Integrand size = 27, antiderivative size = 125 \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {1}{8} \left (4 A+\frac {C d^2}{e^2}\right ) x \sqrt {d^2-e^2 x^2}-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {d^2 \left (C d^2+4 A e^2\right ) \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]
-1/3*B*(-e^2*x^2+d^2)^(3/2)/e^2-1/4*C*x*(-e^2*x^2+d^2)^(3/2)/e^2+1/8*d^2*( 4*A*e^2+C*d^2)*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+1/8*(4*A+C*d^2/e^2)*x* (-e^2*x^2+d^2)^(1/2)
Time = 0.40 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.91 \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {e \sqrt {d^2-e^2 x^2} \left (-8 B d^2-3 C d^2 x+12 A e^2 x+8 B e^2 x^2+6 C e^2 x^3\right )-6 d^2 \left (C d^2+4 A e^2\right ) \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e^3} \]
(e*Sqrt[d^2 - e^2*x^2]*(-8*B*d^2 - 3*C*d^2*x + 12*A*e^2*x + 8*B*e^2*x^2 + 6*C*e^2*x^3) - 6*d^2*(C*d^2 + 4*A*e^2)*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(24*e^3)
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2346, 25, 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.
\(\displaystyle \int \sqrt {d^2-e^2 x^2} \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle -\frac {\int -\left (\left (C d^2+4 A e^2+4 B e^2 x\right ) \sqrt {d^2-e^2 x^2}\right )dx}{4 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \left (C d^2+4 A e^2+4 B e^2 x\right ) \sqrt {d^2-e^2 x^2}dx}{4 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\left (4 A e^2+C d^2\right ) \int \sqrt {d^2-e^2 x^2}dx-\frac {4}{3} B \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\left (4 A e^2+C d^2\right ) \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {4}{3} B \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\left (4 A e^2+C d^2\right ) \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {4}{3} B \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\left (4 A e^2+C d^2\right ) \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {4}{3} B \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}\) |
-1/4*(C*x*(d^2 - e^2*x^2)^(3/2))/e^2 + ((-4*B*(d^2 - e^2*x^2)^(3/2))/3 + ( C*d^2 + 4*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*e^2)
3.1.3.3.1 Defintions of rubi rules used
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 = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\left (6 C \,e^{2} x^{3}+8 B \,e^{2} x^{2}+12 A \,e^{2} x -3 C \,d^{2} x -8 B \,d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e^{2}}+\frac {d^{2} \left (4 A \,e^{2}+C \,d^{2}\right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}\) | \(107\) |
default | \(A \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )+C \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )-\frac {B \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{2}}\) | \(155\) |
1/24*(6*C*e^2*x^3+8*B*e^2*x^2+12*A*e^2*x-3*C*d^2*x-8*B*d^2)/e^2*(-e^2*x^2+ d^2)^(1/2)+1/8*d^2*(4*A*e^2+C*d^2)/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(- e^2*x^2+d^2)^(1/2))
Time = 0.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=-\frac {6 \, {\left (C d^{4} + 4 \, A d^{2} e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, C e^{3} x^{3} + 8 \, B e^{3} x^{2} - 8 \, B d^{2} e - 3 \, {\left (C d^{2} e - 4 \, A e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e^{3}} \]
-1/24*(6*(C*d^4 + 4*A*d^2*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*C*e^3*x^3 + 8*B*e^3*x^2 - 8*B*d^2*e - 3*(C*d^2*e - 4*A*e^3)*x)*sqrt(-e ^2*x^2 + d^2))/e^3
Time = 0.35 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.32 \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {B d^{2}}{3 e^{2}} + \frac {B x^{2}}{3} + \frac {C x^{3}}{4} - \frac {x \left (- A e^{2} + \frac {C d^{2}}{4}\right )}{2 e^{2}}\right ) + \left (A d^{2} + \frac {d^{2} \left (- A e^{2} + \frac {C d^{2}}{4}\right )}{2 e^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: e^{2} \neq 0 \\\left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3}\right ) \sqrt {d^{2}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(d**2 - e**2*x**2)*(-B*d**2/(3*e**2) + B*x**2/3 + C*x**3/4 - x*(-A*e**2 + C*d**2/4)/(2*e**2)) + (A*d**2 + d**2*(-A*e**2 + C*d**2/4)/( 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*x + B*x**2/2 + C*x**3/3)*sqrt(d**2), True))
Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {A d^{2} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}}} + \frac {C d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{2}} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} A x + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C d^{2} x}{8 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C x}{4 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} B}{3 \, e^{2}} \]
1/2*A*d^2*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) + 1/8*C*d^4*arcsin(e^2*x/( d*sqrt(e^2)))/(sqrt(e^2)*e^2) + 1/2*sqrt(-e^2*x^2 + d^2)*A*x + 1/8*sqrt(-e ^2*x^2 + d^2)*C*d^2*x/e^2 - 1/4*(-e^2*x^2 + d^2)^(3/2)*C*x/e^2 - 1/3*(-e^2 *x^2 + d^2)^(3/2)*B/e^2
Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, C x + 4 \, B\right )} x - \frac {3 \, {\left (C d^{2} e^{2} - 4 \, A e^{4}\right )}}{e^{4}}\right )} x - \frac {8 \, B d^{2}}{e^{2}}\right )} + \frac {{\left (C d^{4} + 4 \, A d^{2} e^{2}\right )} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} \]
1/24*sqrt(-e^2*x^2 + d^2)*((2*(3*C*x + 4*B)*x - 3*(C*d^2*e^2 - 4*A*e^4)/e^ 4)*x - 8*B*d^2/e^2) + 1/8*(C*d^4 + 4*A*d^2*e^2)*arcsin(e*x/d)*sgn(d)*sgn(e )/(e^2*abs(e))
Timed out. \[ \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\int \sqrt {d^2-e^2\,x^2}\,\left (C\,x^2+B\,x+A\right ) \,d x \]