\(\int \frac {A+B x}{\sqrt {d+e x} (b x+c x^2)^2} \, dx\) [82]

Optimal result

Integrand size = 26, antiderivative size = 195 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (b x+c x^2\right )^2} \, dx=\frac {c (b B d-2 A c d+A b e) \sqrt {d+e x}}{b^2 d (c d-b e) (b+c x)}-\frac {A \sqrt {d+e x}}{b d x (b+c x)}-\frac {(2 b B d-4 A c d-A b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3 d^{3/2}}+\frac {\sqrt {c} \left (2 b B c d-4 A c^2 d-3 b^2 B e+5 A b c e\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 (c d-b e)^{3/2}} \] Output:

c*(A*b*e-2*A*c*d+B*b*d)*(e*x+d)^(1/2)/b^2/d/(-b*e+c*d)/(c*x+b)-A*(e*x+d)^( 
1/2)/b/d/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^(3/2)+c^(1/2)*(5*A*b*c*e-4*A*c^2*d-3*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/(-b*e+c*d)^(3/2)
 

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (b x+c x^2\right )^2} \, dx=\frac {\frac {b \sqrt {d+e x} \left (b B c d x+A \left (-b c d+b^2 e-2 c^2 d x+b c e x\right )\right )}{d (c d-b e) x (b+c x)}-\frac {\sqrt {c} \left (4 A c^2 d+3 b^2 B e-b c (2 B d+5 A e)\right ) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{(-c d+b e)^{3/2}}+\frac {(-2 b B d+4 A c d+A b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{d^{3/2}}}{b^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1235, 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.

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

\(\Big \downarrow \) 1235

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((Sqrt[d + e*x]*(A*b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x))/(b^2*d 
*(c*d - b*e)*(b*x + c*x^2))) - (e*(((c*d - b*e)*(2*b*B*d - 4*A*c*d - A*b*e 
)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d]*e) - (Sqrt[c]*d*(2*b*B*c*d - 
4*A*c^2*d - 3*b^2*B*e + 5*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c* 
d - b*e]])/(b*e*Sqrt[c*d - b*e])))/(b^2*d*(c*d - b*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 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 1235

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 
*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*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 
*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 
 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* 
m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - 
f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && LtQ[p, -1] && (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 = 1.22 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(2 e^{2} \left (\frac {c \left (\frac {\left (A c -B b \right ) b e \sqrt {e x +d}}{2 \left (b e -c d \right ) \left (\left (e x +d \right ) c +b e -c d \right )}+\frac {\left (5 A c e b -4 A \,c^{2} d -3 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \left (b e -c d \right ) \sqrt {c \left (b e -c d \right )}}\right )}{b^{3} e^{2}}+\frac {-\frac {A b \sqrt {e x +d}}{2 d x}+\frac {\left (A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 d^{\frac {3}{2}}}}{e^{2} b^{3}}\right )\)	\(192\)
default	\(2 e^{2} \left (\frac {c \left (\frac {\left (A c -B b \right ) b e \sqrt {e x +d}}{2 \left (b e -c d \right ) \left (\left (e x +d \right ) c +b e -c d \right )}+\frac {\left (5 A c e b -4 A \,c^{2} d -3 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \left (b e -c d \right ) \sqrt {c \left (b e -c d \right )}}\right )}{b^{3} e^{2}}+\frac {-\frac {A b \sqrt {e x +d}}{2 d x}+\frac {\left (A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 d^{\frac {3}{2}}}}{e^{2} b^{3}}\right )\)	\(192\)
risch	\(-\frac {A \sqrt {e x +d}}{d \,b^{2} x}-\frac {e \left (-\frac {\left (A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e \sqrt {d}}-\frac {2 c d \left (\frac {\left (A c -B b \right ) b e \sqrt {e x +d}}{2 \left (b e -c d \right ) \left (\left (e x +d \right ) c +b e -c d \right )}+\frac {\left (5 A c e b -4 A \,c^{2} d -3 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \left (b e -c d \right ) \sqrt {c \left (b e -c d \right )}}\right )}{b e}\right )}{b^{2} d}\)	\(199\)
pseudoelliptic	\(-\frac {4 \left (c x \left (c x +b \right ) \left (A \,c^{2} d -\frac {5 b \left (A e +\frac {2 B d}{5}\right ) c}{4}+\frac {3 b^{2} B e}{4}\right ) d^{\frac {5}{2}} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )-\frac {\sqrt {c \left (b e -c d \right )}\, \left (d \left (4 A c d +b \left (A e -2 B d \right )\right ) x \left (c x +b \right ) \left (b e -c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )+\left (2 A \,c^{2} d x +\left (d \left (-B x +A \right )-A e x \right ) b c -A \,b^{2} e \right ) b \sqrt {e x +d}\, d^{\frac {3}{2}}\right )}{4}\right )}{\sqrt {c \left (b e -c d \right )}\, d^{\frac {5}{2}} x \,b^{3} \left (b e -c d \right ) \left (c x +b \right )}\)	\(205\)

Input:

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

Output:

2*e^2*(c/b^3/e^2*(1/2*(A*c-B*b)*b*e/(b*e-c*d)*(e*x+d)^(1/2)/((e*x+d)*c+b*e 
-c*d)+1/2*(5*A*b*c*e-4*A*c^2*d-3*B*b^2*e+2*B*b*c*d)/(b*e-c*d)/(c*(b*e-c*d) 
)^(1/2)*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2)))+1/e^2/b^3*(-1/2*A*b/d 
*(e*x+d)^(1/2)/x+1/2*(A*b*e+4*A*c*d-2*B*b*d)/d^(3/2)*arctanh((e*x+d)^(1/2) 
/d^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (175) = 350\).

Time = 1.37 (sec) , antiderivative size = 1496, normalized size of antiderivative = 7.67 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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

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 = 0.27 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.50 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (b x+c x^2\right )^2} \, dx=-\frac {{\left (2 \, B b c^{2} d - 4 \, A c^{3} d - 3 \, B b^{2} c e + 5 \, A b c^{2} e\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{{\left (b^{3} c d - b^{4} e\right )} \sqrt {-c^{2} d + b c e}} + \frac {{\left (e x + d\right )}^{\frac {3}{2}} B b c d e - 2 \, {\left (e x + d\right )}^{\frac {3}{2}} A c^{2} d e - \sqrt {e x + d} B b c d^{2} e + 2 \, \sqrt {e x + d} A c^{2} d^{2} e + {\left (e x + d\right )}^{\frac {3}{2}} A b c e^{2} - 2 \, \sqrt {e x + d} A b c d e^{2} + \sqrt {e x + d} A b^{2} e^{3}}{{\left (b^{2} c d^{2} - b^{3} d e\right )} {\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e\right )}} + \frac {{\left (2 \, B b d - 4 \, A c d - A b e\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b^{3} \sqrt {-d} d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 13.54 (sec) , antiderivative size = 5828, normalized size of antiderivative = 29.89 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(((d + e*x)^(1/2)*(A*b^2*e^3 + 2*A*c^2*d^2*e - 2*A*b*c*d*e^2 - B*b*c*d^2*e 
))/(b^2*(c*d^2 - b*d*e)) + (c*(d + e*x)^(3/2)*(A*b*e^2 - 2*A*c*d*e + B*b*d 
*e))/(b^2*(c*d^2 - b*d*e)))/((b*e - 2*c*d)*(d + e*x) + c*(d + e*x)^2 + c*d 
^2 - b*d*e) + (atan(((((2*(d + e*x)^(1/2)*(A^2*b^4*c^3*e^6 + 32*A^2*c^7*d^ 
4*e^2 + 26*A^2*b^2*c^5*d^2*e^4 + 8*B^2*b^2*c^5*d^4*e^2 - 20*B^2*b^3*c^4*d^ 
3*e^3 + 13*B^2*b^4*c^3*d^2*e^4 - 64*A^2*b*c^6*d^3*e^3 + 6*A^2*b^3*c^4*d*e^ 
5 - 32*A*B*b*c^6*d^4*e^2 - 4*A*B*b^4*c^3*d*e^5 + 72*A*B*b^2*c^5*d^3*e^3 - 
38*A*B*b^3*c^4*d^2*e^4))/(b^4*c^2*d^4 + b^6*d^2*e^2 - 2*b^5*c*d^3*e) - ((( 
4*A*b^9*c^2*d*e^6 + 8*A*b^6*c^5*d^4*e^3 - 16*A*b^7*c^4*d^3*e^4 + 4*A*b^8*c 
^3*d^2*e^5 - 4*B*b^7*c^4*d^4*e^3 + 12*B*b^8*c^3*d^3*e^4 - 8*B*b^9*c^2*d^2* 
e^5)/(b^6*c^2*d^4 + b^8*d^2*e^2 - 2*b^7*c*d^3*e) + ((-c*(b*e - c*d)^3)^(1/ 
2)*(d + e*x)^(1/2)*(8*b^6*c^5*d^5*e^2 - 20*b^7*c^4*d^4*e^3 + 16*b^8*c^3*d^ 
3*e^4 - 4*b^9*c^2*d^2*e^5)*(4*A*c^2*d + 3*B*b^2*e - 5*A*b*c*e - 2*B*b*c*d) 
)/((b^4*c^2*d^4 + b^6*d^2*e^2 - 2*b^5*c*d^3*e)*(b^6*e^3 - b^3*c^3*d^3 + 3* 
b^4*c^2*d^2*e - 3*b^5*c*d*e^2)))*(-c*(b*e - c*d)^3)^(1/2)*(4*A*c^2*d + 3*B 
*b^2*e - 5*A*b*c*e - 2*B*b*c*d))/(2*(b^6*e^3 - b^3*c^3*d^3 + 3*b^4*c^2*d^2 
*e - 3*b^5*c*d*e^2)))*(-c*(b*e - c*d)^3)^(1/2)*(4*A*c^2*d + 3*B*b^2*e - 5* 
A*b*c*e - 2*B*b*c*d)*1i)/(2*(b^6*e^3 - b^3*c^3*d^3 + 3*b^4*c^2*d^2*e - 3*b 
^5*c*d*e^2)) + (((2*(d + e*x)^(1/2)*(A^2*b^4*c^3*e^6 + 32*A^2*c^7*d^4*e^2 
+ 26*A^2*b^2*c^5*d^2*e^4 + 8*B^2*b^2*c^5*d^4*e^2 - 20*B^2*b^3*c^4*d^3*e...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1297, normalized size of antiderivative = 6.65 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(10*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**2*e*x - 8*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**3*x + 10*sqrt(c)*sqrt(b*e - c*d)*ata 
n((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*b*c**2*d**2*e*x**2 - 8*sq 
rt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a* 
c**3*d**3*x**2 - 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c) 
*sqrt(b*e - c*d)))*b**4*d**2*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**3*x - 6*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*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**3*x**2 - 2*sqrt(d + e*x)*a*b**4*d*e**2 + 4*sqrt(d + e*x)* 
a*b**3*c*d**2*e - 2*sqrt(d + e*x)*a*b**3*c*d*e**2*x - 2*sqrt(d + e*x)*a*b* 
*2*c**2*d**3 + 6*sqrt(d + e*x)*a*b**2*c**2*d**2*e*x - 4*sqrt(d + e*x)*a*b* 
c**3*d**3*x - 2*sqrt(d + e*x)*b**4*c*d**2*e*x + 2*sqrt(d + e*x)*b**3*c**2* 
d**3*x - sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*b**4*e**3*x - 2*sqrt(d)*lo 
g(sqrt(d + e*x) - sqrt(d))*a*b**3*c*d*e**2*x - sqrt(d)*log(sqrt(d + e*x) - 
 sqrt(d))*a*b**3*c*e**3*x**2 + 7*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*b* 
*2*c**2*d**2*e*x - 2*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*b**2*c**2*d*e* 
*2*x**2 - 4*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*b*c**3*d**3*x + 7*sqrt( 
d)*log(sqrt(d + e*x) - sqrt(d))*a*b*c**3*d**2*e*x**2 - 4*sqrt(d)*log(sq...