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

Optimal. Leaf size=255 \[ \frac{5 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{c}}-\frac{5 \sqrt{a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{8 x}-\frac{5 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{a}}-\frac{(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac{5 \left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{12 x^2} \]

[Out]

(-5*(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])/(8*x) - (5*(A*b + 2*a*B
- (b*B + 2*A*c)*x)*(a + b*x + c*x^2)^(3/2))/(12*x^2) - ((A - B*x)*(a + b*x + c*x^2)^(5/2))/(3*x^3) - (5*(2*a*
B*(3*b^2 + 4*a*c) + A*(b^3 + 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*Sqrt[a]) +
(5*(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])])/(16*S
qrt[c])

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Rubi [A]  time = 0.321655, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {812, 843, 621, 206, 724} \[ \frac{5 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{c}}-\frac{5 \sqrt{a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{8 x}-\frac{5 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{a}}-\frac{(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac{5 \left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{12 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(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])/(8*x) - (5*(A*b + 2*a*B
- (b*B + 2*A*c)*x)*(a + b*x + c*x^2)^(3/2))/(12*x^2) - ((A - B*x)*(a + b*x + c*x^2)^(5/2))/(3*x^3) - (5*(2*a*
B*(3*b^2 + 4*a*c) + A*(b^3 + 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*Sqrt[a]) +
(5*(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])])/(16*S
qrt[c])

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((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] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && 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 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

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

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]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx &=-\frac{(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac{5}{18} \int \frac{(-3 (A b+2 a B)-3 (b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{x^3} \, dx\\ &=-\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}{48} \int \frac{\left (6 \left (4 a b B+A \left (b^2+4 a c\right )\right )+6 \left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{x^2} \, 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}{96} \int \frac{-6 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right )-6 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right ) x}{x \sqrt{a+b x+c x^2}} \, 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{1}{16} \left (5 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx+\frac{1}{16} \left (5 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right )\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, 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{1}{8} \left (5 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )-\frac{1}{8} \left (5 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\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 )\\ &=-\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 ) \tanh ^{-1}\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 ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.74498, size = 236, normalized size = 0.93 \[ \frac{1}{48} \left (\frac{2 \sqrt{a+x (b+c x)} \left (-4 a^2 (2 A+3 B x)-2 a x (A (13 b+28 c x)+B x (27 b-28 c x))+x^2 \left (A \left (-33 b^2+54 b c x+12 c^2 x^2\right )+B x \left (33 b^2+26 b c x+8 c^2 x^2\right )\right )\right )}{x^3}+\frac{15 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{\sqrt{c}}-\frac{15 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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*(-33*b^2 + 54*b*c*x + 12*c^2*x^2))))/x^3 - (15*(2*a*B*(3*b^2 + 4*a*c) + A*(
b^3 + 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/Sqrt[a] + (15*(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 + x*(b + c*x)])])/Sqrt[c])/48

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Maple [B]  time = 0.011, size = 840, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*A/a*b^2*(c*x^2+b*x+a)^(1/2)*x*c+5/24*A/a^2*b^2*c*(c*x^2+b*x+a)^(3/2)*x+1/8*A/a^3*b^2*c*(c*x^2+b*x+a)^(5/2)
*x+3/4*B/a^2*b*c*(c*x^2+b*x+a)^(5/2)*x+5/4*B/a*b*c*(c*x^2+b*x+a)^(3/2)*x-1/2*B/a/x^2*(c*x^2+b*x+a)^(7/2)-5/2*B
*a^(3/2)*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-15/8*B*a^(1/2)*b^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a
)^(1/2))/x)+5/2*B*a*c*(c*x^2+b*x+a)^(1/2)+1/2*B/a*c*(c*x^2+b*x+a)^(5/2)+5/16*B*b^3/c^(1/2)*ln((1/2*b+c*x)/c^(1
/2)+(c*x^2+b*x+a)^(1/2))+5/4*B/a*b^2*(c*x^2+b*x+a)^(3/2)+3/4*B/a^2*b^2*(c*x^2+b*x+a)^(5/2)-1/3*A/a/x^3*(c*x^2+
b*x+a)^(7/2)+5/2*A*a*c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-5/16*A/a^(1/2)*b^3*ln((2*a+b*x+2*a^(1
/2)*(c*x^2+b*x+a)^(1/2))/x)+5/2*A*c^2*(c*x^2+b*x+a)^(1/2)*x+5/8*A/a*b^3*(c*x^2+b*x+a)^(1/2)+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*A*b*c*(c*x^2+b*x+a)^(1/2)+5/24*A/a^2*b^3*(c*x^2+b*x+a)^(3/2)+1/
8*A/a^3*b^3*(c*x^2+b*x+a)^(5/2)+5/6*B*c*(c*x^2+b*x+a)^(3/2)+25/8*B*b^2*(c*x^2+b*x+a)^(1/2)+4/3*A/a^2*c^2*(c*x^
2+b*x+a)^(5/2)*x-1/8*A/a^3*b^2/x*(c*x^2+b*x+a)^(7/2)+17/12*A/a^2*b*c*(c*x^2+b*x+a)^(5/2)-15/4*A*a^(1/2)*b*c*ln
((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-3/4*B/a^2*b/x*(c*x^2+b*x+a)^(7/2)-4/3*A/a^2*c/x*(c*x^2+b*x+a)^(7/2
)+5/3*A/a*c^2*(c*x^2+b*x+a)^(3/2)*x+25/12*A/a*b*c*(c*x^2+b*x+a)^(3/2)-1/12*A/a^2*b/x^2*(c*x^2+b*x+a)^(7/2)+5/2
*B*b*(c*x^2+b*x+a)^(1/2)*x*c+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))

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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)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 18.2065, size = 2971, normalized size = 11.65 \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)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[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*s
qrt(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)*sqr
t(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)*sqrt(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*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), 1/48*(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*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*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*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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac{5}{2}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.67389, size = 1058, normalized size = 4.15 \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)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="giac")

[Out]

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*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2*sqrt(c) + 33*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))^5*A*b^3*sqrt(c) + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*c^(3/2) + 108*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^5*A*a*b*c^(3/2) + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^2*b*c + 144*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^4*A*a*b^2*c + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^2*c^2 - 96*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^3*B*a^2*b^2*sqrt(c) - 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^3*sqrt(c) - 48*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*b*c^(3/2) - 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*b*c - 1
44*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^2*b^2*c - 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^3*c^2 + 4
2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^3*b^2*sqrt(c) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3*sqr
t(c) - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*c^(3/2) + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b*c
^(3/2) + 96*B*a^4*b*c + 48*A*a^3*b^2*c + 112*A*a^4*c^2)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^3*sqrt(c)
)