Integrand size = 38, antiderivative size = 317 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=-\frac {\left (2 a d \left (4 c d-5 e^2\right )-b \left (12 c d e-7 e^3\right )\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{128 d^4 (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac {\left (32 b c d+50 a d e-35 b e^2-6 d (10 a d-7 b e) x\right ) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{240 d^3 (a+b x)}-\frac {\left (4 c d-e^2\right ) \left (8 a c d^2-12 b c d e-10 a d e^2+7 b e^3\right ) \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{256 d^{9/2} (a+b x)} \]
1/5*b*x^2*(d*x^2+e*x+c)^(3/2)*((b*x+a)^2)^(1/2)/d/(b*x+a)-1/240*(32*b*c*d+ 50*a*d*e-35*b*e^2-6*d*(10*a*d-7*b*e)*x)*(d*x^2+e*x+c)^(3/2)*((b*x+a)^2)^(1 /2)/d^3/(b*x+a)-1/256*(4*c*d-e^2)*(8*a*c*d^2-10*a*d*e^2-12*b*c*d*e+7*b*e^3 )*arctanh(1/2*(2*d*x+e)/d^(1/2)/(d*x^2+e*x+c)^(1/2))*((b*x+a)^2)^(1/2)/d^( 9/2)/(b*x+a)-1/128*(2*a*d*(4*c*d-5*e^2)-b*(12*c*d*e-7*e^3))*(2*d*x+e)*((b* x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/d^4/(b*x+a)
Time = 0.61 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.74 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (2 \sqrt {d} \sqrt {c+x (e+d x)} \left (10 a d \left (15 e^3-10 d e^2 x+8 d^2 e x^2+48 d^3 x^3+4 c d (-13 e+6 d x)\right )+b \left (-256 c^2 d^2-105 e^4+70 d e^3 x-56 d^2 e^2 x^2+48 d^3 e x^3+384 d^4 x^4+4 c d \left (115 e^2-58 d e x+32 d^2 x^2\right )\right )\right )+15 \left (4 c d-e^2\right ) \left (2 a d \left (4 c d-5 e^2\right )+b \left (-12 c d e+7 e^3\right )\right ) \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )}{3840 d^{9/2} (a+b x)} \]
(Sqrt[(a + b*x)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(10*a*d*(15*e^3 - 10*d *e^2*x + 8*d^2*e*x^2 + 48*d^3*x^3 + 4*c*d*(-13*e + 6*d*x)) + b*(-256*c^2*d ^2 - 105*e^4 + 70*d*e^3*x - 56*d^2*e^2*x^2 + 48*d^3*e*x^3 + 384*d^4*x^4 + 4*c*d*(115*e^2 - 58*d*e*x + 32*d^2*x^2))) + 15*(4*c*d - e^2)*(2*a*d*(4*c*d - 5*e^2) + b*(-12*c*d*e + 7*e^3))*Log[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + x*(e + d*x)]]))/(3840*d^(9/2)*(a + b*x))
Time = 0.42 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.71, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1333, 27, 1236, 27, 1225, 1087, 1092, 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.
| \(\displaystyle \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2+e x} \, dx\) |
| \(\Big \downarrow \) 1333 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int 2 b x^2 (a+b x) \sqrt {d x^2+e x+c}dx}{2 b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^2 (a+b x) \sqrt {d x^2+e x+c}dx}{a+b x}\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\int -\frac {1}{2} x (4 b c-(10 a d-7 b e) x) \sqrt {d x^2+e x+c}dx}{5 d}+\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\int x (4 b c-(10 a d-7 b e) x) \sqrt {d x^2+e x+c}dx}{10 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {5 \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \int \sqrt {d x^2+e x+c}dx}{16 d^2}+\frac {\left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{24 d^2}}{10 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {5 \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{\sqrt {d x^2+e x+c}}dx}{8 d}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}+\frac {\left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{24 d^2}}{10 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {5 \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{4 d-\frac {(e+2 d x)^2}{d x^2+e x+c}}d\frac {e+2 d x}{\sqrt {d x^2+e x+c}}}{4 d}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}+\frac {\left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{24 d^2}}{10 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2+e x\right )^{3/2}}{5 d}-\frac {\frac {5 \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \left (\frac {\left (4 c d-e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2}}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}+\frac {\left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{24 d^2}}{10 d}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*x^2*(c + e*x + d*x^2)^(3/2))/(5*d) - (( (32*b*c*d + 50*a*d*e - 35*b*e^2 - 6*d*(10*a*d - 7*b*e)*x)*(c + e*x + d*x^2 )^(3/2))/(24*d^2) + (5*(8*a*c*d^2 - 12*b*c*d*e - 10*a*d*e^2 + 7*b*e^3)*((( e + 2*d*x)*Sqrt[c + e*x + d*x^2])/(4*d) + ((4*c*d - e^2)*ArcTanh[(e + 2*d* x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(8*d^(3/2))))/(16*d^2))/(10*d)))/(a + b*x)
3.1.46.3.1 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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ ) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a + b*x + c*x^2)^Fr acPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])) Int[(g + h*x)^m* (b + 2*c*x)^(2*p)*(d + e*x + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g , h, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.67 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.84
| method | result | size |
| risch | \(\frac {\left (384 b \,x^{4} d^{4}+480 a \,d^{4} x^{3}+48 b \,d^{3} e \,x^{3}+80 a \,d^{3} e \,x^{2}+128 b c \,d^{3} x^{2}-56 b \,d^{2} e^{2} x^{2}+240 a c \,d^{3} x -100 a \,d^{2} e^{2} x -232 x b c \,d^{2} e +70 b d \,e^{3} x -520 a c \,d^{2} e +150 a d \,e^{3}-256 b \,d^{2} c^{2}+460 b c d \,e^{2}-105 b \,e^{4}\right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b x +a \right )^{2}}}{1920 d^{4} \left (b x +a \right )}-\frac {\left (32 a \,c^{2} d^{3}-48 a c \,d^{2} e^{2}+10 a d \,e^{4}-48 b \,c^{2} d^{2} e +40 b c d \,e^{3}-7 b \,e^{5}\right ) \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) \sqrt {\left (b x +a \right )^{2}}}{256 d^{\frac {9}{2}} \left (b x +a \right )}\) | \(266\) |
| default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (768 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {9}{2}} b \,x^{2}+960 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {9}{2}} a x -672 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {7}{2}} b e x -800 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {7}{2}} a e -512 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {7}{2}} b c +560 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {5}{2}} b \,e^{2}-480 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {9}{2}} a c x +600 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} a \,e^{2} x +720 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} b c e x -420 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b \,e^{3} x -240 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} a c e +300 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a \,e^{3}+360 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b c \,e^{2}-210 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b \,e^{4}-480 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,c^{2} d^{4}+720 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{3} e^{2}-150 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{2} e^{4}+720 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{3} e -600 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e^{3}+105 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{5}\right )}{3840 d^{\frac {11}{2}}}\) | \(530\) |
1/1920*(384*b*d^4*x^4+480*a*d^4*x^3+48*b*d^3*e*x^3+80*a*d^3*e*x^2+128*b*c* d^3*x^2-56*b*d^2*e^2*x^2+240*a*c*d^3*x-100*a*d^2*e^2*x-232*b*c*d^2*e*x+70* b*d*e^3*x-520*a*c*d^2*e+150*a*d*e^3-256*b*c^2*d^2+460*b*c*d*e^2-105*b*e^4) *(d*x^2+e*x+c)^(1/2)/d^4*((b*x+a)^2)^(1/2)/(b*x+a)-1/256*(32*a*c^2*d^3-48* a*c*d^2*e^2+10*a*d*e^4-48*b*c^2*d^2*e+40*b*c*d*e^3-7*b*e^5)/d^(9/2)*ln((1/ 2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.35 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.63 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\left [-\frac {15 \, {\left (32 \, a c^{2} d^{3} - 48 \, b c^{2} d^{2} e - 48 \, a c d^{2} e^{2} + 40 \, b c d e^{3} + 10 \, a d e^{4} - 7 \, b e^{5}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) - 4 \, {\left (384 \, b d^{5} x^{4} - 256 \, b c^{2} d^{3} - 520 \, a c d^{3} e + 460 \, b c d^{2} e^{2} + 150 \, a d^{2} e^{3} - 105 \, b d e^{4} + 48 \, {\left (10 \, a d^{5} + b d^{4} e\right )} x^{3} + 8 \, {\left (16 \, b c d^{4} + 10 \, a d^{4} e - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (120 \, a c d^{4} - 116 \, b c d^{3} e - 50 \, a d^{3} e^{2} + 35 \, b d^{2} e^{3}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{7680 \, d^{5}}, \frac {15 \, {\left (32 \, a c^{2} d^{3} - 48 \, b c^{2} d^{2} e - 48 \, a c d^{2} e^{2} + 40 \, b c d e^{3} + 10 \, a d e^{4} - 7 \, b e^{5}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (384 \, b d^{5} x^{4} - 256 \, b c^{2} d^{3} - 520 \, a c d^{3} e + 460 \, b c d^{2} e^{2} + 150 \, a d^{2} e^{3} - 105 \, b d e^{4} + 48 \, {\left (10 \, a d^{5} + b d^{4} e\right )} x^{3} + 8 \, {\left (16 \, b c d^{4} + 10 \, a d^{4} e - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (120 \, a c d^{4} - 116 \, b c d^{3} e - 50 \, a d^{3} e^{2} + 35 \, b d^{2} e^{3}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{3840 \, d^{5}}\right ] \]
[-1/7680*(15*(32*a*c^2*d^3 - 48*b*c^2*d^2*e - 48*a*c*d^2*e^2 + 40*b*c*d*e^ 3 + 10*a*d*e^4 - 7*b*e^5)*sqrt(d)*log(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) - 4*(384*b*d^5*x^4 - 256*b*c^ 2*d^3 - 520*a*c*d^3*e + 460*b*c*d^2*e^2 + 150*a*d^2*e^3 - 105*b*d*e^4 + 48 *(10*a*d^5 + b*d^4*e)*x^3 + 8*(16*b*c*d^4 + 10*a*d^4*e - 7*b*d^3*e^2)*x^2 + 2*(120*a*c*d^4 - 116*b*c*d^3*e - 50*a*d^3*e^2 + 35*b*d^2*e^3)*x)*sqrt(d* x^2 + e*x + c))/d^5, 1/3840*(15*(32*a*c^2*d^3 - 48*b*c^2*d^2*e - 48*a*c*d^ 2*e^2 + 40*b*c*d*e^3 + 10*a*d*e^4 - 7*b*e^5)*sqrt(-d)*arctan(1/2*sqrt(d*x^ 2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(384*b*d^5* x^4 - 256*b*c^2*d^3 - 520*a*c*d^3*e + 460*b*c*d^2*e^2 + 150*a*d^2*e^3 - 10 5*b*d*e^4 + 48*(10*a*d^5 + b*d^4*e)*x^3 + 8*(16*b*c*d^4 + 10*a*d^4*e - 7*b *d^3*e^2)*x^2 + 2*(120*a*c*d^4 - 116*b*c*d^3*e - 50*a*d^3*e^2 + 35*b*d^2*e ^3)*x)*sqrt(d*x^2 + e*x + c))/d^5]
\[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int x^{2} \sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x\right )^{2}}\, dx \]
\[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int { \sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}} x^{2} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.16 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {1}{1920} \, \sqrt {d x^{2} + e x + c} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b x \mathrm {sgn}\left (b x + a\right ) + \frac {10 \, a d^{4} \mathrm {sgn}\left (b x + a\right ) + b d^{3} e \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} x + \frac {16 \, b c d^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a d^{3} e \mathrm {sgn}\left (b x + a\right ) - 7 \, b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} x + \frac {120 \, a c d^{3} \mathrm {sgn}\left (b x + a\right ) - 116 \, b c d^{2} e \mathrm {sgn}\left (b x + a\right ) - 50 \, a d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, b d e^{3} \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} x - \frac {256 \, b c^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 520 \, a c d^{2} e \mathrm {sgn}\left (b x + a\right ) - 460 \, b c d e^{2} \mathrm {sgn}\left (b x + a\right ) - 150 \, a d e^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, b e^{4} \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} + \frac {{\left (32 \, a c^{2} d^{3} \mathrm {sgn}\left (b x + a\right ) - 48 \, b c^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 48 \, a c d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 40 \, b c d e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a d e^{4} \mathrm {sgn}\left (b x + a\right ) - 7 \, b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} + e \right |}\right )}{256 \, d^{\frac {9}{2}}} \]
1/1920*sqrt(d*x^2 + e*x + c)*(2*(4*(6*(8*b*x*sgn(b*x + a) + (10*a*d^4*sgn( b*x + a) + b*d^3*e*sgn(b*x + a))/d^4)*x + (16*b*c*d^3*sgn(b*x + a) + 10*a* d^3*e*sgn(b*x + a) - 7*b*d^2*e^2*sgn(b*x + a))/d^4)*x + (120*a*c*d^3*sgn(b *x + a) - 116*b*c*d^2*e*sgn(b*x + a) - 50*a*d^2*e^2*sgn(b*x + a) + 35*b*d* e^3*sgn(b*x + a))/d^4)*x - (256*b*c^2*d^2*sgn(b*x + a) + 520*a*c*d^2*e*sgn (b*x + a) - 460*b*c*d*e^2*sgn(b*x + a) - 150*a*d*e^3*sgn(b*x + a) + 105*b* e^4*sgn(b*x + a))/d^4) + 1/256*(32*a*c^2*d^3*sgn(b*x + a) - 48*b*c^2*d^2*e *sgn(b*x + a) - 48*a*c*d^2*e^2*sgn(b*x + a) + 40*b*c*d*e^3*sgn(b*x + a) + 10*a*d*e^4*sgn(b*x + a) - 7*b*e^5*sgn(b*x + a))*log(abs(2*(sqrt(d)*x - sqr t(d*x^2 + e*x + c))*sqrt(d) + e))/d^(9/2)
Timed out. \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int x^2\,\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c} \,d x \]