Integrand size = 34, antiderivative size = 236 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt {d^2-e^2 x^2}}{15 e}-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}+\frac {d^3 \left (13 C d^2+15 B d e+20 A e^2\right ) \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]
1/8*d^3*(20*A*e^2+15*B*d*e+13*C*d^2)*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3- 1/15*d^2*(55*A*e^2+45*B*d*e+38*C*d^2)*(-e^2*x^2+d^2)^(1/2)/e^3-1/8*d*(12*A *e^2+15*B*d*e+13*C*d^2)*x*(-e^2*x^2+d^2)^(1/2)/e^2-1/15*(19*C*d^2+5*e*(A*e +3*B*d))*x^2*(-e^2*x^2+d^2)^(1/2)/e-1/4*(B*e+3*C*d)*x^3*(-e^2*x^2+d^2)^(1/ 2)-1/5*C*e*x^4*(-e^2*x^2+d^2)^(1/2)
Time = 0.84 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (C \left (304 d^4+195 d^3 e x+152 d^2 e^2 x^2+90 d e^3 x^3+24 e^4 x^4\right )+5 e \left (4 A e \left (22 d^2+9 d e x+2 e^2 x^2\right )+3 B \left (24 d^3+15 d^2 e x+8 d e^2 x^2+2 e^3 x^3\right )\right )\right )+30 d^3 \left (13 C d^2+5 e (3 B d+4 A e)\right ) \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{120 e^3} \]
-1/120*(Sqrt[d^2 - e^2*x^2]*(C*(304*d^4 + 195*d^3*e*x + 152*d^2*e^2*x^2 + 90*d*e^3*x^3 + 24*e^4*x^4) + 5*e*(4*A*e*(22*d^2 + 9*d*e*x + 2*e^2*x^2) + 3 *B*(24*d^3 + 15*d^2*e*x + 8*d*e^2*x^2 + 2*e^3*x^3))) + 30*d^3*(13*C*d^2 + 5*e*(3*B*d + 4*A*e))*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^3
Time = 1.03 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2346, 25, 2346, 25, 2346, 25, 2346, 25, 27, 455, 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.
| \(\displaystyle \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle -\frac {\int -\frac {5 e^4 (3 C d+B e) x^4+e^3 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3+5 d e^2 \left (C d^2+3 e (B d+A e)\right ) x^2+5 d^2 e^2 (B d+3 A e) x+5 A d^3 e^2}{\sqrt {d^2-e^2 x^2}}dx}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {5 e^4 (3 C d+B e) x^4+e^3 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3+5 d e^2 \left (C d^2+3 e (B d+A e)\right ) x^2+5 d^2 e^2 (B d+3 A e) x+5 A d^3 e^2}{\sqrt {d^2-e^2 x^2}}dx}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {-\frac {\int -\frac {4 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3 e^5+20 A d^3 e^4+5 d \left (13 C d^2+15 B e d+12 A e^2\right ) x^2 e^4+20 d^2 (B d+3 A e) x e^4}{\sqrt {d^2-e^2 x^2}}dx}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {4 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3 e^5+20 A d^3 e^4+5 d \left (13 C d^2+15 B e d+12 A e^2\right ) x^2 e^4+20 d^2 (B d+3 A e) x e^4}{\sqrt {d^2-e^2 x^2}}dx}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {60 A d^3 e^6+15 d \left (13 C d^2+15 B e d+12 A e^2\right ) x^2 e^6+4 d^2 \left (38 C d^2+45 B e d+55 A e^2\right ) x e^5}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {60 A d^3 e^6+15 d \left (13 C d^2+15 B e d+12 A e^2\right ) x^2 e^6+4 d^2 \left (38 C d^2+45 B e d+55 A e^2\right ) x e^5}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int -\frac {d^2 e^6 \left (15 d \left (13 C d^2+15 B e d+20 A e^2\right )+8 e \left (38 C d^2+45 B e d+55 A e^2\right ) x\right )}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {15}{2} d e^4 x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {d^2 e^6 \left (15 d \left (13 C d^2+5 e (3 B d+4 A e)\right )+8 e \left (38 C d^2+45 B e d+55 A e^2\right ) x\right )}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {15}{2} d e^4 x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {1}{2} d^2 e^4 \int \frac {15 d \left (13 C d^2+5 e (3 B d+4 A e)\right )+8 e \left (38 C d^2+45 B e d+55 A e^2\right ) x}{\sqrt {d^2-e^2 x^2}}dx-\frac {15}{2} d e^4 x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {\frac {\frac {\frac {1}{2} d^2 e^4 \left (15 d \left (20 A e^2+15 B d e+13 C d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {8 \sqrt {d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{e}\right )-\frac {15}{2} d e^4 x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {\frac {1}{2} d^2 e^4 \left (15 d \left (20 A e^2+15 B d e+13 C d^2\right ) \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {8 \sqrt {d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{e}\right )-\frac {15}{2} d e^4 x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {1}{2} d^2 e^4 \left (\frac {15 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (20 A e^2+15 B d e+13 C d^2\right )}{e}-\frac {8 \sqrt {d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{e}\right )-\frac {15}{2} d e^4 x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{3 e^2}-\frac {4}{3} e^3 x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{4 e^2}-\frac {5}{4} e^2 x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)}{5 e^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}\) |
-1/5*(C*e*x^4*Sqrt[d^2 - e^2*x^2]) + ((-5*e^2*(3*C*d + B*e)*x^3*Sqrt[d^2 - e^2*x^2])/4 + ((-4*e^3*(19*C*d^2 + 5*e*(3*B*d + A*e))*x^2*Sqrt[d^2 - e^2* x^2])/3 + ((-15*d*e^4*(13*C*d^2 + 15*B*d*e + 12*A*e^2)*x*Sqrt[d^2 - e^2*x^ 2])/2 + (d^2*e^4*((-8*(38*C*d^2 + 45*B*d*e + 55*A*e^2)*Sqrt[d^2 - e^2*x^2] )/e + (15*d*(13*C*d^2 + 15*B*d*e + 20*A*e^2)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x ^2]])/e))/2)/(3*e^2))/(4*e^2))/(5*e^2)
3.1.10.3.1 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]))*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 = 0.79 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.77
| method | result | size |
| risch | \(-\frac {\left (24 e^{4} C \,x^{4}+30 x^{3} B \,e^{4}+90 C d \,e^{3} x^{3}+40 A \,e^{4} x^{2}+120 x^{2} d B \,e^{3}+152 C \,d^{2} e^{2} x^{2}+180 A d \,e^{3} x +225 x B \,d^{2} e^{2}+195 C \,d^{3} x e +440 A \,d^{2} e^{2}+360 B \,d^{3} e +304 C \,d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{120 e^{3}}+\frac {d^{3} \left (20 A \,e^{2}+15 B d e +13 C \,d^{2}\right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}\) | \(182\) |
| default | \(\frac {A \,d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+e^{3} C \left (-\frac {x^{4} \sqrt {-e^{2} x^{2}+d^{2}}}{5 e^{2}}+\frac {4 d^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )}{5 e^{2}}\right )+\left (B \,e^{3}+3 d \,e^{2} C \right ) \left (-\frac {x^{3} \sqrt {-e^{2} x^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )-\frac {\left (3 A \,d^{2} e +B \,d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{e^{2}}+\left (A \,e^{3}+3 B d \,e^{2}+3 d^{2} e C \right ) \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+\left (3 A d \,e^{2}+3 B \,d^{2} e +d^{3} C \right ) \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )\) | \(391\) |
-1/120/e^3*(24*C*e^4*x^4+30*B*e^4*x^3+90*C*d*e^3*x^3+40*A*e^4*x^2+120*B*d* e^3*x^2+152*C*d^2*e^2*x^2+180*A*d*e^3*x+225*B*d^2*e^2*x+195*C*d^3*e*x+440* A*d^2*e^2+360*B*d^3*e+304*C*d^4)*(-e^2*x^2+d^2)^(1/2)+1/8*d^3/e^2*(20*A*e^ 2+15*B*d*e+13*C*d^2)/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2) )
Time = 0.29 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.75 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {30 \, {\left (13 \, C d^{5} + 15 \, B d^{4} e + 20 \, A d^{3} e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (24 \, C e^{4} x^{4} + 304 \, C d^{4} + 360 \, B d^{3} e + 440 \, A d^{2} e^{2} + 30 \, {\left (3 \, C d e^{3} + B e^{4}\right )} x^{3} + 8 \, {\left (19 \, C d^{2} e^{2} + 15 \, B d e^{3} + 5 \, A e^{4}\right )} x^{2} + 15 \, {\left (13 \, C d^{3} e + 15 \, B d^{2} e^{2} + 12 \, A d e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \]
-1/120*(30*(13*C*d^5 + 15*B*d^4*e + 20*A*d^3*e^2)*arctan(-(d - sqrt(-e^2*x ^2 + d^2))/(e*x)) + (24*C*e^4*x^4 + 304*C*d^4 + 360*B*d^3*e + 440*A*d^2*e^ 2 + 30*(3*C*d*e^3 + B*e^4)*x^3 + 8*(19*C*d^2*e^2 + 15*B*d*e^3 + 5*A*e^4)*x ^2 + 15*(13*C*d^3*e + 15*B*d^2*e^2 + 12*A*d*e^3)*x)*sqrt(-e^2*x^2 + d^2))/ e^3
Time = 0.58 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.76 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {C e x^{4}}{5} - \frac {x^{3} \left (B e^{3} + 3 C d e^{2}\right )}{4 e^{2}} - \frac {x^{2} \left (A e^{3} + 3 B d e^{2} + \frac {19 C d^{2} e}{5}\right )}{3 e^{2}} - \frac {x \left (3 A d e^{2} + 3 B d^{2} e + C d^{3} + \frac {3 d^{2} \left (B e^{3} + 3 C d e^{2}\right )}{4 e^{2}}\right )}{2 e^{2}} - \frac {3 A d^{2} e + B d^{3} + \frac {2 d^{2} \left (A e^{3} + 3 B d e^{2} + \frac {19 C d^{2} e}{5}\right )}{3 e^{2}}}{e^{2}}\right ) + \left (A d^{3} + \frac {d^{2} \cdot \left (3 A d e^{2} + 3 B d^{2} e + C d^{3} + \frac {3 d^{2} \left (B e^{3} + 3 C d e^{2}\right )}{4 e^{2}}\right )}{2 e^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {A d^{3} x + \frac {C e^{3} x^{6}}{6} + \frac {x^{5} \left (B e^{3} + 3 C d e^{2}\right )}{5} + \frac {x^{4} \left (A e^{3} + 3 B d e^{2} + 3 C d^{2} e\right )}{4} + \frac {x^{3} \cdot \left (3 A d e^{2} + 3 B d^{2} e + C d^{3}\right )}{3} + \frac {x^{2} \cdot \left (3 A d^{2} e + B d^{3}\right )}{2}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(d**2 - e**2*x**2)*(-C*e*x**4/5 - x**3*(B*e**3 + 3*C*d*e**2 )/(4*e**2) - x**2*(A*e**3 + 3*B*d*e**2 + 19*C*d**2*e/5)/(3*e**2) - x*(3*A* d*e**2 + 3*B*d**2*e + C*d**3 + 3*d**2*(B*e**3 + 3*C*d*e**2)/(4*e**2))/(2*e **2) - (3*A*d**2*e + B*d**3 + 2*d**2*(A*e**3 + 3*B*d*e**2 + 19*C*d**2*e/5) /(3*e**2))/e**2) + (A*d**3 + d**2*(3*A*d*e**2 + 3*B*d**2*e + C*d**3 + 3*d* *2*(B*e**3 + 3*C*d*e**2)/(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)/s qrt(-e**2*x**2), True)), Ne(e**2, 0)), ((A*d**3*x + C*e**3*x**6/6 + x**5*( B*e**3 + 3*C*d*e**2)/5 + x**4*(A*e**3 + 3*B*d*e**2 + 3*C*d**2*e)/4 + x**3* (3*A*d*e**2 + 3*B*d**2*e + C*d**3)/3 + x**2*(3*A*d**2*e + B*d**3)/2)/sqrt( d**2), True))
Time = 0.27 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.79 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{5} \, \sqrt {-e^{2} x^{2} + d^{2}} C e x^{4} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2} x^{2}}{15 \, e} + \frac {A d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{4}}{15 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B d^{3}}{e^{2}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} A d^{2}}{e} - \frac {{\left (3 \, C d e^{2} + B e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} x^{3}}{4 \, e^{2}} - \frac {{\left (3 \, C d^{2} e + 3 \, B d e^{2} + A e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e^{2}} + \frac {3 \, {\left (3 \, C d e^{2} + B e^{3}\right )} d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{4}} + \frac {{\left (C d^{3} + 3 \, B d^{2} e + 3 \, A d e^{2}\right )} d^{2} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}} e^{2}} - \frac {3 \, {\left (3 \, C d e^{2} + B e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{4}} - \frac {{\left (C d^{3} + 3 \, B d^{2} e + 3 \, A d e^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e^{2}} - \frac {2 \, {\left (3 \, C d^{2} e + 3 \, B d e^{2} + A e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{4}} \]
-1/5*sqrt(-e^2*x^2 + d^2)*C*e*x^4 - 4/15*sqrt(-e^2*x^2 + d^2)*C*d^2*x^2/e + A*d^3*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) - 8/15*sqrt(-e^2*x^2 + d^2)* C*d^4/e^3 - sqrt(-e^2*x^2 + d^2)*B*d^3/e^2 - 3*sqrt(-e^2*x^2 + d^2)*A*d^2/ e - 1/4*(3*C*d*e^2 + B*e^3)*sqrt(-e^2*x^2 + d^2)*x^3/e^2 - 1/3*(3*C*d^2*e + 3*B*d*e^2 + A*e^3)*sqrt(-e^2*x^2 + d^2)*x^2/e^2 + 3/8*(3*C*d*e^2 + B*e^3 )*d^4*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^4) + 1/2*(C*d^3 + 3*B*d^2*e + 3*A*d*e^2)*d^2*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) - 3/8*(3*C*d *e^2 + B*e^3)*sqrt(-e^2*x^2 + d^2)*d^2*x/e^4 - 1/2*(C*d^3 + 3*B*d^2*e + 3* A*d*e^2)*sqrt(-e^2*x^2 + d^2)*x/e^2 - 2/3*(3*C*d^2*e + 3*B*d*e^2 + A*e^3)* sqrt(-e^2*x^2 + d^2)*d^2/e^4
Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.79 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{120} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, C e x + \frac {5 \, {\left (3 \, C d e^{6} + B e^{7}\right )}}{e^{6}}\right )} x + \frac {4 \, {\left (19 \, C d^{2} e^{5} + 15 \, B d e^{6} + 5 \, A e^{7}\right )}}{e^{6}}\right )} x + \frac {15 \, {\left (13 \, C d^{3} e^{4} + 15 \, B d^{2} e^{5} + 12 \, A d e^{6}\right )}}{e^{6}}\right )} x + \frac {8 \, {\left (38 \, C d^{4} e^{3} + 45 \, B d^{3} e^{4} + 55 \, A d^{2} e^{5}\right )}}{e^{6}}\right )} + \frac {{\left (13 \, C d^{5} + 15 \, B d^{4} e + 20 \, A d^{3} e^{2}\right )} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} \]
-1/120*sqrt(-e^2*x^2 + d^2)*((2*(3*(4*C*e*x + 5*(3*C*d*e^6 + B*e^7)/e^6)*x + 4*(19*C*d^2*e^5 + 15*B*d*e^6 + 5*A*e^7)/e^6)*x + 15*(13*C*d^3*e^4 + 15* B*d^2*e^5 + 12*A*d*e^6)/e^6)*x + 8*(38*C*d^4*e^3 + 45*B*d^3*e^4 + 55*A*d^2 *e^5)/e^6) + 1/8*(13*C*d^5 + 15*B*d^4*e + 20*A*d^3*e^2)*arcsin(e*x/d)*sgn( d)*sgn(e)/(e^2*abs(e))
Timed out. \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^3\,\left (C\,x^2+B\,x+A\right )}{\sqrt {d^2-e^2\,x^2}} \,d x \]