Integrand size = 32, antiderivative size = 186 \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {d \left (C d^2+e (B d+4 A e)\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {\left (2 C d^2+5 e (B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac {(C d+B e) x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac {d^3 \left (C d^2+e (B d+4 A e)\right ) \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]
-1/15*(2*C*d^2+5*e*(A*e+B*d))*(-e^2*x^2+d^2)^(3/2)/e^3-1/4*(B*e+C*d)*x*(-e ^2*x^2+d^2)^(3/2)/e^2-1/5*C*x^2*(-e^2*x^2+d^2)^(3/2)/e+1/8*d^3*(C*d^2+e*(4 *A*e+B*d))*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+1/8*d*(C*d^2+e*(4*A*e+B*d) )*x*(-e^2*x^2+d^2)^(1/2)/e^2
Time = 0.87 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.97 \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (C \left (-16 d^4-15 d^3 e x-8 d^2 e^2 x^2+30 d e^3 x^3+24 e^4 x^4\right )-5 e \left (-4 A e \left (-2 d^2+3 d e x+2 e^2 x^2\right )+B \left (8 d^3+3 d^2 e x-8 d e^2 x^2-6 e^3 x^3\right )\right )\right )-30 d^3 \left (C d^2+e (B d+4 A e)\right ) \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{120 e^3} \]
(Sqrt[d^2 - e^2*x^2]*(C*(-16*d^4 - 15*d^3*e*x - 8*d^2*e^2*x^2 + 30*d*e^3*x ^3 + 24*e^4*x^4) - 5*e*(-4*A*e*(-2*d^2 + 3*d*e*x + 2*e^2*x^2) + B*(8*d^3 + 3*d^2*e*x - 8*d*e^2*x^2 - 6*e^3*x^3))) - 30*d^3*(C*d^2 + e*(B*d + 4*A*e)) *ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(120*e^3)
Time = 0.47 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {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.
\(\displaystyle \int (d+e x) \sqrt {d^2-e^2 x^2} \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle -\frac {\int -\sqrt {d^2-e^2 x^2} \left (5 (C d+B e) x^2 e^2+5 A d e^2+\left (2 C d^2+5 e (B d+A e)\right ) x e\right )dx}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sqrt {d^2-e^2 x^2} \left (5 (C d+B e) x^2 e^2+5 A d e^2+\left (2 C d^2+5 e (B d+A e)\right ) x e\right )dx}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {-\frac {\int -e^2 \left (5 d \left (C d^2+e (B d+4 A e)\right )+4 e \left (2 C d^2+5 e (B d+A e)\right ) x\right ) \sqrt {d^2-e^2 x^2}dx}{4 e^2}-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int e^2 \left (5 d \left (C d^2+e (B d+4 A e)\right )+4 e \left (2 C d^2+5 e (B d+A e)\right ) x\right ) \sqrt {d^2-e^2 x^2}dx}{4 e^2}-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{4} \int \left (5 d \left (C d^2+e (B d+4 A e)\right )+4 e \left (2 C d^2+5 e (B d+A e)\right ) x\right ) \sqrt {d^2-e^2 x^2}dx-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {1}{4} \left (5 d \left (e (4 A e+B d)+C d^2\right ) \int \sqrt {d^2-e^2 x^2}dx-\frac {4 \left (d^2-e^2 x^2\right )^{3/2} \left (5 e (A e+B d)+2 C d^2\right )}{3 e}\right )-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {1}{4} \left (5 d \left (e (4 A e+B d)+C d^2\right ) \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {4 \left (d^2-e^2 x^2\right )^{3/2} \left (5 e (A e+B d)+2 C d^2\right )}{3 e}\right )-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {1}{4} \left (5 d \left (e (4 A e+B d)+C d^2\right ) \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {4 \left (d^2-e^2 x^2\right )^{3/2} \left (5 e (A e+B d)+2 C d^2\right )}{3 e}\right )-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{4} \left (5 d \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right ) \left (e (4 A e+B d)+C d^2\right )-\frac {4 \left (d^2-e^2 x^2\right )^{3/2} \left (5 e (A e+B d)+2 C d^2\right )}{3 e}\right )-\frac {5}{4} x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{5 e^2}-\frac {C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
-1/5*(C*x^2*(d^2 - e^2*x^2)^(3/2))/e + ((-5*(C*d + B*e)*x*(d^2 - e^2*x^2)^ (3/2))/4 + ((-4*(2*C*d^2 + 5*e*(B*d + A*e))*(d^2 - e^2*x^2)^(3/2))/(3*e) + 5*d*(C*d^2 + e*(B*d + 4*A*e))*((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)
3.1.2.3.1 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 = 0.54 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-\frac {\left (-24 e^{4} C \,x^{4}-30 x^{3} B \,e^{4}-30 C d \,e^{3} x^{3}-40 A \,e^{4} x^{2}-40 x^{2} d B \,e^{3}+8 C \,d^{2} e^{2} x^{2}-60 A d \,e^{3} x +15 x B \,d^{2} e^{2}+15 C \,d^{3} x e +40 A \,d^{2} e^{2}+40 B \,d^{3} e +16 C \,d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{120 e^{3}}+\frac {d^{3} \left (4 A \,e^{2}+B d e +C \,d^{2}\right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}\) | \(180\) |
default | \(d A \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )+e C \left (-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{5 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{15 e^{4}}\right )+\left (B e +C d \right ) \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )-\frac {\left (A e +B d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{2}}\) | \(216\) |
-1/120/e^3*(-24*C*e^4*x^4-30*B*e^4*x^3-30*C*d*e^3*x^3-40*A*e^4*x^2-40*B*d* e^3*x^2+8*C*d^2*e^2*x^2-60*A*d*e^3*x+15*B*d^2*e^2*x+15*C*d^3*e*x+40*A*d^2* e^2+40*B*d^3*e+16*C*d^4)*(-e^2*x^2+d^2)^(1/2)+1/8*d^3/e^2*(4*A*e^2+B*d*e+C *d^2)/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
Time = 0.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.93 \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=-\frac {30 \, {\left (C d^{5} + B d^{4} e + 4 \, A d^{3} e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (24 \, C e^{4} x^{4} - 16 \, C d^{4} - 40 \, B d^{3} e - 40 \, A d^{2} e^{2} + 30 \, {\left (C d e^{3} + B e^{4}\right )} x^{3} - 8 \, {\left (C d^{2} e^{2} - 5 \, B d e^{3} - 5 \, A e^{4}\right )} x^{2} - 15 \, {\left (C d^{3} e + B d^{2} e^{2} - 4 \, A d e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \]
-1/120*(30*(C*d^5 + B*d^4*e + 4*A*d^3*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^ 2))/(e*x)) - (24*C*e^4*x^4 - 16*C*d^4 - 40*B*d^3*e - 40*A*d^2*e^2 + 30*(C* d*e^3 + B*e^4)*x^3 - 8*(C*d^2*e^2 - 5*B*d*e^3 - 5*A*e^4)*x^2 - 15*(C*d^3*e + B*d^2*e^2 - 4*A*d*e^3)*x)*sqrt(-e^2*x^2 + d^2))/e^3
Time = 0.58 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.76 \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (\frac {C e x^{4}}{5} - \frac {x^{3} \left (- B e^{3} - C d e^{2}\right )}{4 e^{2}} - \frac {x^{2} \left (- A e^{3} - B d e^{2} + \frac {C d^{2} e}{5}\right )}{3 e^{2}} - \frac {x \left (- A d e^{2} + B d^{2} e + C d^{3} + \frac {3 d^{2} \left (- B e^{3} - C d e^{2}\right )}{4 e^{2}}\right )}{2 e^{2}} - \frac {A d^{2} e + B d^{3} + \frac {2 d^{2} \left (- A e^{3} - B d e^{2} + \frac {C d^{2} e}{5}\right )}{3 e^{2}}}{e^{2}}\right ) + \left (A d^{3} + \frac {d^{2} \left (- A d e^{2} + B d^{2} e + C d^{3} + \frac {3 d^{2} \left (- B e^{3} - C d e^{2}\right )}{4 e^{2}}\right )}{2 e^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: e^{2} \neq 0 \\\left (A d x + \frac {C e x^{4}}{4} + \frac {x^{3} \left (B e + C d\right )}{3} + \frac {x^{2} \left (A e + B d\right )}{2}\right ) \sqrt {d^{2}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(d**2 - e**2*x**2)*(C*e*x**4/5 - x**3*(-B*e**3 - C*d*e**2)/ (4*e**2) - x**2*(-A*e**3 - B*d*e**2 + C*d**2*e/5)/(3*e**2) - x*(-A*d*e**2 + B*d**2*e + C*d**3 + 3*d**2*(-B*e**3 - C*d*e**2)/(4*e**2))/(2*e**2) - (A* d**2*e + B*d**3 + 2*d**2*(-A*e**3 - B*d*e**2 + C*d**2*e/5)/(3*e**2))/e**2) + (A*d**3 + d**2*(-A*d*e**2 + B*d**2*e + C*d**3 + 3*d**2*(-B*e**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)/sqrt(-e**2*x**2), Tru e)), Ne(e**2, 0)), ((A*d*x + C*e*x**4/4 + x**3*(B*e + C*d)/3 + x**2*(A*e + B*d)/2)*sqrt(d**2), True))
Time = 0.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.20 \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {A d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}}} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} A d x - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C x^{2}}{5 \, e} + \frac {{\left (C d + B e\right )} d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} {\left (C d + B e\right )} d^{2} x}{8 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C d^{2}}{15 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} B d}{3 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} A}{3 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (C d + B e\right )} x}{4 \, e^{2}} \]
1/2*A*d^3*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) + 1/2*sqrt(-e^2*x^2 + d^2) *A*d*x - 1/5*(-e^2*x^2 + d^2)^(3/2)*C*x^2/e + 1/8*(C*d + B*e)*d^4*arcsin(e ^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) + 1/8*sqrt(-e^2*x^2 + d^2)*(C*d + B*e) *d^2*x/e^2 - 2/15*(-e^2*x^2 + d^2)^(3/2)*C*d^2/e^3 - 1/3*(-e^2*x^2 + d^2)^ (3/2)*B*d/e^2 - 1/3*(-e^2*x^2 + d^2)^(3/2)*A/e - 1/4*(-e^2*x^2 + d^2)^(3/2 )*(C*d + B*e)*x/e^2
Time = 0.31 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.97 \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\frac {1}{120} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, C e x + \frac {5 \, {\left (C d e^{6} + B e^{7}\right )}}{e^{6}}\right )} x - \frac {4 \, {\left (C d^{2} e^{5} - 5 \, B d e^{6} - 5 \, A e^{7}\right )}}{e^{6}}\right )} x - \frac {15 \, {\left (C d^{3} e^{4} + B d^{2} e^{5} - 4 \, A d e^{6}\right )}}{e^{6}}\right )} x - \frac {8 \, {\left (2 \, C d^{4} e^{3} + 5 \, B d^{3} e^{4} + 5 \, A d^{2} e^{5}\right )}}{e^{6}}\right )} + \frac {{\left (C d^{5} + B d^{4} e + 4 \, A d^{3} e^{2}\right )} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} \]
1/120*sqrt(-e^2*x^2 + d^2)*((2*(3*(4*C*e*x + 5*(C*d*e^6 + B*e^7)/e^6)*x - 4*(C*d^2*e^5 - 5*B*d*e^6 - 5*A*e^7)/e^6)*x - 15*(C*d^3*e^4 + B*d^2*e^5 - 4 *A*d*e^6)/e^6)*x - 8*(2*C*d^4*e^3 + 5*B*d^3*e^4 + 5*A*d^2*e^5)/e^6) + 1/8* (C*d^5 + B*d^4*e + 4*A*d^3*e^2)*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e))
Timed out. \[ \int (d+e x) \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx=\int \sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )\,\left (C\,x^2+B\,x+A\right ) \,d x \]