Integrand size = 25, antiderivative size = 132 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]
-1/3*d^2*(-e^2*x^2+d^2)^(3/2)/e^3-1/4*d*x*(-e^2*x^2+d^2)^(3/2)/e^2+1/5*(-e ^2*x^2+d^2)^(5/2)/e^3+1/8*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+1/8*d^3 *x*(-e^2*x^2+d^2)^(1/2)/e^2
Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.78 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-16 d^4-15 d^3 e x-8 d^2 e^2 x^2+30 d e^3 x^3+24 e^4 x^4\right )-30 d^5 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{120 e^3} \]
(Sqrt[d^2 - e^2*x^2]*(-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) - 30*d^5*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(1 20*e^3)
Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {533, 27, 533, 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.
\(\displaystyle \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\int d e x (2 d+5 e x) \sqrt {d^2-e^2 x^2}dx}{5 e^2}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int x (2 d+5 e x) \sqrt {d^2-e^2 x^2}dx}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {d \left (\frac {\int d e (5 d+8 e x) \sqrt {d^2-e^2 x^2}dx}{4 e^2}-\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \left (\frac {d \int (5 d+8 e x) \sqrt {d^2-e^2 x^2}dx}{4 e}-\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {d \left (\frac {d \left (5 d \int \sqrt {d^2-e^2 x^2}dx-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{4 e}-\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {d \left (\frac {d \left (5 d \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{4 e}-\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {d \left (\frac {d \left (5 d \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{4 e}-\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {d \left (\frac {d \left (5 d \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{4 e}-\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )}{5 e}-\frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}\) |
-1/5*(x^2*(d^2 - e^2*x^2)^(3/2))/e + (d*((-5*x*(d^2 - e^2*x^2)^(3/2))/(4*e ) + (d*((-8*(d^2 - e^2*x^2)^(3/2))/(3*e) + 5*d*((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)))/(5*e)
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[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {\left (-24 e^{4} x^{4}-30 d \,e^{3} x^{3}+8 d^{2} e^{2} x^{2}+15 d^{3} e x +16 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{120 e^{3}}+\frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}\) | \(97\) |
default | \(e \left (-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{5 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{15 e^{4}}\right )+d \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )\) | \(130\) |
-1/120*(-24*e^4*x^4-30*d*e^3*x^3+8*d^2*e^2*x^2+15*d^3*e*x+16*d^4)/e^3*(-e^ 2*x^2+d^2)^(1/2)+1/8*d^5/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^ 2)^(1/2))
Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.72 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=-\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (24 \, e^{4} x^{4} + 30 \, d e^{3} x^{3} - 8 \, d^{2} e^{2} x^{2} - 15 \, d^{3} e x - 16 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \]
-1/120*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (24*e^4*x^4 + 3 0*d*e^3*x^3 - 8*d^2*e^2*x^2 - 15*d^3*e*x - 16*d^4)*sqrt(-e^2*x^2 + d^2))/e ^3
Time = 0.46 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\begin {cases} \frac {d^{5} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{3}} - \frac {d^{3} x}{8 e^{2}} - \frac {d^{2} x^{2}}{15 e} + \frac {d x^{3}}{4} + \frac {e x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{3}}{3} + \frac {e x^{4}}{4}\right ) \sqrt {d^{2}} & \text {otherwise} \end {cases} \]
Piecewise((d**5*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))/(8*e* *2) + sqrt(d**2 - e**2*x**2)*(-2*d**4/(15*e**3) - d**3*x/(8*e**2) - d**2*x **2/(15*e) + d*x**3/4 + e*x**4/5), Ne(e**2, 0)), ((d*x**3/3 + e*x**4/4)*sq rt(d**2), True))
Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.88 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^{5} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{15 \, e^{3}} \]
1/8*d^5*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) + 1/8*sqrt(-e^2*x^2 + d^2)*d^3*x/e^2 - 1/5*(-e^2*x^2 + d^2)^(3/2)*x^2/e - 1/4*(-e^2*x^2 + d^2)^( 3/2)*d*x/e^2 - 2/15*(-e^2*x^2 + d^2)^(3/2)*d^2/e^3
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.63 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} + \frac {1}{120} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, e x + 5 \, d\right )} x - \frac {4 \, d^{2}}{e}\right )} x - \frac {15 \, d^{3}}{e^{2}}\right )} x - \frac {16 \, d^{4}}{e^{3}}\right )} \]
1/8*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) + 1/120*sqrt(-e^2*x^2 + d ^2)*((2*(3*(4*e*x + 5*d)*x - 4*d^2/e)*x - 15*d^3/e^2)*x - 16*d^4/e^3)
Timed out. \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\int x^2\,\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right ) \,d x \]