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

Optimal. Leaf size=231 \[ -\frac {3 \left (-4 a A c-4 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{7/2}}-\frac {\sqrt {a+b x+c x^2} \left (4 a B \left (3 b^2-8 a c\right )-A \left (15 b^3-52 a b c\right )\right )}{4 a^3 x \left (b^2-4 a c\right )}-\frac {\sqrt {a+b x+c x^2} \left (-12 a A c-4 a b B+5 A b^2\right )}{2 a^2 x^2 \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a x^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[Out]

-3/8*(-4*A*a*c+5*A*b^2-4*B*a*b)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(7/2)+2*(A*b^2-a*b*B-2*a*
A*c+(A*b-2*B*a)*c*x)/a/(-4*a*c+b^2)/x^2/(c*x^2+b*x+a)^(1/2)-1/2*(-12*A*a*c+5*A*b^2-4*B*a*b)*(c*x^2+b*x+a)^(1/2
)/a^2/(-4*a*c+b^2)/x^2-1/4*(4*a*B*(-8*a*c+3*b^2)-A*(-52*a*b*c+15*b^3))*(c*x^2+b*x+a)^(1/2)/a^3/(-4*a*c+b^2)/x

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {822, 834, 806, 724, 206} \[ -\frac {\sqrt {a+b x+c x^2} \left (-12 a A c-4 a b B+5 A b^2\right )}{2 a^2 x^2 \left (b^2-4 a c\right )}-\frac {\sqrt {a+b x+c x^2} \left (4 a B \left (3 b^2-8 a c\right )-A \left (15 b^3-52 a b c\right )\right )}{4 a^3 x \left (b^2-4 a c\right )}-\frac {3 \left (-4 a A c-4 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{7/2}}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a x^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x))/(a*(b^2 - 4*a*c)*x^2*Sqrt[a + b*x + c*x^2]) - ((5*A*b^2 - 4*
a*b*B - 12*a*A*c)*Sqrt[a + b*x + c*x^2])/(2*a^2*(b^2 - 4*a*c)*x^2) - ((4*a*B*(3*b^2 - 8*a*c) - A*(15*b^3 - 52*
a*b*c))*Sqrt[a + b*x + c*x^2])/(4*a^3*(b^2 - 4*a*c)*x) - (3*(5*A*b^2 - 4*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b*x)/
(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(8*a^(7/2))

Rule 206

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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] - Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 822

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] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((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] + Dist[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] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{x^3 \left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} \left (-5 A b^2+4 a b B+12 a A c\right )-2 (A b-2 a B) c x}{x^3 \sqrt {a+b x+c x^2}} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x+c x^2}}-\frac {\left (5 A b^2-4 a b B-12 a A c\right ) \sqrt {a+b x+c x^2}}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {\int \frac {\frac {1}{4} \left (4 a B \left (3 b^2-8 a c\right )-4 A \left (\frac {15 b^3}{4}-13 a b c\right )\right )-\frac {1}{2} c \left (5 A b^2-4 a b B-12 a A c\right ) x}{x^2 \sqrt {a+b x+c x^2}} \, dx}{a^2 \left (b^2-4 a c\right )}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x+c x^2}}-\frac {\left (5 A b^2-4 a b B-12 a A c\right ) \sqrt {a+b x+c x^2}}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac {\left (4 a B \left (3 b^2-8 a c\right )-A \left (15 b^3-52 a b c\right )\right ) \sqrt {a+b x+c x^2}}{4 a^3 \left (b^2-4 a c\right ) x}+\frac {\left (3 \left (5 A b^2-4 a b B-4 a A c\right )\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{8 a^3}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x+c x^2}}-\frac {\left (5 A b^2-4 a b B-12 a A c\right ) \sqrt {a+b x+c x^2}}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac {\left (4 a B \left (3 b^2-8 a c\right )-A \left (15 b^3-52 a b c\right )\right ) \sqrt {a+b x+c x^2}}{4 a^3 \left (b^2-4 a c\right ) x}-\frac {\left (3 \left (5 A b^2-4 a b B-4 a A c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x}{\sqrt {a+b x+c x^2}}\right )}{4 a^3}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x+c x^2}}-\frac {\left (5 A b^2-4 a b B-12 a A c\right ) \sqrt {a+b x+c x^2}}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac {\left (4 a B \left (3 b^2-8 a c\right )-A \left (15 b^3-52 a b c\right )\right ) \sqrt {a+b x+c x^2}}{4 a^3 \left (b^2-4 a c\right ) x}-\frac {3 \left (5 A b^2-4 a b B-4 a A c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 214, normalized size = 0.93 \[ \frac {3 \left (b^2-4 a c\right ) \left (-4 a A c-4 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )-\frac {2 \sqrt {a} \left (8 a^3 c (A+2 B x)+a^2 \left (4 B x \left (-b^2+10 b c x+8 c^2 x^2\right )-2 A \left (b^2+10 b c x-12 c^2 x^2\right )\right )-a b x \left (A \left (-5 b^2+62 b c x+52 c^2 x^2\right )+12 b B x (b+c x)\right )+15 A b^3 x^2 (b+c x)\right )}{x^2 \sqrt {a+x (b+c x)}}}{8 a^{7/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[a]*(8*a^3*c*(A + 2*B*x) + 15*A*b^3*x^2*(b + c*x) + a^2*(-2*A*(b^2 + 10*b*c*x - 12*c^2*x^2) + 4*B*x*(
-b^2 + 10*b*c*x + 8*c^2*x^2)) - a*b*x*(12*b*B*x*(b + c*x) + A*(-5*b^2 + 62*b*c*x + 52*c^2*x^2))))/(x^2*Sqrt[a
+ x*(b + c*x)]) + 3*(b^2 - 4*a*c)*(5*A*b^2 - 4*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b +
c*x)])])/(8*a^(7/2)*(-b^2 + 4*a*c))

________________________________________________________________________________________

fricas [B]  time = 2.71, size = 869, normalized size = 3.76 \[ \left [-\frac {3 \, {\left ({\left (16 \, A a^{2} c^{3} + 8 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} c^{2} - {\left (4 \, B a b^{3} - 5 \, A b^{4}\right )} c\right )} x^{4} - {\left (4 \, B a b^{4} - 5 \, A b^{5} - 16 \, A a^{2} b c^{2} - 8 \, {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} c\right )} x^{3} - {\left (4 \, B a^{2} b^{3} - 5 \, A a b^{4} - 16 \, A a^{3} c^{2} - 8 \, {\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} c\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) + 4 \, {\left (2 \, A a^{3} b^{2} - 8 \, A a^{4} c - {\left (4 \, {\left (8 \, B a^{3} - 13 \, A a^{2} b\right )} c^{2} - 3 \, {\left (4 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} c\right )} x^{3} + {\left (12 \, B a^{2} b^{3} - 15 \, A a b^{4} - 24 \, A a^{3} c^{2} - 2 \, {\left (20 \, B a^{3} b - 31 \, A a^{2} b^{2}\right )} c\right )} x^{2} + {\left (4 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} - 4 \, {\left (4 \, B a^{4} - 5 \, A a^{3} b\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{16 \, {\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{4} + {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{3} + {\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{2}\right )}}, \frac {3 \, {\left ({\left (16 \, A a^{2} c^{3} + 8 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} c^{2} - {\left (4 \, B a b^{3} - 5 \, A b^{4}\right )} c\right )} x^{4} - {\left (4 \, B a b^{4} - 5 \, A b^{5} - 16 \, A a^{2} b c^{2} - 8 \, {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} c\right )} x^{3} - {\left (4 \, B a^{2} b^{3} - 5 \, A a b^{4} - 16 \, A a^{3} c^{2} - 8 \, {\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} c\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) - 2 \, {\left (2 \, A a^{3} b^{2} - 8 \, A a^{4} c - {\left (4 \, {\left (8 \, B a^{3} - 13 \, A a^{2} b\right )} c^{2} - 3 \, {\left (4 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} c\right )} x^{3} + {\left (12 \, B a^{2} b^{3} - 15 \, A a b^{4} - 24 \, A a^{3} c^{2} - 2 \, {\left (20 \, B a^{3} b - 31 \, A a^{2} b^{2}\right )} c\right )} x^{2} + {\left (4 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} - 4 \, {\left (4 \, B a^{4} - 5 \, A a^{3} b\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{8 \, {\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{4} + {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{3} + {\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*((16*A*a^2*c^3 + 8*(2*B*a^2*b - 3*A*a*b^2)*c^2 - (4*B*a*b^3 - 5*A*b^4)*c)*x^4 - (4*B*a*b^4 - 5*A*b^5
- 16*A*a^2*b*c^2 - 8*(2*B*a^2*b^2 - 3*A*a*b^3)*c)*x^3 - (4*B*a^2*b^3 - 5*A*a*b^4 - 16*A*a^3*c^2 - 8*(2*B*a^3*
b - 3*A*a^2*b^2)*c)*x^2)*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*(2*A*a^3*b^2 - 8*A*a^4*c - (4*(8*B*a^3 - 13*A*a^2*b)*c^2 - 3*(4*B*a^2*b^2 - 5*A*a*b^3)*c)
*x^3 + (12*B*a^2*b^3 - 15*A*a*b^4 - 24*A*a^3*c^2 - 2*(20*B*a^3*b - 31*A*a^2*b^2)*c)*x^2 + (4*B*a^3*b^2 - 5*A*a
^2*b^3 - 4*(4*B*a^4 - 5*A*a^3*b)*c)*x)*sqrt(c*x^2 + b*x + a))/((a^4*b^2*c - 4*a^5*c^2)*x^4 + (a^4*b^3 - 4*a^5*
b*c)*x^3 + (a^5*b^2 - 4*a^6*c)*x^2), 1/8*(3*((16*A*a^2*c^3 + 8*(2*B*a^2*b - 3*A*a*b^2)*c^2 - (4*B*a*b^3 - 5*A*
b^4)*c)*x^4 - (4*B*a*b^4 - 5*A*b^5 - 16*A*a^2*b*c^2 - 8*(2*B*a^2*b^2 - 3*A*a*b^3)*c)*x^3 - (4*B*a^2*b^3 - 5*A*
a*b^4 - 16*A*a^3*c^2 - 8*(2*B*a^3*b - 3*A*a^2*b^2)*c)*x^2)*sqrt(-a)*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)) - 2*(2*A*a^3*b^2 - 8*A*a^4*c - (4*(8*B*a^3 - 13*A*a^2*b)*c^2 - 3*(4*B*a^2
*b^2 - 5*A*a*b^3)*c)*x^3 + (12*B*a^2*b^3 - 15*A*a*b^4 - 24*A*a^3*c^2 - 2*(20*B*a^3*b - 31*A*a^2*b^2)*c)*x^2 +
(4*B*a^3*b^2 - 5*A*a^2*b^3 - 4*(4*B*a^4 - 5*A*a^3*b)*c)*x)*sqrt(c*x^2 + b*x + a))/((a^4*b^2*c - 4*a^5*c^2)*x^4
+ (a^4*b^3 - 4*a^5*b*c)*x^3 + (a^5*b^2 - 4*a^6*c)*x^2)]

________________________________________________________________________________________

giac [B]  time = 0.26, size = 467, normalized size = 2.02 \[ -\frac {2 \, {\left (\frac {{\left (B a^{4} b^{2} c - A a^{3} b^{3} c - 2 \, B a^{5} c^{2} + 3 \, A a^{4} b c^{2}\right )} x}{a^{6} b^{2} - 4 \, a^{7} c} + \frac {B a^{4} b^{3} - A a^{3} b^{4} - 3 \, B a^{5} b c + 4 \, A a^{4} b^{2} c - 2 \, A a^{5} c^{2}}{a^{6} b^{2} - 4 \, a^{7} c}\right )}}{\sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (4 \, B a b - 5 \, A b^{2} + 4 \, A a c\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a b - 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A b^{2} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a c + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{2} \sqrt {c} - 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a b \sqrt {c} - 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{2} b + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a b^{2} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{2} c - 8 \, B a^{3} \sqrt {c} + 16 \, A a^{2} b \sqrt {c}}{4 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((B*a^4*b^2*c - A*a^3*b^3*c - 2*B*a^5*c^2 + 3*A*a^4*b*c^2)*x/(a^6*b^2 - 4*a^7*c) + (B*a^4*b^3 - A*a^3*b^4 -
3*B*a^5*b*c + 4*A*a^4*b^2*c - 2*A*a^5*c^2)/(a^6*b^2 - 4*a^7*c))/sqrt(c*x^2 + b*x + a) - 3/4*(4*B*a*b - 5*A*b^
2 + 4*A*a*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^3) + 1/4*(4*(sqrt(c)*x - sqrt(c
*x^2 + b*x + a))^3*B*a*b - 7*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))^3*A*a*c + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^2*sqrt(c) - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A
*a*b*sqrt(c) - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^2*b + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a*b^2 +
4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*c - 8*B*a^3*sqrt(c) + 16*A*a^2*b*sqrt(c))/(((sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^2 - a)^2*a^3)

________________________________________________________________________________________

maple [B]  time = 0.07, size = 506, normalized size = 2.19 \[ \frac {13 A b \,c^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}-\frac {15 A \,b^{3} c x}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{3}}-\frac {8 B \,c^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}+\frac {3 B \,b^{2} c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {13 A \,b^{2} c}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}-\frac {15 A \,b^{4}}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{3}}-\frac {4 B b c}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}+\frac {3 B \,b^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {3 A c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {5}{2}}}-\frac {15 A \,b^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {7}{2}}}+\frac {3 B b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {5}{2}}}-\frac {3 A c}{2 \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {15 A \,b^{2}}{8 \sqrt {c \,x^{2}+b x +a}\, a^{3}}-\frac {3 B b}{2 \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {5 A b}{4 \sqrt {c \,x^{2}+b x +a}\, a^{2} x}-\frac {B}{\sqrt {c \,x^{2}+b x +a}\, a x}-\frac {A}{2 \sqrt {c \,x^{2}+b x +a}\, a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*A/a/x^2/(c*x^2+b*x+a)^(1/2)+5/4*A/a^2*b/x/(c*x^2+b*x+a)^(1/2)+15/8*A/a^3*b^2/(c*x^2+b*x+a)^(1/2)-15/4*A/a
^3*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*c*x-15/8*A/a^3*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-15/8*A/a^(7/2)*b^2*l
n((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+13*A/a^2*b*c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+13/2*A/a^2*b^2*c
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3/2*A*c/a^2/(c*x^2+b*x+a)^(1/2)+3/2*A*c/a^(5/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(
1/2)*a^(1/2))/x)-B/a/x/(c*x^2+b*x+a)^(1/2)-3/2*B/a^2*b/(c*x^2+b*x+a)^(1/2)+3*B/a^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+
a)^(1/2)*c*x+3/2*B/a^2*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+3/2*B/a^(5/2)*b*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a
^(1/2))/x)-8*B*c^2/a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-4*B*c/a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*b

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{x^3\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{x^{3} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________