5 + sqrt(2)*sqrt(-A*B*e)*A^3*B^2*e^5)*abs(B)*abs(e)) + (2*A 
*B^5*d^2*e^3 - A^2*B^4*d*e^4 + (4*A*B^3*d^2*e - 4*A^2*B^2*d*e^2 + A^3*B*e^ 
3 - (2*A*B^2*d*e - A^2*B*e^2)*sqrt(2*A*B*d*e - A^2*e^2))*B^2*e^2*sgn(e) - 
(4*A*B^3*d^2*e - 4*A^2*B^2*d*e^2 + A^3*B*e^3 + (4*B^3*d^2 - 4*A*B^2*d*e + 
A^2*B*e^2)*sqrt(2*A*B*d*e - A^2*e^2))*B^2*e^2 + (2*A*B^4*d^2*e^2 - 3*A^...
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 310, normalized size of antiderivative = 4.08 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\frac {\left (A\,e^2\,\sqrt {A\,B\,e-2\,B^2\,d}-2\,B\,d\,e\,\sqrt {A\,B\,e-2\,B^2\,d}\right )\,\left (\left (\frac {\sqrt {2}\,\left (\frac {2\,B^4\,d-2\,A\,B^3\,e}{e^2}-\frac {4\,A^2\,B^4\,e^4-8\,A\,B^5\,d\,e^3+4\,B^6\,d^2\,e^2}{e^4\,\left (2\,B^2\,d-A\,B\,e\right )}\right )}{\left (A\,e-B\,d\right )\,\left (A\,e-2\,B\,d\right )}+\frac {4\,\sqrt {2}\,A\,B^4}{e\,\left (2\,B^2\,d-A\,B\,e\right )\,\left (A\,e-B\,d\right )}\right )\,\sqrt {d+e\,x}-\frac {\sqrt {2}\,\left (\frac {2\,B^4}{e^2}-\frac {4\,B^6\,d}{e^2\,\left (2\,B^2\,d-A\,B\,e\right )}\right )\,{\left (d+e\,x\right )}^{3/2}}{\left (A\,e-B\,d\right )\,\left (A\,e-2\,B\,d\right )}\right )}{4\,A\,B^3}\right )-\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {A\,B\,e-2\,B^2\,d}\,\sqrt {d+e\,x}}{2\,\left (A\,e-2\,B\,d\right )}\right )\right )}{e\,\sqrt {A\,B\,e-2\,B^2\,d}} \] Input:

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

Output:

(2^(1/2)*(atan(((A*e^2*(A*B*e - 2*B^2*d)^(1/2) - 2*B*d*e*(A*B*e - 2*B^2*d) 
^(1/2))*(((2^(1/2)*((2*B^4*d - 2*A*B^3*e)/e^2 - (4*A^2*B^4*e^4 + 4*B^6*d^2 
*e^2 - 8*A*B^5*d*e^3)/(e^4*(2*B^2*d - A*B*e))))/((A*e - B*d)*(A*e - 2*B*d) 
) + (4*2^(1/2)*A*B^4)/(e*(2*B^2*d - A*B*e)*(A*e - B*d)))*(d + e*x)^(1/2) - 
 (2^(1/2)*((2*B^4)/e^2 - (4*B^6*d)/(e^2*(2*B^2*d - A*B*e)))*(d + e*x)^(3/2 
))/((A*e - B*d)*(A*e - 2*B*d))))/(4*A*B^3)) - atan((2^(1/2)*(A*B*e - 2*B^2 
*d)^(1/2)*(d + e*x)^(1/2))/(2*(A*e - 2*B*d)))))/(e*(A*B*e - 2*B^2*d)^(1/2) 
)
 

Reduce [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.49 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (2 A B d-A^2 e-B^2 e x^2\right )} \, dx=-\frac {\sqrt {b}\, \sqrt {a e -2 b d}\, \sqrt {2}\, \left (\mathit {atan} \left (\frac {2 \sqrt {e x +d}\, b -\sqrt {e}\, \sqrt {b}\, \sqrt {a}\, \sqrt {2}}{\sqrt {b}\, \sqrt {a e -2 b d}\, \sqrt {2}}\right )+\mathit {atan} \left (\frac {2 \sqrt {e x +d}\, b +\sqrt {e}\, \sqrt {b}\, \sqrt {a}\, \sqrt {2}}{\sqrt {b}\, \sqrt {a e -2 b d}\, \sqrt {2}}\right )\right )}{b e \left (a e -2 b d \right )} \] Input:

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

Output:

( - sqrt(b)*sqrt(a*e - 2*b*d)*sqrt(2)*(atan((2*sqrt(d + e*x)*b - sqrt(e)*s 
qrt(b)*sqrt(a)*sqrt(2))/(sqrt(b)*sqrt(a*e - 2*b*d)*sqrt(2))) + atan((2*sqr 
t(d + e*x)*b + sqrt(e)*sqrt(b)*sqrt(a)*sqrt(2))/(sqrt(b)*sqrt(a*e - 2*b*d) 
*sqrt(2)))))/(b*e*(a*e - 2*b*d))