\(\int \frac {(A+B x) (a+b x+c x^2)^{3/2}}{x^2} \, dx\) [122]

Optimal result

Integrand size = 23, antiderivative size = 193 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {\left (b^2 B+18 A b c+8 a B c+2 c (b B+6 A c) x\right ) \sqrt {a+b x+c x^2}}{8 c}-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}-\frac {1}{2} \sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )-\frac {\left (b^3 B-6 A b^2 c-12 a b B c-24 a A c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2}} \] Output:

1/8*(B*b^2+18*A*b*c+8*B*a*c+2*c*(6*A*c+B*b)*x)*(c*x^2+b*x+a)^(1/2)/c-1/3*( 
-B*x+3*A)*(c*x^2+b*x+a)^(3/2)/x-1/2*a^(1/2)*(3*A*b+2*B*a)*arctanh(1/2*(b*x 
+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))-1/16*(-24*A*a*c^2-6*A*b^2*c-12*B*a*b*c+ 
B*b^3)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {a+x (b+c x)} \left (-8 a c (3 A-4 B x)+x \left (3 b^2 B+4 c^2 x (3 A+2 B x)+2 b c (15 A+7 B x)\right )\right )}{24 c x}+\frac {\left (-b^3 B+6 A b^2 c+12 a b B c+24 a A c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{16 c^{3/2}}-\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right ) \] Input:

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

Output:

(Sqrt[a + x*(b + c*x)]*(-8*a*c*(3*A - 4*B*x) + x*(3*b^2*B + 4*c^2*x*(3*A + 
 2*B*x) + 2*b*c*(15*A + 7*B*x))))/(24*c*x) + ((-(b^3*B) + 6*A*b^2*c + 12*a 
*b*B*c + 24*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]) 
])/(16*c^(3/2)) - Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + 
 x*(b + c*x)])/Sqrt[a]]
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {1230, 25, 1231, 27, 1269, 1092, 219, 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.

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

\(\Big \downarrow \) 1230

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1231

\(\displaystyle \frac {1}{2} \left (\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}-\frac {\int -\frac {8 a (3 A b+2 a B) c-\left (B b^3-6 A c b^2-12 a B c b-24 a A c^2\right ) x}{2 x \sqrt {c x^2+b x+a}}dx}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {8 a (3 A b+2 a B) c-\left (B b^3-6 A c b^2-12 a B c b-24 a A c^2\right ) x}{x \sqrt {c x^2+b x+a}}dx}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{2} \left (\frac {8 a c (2 a B+3 A b) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-\left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (\frac {8 a c (2 a B+3 A b) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-2 \left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {8 a c (2 a B+3 A b) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-\frac {\left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (\frac {-16 a c (2 a B+3 A b) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}-\frac {\left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-8 \sqrt {a} c (2 a B+3 A b) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{4 c}\right )-\frac {(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}\)

Input:

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

Output:

-1/3*((3*A - B*x)*(a + b*x + c*x^2)^(3/2))/x + (((b^2*B + 18*A*b*c + 8*a*B 
*c + 2*c*(b*B + 6*A*c)*x)*Sqrt[a + b*x + c*x^2])/(4*c) + (-8*Sqrt[a]*(3*A* 
b + 2*a*B)*c*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])] - ((b^ 
3*B - 6*A*b^2*c - 12*a*b*B*c - 24*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]* 
Sqrt[a + b*x + c*x^2])])/Sqrt[c])/(8*c))/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 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 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

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 1230

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

rule 1231

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

rule 1269

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

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.72

method	result	size
risch	\(-\frac {a A \sqrt {c \,x^{2}+b x +a}}{x}+\frac {3 A a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+\frac {B \,b^{2} \sqrt {c \,x^{2}+b x +a}}{8 c}-\frac {B \,b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {B c \,x^{2} \sqrt {c \,x^{2}+b x +a}}{3}+\frac {7 B b x \sqrt {c \,x^{2}+b x +a}}{12}+\frac {3 a b B \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 \sqrt {c}}+\frac {5 A b \sqrt {c \,x^{2}+b x +a}}{4}-\frac {3 A \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b}{2}-B \,a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )+\frac {3 b^{2} A \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}+\frac {A c x \sqrt {c \,x^{2}+b x +a}}{2}+\frac {4 B a \sqrt {c \,x^{2}+b x +a}}{3}\)	\(331\)
default	\(A \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{a x}+\frac {3 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2}+a \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )\right )}{2 a}+\frac {4 c \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{a}\right )+B \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2}+a \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )\right )\)	\(466\)

Input:

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

Output:

-a*A*(c*x^2+b*x+a)^(1/2)/x+3/2*A*a*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b 
*x+a)^(1/2))+1/8*B*b^2/c*(c*x^2+b*x+a)^(1/2)-1/16*B*b^3/c^(3/2)*ln((1/2*b+ 
c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*B*c*x^2*(c*x^2+b*x+a)^(1/2)+7/12*B*b 
*x*(c*x^2+b*x+a)^(1/2)+3/4*a*b*B*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2 
))/c^(1/2)+5/4*A*b*(c*x^2+b*x+a)^(1/2)-3/2*A*a^(1/2)*ln((2*a+b*x+2*a^(1/2) 
*(c*x^2+b*x+a)^(1/2))/x)*b-B*a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^( 
1/2))/x)+3/8*b^2*A*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+1/2 
*A*c*x*(c*x^2+b*x+a)^(1/2)+4/3*B*a*(c*x^2+b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 917, normalized size of antiderivative = 4.75 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/96*(24*(2*B*a + 3*A*b)*sqrt(a)*c^2*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 
- 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 3*(B*b^3 - 2 
4*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(c)*x*log(-8*c^2*x^2 - 8*b*c*x - b^ 
2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*B*c^3*x^3 
- 24*A*a*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2 + (3*B*b^2*c + 2*(16*B*a + 15*A 
*b)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^2*x), 1/48*(12*(2*B*a + 3*A*b)*sqrt( 
a)*c^2*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x 
+ 2*a)*sqrt(a) + 8*a^2)/x^2) + 3*(B*b^3 - 24*A*a*c^2 - 6*(2*B*a*b + A*b^2) 
*c)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2* 
x^2 + b*c*x + a*c)) + 2*(8*B*c^3*x^3 - 24*A*a*c^2 + 2*(7*B*b*c^2 + 6*A*c^3 
)*x^2 + (3*B*b^2*c + 2*(16*B*a + 15*A*b)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c 
^2*x), 1/96*(48*(2*B*a + 3*A*b)*sqrt(-a)*c^2*x*arctan(1/2*sqrt(c*x^2 + b*x 
 + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 3*(B*b^3 - 24*A*a*c^ 
2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(c)*x*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sq 
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*B*c^3*x^3 - 24*A*a 
*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2 + (3*B*b^2*c + 2*(16*B*a + 15*A*b)*c^2) 
*x)*sqrt(c*x^2 + b*x + a))/(c^2*x), 1/48*(24*(2*B*a + 3*A*b)*sqrt(-a)*c^2* 
x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + 
 a^2)) + 3*(B*b^3 - 24*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(-c)*x*arctan( 
1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c))...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((B*x+A)*(c*x^2+b*x+a)^(3/2)/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(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, B c x + \frac {7 \, B b c^{2} + 6 \, A c^{3}}{c^{2}}\right )} x + \frac {3 \, B b^{2} c + 32 \, B a c^{2} + 30 \, A b c^{2}}{c^{2}}\right )} + \frac {{\left (2 \, B a^{2} + 3 \, A a b\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {{\left (B b^{3} - 12 \, B a b c - 6 \, A b^{2} c - 24 \, A a c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {3}{2}}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a b + 2 \, A a^{2} \sqrt {c}}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a} \] Input:

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

Output:

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*B*c*x + (7*B*b*c^2 + 6*A*c^3)/c^2)*x + (3 
*B*b^2*c + 32*B*a*c^2 + 30*A*b*c^2)/c^2) + (2*B*a^2 + 3*A*a*b)*arctan(-(sq 
rt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/sqrt(-a) + 1/16*(B*b^3 - 12*B*a 
*b*c - 6*A*b^2*c - 24*A*a*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a 
))*sqrt(c) + b))/c^(3/2) + ((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a*b + 2* 
A*a^2*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {-48 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{2}+124 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2} x +24 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x^{2}+6 \sqrt {c \,x^{2}+b x +a}\, b^{3} c x +28 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x^{2}+16 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} x^{3}+120 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a b \,c^{2} x -120 \sqrt {a}\, \mathrm {log}\left (x \right ) a b \,c^{2} x +72 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a^{2} c^{2} x +54 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a \,b^{2} c x -3 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) b^{4} x}{48 c^{2} x} \] Input:

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

Output:

( - 48*sqrt(a + b*x + c*x**2)*a**2*c**2 + 124*sqrt(a + b*x + c*x**2)*a*b*c 
**2*x + 24*sqrt(a + b*x + c*x**2)*a*c**3*x**2 + 6*sqrt(a + b*x + c*x**2)*b 
**3*c*x + 28*sqrt(a + b*x + c*x**2)*b**2*c**2*x**2 + 16*sqrt(a + b*x + c*x 
**2)*b*c**3*x**3 + 120*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a 
- b*x)*a*b*c**2*x - 120*sqrt(a)*log(x)*a*b*c**2*x + 72*sqrt(c)*log( - 2*sq 
rt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*a**2*c**2*x + 54*sqrt(c)*log( - 
2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*a*b**2*c*x - 3*sqrt(c)*log( 
- 2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*b**4*x)/(48*c**2*x)