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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.200162, antiderivative size = 231, normalized size of antiderivative = 1., 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 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])

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 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 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])

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 = 0.303499, size = 214, normalized size = 0.93 \[ \frac{3 \left (b^2-4 a c\right ) \left (-4 a A c-4 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )-\frac{2 \sqrt{a} \left (a^2 \left (4 B x \left (-b^2+10 b c x+8 c^2 x^2\right )-2 A \left (b^2+10 b c x-12 c^2 x^2\right )\right )+8 a^3 c (A+2 B x)-a b x \left (A \left (-5 b^2+62 b c x+52 c^2 x^2\right )+12 b B x (b+c x)\right )+15 A b^3 x^2 (b+c x)\right )}{x^2 \sqrt{a+x (b+c x)}}}{8 a^{7/2} \left (4 a c-b^2\right )} \]

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))

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 506, normalized size = 2.2 \begin{align*} -{\frac{A}{2\,a{x}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{5\,Ab}{4\,{a}^{2}x}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{15\,A{b}^{2}}{8\,{a}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{\frac{15\,A{b}^{3}xc}{4\,{a}^{3} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{\frac{15\,A{b}^{4}}{8\,{a}^{3} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{\frac{15\,A{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{7}{2}}}}+13\,{\frac{Ab{c}^{2}x}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+{\frac{13\,A{b}^{2}c}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{\frac{3\,Ac}{2\,{a}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{3\,Ac}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{B}{ax}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{\frac{3\,bB}{2\,{a}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+3\,{\frac{B{b}^{2}cx}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+{\frac{3\,{b}^{3}B}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{3\,bB}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-8\,{\frac{B{c}^{2}x}{a \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-4\,{\frac{Bcb}{a \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}} \end{align*}

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)*x*c-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((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+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/a^2*c/(c*x^2+b*x+a)^(1/2)+3/2*A/a^(5/2)*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+
b*x+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)*x*c+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((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

________________________________________________________________________________________

Fricas [B]  time = 6.99678, size = 1863, normalized size = 8.06 \begin{align*} \text{result too large to display} \end{align*}

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)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.47584, size = 630, normalized size = 2.73 \begin{align*} -\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}} \end{align*}

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)