\(\int \frac {A+B x}{x^2 (a+b x+c x^2)^{5/2}} \, dx\) [168]

Optimal result

Integrand size = 23, antiderivative size = 277 \[ \int \frac {A+B x}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {A}{a x \left (a+b x+c x^2\right )^{3/2}}+\frac {2 a B \left (b^2-2 a c\right )-A \left (5 b^3-18 a b c\right )-c \left (5 A b^2-2 a b B-16 a A c\right ) x}{3 a^2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {3 (5 A b-2 a B) \left (b^2-4 a c\right ) \left (b^2-2 a c\right )-4 a b c \left (5 A b^2-2 a b B-16 a A c\right )-c \left (2 a b B \left (3 b^2-20 a c\right )-A \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )\right ) x}{3 a^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{7/2}} \] Output:

-A/a/x/(c*x^2+b*x+a)^(3/2)+1/3*(2*a*B*(-2*a*c+b^2)-A*(-18*a*b*c+5*b^3)-c*( 
-16*A*a*c+5*A*b^2-2*B*a*b)*x)/a^2/(-4*a*c+b^2)/(c*x^2+b*x+a)^(3/2)-1/3*(3* 
(5*A*b-2*B*a)*(-4*a*c+b^2)*(-2*a*c+b^2)-4*a*b*c*(-16*A*a*c+5*A*b^2-2*B*a*b 
)-c*(2*a*b*B*(-20*a*c+3*b^2)-A*(128*a^2*c^2-100*a*b^2*c+15*b^4))*x)/a^3/(- 
4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)+1/2*(5*A*b-2*B*a)*arctanh(1/2*(b*x+2*a)/a 
^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {16 a^4 c^2 (-3 A+4 B x)-15 A b^4 x^2 (b+c x)^2+8 a^3 c \left (-7 b^2 B x+6 B c^2 x^3+A \left (3 b^2-32 b c x-24 c^2 x^2\right )\right )+2 a b^2 x (b+c x) \left (3 b B x (b+c x)+A \left (-10 b^2+55 b c x+50 c^2 x^2\right )\right )-a^2 \left (4 b B x \left (-2 b^3+9 b^2 c x+21 b c^2 x^2+10 c^3 x^3\right )+A \left (3 b^4-148 b^3 c x+48 b^2 c^2 x^2+312 b c^3 x^3+128 c^4 x^4\right )\right )}{3 a^3 \left (b^2-4 a c\right )^2 x (a+x (b+c x))^{3/2}}+\frac {(-5 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{a^{7/2}} \] Input:

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

Output:

(16*a^4*c^2*(-3*A + 4*B*x) - 15*A*b^4*x^2*(b + c*x)^2 + 8*a^3*c*(-7*b^2*B* 
x + 6*B*c^2*x^3 + A*(3*b^2 - 32*b*c*x - 24*c^2*x^2)) + 2*a*b^2*x*(b + c*x) 
*(3*b*B*x*(b + c*x) + A*(-10*b^2 + 55*b*c*x + 50*c^2*x^2)) - a^2*(4*b*B*x* 
(-2*b^3 + 9*b^2*c*x + 21*b*c^2*x^2 + 10*c^3*x^3) + A*(3*b^4 - 148*b^3*c*x 
+ 48*b^2*c^2*x^2 + 312*b*c^3*x^3 + 128*c^4*x^4)))/(3*a^3*(b^2 - 4*a*c)^2*x 
*(a + x*(b + c*x))^(3/2)) + ((-5*A*b + 2*a*B)*ArcTanh[(Sqrt[c]*x - Sqrt[a 
+ x*(b + c*x)])/Sqrt[a]])/a^(7/2)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1235, 27, 1235, 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.

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

\(\Big \downarrow \) 1235

\(\displaystyle \frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int -\frac {5 A b^2-2 a B b-16 a A c+6 (A b-2 a B) c x}{2 x^2 \left (c x^2+b x+a\right )^{3/2}}dx}{3 a \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 A b^2-2 a B b-16 a A c+6 (A b-2 a B) c x}{x^2 \left (c x^2+b x+a\right )^{3/2}}dx}{3 a \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1235

\(\displaystyle \frac {-\frac {2 \int \frac {2 a b B \left (3 b^2-20 a c\right )-2 A \left (\frac {15 b^4}{2}-50 a c b^2+64 a^2 c^2\right )-2 c \left (5 A b^3-2 a B b^2-28 a A c b+24 a^2 B c\right ) x}{2 x^2 \sqrt {c x^2+b x+a}}dx}{a \left (b^2-4 a c\right )}-\frac {2 \left (-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )+c x \left (2 a B \left (b^2-12 a c\right )-A \left (5 b^3-28 a b c\right )\right )+2 a b B \left (b^2-8 a c\right )\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 a \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {2 a b B \left (3 b^2-20 a c\right )-A \left (15 b^4-100 a c b^2+128 a^2 c^2\right )-2 c \left (5 A b^3-2 a B b^2-28 a A c b+24 a^2 B c\right ) x}{x^2 \sqrt {c x^2+b x+a}}dx}{a \left (b^2-4 a c\right )}-\frac {2 \left (-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )+c x \left (2 a B \left (b^2-12 a c\right )-A \left (5 b^3-28 a b c\right )\right )+2 a b B \left (b^2-8 a c\right )\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 a \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1228

\(\displaystyle \frac {-\frac {\frac {3 \left (b^2-4 a c\right )^2 (5 A b-2 a B) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx}{2 a}-\frac {\sqrt {a+b x+c x^2} \left (2 a b B \left (3 b^2-20 a c\right )-A \left (128 a^2 c^2-100 a b^2 c+15 b^4\right )\right )}{a x}}{a \left (b^2-4 a c\right )}-\frac {2 \left (-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )+c x \left (2 a B \left (b^2-12 a c\right )-A \left (5 b^3-28 a b c\right )\right )+2 a b B \left (b^2-8 a c\right )\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 a \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {-\frac {-\frac {3 \left (b^2-4 a c\right )^2 (5 A b-2 a B) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}}{a}-\frac {\sqrt {a+b x+c x^2} \left (2 a b B \left (3 b^2-20 a c\right )-A \left (128 a^2 c^2-100 a b^2 c+15 b^4\right )\right )}{a x}}{a \left (b^2-4 a c\right )}-\frac {2 \left (-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )+c x \left (2 a B \left (b^2-12 a c\right )-A \left (5 b^3-28 a b c\right )\right )+2 a b B \left (b^2-8 a c\right )\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 a \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {2 \left (-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )+c x \left (2 a B \left (b^2-12 a c\right )-A \left (5 b^3-28 a b c\right )\right )+2 a b B \left (b^2-8 a c\right )\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {-\frac {3 \left (b^2-4 a c\right )^2 (5 A b-2 a B) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2}}-\frac {\sqrt {a+b x+c x^2} \left (2 a b B \left (3 b^2-20 a c\right )-A \left (128 a^2 c^2-100 a b^2 c+15 b^4\right )\right )}{a x}}{a \left (b^2-4 a c\right )}}{3 a \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

Input:

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

Output:

(2*(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x))/(3*a*(b^2 - 4*a*c)*x*(a 
+ b*x + c*x^2)^(3/2)) + ((-2*(2*a*b*B*(b^2 - 8*a*c) - A*(5*b^4 - 32*a*b^2* 
c + 32*a^2*c^2) + c*(2*a*B*(b^2 - 12*a*c) - A*(5*b^3 - 28*a*b*c))*x))/(a*( 
b^2 - 4*a*c)*x*Sqrt[a + b*x + c*x^2]) - (-(((2*a*b*B*(3*b^2 - 20*a*c) - A* 
(15*b^4 - 100*a*b^2*c + 128*a^2*c^2))*Sqrt[a + b*x + c*x^2])/(a*x)) - (3*( 
5*A*b - 2*a*B)*(b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x 
 + c*x^2])])/(2*a^(3/2)))/(a*(b^2 - 4*a*c)))/(3*a*(b^2 - 4*a*c))
 

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] 
)
 

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.69

method	result	size
default	\(A \left (-\frac {1}{a x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {5 b \left (\frac {1}{3 a \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 a}+\frac {\frac {1}{a \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )}{2 a}-\frac {4 c \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{a}\right )+B \left (\frac {1}{3 a \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 a}+\frac {\frac {1}{a \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )\)	\(467\)
risch	\(\text {Expression too large to display}\)	\(4767\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 823 vs. \(2 (256) = 512\).

Time = 1.61 (sec) , antiderivative size = 1655, normalized size of antiderivative = 5.97 \[ \int \frac {A+B x}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/12*(3*((16*(2*B*a^3 - 5*A*a^2*b)*c^4 - 8*(2*B*a^2*b^2 - 5*A*a*b^3)*c^3 
 + (2*B*a*b^4 - 5*A*b^5)*c^2)*x^5 + 2*(16*(2*B*a^3*b - 5*A*a^2*b^2)*c^3 - 
8*(2*B*a^2*b^3 - 5*A*a*b^4)*c^2 + (2*B*a*b^5 - 5*A*b^6)*c)*x^4 + (2*B*a*b^ 
6 - 5*A*b^7 + 32*(2*B*a^4 - 5*A*a^3*b)*c^3 - 6*(2*B*a^2*b^4 - 5*A*a*b^5)*c 
)*x^3 + 2*(2*B*a^2*b^5 - 5*A*a*b^6 + 16*(2*B*a^4*b - 5*A*a^3*b^2)*c^2 - 8* 
(2*B*a^3*b^3 - 5*A*a^2*b^4)*c)*x^2 + (2*B*a^3*b^4 - 5*A*a^2*b^5 + 16*(2*B* 
a^5 - 5*A*a^4*b)*c^2 - 8*(2*B*a^4*b^2 - 5*A*a^3*b^3)*c)*x)*sqrt(a)*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*(3*A*a^3*b^4 - 24*A*a^4*b^2*c + 48*A*a^5*c^2 + (128*A*a^3 
*c^4 + 20*(2*B*a^3*b - 5*A*a^2*b^2)*c^3 - 3*(2*B*a^2*b^3 - 5*A*a*b^4)*c^2) 
*x^4 - 6*(4*(2*B*a^4 - 13*A*a^3*b)*c^3 - 7*(2*B*a^3*b^2 - 5*A*a^2*b^3)*c^2 
 + (2*B*a^2*b^4 - 5*A*a*b^5)*c)*x^3 - 3*(2*B*a^2*b^5 - 5*A*a*b^6 - 16*A*a^ 
3*b^2*c^2 - 64*A*a^4*c^3 - 6*(2*B*a^3*b^3 - 5*A*a^2*b^4)*c)*x^2 - 4*(2*B*a 
^3*b^4 - 5*A*a^2*b^5 + 16*(B*a^5 - 4*A*a^4*b)*c^2 - (14*B*a^4*b^2 - 37*A*a 
^3*b^3)*c)*x)*sqrt(c*x^2 + b*x + a))/((a^4*b^4*c^2 - 8*a^5*b^2*c^3 + 16*a^ 
6*c^4)*x^5 + 2*(a^4*b^5*c - 8*a^5*b^3*c^2 + 16*a^6*b*c^3)*x^4 + (a^4*b^6 - 
 6*a^5*b^4*c + 32*a^7*c^3)*x^3 + 2*(a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)* 
x^2 + (a^6*b^4 - 8*a^7*b^2*c + 16*a^8*c^2)*x), 1/6*(3*((16*(2*B*a^3 - 5*A* 
a^2*b)*c^4 - 8*(2*B*a^2*b^2 - 5*A*a*b^3)*c^3 + (2*B*a*b^4 - 5*A*b^5)*c^2)* 
x^5 + 2*(16*(2*B*a^3*b - 5*A*a^2*b^2)*c^3 - 8*(2*B*a^2*b^3 - 5*A*a*b^4)...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)/x^2/(c*x^2+b*x+a)^(5/2),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 = 0.26 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.74 \[ \int \frac {A+B x}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (\frac {{\left (3 \, B a^{9} b^{3} c^{2} - 6 \, A a^{8} b^{4} c^{2} - 20 \, B a^{10} b c^{3} + 38 \, A a^{9} b^{2} c^{3} - 40 \, A a^{10} c^{4}\right )} x}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}} + \frac {3 \, {\left (2 \, B a^{9} b^{4} c - 4 \, A a^{8} b^{5} c - 14 \, B a^{10} b^{2} c^{2} + 27 \, A a^{9} b^{3} c^{2} + 8 \, B a^{11} c^{3} - 36 \, A a^{10} b c^{3}\right )}}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}}\right )} x + \frac {3 \, {\left (B a^{9} b^{5} - 2 \, A a^{8} b^{6} - 6 \, B a^{10} b^{3} c + 12 \, A a^{9} b^{4} c - 8 \, A a^{10} b^{2} c^{2} - 16 \, A a^{11} c^{3}\right )}}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}}\right )} x + \frac {4 \, B a^{10} b^{4} - 7 \, A a^{9} b^{5} - 28 \, B a^{11} b^{2} c + 50 \, A a^{10} b^{3} c + 32 \, B a^{12} c^{2} - 80 \, A a^{11} b c^{2}}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} + \frac {{\left (2 \, B a - 5 \, A b\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A b + 2 \, A a \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )} a^{3}} \] Input:

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

Output:

2/3*((((3*B*a^9*b^3*c^2 - 6*A*a^8*b^4*c^2 - 20*B*a^10*b*c^3 + 38*A*a^9*b^2 
*c^3 - 40*A*a^10*c^4)*x/(a^11*b^4 - 8*a^12*b^2*c + 16*a^13*c^2) + 3*(2*B*a 
^9*b^4*c - 4*A*a^8*b^5*c - 14*B*a^10*b^2*c^2 + 27*A*a^9*b^3*c^2 + 8*B*a^11 
*c^3 - 36*A*a^10*b*c^3)/(a^11*b^4 - 8*a^12*b^2*c + 16*a^13*c^2))*x + 3*(B* 
a^9*b^5 - 2*A*a^8*b^6 - 6*B*a^10*b^3*c + 12*A*a^9*b^4*c - 8*A*a^10*b^2*c^2 
 - 16*A*a^11*c^3)/(a^11*b^4 - 8*a^12*b^2*c + 16*a^13*c^2))*x + (4*B*a^10*b 
^4 - 7*A*a^9*b^5 - 28*B*a^11*b^2*c + 50*A*a^10*b^3*c + 32*B*a^12*c^2 - 80* 
A*a^11*b*c^2)/(a^11*b^4 - 8*a^12*b^2*c + 16*a^13*c^2))/(c*x^2 + b*x + a)^( 
3/2) + (2*B*a - 5*A*b)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a 
))/(sqrt(-a)*a^3) + ((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*b + 2*A*a*sqrt( 
c))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)*a^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 1284, normalized size of antiderivative = 4.64 \[ \int \frac {A+B x}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 96*sqrt(a + b*x + c*x**2)*a**5*c**2 + 48*sqrt(a + b*x + c*x**2)*a**4*b 
**2*c - 384*sqrt(a + b*x + c*x**2)*a**4*b*c**2*x - 384*sqrt(a + b*x + c*x* 
*2)*a**4*c**3*x**2 - 6*sqrt(a + b*x + c*x**2)*a**3*b**4 + 184*sqrt(a + b*x 
 + c*x**2)*a**3*b**3*c*x - 96*sqrt(a + b*x + c*x**2)*a**3*b**2*c**2*x**2 - 
 528*sqrt(a + b*x + c*x**2)*a**3*b*c**3*x**3 - 256*sqrt(a + b*x + c*x**2)* 
a**3*c**4*x**4 - 24*sqrt(a + b*x + c*x**2)*a**2*b**5*x + 108*sqrt(a + b*x 
+ c*x**2)*a**2*b**4*c*x**2 + 252*sqrt(a + b*x + c*x**2)*a**2*b**3*c**2*x** 
3 + 120*sqrt(a + b*x + c*x**2)*a**2*b**2*c**3*x**4 - 18*sqrt(a + b*x + c*x 
**2)*a*b**6*x**2 - 36*sqrt(a + b*x + c*x**2)*a*b**5*c*x**3 - 18*sqrt(a + b 
*x + c*x**2)*a*b**4*c**2*x**4 + 144*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x 
+ c*x**2) - 2*a - b*x)*a**4*b*c**2*x - 72*sqrt(a)*log( - 2*sqrt(a)*sqrt(a 
+ b*x + c*x**2) - 2*a - b*x)*a**3*b**3*c*x + 288*sqrt(a)*log( - 2*sqrt(a)* 
sqrt(a + b*x + c*x**2) - 2*a - b*x)*a**3*b**2*c**2*x**2 + 288*sqrt(a)*log( 
 - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a**3*b*c**3*x**3 + 9*sqrt 
(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a**2*b**5*x - 144 
*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a**2*b**4*c* 
x**2 + 288*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a* 
*2*b**2*c**3*x**4 + 144*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 
2*a - b*x)*a**2*b*c**4*x**5 + 18*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c 
*x**2) - 2*a - b*x)*a*b**6*x**2 - 54*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + ...