Integrand size = 24, antiderivative size = 189 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {1}{2} (a-b+c)^{3/2} \text {arctanh}\left (\frac {2 a-b+(b-2 c) x}{2 \sqrt {a-b+c} \sqrt {a+b x+c x^2}}\right )-\frac {\left (3 b^2+12 a c+8 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}+\frac {1}{2} (a+b+c)^{3/2} \text {arctanh}\left (\frac {2 a+b+(b+2 c) x}{2 \sqrt {a+b+c} \sqrt {a+b x+c x^2}}\right ) \]
-1/2*(a-b+c)^(3/2)*arctanh(1/2*(2*a-b+(b-2*c)*x)/(a-b+c)^(1/2)/(c*x^2+b*x+ a)^(1/2))+1/2*(a+b+c)^(3/2)*arctanh(1/2*(2*a+b+(b+2*c)*x)/(a+b+c)^(1/2)/(c *x^2+b*x+a)^(1/2))-1/8*(12*a*c+3*b^2+8*c^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/ (c*x^2+b*x+a)^(1/2))/c^(1/2)-1/4*(2*c*x+5*b)*(c*x^2+b*x+a)^(1/2)
Time = 1.53 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=\frac {1}{4} \left (-\left ((5 b+2 c x) \sqrt {a+x (b+c x)}\right )-4 (-a+b-c)^{3/2} \arctan \left (\frac {\sqrt {-a+b-c} x}{\sqrt {a} (1+x)-\sqrt {a+x (b+c x)}}\right )+4 (-a-b-c)^{3/2} \arctan \left (\frac {\sqrt {-a-b-c} x}{\sqrt {a} (-1+x)+\sqrt {a+x (b+c x)}}\right )-\frac {\left (3 b^2+4 c (3 a+2 c)\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}\right ) \]
(-((5*b + 2*c*x)*Sqrt[a + x*(b + c*x)]) - 4*(-a + b - c)^(3/2)*ArcTan[(Sqr t[-a + b - c]*x)/(Sqrt[a]*(1 + x) - Sqrt[a + x*(b + c*x)])] + 4*(-a - b - c)^(3/2)*ArcTan[(Sqrt[-a - b - c]*x)/(Sqrt[a]*(-1 + x) + Sqrt[a + x*(b + c *x)])] - ((3*b^2 + 4*c*(3*a + 2*c))*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/Sqrt[c])/4
Time = 0.61 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1309, 27, 2144, 27, 1092, 219, 1366, 25, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx\) |
| \(\Big \downarrow \) 1309 |
| \(\displaystyle \frac {1}{2} \int \frac {8 a^2+4 c a+5 b^2+\left (3 b^2+8 c^2+12 a c\right ) x^2+16 b (a+c) x}{4 \left (1-x^2\right ) \sqrt {c x^2+b x+a}}dx-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \frac {8 a^2+4 c a+5 b^2+\left (3 b^2+8 c^2+12 a c\right ) x^2+16 b (a+c) x}{\left (1-x^2\right ) \sqrt {c x^2+b x+a}}dx-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 2144 |
| \(\displaystyle \frac {1}{8} \left (-\int -\frac {8 \left (a^2+2 c a+b^2+c^2+2 b (a+c) x\right )}{\left (1-x^2\right ) \sqrt {c x^2+b x+a}}dx-\left (\left (12 a c+3 b^2+8 c^2\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx\right )\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (8 \int \frac {a^2+2 c a+b^2+c^2+2 b (a+c) x}{\left (1-x^2\right ) \sqrt {c x^2+b x+a}}dx-\left (12 a c+3 b^2+8 c^2\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {1}{8} \left (8 \int \frac {a^2+2 c a+b^2+c^2+2 b (a+c) x}{\left (1-x^2\right ) \sqrt {c x^2+b x+a}}dx-2 \left (12 a c+3 b^2+8 c^2\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{8} \left (8 \int \frac {a^2+2 c a+b^2+c^2+2 b (a+c) x}{\left (1-x^2\right ) \sqrt {c x^2+b x+a}}dx-\frac {\left (12 a c+3 b^2+8 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 1366 |
| \(\displaystyle \frac {1}{8} \left (8 \left (\frac {1}{2} (a+b+c)^2 \int \frac {1}{(1-x) \sqrt {c x^2+b x+a}}dx-\frac {1}{2} (a-b+c)^2 \int -\frac {1}{(x+1) \sqrt {c x^2+b x+a}}dx\right )-\frac {\left (12 a c+3 b^2+8 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{8} \left (8 \left (\frac {1}{2} (a-b+c)^2 \int \frac {1}{(x+1) \sqrt {c x^2+b x+a}}dx+\frac {1}{2} (a+b+c)^2 \int \frac {1}{(1-x) \sqrt {c x^2+b x+a}}dx\right )-\frac {\left (12 a c+3 b^2+8 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {1}{8} \left (8 \left ((a-b+c)^2 \left (-\int \frac {1}{4 (a-b+c)-\frac {(2 a-b+(b-2 c) x)^2}{c x^2+b x+a}}d\frac {2 a-b+(b-2 c) x}{\sqrt {c x^2+b x+a}}\right )-(a+b+c)^2 \int \frac {1}{4 (a+b+c)-\frac {(2 a+b+(b+2 c) x)^2}{c x^2+b x+a}}d\left (-\frac {2 a+b+(b+2 c) x}{\sqrt {c x^2+b x+a}}\right )\right )-\frac {\left (12 a c+3 b^2+8 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{8} \left (8 \left (\frac {1}{2} (a+b+c)^{3/2} \text {arctanh}\left (\frac {2 a+x (b+2 c)+b}{2 \sqrt {a+b+c} \sqrt {a+b x+c x^2}}\right )-\frac {1}{2} (a-b+c)^{3/2} \text {arctanh}\left (\frac {2 a+x (b-2 c)-b}{2 \sqrt {a-b+c} \sqrt {a+b x+c x^2}}\right )\right )-\frac {\left (12 a c+3 b^2+8 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}\) |
-1/4*((5*b + 2*c*x)*Sqrt[a + b*x + c*x^2]) + (-(((3*b^2 + 12*a*c + 8*c^2)* ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[c]) + 8*(-1/2 *((a - b + c)^(3/2)*ArcTanh[(2*a - b + (b - 2*c)*x)/(2*Sqrt[a - b + c]*Sqr t[a + b*x + c*x^2])]) + ((a + b + c)^(3/2)*ArcTanh[(2*a + b + (b + 2*c)*x) /(2*Sqrt[a + b + c]*Sqrt[a + b*x + c*x^2])])/2))/8
3.1.91.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x _Symbol] :> Simp[(b*(3*p + 2*q) + 2*c*(p + q)*x)*(a + b*x + c*x^2)^(p - 1)* ((d + f*x^2)^(q + 1)/(2*f*(p + q)*(2*p + 2*q + 1))), x] - Simp[1/(2*f*(p + q)*(2*p + 2*q + 1)) Int[(a + b*x + c*x^2)^(p - 2)*(d + f*x^2)^q*Simp[b^2* d*(p - 1)*(2*p + q) - (p + q)*(b^2*d*(1 - p) - 2*a*(c*d - a*f*(2*p + 2*q + 1))) - (2*b*(c*d - a*f)*(1 - p)*(2*p + q) - 2*(p + q)*b*(2*c*d*(2*p + q) - (c*d + a*f)*(2*p + 2*q + 1)))*x + (b^2*f*p*(1 - p) + 2*c*(p + q)*(c*d*(2*p - 1) - a*f*(4*p + 2*q - 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x ] && NeQ[b^2 - 4*a*c, 0] && GtQ[p, 1] && NeQ[p + q, 0] && NeQ[2*p + 2*q + 1 , 0] && !IGtQ[p, 0] && !IGtQ[q, 0]
Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + ( f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[(h/2 + c*(g/(2*q ))) Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[(h/2 - c*(g/( 2*q))) Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d , e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && PosQ[(-a)*c]
Int[(Px_)/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2]}, Simp[C/c Int[1/Sqrt[d + e*x + f*x^2], x], x] + Simp[1/c Int[(A* c - a*C + B*c*x)/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f}, x] && PolyQ[Px, x, 2]
Time = 0.71 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.63
| method | result | size |
| risch | \(-\frac {\left (2 c x +5 b \right ) \sqrt {c \,x^{2}+b x +a}}{4}-\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )-\frac {3 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+\frac {\left (-4 a^{2}+8 b a -8 a c -4 b^{2}+8 b c -4 c^{2}\right ) \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+x \right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c}}{1+x}\right )}{8 \sqrt {a -b +c}}+\frac {\left (4 a^{2}+8 b a +8 a c +4 b^{2}+8 b c +4 c^{2}\right ) \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (-1+x \right )+2 \sqrt {a +b +c}\, \sqrt {\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c}}{-1+x}\right )}{8 \sqrt {a +b +c}}\) | \(308\) |
| default | \(-\frac {\left (\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c \right )^{\frac {3}{2}}}{6}-\frac {\left (b +2 c \right ) \left (\frac {\left (2 c \left (-1+x \right )+b +2 c \right ) \sqrt {\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c}}{4 c}+\frac {\left (4 c \left (a +b +c \right )-\left (b +2 c \right )^{2}\right ) \ln \left (\frac {\frac {b}{2}+c +c \left (-1+x \right )}{\sqrt {c}}+\sqrt {\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c}\right )}{8 c^{\frac {3}{2}}}\right )}{4}-\frac {\left (a +b +c \right ) \left (\sqrt {\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c}+\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (-1+x \right )}{\sqrt {c}}+\sqrt {\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c}\right )}{2 \sqrt {c}}-\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (-1+x \right )+2 \sqrt {a +b +c}\, \sqrt {\left (-1+x \right )^{2} c +\left (b +2 c \right ) \left (-1+x \right )+a +b +c}}{-1+x}\right )\right )}{2}+\frac {\left (\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c \right )^{\frac {3}{2}}}{6}+\frac {\left (b -2 c \right ) \left (\frac {\left (2 c \left (1+x \right )+b -2 c \right ) \sqrt {\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c}}{4 c}+\frac {\left (4 c \left (a -b +c \right )-\left (b -2 c \right )^{2}\right ) \ln \left (\frac {\frac {b}{2}-c +c \left (1+x \right )}{\sqrt {c}}+\sqrt {\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c}\right )}{8 c^{\frac {3}{2}}}\right )}{4}+\frac {\left (a -b +c \right ) \left (\sqrt {\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c}+\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +c \left (1+x \right )}{\sqrt {c}}+\sqrt {\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c}\right )}{2 \sqrt {c}}-\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+x \right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+x \right )^{2} c +\left (b -2 c \right ) \left (1+x \right )+a -b +c}}{1+x}\right )\right )}{2}\) | \(568\) |
-1/4*(2*c*x+5*b)*(c*x^2+b*x+a)^(1/2)-3/8*b^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2 +b*x+a)^(1/2))/c^(1/2)-c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)) -3/2*a*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/8*(-4*a^2+8*a *b-8*a*c-4*b^2+8*b*c-4*c^2)/(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(1+x)+2* (a-b+c)^(1/2)*((1+x)^2*c+(b-2*c)*(1+x)+a-b+c)^(1/2))/(1+x))+1/8*(4*a^2+8*a *b+8*a*c+4*b^2+8*b*c+4*c^2)/(a+b+c)^(1/2)*ln((2*a+2*b+2*c+(b+2*c)*(-1+x)+2 *(a+b+c)^(1/2)*((-1+x)^2*c+(b+2*c)*(-1+x)+a+b+c)^(1/2))/(-1+x))
Time = 116.63 (sec) , antiderivative size = 2579, normalized size of antiderivative = 13.65 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=\text {Too large to display} \]
[1/16*((3*b^2 + 12*a*c + 8*c^2)*sqrt(c)*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*((a - b)*c + c^2)* sqrt(a - b + c)*log(-((b^2 + 4*(a - 2*b)*c + 8*c^2)*x^2 - 4*sqrt(c*x^2 + b *x + a)*((b - 2*c)*x + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b^2 + 4* a*c + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*x)/(x^2 + 2*x + 1)) + 4*((a + b)*c + c^2)*sqrt(a + b + c)*log(-((b^2 + 4*(a + 2*b)*c + 8*c^2)*x^2 + 4*sqrt(c*x ^2 + b*x + a)*((b + 2*c)*x + 2*a + b)*sqrt(a + b + c) + 8*a^2 + 8*a*b + b^ 2 + 4*a*c + 2*(4*a*b + 3*b^2 + 4*(a + b)*c)*x)/(x^2 - 2*x + 1)) - 4*(2*c^2 *x + 5*b*c)*sqrt(c*x^2 + b*x + a))/c, -1/16*(8*((a - b)*c + c^2)*sqrt(-a + b - c)*arctan(-1/2*sqrt(c*x^2 + b*x + a)*((b - 2*c)*x + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^2)*x^2 + a^2 - a*b + a*c + (a*b - b^2 + b*c)*x)) - (3*b^2 + 12*a*c + 8*c^2)*sqrt(c)*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*((a + b)*c + c^2)*sqrt( a + b + c)*log(-((b^2 + 4*(a + 2*b)*c + 8*c^2)*x^2 + 4*sqrt(c*x^2 + b*x + a)*((b + 2*c)*x + 2*a + b)*sqrt(a + b + c) + 8*a^2 + 8*a*b + b^2 + 4*a*c + 2*(4*a*b + 3*b^2 + 4*(a + b)*c)*x)/(x^2 - 2*x + 1)) + 4*(2*c^2*x + 5*b*c) *sqrt(c*x^2 + b*x + a))/c, -1/16*(8*((a + b)*c + c^2)*sqrt(-a - b - c)*arc tan(1/2*sqrt(c*x^2 + b*x + a)*((b + 2*c)*x + 2*a + b)*sqrt(-a - b - c)/((( a + b)*c + c^2)*x^2 + a^2 + a*b + a*c + (a*b + b^2 + b*c)*x)) - (3*b^2 + 1 2*a*c + 8*c^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + ...
\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=- \int \frac {a \sqrt {a + b x + c x^{2}}}{x^{2} - 1}\, dx - \int \frac {b x \sqrt {a + b x + c x^{2}}}{x^{2} - 1}\, dx - \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{x^{2} - 1}\, dx \]
-Integral(a*sqrt(a + b*x + c*x**2)/(x**2 - 1), x) - Integral(b*x*sqrt(a + b*x + c*x**2)/(x**2 - 1), x) - Integral(c*x**2*sqrt(a + b*x + c*x**2)/(x** 2 - 1), x)
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=\text {Exception raised: ValueError} \]
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
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx=-\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x^2-1} \,d x \]