Integrand size = 25, antiderivative size = 103 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]
1/3*(-e^2*x^2+d^2)^(3/2)/e^3+1/2*d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3- d^2*(-e^2*x^2+d^2)^(1/2)/e^3-1/2*d*x*(-e^2*x^2+d^2)^(1/2)/e^2
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {\left (-4 d^2-3 d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{6 e^3}-\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]
((-4*d^2 - 3*d*e*x - 2*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(6*e^3) - (d^3*ArcTan [(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^3
Time = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {533, 27, 533, 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.
\(\displaystyle \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\int \frac {d e x (2 d+3 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int \frac {x (2 d+3 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {d \left (\frac {\int \frac {d e (3 d+4 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {3 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{3 e}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \left (\frac {d \int \frac {3 d+4 e x}{\sqrt {d^2-e^2 x^2}}dx}{2 e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{3 e}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {d \left (\frac {d \left (3 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {4 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{3 e}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {d \left (\frac {d \left (3 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {4 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{3 e}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {d \left (\frac {d \left (\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {4 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{3 e}-\frac {x^2 \sqrt {d^2-e^2 x^2}}{3 e}\) |
-1/3*(x^2*Sqrt[d^2 - e^2*x^2])/e + (d*((-3*x*Sqrt[d^2 - e^2*x^2])/(2*e) + (d*((-4*Sqrt[d^2 - e^2*x^2])/e + (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e ))/(2*e)))/(3*e)
3.1.16.3.1 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[(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 = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {\left (2 e^{2} x^{2}+3 d e x +4 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{3}}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\) | \(75\) |
default | \(e \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+d \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )\) | \(107\) |
-1/6*(2*e^2*x^2+3*d*e*x+4*d^2)/e^3*(-e^2*x^2+d^2)^(1/2)+1/2*d^3/e^2/(e^2)^ (1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{2} x^{2} + 3 \, d e x + 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]
-1/6*(6*d^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^2*x^2 + 3*d*e *x + 4*d^2)*sqrt(-e^2*x^2 + d^2))/e^3
Time = 0.43 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.25 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {d^{3} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{2}}{3 e^{3}} - \frac {d x}{2 e^{2}} - \frac {x^{2}}{3 e}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d x^{3}}{3} + \frac {e x^{4}}{4}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]
Piecewise((d**3*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))/(2*e* *2) + sqrt(d**2 - e**2*x**2)*(-2*d**2/(3*e**3) - d*x/(2*e**2) - x**2/(3*e) ), Ne(e**2, 0)), ((d*x**3/3 + e*x**4/4)/sqrt(d**2), True))
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e} + \frac {d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}} e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{2}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{3}} \]
-1/3*sqrt(-e^2*x^2 + d^2)*x^2/e + 1/2*d^3*arcsin(e^2*x/(d*sqrt(e^2)))/(sqr t(e^2)*e^2) - 1/2*sqrt(-e^2*x^2 + d^2)*d*x/e^2 - 2/3*sqrt(-e^2*x^2 + d^2)* d^2/e^3
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{2} {\left | e \right |}} - \frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x {\left (\frac {2 \, x}{e} + \frac {3 \, d}{e^{2}}\right )} + \frac {4 \, d^{2}}{e^{3}}\right )} \]
1/2*d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) - 1/6*sqrt(-e^2*x^2 + d^2 )*(x*(2*x/e + 3*d/e^2) + 4*d^2/e^3)
Time = 11.93 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\left \{\begin {array}{cl} \frac {d\,x^3}{3\,\sqrt {d^2}} & \text {\ if\ \ }e=0\\ -\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+e^2\,x^2\right )}{3\,e^3}-\frac {d^3\,\ln \left (2\,x\,\sqrt {-e^2}+2\,\sqrt {d^2-e^2\,x^2}\right )}{2\,{\left (-e^2\right )}^{3/2}}-\frac {d\,x\,\sqrt {d^2-e^2\,x^2}}{2\,e^2} & \text {\ if\ \ }e\neq 0 \end {array}\right . \]