Integrand size = 34, antiderivative size = 236 \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^2 \left (3 C d^2+4 B d e+10 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{16 e^2}-\frac {d \left (4 C d^2+e (7 B d+10 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac {\left (3 C d^2+2 e (2 B d+A e)\right ) x \left (d^2-e^2 x^2\right )^{3/2}}{8 e^2}-\frac {(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^4 \left (3 C d^2+4 B d e+10 A e^2\right ) \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \]
-1/15*d*(4*C*d^2+e*(10*A*e+7*B*d))*(-e^2*x^2+d^2)^(3/2)/e^3-1/8*(3*C*d^2+2 *e*(A*e+2*B*d))*x*(-e^2*x^2+d^2)^(3/2)/e^2-1/5*(B*e+2*C*d)*x^2*(-e^2*x^2+d ^2)^(3/2)/e-1/6*C*x^3*(-e^2*x^2+d^2)^(3/2)+1/16*d^4*(10*A*e^2+4*B*d*e+3*C* d^2)*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+1/16*d^2*(10*A*e^2+4*B*d*e+3*C*d ^2)*x*(-e^2*x^2+d^2)^(1/2)/e^2
Time = 1.07 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.92 \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (C \left (-64 d^5-45 d^4 e x-32 d^3 e^2 x^2+50 d^2 e^3 x^3+96 d e^4 x^4+40 e^5 x^5\right )+2 e \left (5 A e \left (-16 d^3+9 d^2 e x+16 d e^2 x^2+6 e^3 x^3\right )+B \left (-56 d^4-30 d^3 e x+32 d^2 e^2 x^2+60 d e^3 x^3+24 e^4 x^4\right )\right )\right )-30 d^4 \left (3 C d^2+2 e (2 B d+5 A e)\right ) \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^3} \]
(Sqrt[d^2 - e^2*x^2]*(C*(-64*d^5 - 45*d^4*e*x - 32*d^3*e^2*x^2 + 50*d^2*e^ 3*x^3 + 96*d*e^4*x^4 + 40*e^5*x^5) + 2*e*(5*A*e*(-16*d^3 + 9*d^2*e*x + 16* d*e^2*x^2 + 6*e^3*x^3) + B*(-56*d^4 - 30*d^3*e*x + 32*d^2*e^2*x^2 + 60*d*e ^3*x^3 + 24*e^4*x^4))) - 30*d^4*(3*C*d^2 + 2*e*(2*B*d + 5*A*e))*ArcTan[(e* x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(240*e^3)
Time = 0.73 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {2346, 27, 2346, 25, 2346, 25, 27, 455, 211, 224, 216}
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.
| \(\displaystyle \int (d+e x)^2 \sqrt {d^2-e^2 x^2} \left (A+B x+C x^2\right ) \, dx\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle -\frac {\int -3 \sqrt {d^2-e^2 x^2} \left (2 e^3 (2 C d+B e) x^3+e^2 \left (3 C d^2+2 e (2 B d+A e)\right ) x^2+2 d e^2 (B d+2 A e) x+2 A d^2 e^2\right )dx}{6 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {d^2-e^2 x^2} \left (2 e^3 (2 C d+B e) x^3+e^2 \left (3 C d^2+2 e (2 B d+A e)\right ) x^2+2 d e^2 (B d+2 A e) x+2 A d^2 e^2\right )dx}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {-\frac {\int -\sqrt {d^2-e^2 x^2} \left (10 A d^2 e^4+5 \left (3 C d^2+2 e (2 B d+A e)\right ) x^2 e^4+2 d \left (4 C d^2+e (7 B d+10 A e)\right ) x e^3\right )dx}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \sqrt {d^2-e^2 x^2} \left (10 A d^2 e^4+5 \left (3 C d^2+2 e (2 B d+A e)\right ) x^2 e^4+2 d \left (4 C d^2+e (7 B d+10 A e)\right ) x e^3\right )dx}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {-\frac {\int -d e^4 \left (5 d \left (3 C d^2+4 B e d+10 A e^2\right )+8 e \left (4 C d^2+e (7 B d+10 A e)\right ) x\right ) \sqrt {d^2-e^2 x^2}dx}{4 e^2}-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int d e^4 \left (5 d \left (3 C d^2+4 B e d+10 A e^2\right )+8 e \left (4 C d^2+e (7 B d+10 A e)\right ) x\right ) \sqrt {d^2-e^2 x^2}dx}{4 e^2}-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{4} d e^2 \int \left (5 d \left (3 C d^2+4 B e d+10 A e^2\right )+8 e \left (4 C d^2+e (7 B d+10 A e)\right ) x\right ) \sqrt {d^2-e^2 x^2}dx-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {\frac {\frac {1}{4} d e^2 \left (5 d \left (10 A e^2+4 B d e+3 C d^2\right ) \int \sqrt {d^2-e^2 x^2}dx-\frac {8 \left (d^2-e^2 x^2\right )^{3/2} \left (e (10 A e+7 B d)+4 C d^2\right )}{3 e}\right )-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\frac {1}{4} d e^2 \left (5 d \left (10 A e^2+4 B d e+3 C d^2\right ) \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2} \left (e (10 A e+7 B d)+4 C d^2\right )}{3 e}\right )-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {1}{4} d e^2 \left (5 d \left (10 A e^2+4 B d e+3 C d^2\right ) \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2} \left (e (10 A e+7 B d)+4 C d^2\right )}{3 e}\right )-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {1}{4} d e^2 \left (5 d \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right ) \left (10 A e^2+4 B d e+3 C d^2\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2} \left (e (10 A e+7 B d)+4 C d^2\right )}{3 e}\right )-\frac {5}{4} e^2 x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{5 e^2}-\frac {2}{5} e x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{2 e^2}-\frac {1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}\) |
-1/6*(C*x^3*(d^2 - e^2*x^2)^(3/2)) + ((-2*e*(2*C*d + B*e)*x^2*(d^2 - e^2*x ^2)^(3/2))/5 + ((-5*e^2*(3*C*d^2 + 2*e*(2*B*d + A*e))*x*(d^2 - e^2*x^2)^(3 /2))/4 + (d*e^2*((-8*(4*C*d^2 + e*(7*B*d + 10*A*e))*(d^2 - e^2*x^2)^(3/2)) /(3*e) + 5*d*(3*C*d^2 + 4*B*d*e + 10*A*e^2)*((x*Sqrt[d^2 - e^2*x^2])/2 + ( d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e))))/4)/(5*e^2))/(2*e^2)
3.1.1.3.1 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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 0.62 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(-\frac {\left (-40 e^{5} C \,x^{5}-48 B \,x^{4} e^{5}-96 C d \,e^{4} x^{4}-60 A \,e^{5} x^{3}-120 x^{3} d B \,e^{4}-50 C \,d^{2} e^{3} x^{3}-160 A d \,e^{4} x^{2}-64 x^{2} d^{2} B \,e^{3}+32 C \,d^{3} e^{2} x^{2}-90 A \,d^{2} e^{3} x +60 x \,d^{3} B \,e^{2}+45 C \,d^{4} e x +160 A \,d^{3} e^{2}+112 B \,d^{4} e +64 C \,d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{240 e^{3}}+\frac {d^{4} \left (10 A \,e^{2}+4 B d e +3 C \,d^{2}\right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e^{2} \sqrt {e^{2}}}\) | \(218\) |
| default | \(A \,d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )+e^{2} C \left (-\frac {x^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{6 e^{2}}+\frac {d^{2} \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+\left (B \,e^{2}+2 d e C \right ) \left (-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{5 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{15 e^{4}}\right )-\frac {\left (2 A d e +B \,d^{2}\right ) \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{2}}+\left (A \,e^{2}+2 B d e +C \,d^{2}\right ) \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )\) | \(355\) |
-1/240/e^3*(-40*C*e^5*x^5-48*B*e^5*x^4-96*C*d*e^4*x^4-60*A*e^5*x^3-120*B*d *e^4*x^3-50*C*d^2*e^3*x^3-160*A*d*e^4*x^2-64*B*d^2*e^3*x^2+32*C*d^3*e^2*x^ 2-90*A*d^2*e^3*x+60*B*d^3*e^2*x+45*C*d^4*e*x+160*A*d^3*e^2+112*B*d^4*e+64* C*d^5)*(-e^2*x^2+d^2)^(1/2)+1/16*d^4/e^2*(10*A*e^2+4*B*d*e+3*C*d^2)/(e^2)^ (1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
Time = 0.45 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.89 \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=-\frac {30 \, {\left (3 \, C d^{6} + 4 \, B d^{5} e + 10 \, A d^{4} e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (40 \, C e^{5} x^{5} - 64 \, C d^{5} - 112 \, B d^{4} e - 160 \, A d^{3} e^{2} + 48 \, {\left (2 \, C d e^{4} + B e^{5}\right )} x^{4} + 10 \, {\left (5 \, C d^{2} e^{3} + 12 \, B d e^{4} + 6 \, A e^{5}\right )} x^{3} - 32 \, {\left (C d^{3} e^{2} - 2 \, B d^{2} e^{3} - 5 \, A d e^{4}\right )} x^{2} - 15 \, {\left (3 \, C d^{4} e + 4 \, B d^{3} e^{2} - 6 \, A d^{2} e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{3}} \]
-1/240*(30*(3*C*d^6 + 4*B*d^5*e + 10*A*d^4*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (40*C*e^5*x^5 - 64*C*d^5 - 112*B*d^4*e - 160*A*d^3*e^2 + 48*(2*C*d*e^4 + B*e^5)*x^4 + 10*(5*C*d^2*e^3 + 12*B*d*e^4 + 6*A*e^5)*x^3 - 32*(C*d^3*e^2 - 2*B*d^2*e^3 - 5*A*d*e^4)*x^2 - 15*(3*C*d^4*e + 4*B*d^3*e ^2 - 6*A*d^2*e^3)*x)*sqrt(-e^2*x^2 + d^2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (219) = 438\).
Time = 0.56 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.97 \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (\frac {C e^{2} x^{5}}{6} - \frac {x^{4} \left (- B e^{4} - 2 C d e^{3}\right )}{5 e^{2}} - \frac {x^{3} \left (- A e^{4} - 2 B d e^{3} - \frac {5 C d^{2} e^{2}}{6}\right )}{4 e^{2}} - \frac {x^{2} \left (- 2 A d e^{3} + 2 C d^{3} e + \frac {4 d^{2} \left (- B e^{4} - 2 C d e^{3}\right )}{5 e^{2}}\right )}{3 e^{2}} - \frac {x \left (2 B d^{3} e + C d^{4} + \frac {3 d^{2} \left (- A e^{4} - 2 B d e^{3} - \frac {5 C d^{2} e^{2}}{6}\right )}{4 e^{2}}\right )}{2 e^{2}} - \frac {2 A d^{3} e + B d^{4} + \frac {2 d^{2} \left (- 2 A d e^{3} + 2 C d^{3} e + \frac {4 d^{2} \left (- B e^{4} - 2 C d e^{3}\right )}{5 e^{2}}\right )}{3 e^{2}}}{e^{2}}\right ) + \left (A d^{4} + \frac {d^{2} \cdot \left (2 B d^{3} e + C d^{4} + \frac {3 d^{2} \left (- A e^{4} - 2 B d e^{3} - \frac {5 C d^{2} e^{2}}{6}\right )}{4 e^{2}}\right )}{2 e^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: e^{2} \neq 0 \\\left (A d^{2} x + \frac {C e^{2} x^{5}}{5} + \frac {x^{4} \left (B e^{2} + 2 C d e\right )}{4} + \frac {x^{3} \left (A e^{2} + 2 B d e + C d^{2}\right )}{3} + \frac {x^{2} \cdot \left (2 A d e + B d^{2}\right )}{2}\right ) \sqrt {d^{2}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(d**2 - e**2*x**2)*(C*e**2*x**5/6 - x**4*(-B*e**4 - 2*C*d*e **3)/(5*e**2) - x**3*(-A*e**4 - 2*B*d*e**3 - 5*C*d**2*e**2/6)/(4*e**2) - x **2*(-2*A*d*e**3 + 2*C*d**3*e + 4*d**2*(-B*e**4 - 2*C*d*e**3)/(5*e**2))/(3 *e**2) - x*(2*B*d**3*e + C*d**4 + 3*d**2*(-A*e**4 - 2*B*d*e**3 - 5*C*d**2* e**2/6)/(4*e**2))/(2*e**2) - (2*A*d**3*e + B*d**4 + 2*d**2*(-2*A*d*e**3 + 2*C*d**3*e + 4*d**2*(-B*e**4 - 2*C*d*e**3)/(5*e**2))/(3*e**2))/e**2) + (A* d**4 + d**2*(2*B*d**3*e + C*d**4 + 3*d**2*(-A*e**4 - 2*B*d*e**3 - 5*C*d**2 *e**2/6)/(4*e**2))/(2*e**2))*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt (d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True)), Ne(e**2, 0)), ((A*d**2*x + C*e**2*x**5/5 + x**4*(B*e**2 + 2*C*d*e )/4 + x**3*(A*e**2 + 2*B*d*e + C*d**2)/3 + x**2*(2*A*d*e + B*d**2)/2)*sqrt (d**2), True))
Time = 0.28 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.57 \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=-\frac {1}{6} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C x^{3} + \frac {A d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}}} + \frac {C d^{6} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e^{2}} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} A d^{2} x + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C d^{4} x}{16 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C d^{2} x}{8 \, e^{2}} + \frac {{\left (C d^{2} + 2 \, B d e + A e^{2}\right )} d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} B d^{2}}{3 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} A d}{3 \, e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} {\left (C d^{2} + 2 \, B d e + A e^{2}\right )} d^{2} x}{8 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (2 \, C d e + B e^{2}\right )} x^{2}}{5 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (C d^{2} + 2 \, B d e + A e^{2}\right )} x}{4 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (2 \, C d e + B e^{2}\right )} d^{2}}{15 \, e^{4}} \]
-1/6*(-e^2*x^2 + d^2)^(3/2)*C*x^3 + 1/2*A*d^4*arcsin(e^2*x/(d*sqrt(e^2)))/ sqrt(e^2) + 1/16*C*d^6*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) + 1/2*s qrt(-e^2*x^2 + d^2)*A*d^2*x + 1/16*sqrt(-e^2*x^2 + d^2)*C*d^4*x/e^2 - 1/8* (-e^2*x^2 + d^2)^(3/2)*C*d^2*x/e^2 + 1/8*(C*d^2 + 2*B*d*e + A*e^2)*d^4*arc sin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) - 1/3*(-e^2*x^2 + d^2)^(3/2)*B*d^ 2/e^2 - 2/3*(-e^2*x^2 + d^2)^(3/2)*A*d/e + 1/8*sqrt(-e^2*x^2 + d^2)*(C*d^2 + 2*B*d*e + A*e^2)*d^2*x/e^2 - 1/5*(-e^2*x^2 + d^2)^(3/2)*(2*C*d*e + B*e^ 2)*x^2/e^2 - 1/4*(-e^2*x^2 + d^2)^(3/2)*(C*d^2 + 2*B*d*e + A*e^2)*x/e^2 - 2/15*(-e^2*x^2 + d^2)^(3/2)*(2*C*d*e + B*e^2)*d^2/e^4
Time = 0.30 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.95 \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {1}{240} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, C e^{2} x + \frac {6 \, {\left (2 \, C d e^{9} + B e^{10}\right )}}{e^{8}}\right )} x + \frac {5 \, {\left (5 \, C d^{2} e^{8} + 12 \, B d e^{9} + 6 \, A e^{10}\right )}}{e^{8}}\right )} x - \frac {16 \, {\left (C d^{3} e^{7} - 2 \, B d^{2} e^{8} - 5 \, A d e^{9}\right )}}{e^{8}}\right )} x - \frac {15 \, {\left (3 \, C d^{4} e^{6} + 4 \, B d^{3} e^{7} - 6 \, A d^{2} e^{8}\right )}}{e^{8}}\right )} x - \frac {16 \, {\left (4 \, C d^{5} e^{5} + 7 \, B d^{4} e^{6} + 10 \, A d^{3} e^{7}\right )}}{e^{8}}\right )} + \frac {{\left (3 \, C d^{6} + 4 \, B d^{5} e + 10 \, A d^{4} e^{2}\right )} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e^{2} {\left | e \right |}} \]
1/240*sqrt(-e^2*x^2 + d^2)*((2*((4*(5*C*e^2*x + 6*(2*C*d*e^9 + B*e^10)/e^8 )*x + 5*(5*C*d^2*e^8 + 12*B*d*e^9 + 6*A*e^10)/e^8)*x - 16*(C*d^3*e^7 - 2*B *d^2*e^8 - 5*A*d*e^9)/e^8)*x - 15*(3*C*d^4*e^6 + 4*B*d^3*e^7 - 6*A*d^2*e^8 )/e^8)*x - 16*(4*C*d^5*e^5 + 7*B*d^4*e^6 + 10*A*d^3*e^7)/e^8) + 1/16*(3*C* d^6 + 4*B*d^5*e + 10*A*d^4*e^2)*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e))
Timed out. \[ \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\int \sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^2\,\left (C\,x^2+B\,x+A\right ) \,d x \]