Integrand size = 25, antiderivative size = 83 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]
d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^2-1/3*x^2*(-e^2*x^2+d^2)^(1/2)-1/3* d*(3*e*x+5*d)*(-e^2*x^2+d^2)^(1/2)/e^2
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (5 d^2+3 d e x+e^2 x^2\right )+6 d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{3 e^2} \]
-1/3*(Sqrt[d^2 - e^2*x^2]*(5*d^2 + 3*d*e*x + e^2*x^2) + 6*d^3*ArcTan[(e*x) /(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^2
Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {541, 25, 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.
\(\displaystyle \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 541 |
\(\displaystyle -\frac {\int -\frac {d e^2 x (5 d+6 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {d e^2 x (5 d+6 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} d \int \frac {x (5 d+6 e x)}{\sqrt {d^2-e^2 x^2}}dx-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {1}{3} d \left (\frac {\int \frac {2 d e (3 d+5 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {3 x \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} d \left (\frac {d \int \frac {3 d+5 e x}{\sqrt {d^2-e^2 x^2}}dx}{e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {1}{3} d \left (\frac {d \left (3 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {5 \sqrt {d^2-e^2 x^2}}{e}\right )}{e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{3} d \left (\frac {d \left (3 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}\right )}{e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} d \left (\frac {d \left (\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}\right )}{e}-\frac {3 x \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}\) |
-1/3*(x^2*Sqrt[d^2 - e^2*x^2]) + (d*((-3*x*Sqrt[d^2 - e^2*x^2])/e + (d*((- 5*Sqrt[d^2 - e^2*x^2])/e + (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/e)) /3
3.1.36.3.1 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]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x ] + Simp[1/(b*(m + n + 2*p + 1)) Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) *x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {\left (e^{2} x^{2}+3 d e x +5 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e \sqrt {e^{2}}}\) | \(73\) |
default | \(e^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )-\frac {d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e^{2}}+2 d e \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )\) | \(133\) |
-1/3*(e^2*x^2+3*d*e*x+5*d^2)/e^2*(-e^2*x^2+d^2)^(1/2)+d^3/e/(e^2)^(1/2)*ar ctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (e^{2} x^{2} + 3 \, d e x + 5 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, e^{2}} \]
-1/3*(6*d^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (e^2*x^2 + 3*d*e*x + 5*d^2)*sqrt(-e^2*x^2 + d^2))/e^2
Time = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.61 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {d^{3} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{e} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {5 d^{2}}{3 e^{2}} - \frac {d x}{e} - \frac {x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d^{2} x^{2}}{2} + \frac {2 d e x^{3}}{3} + \frac {e^{2} x^{4}}{4}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]
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))/e + s qrt(d**2 - e**2*x**2)*(-5*d**2/(3*e**2) - d*x/e - x**2/3), Ne(e**2, 0)), ( (d**2*x**2/2 + 2*d*e*x**3/3 + e**2*x**4/4)/sqrt(d**2), True))
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{2} + \frac {d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{e} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{2}} \]
-1/3*sqrt(-e^2*x^2 + d^2)*x^2 + d^3*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2) *e) - sqrt(-e^2*x^2 + d^2)*d*x/e - 5/3*sqrt(-e^2*x^2 + d^2)*d^2/e^2
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e {\left | e \right |}} - \frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (x + \frac {3 \, d}{e}\right )} x + \frac {5 \, d^{2}}{e^{2}}\right )} \]
d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/(e*abs(e)) - 1/3*sqrt(-e^2*x^2 + d^2)*((x + 3*d/e)*x + 5*d^2/e^2)
Timed out. \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x\,{\left (d+e\,x\right )}^2}{\sqrt {d^2-e^2\,x^2}} \,d x \]