\(\int \frac {A+B x}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^2} \, dx\) [364]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 33, antiderivative size = 209 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {(A b-a B) \sqrt {d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac {(6 b B d-5 A b e-a B e) \sqrt {d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac {e (6 b B d-5 A b e-a B e) \sqrt {d+e x}}{8 b (b d-a e)^3 (a+b x)}-\frac {e^2 (6 b B d-5 A b e-a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}} \] Output:

-1/3*(A*b-B*a)*(e*x+d)^(1/2)/b/(-a*e+b*d)/(b*x+a)^3-1/12*(-5*A*b*e-B*a*e+6 
*B*b*d)*(e*x+d)^(1/2)/b/(-a*e+b*d)^2/(b*x+a)^2+1/8*e*(-5*A*b*e-B*a*e+6*B*b 
*d)*(e*x+d)^(1/2)/b/(-a*e+b*d)^3/(b*x+a)-1/8*e^2*(-5*A*b*e-B*a*e+6*B*b*d)* 
arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(3/2)/(-a*e+b*d)^(7/2)
 

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {d+e x} \left (B \left (-3 a^3 e^2+6 b^3 d x (2 d-3 e x)+8 a^2 b e (-2 d+e x)+a b^2 \left (4 d^2-50 d e x+3 e^2 x^2\right )\right )+A b \left (33 a^2 e^2+2 a b e (-13 d+20 e x)+b^2 \left (8 d^2-10 d e x+15 e^2 x^2\right )\right )\right )}{24 b (-b d+a e)^3 (a+b x)^3}+\frac {e^2 (-6 b B d+5 A b e+a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{3/2} (-b d+a e)^{7/2}} \] Input:

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

Output:

(Sqrt[d + e*x]*(B*(-3*a^3*e^2 + 6*b^3*d*x*(2*d - 3*e*x) + 8*a^2*b*e*(-2*d 
+ e*x) + a*b^2*(4*d^2 - 50*d*e*x + 3*e^2*x^2)) + A*b*(33*a^2*e^2 + 2*a*b*e 
*(-13*d + 20*e*x) + b^2*(8*d^2 - 10*d*e*x + 15*e^2*x^2))))/(24*b*(-(b*d) + 
 a*e)^3*(a + b*x)^3) + (e^2*(-6*b*B*d + 5*A*b*e + a*B*e)*ArcTan[(Sqrt[b]*S 
qrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(3/2)*(-(b*d) + a*e)^(7/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1184, 27, 87, 52, 52, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^4 \int \frac {A+B x}{b^4 (a+b x)^4 \sqrt {d+e x}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {A+B x}{(a+b x)^4 \sqrt {d+e x}}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(-a B e-5 A b e+6 b B d) \int \frac {1}{(a+b x)^3 \sqrt {d+e x}}dx}{6 b (b d-a e)}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(-a B e-5 A b e+6 b B d) \left (-\frac {3 e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b (b d-a e)}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(-a B e-5 A b e+6 b B d) \left (-\frac {3 e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b (b d-a e)}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(-a B e-5 A b e+6 b B d) \left (-\frac {3 e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b (b d-a e)}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(-a B e-5 A b e+6 b B d) \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b (b d-a e)}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\)

Input:

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

Output:

-1/3*((A*b - a*B)*Sqrt[d + e*x])/(b*(b*d - a*e)*(a + b*x)^3) + ((6*b*B*d - 
 5*A*b*e - a*B*e)*(-1/2*Sqrt[d + e*x]/((b*d - a*e)*(a + b*x)^2) - (3*e*(-( 
Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) + (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x] 
)/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2))))/(4*(b*d - a*e))))/(6*b*( 
b*d - a*e))
 

Defintions of rubi rules used

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 52
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

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 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 1184
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 
Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00

method result size
pseudoelliptic \(\frac {\frac {11 \sqrt {b \left (a e -b d \right )}\, \left (\frac {4 \left (\left (-\frac {3 e x}{2}+d \right ) d x B +\frac {2 \left (\frac {15}{8} e^{2} x^{2}-\frac {5}{4} d e x +d^{2}\right ) A}{3}\right ) b^{3}}{11}-\frac {26 \left (\frac {\left (-\frac {3}{2} e^{2} x^{2}+25 d e x -2 d^{2}\right ) B}{13}+A e \left (-\frac {20 e x}{13}+d \right )\right ) a \,b^{2}}{33}+e \left (\frac {8 \left (e x -2 d \right ) B}{33}+A e \right ) a^{2} b -\frac {B \,e^{2} a^{3}}{11}\right ) \sqrt {e x +d}}{8}+\frac {5 \left (\left (A e -\frac {6 B d}{5}\right ) b +\frac {B a e}{5}\right ) e^{2} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{8}}{\sqrt {b \left (a e -b d \right )}\, \left (b x +a \right )^{3} \left (a e -b d \right )^{3} b}\) \(210\)
derivativedivides \(2 e^{2} \left (\frac {\frac {\left (5 A b e +B a e -6 B b d \right ) b \left (e x +d \right )^{\frac {5}{2}}}{16 e^{3} a^{3}-48 a^{2} b d \,e^{2}+48 a \,b^{2} d^{2} e -16 b^{3} d^{3}}+\frac {\left (5 A b e +B a e -6 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{6 e^{2} a^{2}-12 a b d e +6 b^{2} d^{2}}+\frac {\left (11 A b e -B a e -10 B b d \right ) \sqrt {e x +d}}{16 b \left (a e -b d \right )}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {\left (5 A b e +B a e -6 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 b \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {b \left (a e -b d \right )}}\right )\) \(266\)
default \(2 e^{2} \left (\frac {\frac {\left (5 A b e +B a e -6 B b d \right ) b \left (e x +d \right )^{\frac {5}{2}}}{16 e^{3} a^{3}-48 a^{2} b d \,e^{2}+48 a \,b^{2} d^{2} e -16 b^{3} d^{3}}+\frac {\left (5 A b e +B a e -6 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{6 e^{2} a^{2}-12 a b d e +6 b^{2} d^{2}}+\frac {\left (11 A b e -B a e -10 B b d \right ) \sqrt {e x +d}}{16 b \left (a e -b d \right )}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {\left (5 A b e +B a e -6 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 b \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {b \left (a e -b d \right )}}\right )\) \(266\)

Input:

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

Output:

11/8/(b*(a*e-b*d))^(1/2)*((b*(a*e-b*d))^(1/2)*(4/11*((-3/2*e*x+d)*d*x*B+2/ 
3*(15/8*e^2*x^2-5/4*d*e*x+d^2)*A)*b^3-26/33*(1/13*(-3/2*e^2*x^2+25*d*e*x-2 
*d^2)*B+A*e*(-20/13*e*x+d))*a*b^2+e*(8/33*(e*x-2*d)*B+A*e)*a^2*b-1/11*B*e^ 
2*a^3)*(e*x+d)^(1/2)+5/11*((A*e-6/5*B*d)*b+1/5*B*a*e)*e^2*(b*x+a)^3*arctan 
(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2)))/(b*x+a)^3/(a*e-b*d)^3/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 662 vs. \(2 (185) = 370\).

Time = 0.13 (sec) , antiderivative size = 1337, normalized size of antiderivative = 6.40 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fric 
as")
 

Output:

[1/48*(3*(6*B*a^3*b*d*e^2 - (B*a^4 + 5*A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (B* 
a*b^3 + 5*A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (B*a^2*b^2 + 5*A*a*b^3)*e 
^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (B*a^3*b + 5*A*a^2*b^2)*e^3)*x)*sqrt(b^2* 
d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d)) 
/(b*x + a)) - 2*(4*(B*a*b^4 + 2*A*b^5)*d^3 - 2*(10*B*a^2*b^3 + 17*A*a*b^4) 
*d^2*e + (13*B*a^3*b^2 + 59*A*a^2*b^3)*d*e^2 + 3*(B*a^4*b - 11*A*a^3*b^2)* 
e^3 - 3*(6*B*b^5*d^2*e - (7*B*a*b^4 + 5*A*b^5)*d*e^2 + (B*a^2*b^3 + 5*A*a* 
b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 - (31*B*a*b^4 + 5*A*b^5)*d^2*e + (29*B*a^2* 
b^3 + 25*A*a*b^4)*d*e^2 - 4*(B*a^3*b^2 + 5*A*a^2*b^3)*e^3)*x)*sqrt(e*x + d 
))/(a^3*b^6*d^4 - 4*a^4*b^5*d^3*e + 6*a^5*b^4*d^2*e^2 - 4*a^6*b^3*d*e^3 + 
a^7*b^2*e^4 + (b^9*d^4 - 4*a*b^8*d^3*e + 6*a^2*b^7*d^2*e^2 - 4*a^3*b^6*d*e 
^3 + a^4*b^5*e^4)*x^3 + 3*(a*b^8*d^4 - 4*a^2*b^7*d^3*e + 6*a^3*b^6*d^2*e^2 
 - 4*a^4*b^5*d*e^3 + a^5*b^4*e^4)*x^2 + 3*(a^2*b^7*d^4 - 4*a^3*b^6*d^3*e + 
 6*a^4*b^5*d^2*e^2 - 4*a^5*b^4*d*e^3 + a^6*b^3*e^4)*x), 1/24*(3*(6*B*a^3*b 
*d*e^2 - (B*a^4 + 5*A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (B*a*b^3 + 5*A*b^4)*e^ 
3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (B*a^2*b^2 + 5*A*a*b^3)*e^3)*x^2 + 3*(6*B*a^ 
2*b^2*d*e^2 - (B*a^3*b + 5*A*a^2*b^2)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan( 
sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (4*(B*a*b^4 + 2*A*b^5) 
*d^3 - 2*(10*B*a^2*b^3 + 17*A*a*b^4)*d^2*e + (13*B*a^3*b^2 + 59*A*a^2*b^3) 
*d*e^2 + 3*(B*a^4*b - 11*A*a^3*b^2)*e^3 - 3*(6*B*b^5*d^2*e - (7*B*a*b^4...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxi 
ma")
 

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*e-b*d>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. 429 vs. \(2 (185) = 370\).

Time = 0.21 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.05 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {{\left (6 \, B b d e^{2} - B a e^{3} - 5 \, A b e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {18 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} d e^{2} - 48 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e^{2} + 30 \, \sqrt {e x + d} B b^{3} d^{3} e^{2} - 3 \, {\left (e x + d\right )}^{\frac {5}{2}} B a b^{2} e^{3} - 15 \, {\left (e x + d\right )}^{\frac {5}{2}} A b^{3} e^{3} + 56 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} d e^{3} + 40 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} d e^{3} - 57 \, \sqrt {e x + d} B a b^{2} d^{2} e^{3} - 33 \, \sqrt {e x + d} A b^{3} d^{2} e^{3} - 8 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b e^{4} - 40 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{2} e^{4} + 24 \, \sqrt {e x + d} B a^{2} b d e^{4} + 66 \, \sqrt {e x + d} A a b^{2} d e^{4} + 3 \, \sqrt {e x + d} B a^{3} e^{5} - 33 \, \sqrt {e x + d} A a^{2} b e^{5}}{24 \, {\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \] Input:

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

Output:

1/8*(6*B*b*d*e^2 - B*a*e^3 - 5*A*b*e^3)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d 
 + a*b*e))/((b^4*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*sqrt(- 
b^2*d + a*b*e)) + 1/24*(18*(e*x + d)^(5/2)*B*b^3*d*e^2 - 48*(e*x + d)^(3/2 
)*B*b^3*d^2*e^2 + 30*sqrt(e*x + d)*B*b^3*d^3*e^2 - 3*(e*x + d)^(5/2)*B*a*b 
^2*e^3 - 15*(e*x + d)^(5/2)*A*b^3*e^3 + 56*(e*x + d)^(3/2)*B*a*b^2*d*e^3 + 
 40*(e*x + d)^(3/2)*A*b^3*d*e^3 - 57*sqrt(e*x + d)*B*a*b^2*d^2*e^3 - 33*sq 
rt(e*x + d)*A*b^3*d^2*e^3 - 8*(e*x + d)^(3/2)*B*a^2*b*e^4 - 40*(e*x + d)^( 
3/2)*A*a*b^2*e^4 + 24*sqrt(e*x + d)*B*a^2*b*d*e^4 + 66*sqrt(e*x + d)*A*a*b 
^2*d*e^4 + 3*sqrt(e*x + d)*B*a^3*e^5 - 33*sqrt(e*x + d)*A*a^2*b*e^5)/((b^4 
*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*((e*x + d)*b - b*d + a 
*e)^3)
 

Mupad [B] (verification not implemented)

Time = 11.14 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {{\left (d+e\,x\right )}^{3/2}\,\left (5\,A\,b\,e^3+B\,a\,e^3-6\,B\,b\,d\,e^2\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {b\,{\left (d+e\,x\right )}^{5/2}\,\left (5\,A\,b\,e^3+B\,a\,e^3-6\,B\,b\,d\,e^2\right )}{8\,{\left (a\,e-b\,d\right )}^3}-\frac {\sqrt {d+e\,x}\,\left (B\,a\,e^3-11\,A\,b\,e^3+10\,B\,b\,d\,e^2\right )}{8\,b\,\left (a\,e-b\,d\right )}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2}+\frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {d+e\,x}\,\left (5\,A\,b\,e+B\,a\,e-6\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (5\,A\,b\,e^3+B\,a\,e^3-6\,B\,b\,d\,e^2\right )}\right )\,\left (5\,A\,b\,e+B\,a\,e-6\,B\,b\,d\right )}{8\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{7/2}} \] Input:

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

Output:

(((d + e*x)^(3/2)*(5*A*b*e^3 + B*a*e^3 - 6*B*b*d*e^2))/(3*(a*e - b*d)^2) + 
 (b*(d + e*x)^(5/2)*(5*A*b*e^3 + B*a*e^3 - 6*B*b*d*e^2))/(8*(a*e - b*d)^3) 
 - ((d + e*x)^(1/2)*(B*a*e^3 - 11*A*b*e^3 + 10*B*b*d*e^2))/(8*b*(a*e - b*d 
)))/((d + e*x)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e) + b^3*(d + e*x)^3 - 
 (3*b^3*d - 3*a*b^2*e)*(d + e*x)^2 + a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3 
*a^2*b*d*e^2) + (e^2*atan((b^(1/2)*e^2*(d + e*x)^(1/2)*(5*A*b*e + B*a*e - 
6*B*b*d))/((a*e - b*d)^(1/2)*(5*A*b*e^3 + B*a*e^3 - 6*B*b*d*e^2)))*(5*A*b* 
e + B*a*e - 6*B*b*d))/(8*b^(3/2)*(a*e - b*d)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.68 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} e^{2}+6 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a b \,e^{2} x +3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{2} e^{2} x^{2}+5 \sqrt {e x +d}\, a^{2} b \,e^{2}-7 \sqrt {e x +d}\, a \,b^{2} d e +3 \sqrt {e x +d}\, a \,b^{2} e^{2} x +2 \sqrt {e x +d}\, b^{3} d^{2}-3 \sqrt {e x +d}\, b^{3} d e x}{4 b \left (a^{3} b^{2} e^{3} x^{2}-3 a^{2} b^{3} d \,e^{2} x^{2}+3 a \,b^{4} d^{2} e \,x^{2}-b^{5} d^{3} x^{2}+2 a^{4} b \,e^{3} x -6 a^{3} b^{2} d \,e^{2} x +6 a^{2} b^{3} d^{2} e x -2 a \,b^{4} d^{3} x +a^{5} e^{3}-3 a^{4} b d \,e^{2}+3 a^{3} b^{2} d^{2} e -a^{2} b^{3} d^{3}\right )} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d) 
))*a**2*e**2 + 6*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*s 
qrt(a*e - b*d)))*a*b*e**2*x + 3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x 
)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**2*e**2*x**2 + 5*sqrt(d + e*x)*a**2*b*e* 
*2 - 7*sqrt(d + e*x)*a*b**2*d*e + 3*sqrt(d + e*x)*a*b**2*e**2*x + 2*sqrt(d 
 + e*x)*b**3*d**2 - 3*sqrt(d + e*x)*b**3*d*e*x)/(4*b*(a**5*e**3 - 3*a**4*b 
*d*e**2 + 2*a**4*b*e**3*x + 3*a**3*b**2*d**2*e - 6*a**3*b**2*d*e**2*x + a* 
*3*b**2*e**3*x**2 - a**2*b**3*d**3 + 6*a**2*b**3*d**2*e*x - 3*a**2*b**3*d* 
e**2*x**2 - 2*a*b**4*d**3*x + 3*a*b**4*d**2*e*x**2 - b**5*d**3*x**2))