\(\int \frac {A+B x}{(d+e x)^3 (b x+c x^2)^{3/2}} \, dx\) [131]

Optimal result

Integrand size = 26, antiderivative size = 384 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {8 A c^2 d^2+4 b c d (2 B d-7 A e)-3 b^2 e (B d-5 A e)}{4 b d^3 (c d-b e)^2 \sqrt {b x+c x^2}}-\frac {c \left (16 A c^3 d^3-2 b^2 c d e (5 B d-19 A e)+3 b^3 e^2 (B d-5 A e)-8 b c^2 d^2 (B d+3 A e)\right ) x}{4 b^2 d^3 (c d-b e)^3 \sqrt {b x+c x^2}}+\frac {B d-A e}{2 d (c d-b e) (d+e x)^2 \sqrt {b x+c x^2}}-\frac {5 A e (2 c d-b e)-B d (6 c d-b e)}{4 d^2 (c d-b e)^2 (d+e x) \sqrt {b x+c x^2}}-\frac {3 e \left (B d \left (8 c^2 d^2-4 b c d e+b^2 e^2\right )-A e \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c d-b e} x}{\sqrt {d} \sqrt {b x+c x^2}}\right )}{4 d^{7/2} (c d-b e)^{7/2}} \] Output:

-1/4*(8*A*c^2*d^2+4*b*c*d*(-7*A*e+2*B*d)-3*b^2*e*(-5*A*e+B*d))/b/d^3/(-b*e 
+c*d)^2/(c*x^2+b*x)^(1/2)-1/4*c*(16*A*c^3*d^3-2*b^2*c*d*e*(-19*A*e+5*B*d)+ 
3*b^3*e^2*(-5*A*e+B*d)-8*b*c^2*d^2*(3*A*e+B*d))*x/b^2/d^3/(-b*e+c*d)^3/(c* 
x^2+b*x)^(1/2)+1/2*(-A*e+B*d)/d/(-b*e+c*d)/(e*x+d)^2/(c*x^2+b*x)^(1/2)-1/4 
*(5*A*e*(-b*e+2*c*d)-B*d*(-b*e+6*c*d))/d^2/(-b*e+c*d)^2/(e*x+d)/(c*x^2+b*x 
)^(1/2)-3/4*e*(B*d*(b^2*e^2-4*b*c*d*e+8*c^2*d^2)-A*e*(5*b^2*e^2-16*b*c*d*e 
+16*c^2*d^2))*arctanh((-b*e+c*d)^(1/2)*x/d^(1/2)/(c*x^2+b*x)^(1/2))/d^(7/2 
)/(-b*e+c*d)^(7/2)
 

Mathematica [A] (verified)

Time = 11.54 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 b^2 d^{5/2} (B d-A e) (c d-b e)^{5/2}+(d+e x) \left (b^2 d^{3/2} (c d-b e)^{3/2} (B d (6 c d-b e)+5 A e (-2 c d+b e))-(d+e x) \left (b \sqrt {d} (c d-b e)^{3/2} \left (8 A c^2 d^2+4 b c d (2 B d-7 A e)+3 b^2 e (-B d+5 A e)\right )+c \sqrt {d} \sqrt {c d-b e} \left (16 A c^3 d^3+3 b^3 e^2 (B d-5 A e)-8 b c^2 d^2 (B d+3 A e)+2 b^2 c d e (-5 B d+19 A e)\right ) x+3 b^2 e \left (A e \left (-16 c^2 d^2+16 b c d e-5 b^2 e^2\right )+B d \left (8 c^2 d^2-4 b c d e+b^2 e^2\right )\right ) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )\right )\right )}{4 b^2 d^{7/2} (c d-b e)^{7/2} \sqrt {x (b+c x)} (d+e x)^2} \] Input:

Integrate[(A + B*x)/((d + e*x)^3*(b*x + c*x^2)^(3/2)),x]
 

Output:

(2*b^2*d^(5/2)*(B*d - A*e)*(c*d - b*e)^(5/2) + (d + e*x)*(b^2*d^(3/2)*(c*d 
 - b*e)^(3/2)*(B*d*(6*c*d - b*e) + 5*A*e*(-2*c*d + b*e)) - (d + e*x)*(b*Sq 
rt[d]*(c*d - b*e)^(3/2)*(8*A*c^2*d^2 + 4*b*c*d*(2*B*d - 7*A*e) + 3*b^2*e*( 
-(B*d) + 5*A*e)) + c*Sqrt[d]*Sqrt[c*d - b*e]*(16*A*c^3*d^3 + 3*b^3*e^2*(B* 
d - 5*A*e) - 8*b*c^2*d^2*(B*d + 3*A*e) + 2*b^2*c*d*e*(-5*B*d + 19*A*e))*x 
+ 3*b^2*e*(A*e*(-16*c^2*d^2 + 16*b*c*d*e - 5*b^2*e^2) + B*d*(8*c^2*d^2 - 4 
*b*c*d*e + b^2*e^2))*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x 
])/(Sqrt[d]*Sqrt[b + c*x])])))/(4*b^2*d^(7/2)*(c*d - b*e)^(7/2)*Sqrt[x*(b 
+ c*x)]*(d + e*x)^2)
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1235, 27, 1237, 27, 1228, 1154, 219}

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.

\(\displaystyle \int \frac {A+B x}{\left (b x+c x^2\right )^{3/2} (d+e x)^3} \, dx\)

\(\Big \downarrow \) 1235

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1237

\(\displaystyle -\frac {e \left (\frac {\sqrt {b x+c x^2} \left (b^2 (-e) (B d-5 A e)-4 b c d (2 A e+B d)+8 A c^2 d^2\right )}{2 d (d+e x)^2 (c d-b e)}-\frac {\int -\frac {b \left (-3 e (B d-5 A e) b^2+4 c d (2 B d-7 A e) b+8 A c^2 d^2\right )+2 c \left (-e (B d-5 A e) b^2-4 c d (B d+2 A e) b+8 A c^2 d^2\right ) x}{2 (d+e x)^2 \sqrt {c x^2+b x}}dx}{2 d (c d-b e)}\right )}{b^2 d (c d-b e)}-\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (d+e x)^2 (c d-b e)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {e \left (\frac {\int \frac {b \left (-3 e (B d-5 A e) b^2+4 c d (2 B d-7 A e) b+8 A c^2 d^2\right )+2 c \left (-e (B d-5 A e) b^2-4 c d (B d+2 A e) b+8 A c^2 d^2\right ) x}{(d+e x)^2 \sqrt {c x^2+b x}}dx}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} \left (b^2 (-e) (B d-5 A e)-4 b c d (2 A e+B d)+8 A c^2 d^2\right )}{2 d (d+e x)^2 (c d-b e)}\right )}{b^2 d (c d-b e)}-\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (d+e x)^2 (c d-b e)}\)

\(\Big \downarrow \) 1228

\(\displaystyle -\frac {e \left (\frac {\frac {3 b^2 \left (B d \left (b^2 e^2-4 b c d e+8 c^2 d^2\right )-A e \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{2 d (c d-b e)}+\frac {\sqrt {b x+c x^2} \left (3 b^3 e^2 (B d-5 A e)-2 b^2 c d e (5 B d-19 A e)-8 b c^2 d^2 (3 A e+B d)+16 A c^3 d^3\right )}{d (d+e x) (c d-b e)}}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} \left (b^2 (-e) (B d-5 A e)-4 b c d (2 A e+B d)+8 A c^2 d^2\right )}{2 d (d+e x)^2 (c d-b e)}\right )}{b^2 d (c d-b e)}-\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (d+e x)^2 (c d-b e)}\)

\(\Big \downarrow \) 1154

\(\displaystyle -\frac {e \left (\frac {\frac {\sqrt {b x+c x^2} \left (3 b^3 e^2 (B d-5 A e)-2 b^2 c d e (5 B d-19 A e)-8 b c^2 d^2 (3 A e+B d)+16 A c^3 d^3\right )}{d (d+e x) (c d-b e)}-\frac {3 b^2 \left (B d \left (b^2 e^2-4 b c d e+8 c^2 d^2\right )-A e \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )\right ) \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{d (c d-b e)}}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} \left (b^2 (-e) (B d-5 A e)-4 b c d (2 A e+B d)+8 A c^2 d^2\right )}{2 d (d+e x)^2 (c d-b e)}\right )}{b^2 d (c d-b e)}-\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (d+e x)^2 (c d-b e)}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {e \left (\frac {\frac {3 b^2 \left (B d \left (b^2 e^2-4 b c d e+8 c^2 d^2\right )-A e \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )\right ) \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}}+\frac {\sqrt {b x+c x^2} \left (3 b^3 e^2 (B d-5 A e)-2 b^2 c d e (5 B d-19 A e)-8 b c^2 d^2 (3 A e+B d)+16 A c^3 d^3\right )}{d (d+e x) (c d-b e)}}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} \left (b^2 (-e) (B d-5 A e)-4 b c d (2 A e+B d)+8 A c^2 d^2\right )}{2 d (d+e x)^2 (c d-b e)}\right )}{b^2 d (c d-b e)}-\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (d+e x)^2 (c d-b e)}\)

Input:

Int[(A + B*x)/((d + e*x)^3*(b*x + c*x^2)^(3/2)),x]
 

Output:

(-2*(A*b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x))/(b^2*d*(c*d - b*e)* 
(d + e*x)^2*Sqrt[b*x + c*x^2]) - (e*(((8*A*c^2*d^2 - b^2*e*(B*d - 5*A*e) - 
 4*b*c*d*(B*d + 2*A*e))*Sqrt[b*x + c*x^2])/(2*d*(c*d - b*e)*(d + e*x)^2) + 
 (((16*A*c^3*d^3 - 2*b^2*c*d*e*(5*B*d - 19*A*e) + 3*b^3*e^2*(B*d - 5*A*e) 
- 8*b*c^2*d^2*(B*d + 3*A*e))*Sqrt[b*x + c*x^2])/(d*(c*d - b*e)*(d + e*x)) 
+ (3*b^2*(B*d*(8*c^2*d^2 - 4*b*c*d*e + b^2*e^2) - A*e*(16*c^2*d^2 - 16*b*c 
*d*e + 5*b^2*e^2))*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b 
*e]*Sqrt[b*x + c*x^2])])/(2*d^(3/2)*(c*d - b*e)^(3/2)))/(4*d*(c*d - b*e))) 
)/(b^2*d*(c*d - b*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1228

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

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 1237

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

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 363, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\frac {15 e \left (e x +d \right )^{2} b^{2} \left (-\frac {8 B \,c^{2} d^{3}}{5}+\frac {16 c \left (A c +\frac {B b}{4}\right ) e \,d^{2}}{5}-\frac {16 e^{2} b \left (A c +\frac {B b}{16}\right ) d}{5}+A \,b^{2} e^{3}\right ) \sqrt {x \left (c x +b \right )}\, \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )}{8}+\sqrt {d \left (b e -c d \right )}\, \left (-c^{3} \left (2 A c x +b \left (-B x +A \right )\right ) d^{5}+3 \left (b^{2} A +\frac {c x \left (2 B x +A \right ) b}{3}-\frac {4 A \,c^{2} x^{2}}{3}\right ) c^{2} e \,d^{4}-3 c \,e^{2} \left (\left (-\frac {B x}{2}+A \right ) b^{3}-c x \left (\frac {B x}{2}+A \right ) b^{2}-\frac {5 c^{2} x^{2} \left (\frac {B x}{5}+A \right ) b}{3}+\frac {2 A \,c^{3} x^{3}}{3}\right ) d^{3}+e^{3} \left (c x +b \right ) b \left (\left (-\frac {5 B x}{8}+A \right ) b^{2}-8 c \left (-\frac {5 B x}{32}+A \right ) x b +3 A \,c^{2} x^{2}\right ) d^{2}+\frac {25 e^{4} \left (\left (-\frac {3 B x}{25}+A \right ) b -\frac {38 A c x}{25}\right ) x \left (c x +b \right ) b^{2} d}{8}+\frac {15 A \,b^{3} e^{5} x^{2} \left (c x +b \right )}{8}\right )\right )}{\sqrt {x \left (c x +b \right )}\, \sqrt {d \left (b e -c d \right )}\, d^{3} \left (e x +d \right )^{2} \left (b e -c d \right )^{3} b^{2}}\)	\(363\)
risch	\(\text {Expression too large to display}\)	\(1025\)
default	\(\text {Expression too large to display}\)	\(1582\)

Input:

int((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/(x*(c*x+b))^(1/2)*(-15/8*e*(e*x+d)^2*b^2*(-8/5*B*c^2*d^3+16/5*c*(A*c+1/ 
4*B*b)*e*d^2-16/5*e^2*b*(A*c+1/16*B*b)*d+A*b^2*e^3)*(x*(c*x+b))^(1/2)*arct 
an((x*(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1/2))+(d*(b*e-c*d))^(1/2)*(-c^3*(2 
*A*c*x+b*(-B*x+A))*d^5+3*(b^2*A+1/3*c*x*(2*B*x+A)*b-4/3*A*c^2*x^2)*c^2*e*d 
^4-3*c*e^2*((-1/2*B*x+A)*b^3-c*x*(1/2*B*x+A)*b^2-5/3*c^2*x^2*(1/5*B*x+A)*b 
+2/3*A*c^3*x^3)*d^3+e^3*(c*x+b)*b*((-5/8*B*x+A)*b^2-8*c*(-5/32*B*x+A)*x*b+ 
3*A*c^2*x^2)*d^2+25/8*e^4*((-3/25*B*x+A)*b-38/25*A*c*x)*x*(c*x+b)*b^2*d+15 
/8*A*b^3*e^5*x^2*(c*x+b)))/(d*(b*e-c*d))^(1/2)/d^3/(e*x+d)^2/(b*e-c*d)^3/b 
^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1132 vs. \(2 (355) = 710\).

Time = 0.24 (sec) , antiderivative size = 2279, normalized size of antiderivative = 5.93 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*(3*((8*B*b^2*c^3*d^3*e^3 - 5*A*b^4*c*e^6 - 4*(B*b^3*c^2 + 4*A*b^2*c^3 
)*d^2*e^4 + (B*b^4*c + 16*A*b^3*c^2)*d*e^5)*x^4 + (16*B*b^2*c^3*d^4*e^2 - 
32*A*b^2*c^3*d^3*e^3 - 5*A*b^5*e^6 - 2*(B*b^4*c - 8*A*b^3*c^2)*d^2*e^4 + ( 
B*b^5 + 6*A*b^4*c)*d*e^5)*x^3 + (8*B*b^2*c^3*d^5*e - 10*A*b^5*d*e^5 + 4*(3 
*B*b^3*c^2 - 4*A*b^2*c^3)*d^4*e^2 - (7*B*b^4*c + 16*A*b^3*c^2)*d^3*e^3 + ( 
2*B*b^5 + 27*A*b^4*c)*d^2*e^4)*x^2 + (8*B*b^3*c^2*d^5*e - 5*A*b^5*d^2*e^4 
- 4*(B*b^4*c + 4*A*b^3*c^2)*d^4*e^2 + (B*b^5 + 16*A*b^4*c)*d^3*e^3)*x)*sqr 
t(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x - 2*sqrt(c*d^2 - b*d*e)*sqrt(c 
*x^2 + b*x))/(e*x + d)) - 2*(8*A*b*c^4*d^7 - 32*A*b^2*c^3*d^6*e + 48*A*b^3 
*c^2*d^5*e^2 - 32*A*b^4*c*d^4*e^3 + 8*A*b^5*d^3*e^4 + (15*A*b^4*c*d*e^6 - 
8*(B*b*c^4 - 2*A*c^5)*d^5*e^2 - 2*(B*b^2*c^3 + 20*A*b*c^4)*d^4*e^3 + (13*B 
*b^3*c^2 + 62*A*b^2*c^3)*d^3*e^4 - (3*B*b^4*c + 53*A*b^3*c^2)*d^2*e^5)*x^3 
 + (15*A*b^5*d*e^6 - 16*(B*b*c^4 - 2*A*c^5)*d^6*e + 4*(B*b^2*c^3 - 18*A*b* 
c^4)*d^5*e^2 + (7*B*b^3*c^2 + 80*A*b^2*c^3)*d^4*e^3 + (8*B*b^4*c - 27*A*b^ 
3*c^2)*d^3*e^4 - (3*B*b^5 + 28*A*b^4*c)*d^2*e^5)*x^2 + (25*A*b^5*d^2*e^5 - 
 8*(B*b*c^4 - 2*A*c^5)*d^7 + 8*(B*b^2*c^3 - 3*A*b*c^4)*d^6*e - 4*(3*B*b^3* 
c^2 + 4*A*b^2*c^3)*d^5*e^2 + (17*B*b^4*c + 80*A*b^3*c^2)*d^4*e^3 - (5*B*b^ 
5 + 81*A*b^4*c)*d^3*e^4)*x)*sqrt(c*x^2 + b*x))/((b^2*c^5*d^8*e^2 - 4*b^3*c 
^4*d^7*e^3 + 6*b^4*c^3*d^6*e^4 - 4*b^5*c^2*d^5*e^5 + b^6*c*d^4*e^6)*x^4 + 
(2*b^2*c^5*d^9*e - 7*b^3*c^4*d^8*e^2 + 8*b^4*c^3*d^7*e^3 - 2*b^5*c^2*d^...
 

Sympy [F]

\[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{3}}\, dx \] Input:

integrate((B*x+A)/(e*x+d)**3/(c*x**2+b*x)**(3/2),x)
 

Output:

Integral((A + B*x)/((x*(b + c*x))**(3/2)*(d + e*x)**3), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(3/2),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1117 vs. \(2 (355) = 710\).

Time = 0.15 (sec) , antiderivative size = 1117, normalized size of antiderivative = 2.91 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(3/2),x, algorithm="giac")
 

Output:

2*((B*b*c^3*d^6 - 2*A*c^4*d^6 + 3*A*b*c^3*d^5*e - 3*A*b^2*c^2*d^4*e^2 + A* 
b^3*c*d^3*e^3)*x/(b^2*c^3*d^9 - 3*b^3*c^2*d^8*e + 3*b^4*c*d^7*e^2 - b^5*d^ 
6*e^3) - (A*b*c^3*d^6 - 3*A*b^2*c^2*d^5*e + 3*A*b^3*c*d^4*e^2 - A*b^4*d^3* 
e^3)/(b^2*c^3*d^9 - 3*b^3*c^2*d^8*e + 3*b^4*c*d^7*e^2 - b^5*d^6*e^3))/sqrt 
(c*x^2 + b*x) - 3/4*(8*B*c^2*d^3*e - 4*B*b*c*d^2*e^2 - 16*A*c^2*d^2*e^2 + 
B*b^2*d*e^3 + 16*A*b*c*d*e^3 - 5*A*b^2*e^4)*arctan(-((sqrt(c)*x - sqrt(c*x 
^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c^3*d^6 - 3*b*c^2*d^5*e 
+ 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*sqrt(-c*d^2 + b*d*e)) + 1/4*(16*(sqrt(c)* 
x - sqrt(c*x^2 + b*x))^3*B*c^2*d^3*e^2 - 12*(sqrt(c)*x - sqrt(c*x^2 + b*x) 
)^3*B*b*c*d^2*e^3 - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*c^2*d^2*e^3 + 3 
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*d*e^4 + 24*(sqrt(c)*x - sqrt(c*x^ 
2 + b*x))^3*A*b*c*d*e^4 - 7*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*e^5 + 
40*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*c^(5/2)*d^4*e - 28*(sqrt(c)*x - sqr 
t(c*x^2 + b*x))^2*B*b*c^(3/2)*d^3*e^2 - 56*(sqrt(c)*x - sqrt(c*x^2 + b*x)) 
^2*A*c^(5/2)*d^3*e^2 + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2*sqrt(c)*d 
^2*e^3 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b*c^(3/2)*d^2*e^3 - 13*(sq 
rt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*sqrt(c)*d*e^4 + 40*(sqrt(c)*x - sqrt( 
c*x^2 + b*x))*B*b*c^2*d^4*e - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b^2*c*d 
^3*e^2 - 56*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b*c^2*d^3*e^2 + 5*(sqrt(c)*x 
 - sqrt(c*x^2 + b*x))*B*b^3*d^2*e^3 + 44*(sqrt(c)*x - sqrt(c*x^2 + b*x)...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{{\left (c\,x^2+b\,x\right )}^{3/2}\,{\left (d+e\,x\right )}^3} \,d x \] Input:

int((A + B*x)/((b*x + c*x^2)^(3/2)*(d + e*x)^3),x)
 

Output:

int((A + B*x)/((b*x + c*x^2)^(3/2)*(d + e*x)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.98 (sec) , antiderivative size = 5061, normalized size of antiderivative = 13.18 \[ \int \frac {A+B x}{(d+e x)^3 \left (b x+c x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(3/2),x)
 

Output:

(30*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sqrt(e)* 
sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*a*b**5*d**2*e* 
*5*x + 60*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sq 
rt(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*a*b**5*d 
*e**6*x**2 + 30*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d 
) - sqrt(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*a* 
b**5*e**7*x**3 - 156*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e 
- c*d) - sqrt(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c) 
))*a*b**4*c*d**3*e**4*x - 312*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan(( 
sqrt(b*e - c*d) - sqrt(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d 
)*sqrt(c)))*a*b**4*c*d**2*e**5*x**2 - 156*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - 
 c*d)*atan((sqrt(b*e - c*d) - sqrt(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt 
(c))/(sqrt(d)*sqrt(c)))*a*b**4*c*d*e**6*x**3 + 288*sqrt(d)*sqrt(b + c*x)*s 
qrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sqrt(e)*sqrt(b + c*x) - sqrt(x)*sqr 
t(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*a*b**3*c**2*d**4*e**3*x + 576*sqrt(d)*sqr 
t(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sqrt(e)*sqrt(b + c*x) - 
 sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*a*b**3*c**2*d**3*e**4*x**2 + 
288*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sqrt(e)* 
sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*a*b**3*c**2*d* 
*2*e**5*x**3 - 192*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e...