\(\int \frac {(A+B x) (a+b x+c x^2)^{5/2}}{x^4} \, dx\) [137]

Optimal result

Integrand size = 23, antiderivative size = 255 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=-\frac {5 \left (4 a b B+A \left (b^2+4 a c\right )-\left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{8 x}-\frac {5 (A b+2 a B-(b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac {5 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {a}}+\frac {5 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 2.70 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{48} \left (-\frac {2 \sqrt {a+x (b+c x)} \left (4 a^2 (2 A+3 B x)+2 a x (B x (27 b-28 c x)+A (13 b+28 c x))-x^2 \left (B x \left (33 b^2+26 b c x+8 c^2 x^2\right )+A \left (-33 b^2+54 b c x+12 c^2 x^2\right )\right )\right )}{x^3}-\frac {30 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right ) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {15 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{\sqrt {c}}\right ) \] Input:

Integrate[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^4,x]
 

Output:

((-2*Sqrt[a + x*(b + c*x)]*(4*a^2*(2*A + 3*B*x) + 2*a*x*(B*x*(27*b - 28*c* 
x) + A*(13*b + 28*c*x)) - x^2*(B*x*(33*b^2 + 26*b*c*x + 8*c^2*x^2) + A*(-3 
3*b^2 + 54*b*c*x + 12*c^2*x^2))))/x^3 - (30*(2*a*B*(3*b^2 + 4*a*c) + A*(b^ 
3 + 12*a*b*c))*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + x*(b + c*x)])/Sqrt[a]])/Sq 
rt[a] - (15*(b^3*B + 6*A*b^2*c + 12*a*b*B*c + 8*a*A*c^2)*Log[b + 2*c*x - 2 
*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/Sqrt[c])/48
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {1230, 27, 1230, 27, 1230, 25, 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.

\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx\)

\(\Big \downarrow \) 1230

\(\displaystyle -\frac {5}{18} \int -\frac {3 (A b+2 a B+(b B+2 A c) x) \left (c x^2+b x+a\right )^{3/2}}{x^3}dx-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{6} \int \frac {(A b+2 a B+(b B+2 A c) x) \left (c x^2+b x+a\right )^{3/2}}{x^3}dx-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 1230

\(\displaystyle \frac {5}{6} \left (-\frac {3}{8} \int -\frac {2 \left (4 a b B+A \left (b^2+4 a c\right )+\left (B b^2+4 A c b+4 a B c\right ) x\right ) \sqrt {c x^2+b x+a}}{x^2}dx-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1230

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (-\frac {1}{2} \int -\frac {2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a c b\right )+\left (B b^3+6 A c b^2+12 a B c b+8 a A c^2\right ) x}{x \sqrt {c x^2+b x+a}}dx-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a c b\right )+\left (B b^3+6 A c b^2+12 a B c b+8 a A c^2\right ) x}{x \sqrt {c x^2+b x+a}}dx-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx+\left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx\right )-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (2 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}+\left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx\right )-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+\frac {\left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-2 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-\frac {\left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{x}\right )-\frac {\left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{2 x^2}\right )-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}\)

Input:

Int[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^4,x]
 

Output:

-1/3*((A - B*x)*(a + b*x + c*x^2)^(5/2))/x^3 + (5*(-1/2*((A*b + 2*a*B - (b 
*B + 2*A*c)*x)*(a + b*x + c*x^2)^(3/2))/x^2 + (3*(-(((4*a*b*B + A*(b^2 + 4 
*a*c) - (b^2*B + 4*A*b*c + 4*a*B*c)*x)*Sqrt[a + b*x + c*x^2])/x) + (-(((2* 
a*B*(3*b^2 + 4*a*c) + A*(b^3 + 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*S 
qrt[a + b*x + c*x^2])])/Sqrt[a]) + ((b^3*B + 6*A*b^2*c + 12*a*b*B*c + 8*a* 
A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[c])/2) 
)/4))/6
 

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(455\) vs. \(2(223)=446\).

Time = 1.32 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.79

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (56 A a c \,x^{2}+33 x^{2} b^{2} A +54 B a \,x^{2} b +26 a b A x +12 a^{2} B x +8 a^{2} A \right )}{24 x^{3}}-\frac {15 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A b c}{4}+\frac {15 A \,b^{2} \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8}+\frac {5 A a \,c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+\frac {c^{2} x \sqrt {c \,x^{2}+b x +a}\, A}{2}+\frac {9 c b \sqrt {c \,x^{2}+b x +a}\, A}{4}-\frac {5 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A \,b^{3}}{16 \sqrt {a}}+\frac {5 B \,b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 \sqrt {c}}+\frac {11 b^{2} \sqrt {c \,x^{2}+b x +a}\, B}{8}-\frac {5 a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) B c}{2}-\frac {15 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) B \,b^{2}}{8}+\frac {13 c x \sqrt {c \,x^{2}+b x +a}\, B b}{12}+\frac {15 B a b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4}+\frac {B \,c^{2} x^{2} \sqrt {c \,x^{2}+b x +a}}{3}+\frac {7 c \sqrt {c \,x^{2}+b x +a}\, a B}{3}\)	\(456\)
default	\(\text {Expression too large to display}\)	\(2055\)

Input:

int((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(c*x^2+b*x+a)^(1/2)*(56*A*a*c*x^2+33*A*b^2*x^2+54*B*a*b*x^2+26*A*a*b 
*x+12*B*a^2*x+8*A*a^2)/x^3-15/4*a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a 
)^(1/2))/x)*A*b*c+15/8*A*b^2*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^ 
(1/2))+5/2*A*a*c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*c^2 
*x*(c*x^2+b*x+a)^(1/2)*A+9/4*c*b*(c*x^2+b*x+a)^(1/2)*A-5/16/a^(1/2)*ln((2* 
a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*A*b^3+5/16*B*b^3*ln((1/2*b+c*x)/c^ 
(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+11/8*b^2*(c*x^2+b*x+a)^(1/2)*B-5/2*a^(3 
/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*B*c-15/8*a^(1/2)*ln((2*a 
+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*B*b^2+13/12*c*x*(c*x^2+b*x+a)^(1/2) 
*B*b+15/4*B*a*b*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*B* 
c^2*x^2*(c*x^2+b*x+a)^(1/2)+7/3*c*(c*x^2+b*x+a)^(1/2)*a*B
 

Fricas [A] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 1293, normalized size of antiderivative = 5.07 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="fricas")
 

Output:

[1/96*(15*(B*a*b^3 + 8*A*a^2*c^2 + 6*(2*B*a^2*b + A*a*b^2)*c)*sqrt(c)*x^3* 
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) + 15*(4*(2*B*a^2 + 3*A*a*b)*c^2 + (6*B*a*b^2 + A*b^3)*c)*sqrt( 
a)*x^3*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) + 4*(8*B*a*c^3*x^5 - 8*A*a^3*c + 2*(13*B*a*b*c^ 
2 + 6*A*a*c^3)*x^4 + (33*B*a*b^2*c + 2*(28*B*a^2 + 27*A*a*b)*c^2)*x^3 - 2* 
(6*B*a^3 + 13*A*a^2*b)*c*x - (56*A*a^2*c^2 + 3*(18*B*a^2*b + 11*A*a*b^2)*c 
)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^3), -1/96*(30*(B*a*b^3 + 8*A*a^2*c^2 
+ 6*(2*B*a^2*b + A*a*b^2)*c)*sqrt(-c)*x^3*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)) - 15*(4*(2*B*a^2 + 3*A*a*b) 
*c^2 + (6*B*a*b^2 + A*b^3)*c)*sqrt(a)*x^3*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) - 4*(8*B*a*c 
^3*x^5 - 8*A*a^3*c + 2*(13*B*a*b*c^2 + 6*A*a*c^3)*x^4 + (33*B*a*b^2*c + 2* 
(28*B*a^2 + 27*A*a*b)*c^2)*x^3 - 2*(6*B*a^3 + 13*A*a^2*b)*c*x - (56*A*a^2* 
c^2 + 3*(18*B*a^2*b + 11*A*a*b^2)*c)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^3) 
, 1/96*(30*(4*(2*B*a^2 + 3*A*a*b)*c^2 + (6*B*a*b^2 + A*b^3)*c)*sqrt(-a)*x^ 
3*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)) + 15*(B*a*b^3 + 8*A*a^2*c^2 + 6*(2*B*a^2*b + A*a*b^2)*c)*sqrt(c)*x^ 
3*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqr 
t(c) - 4*a*c) + 4*(8*B*a*c^3*x^5 - 8*A*a^3*c + 2*(13*B*a*b*c^2 + 6*A*a*...
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{x^{4}}\, dx \] Input:

integrate((B*x+A)*(c*x**2+b*x+a)**(5/2)/x**4,x)
 

Output:

Integral((A + B*x)*(a + b*x + c*x**2)**(5/2)/x**4, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 765 vs. \(2 (224) = 448\).

Time = 0.33 (sec) , antiderivative size = 765, normalized size of antiderivative = 3.00 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="giac")
 

Output:

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*B*c^2*x + (13*B*b*c^3 + 6*A*c^4)/c^2)*x + 
 (33*B*b^2*c^2 + 56*B*a*c^3 + 54*A*b*c^3)/c^2) + 5/8*(6*B*a*b^2 + A*b^3 + 
8*B*a^2*c + 12*A*a*b*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(- 
a))/sqrt(-a) - 5/16*(B*b^3 + 12*B*a*b*c + 6*A*b^2*c + 8*A*a*c^2)*log(abs(2 
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c) + 1/24*(54*(sqr 
t(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2 + 33*(sqrt(c)*x - sqrt(c*x^2 + b 
*x + a))^5*A*b^3 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*c + 108* 
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a*b*c + 144*(sqrt(c)*x - sqrt(c*x^ 
2 + b*x + a))^4*B*a^2*b*sqrt(c) + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^ 
4*A*a*b^2*sqrt(c) + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^2*c^(3/2 
) - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^2*b^2 - 40*(sqrt(c)*x - s 
qrt(c*x^2 + b*x + a))^3*A*a*b^3 - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3 
*A*a^2*b*c - 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*b*sqrt(c) - 1 
44*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^2*b^2*sqrt(c) - 192*(sqrt(c)* 
x - sqrt(c*x^2 + b*x + a))^2*A*a^3*c^(3/2) + 42*(sqrt(c)*x - sqrt(c*x^2 + 
b*x + a))*B*a^3*b^2 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3 - 2 
4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*c + 36*(sqrt(c)*x - sqrt(c*x^2 
 + b*x + a))*A*a^3*b*c + 96*B*a^4*b*sqrt(c) + 48*A*a^3*b^2*sqrt(c) + 112*A 
*a^4*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{x^4} \,d x \] Input:

int(((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^4,x)
                                                                                    
                                                                                    
 

Output:

int(((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.55 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx=\frac {-16 \sqrt {c \,x^{2}+b x +a}\, a^{3} c -76 \sqrt {c \,x^{2}+b x +a}\, a^{2} b c x -112 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{2} x^{2}-174 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c \,x^{2}+220 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2} x^{3}+24 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x^{4}+66 \sqrt {c \,x^{2}+b x +a}\, b^{3} c \,x^{3}+52 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x^{4}+16 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} x^{5}+300 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a b \,c^{2} x^{3}+105 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b^{3} c \,x^{3}-300 \sqrt {a}\, \mathrm {log}\left (x \right ) a b \,c^{2} x^{3}-105 \sqrt {a}\, \mathrm {log}\left (x \right ) b^{3} c \,x^{3}+120 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a^{2} c^{2} x^{3}+270 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a \,b^{2} c \,x^{3}+15 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) b^{4} x^{3}}{48 c \,x^{3}} \] Input:

int((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x)
 

Output:

( - 16*sqrt(a + b*x + c*x**2)*a**3*c - 76*sqrt(a + b*x + c*x**2)*a**2*b*c* 
x - 112*sqrt(a + b*x + c*x**2)*a**2*c**2*x**2 - 174*sqrt(a + b*x + c*x**2) 
*a*b**2*c*x**2 + 220*sqrt(a + b*x + c*x**2)*a*b*c**2*x**3 + 24*sqrt(a + b* 
x + c*x**2)*a*c**3*x**4 + 66*sqrt(a + b*x + c*x**2)*b**3*c*x**3 + 52*sqrt( 
a + b*x + c*x**2)*b**2*c**2*x**4 + 16*sqrt(a + b*x + c*x**2)*b*c**3*x**5 + 
 300*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a*b*c**2*x* 
*3 + 105*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b**3*c* 
x**3 - 300*sqrt(a)*log(x)*a*b*c**2*x**3 - 105*sqrt(a)*log(x)*b**3*c*x**3 + 
 120*sqrt(c)*log( - 2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*a**2*c** 
2*x**3 + 270*sqrt(c)*log( - 2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)* 
a*b**2*c*x**3 + 15*sqrt(c)*log( - 2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2 
*c*x)*b**4*x**3)/(48*c*x**3)