Integrand size = 27, antiderivative size = 144 \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {3 d^3 (8 d+5 e x) \sqrt {d^2-e^2 x^2}}{20 e^4}+\frac {3 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^4} \]
3/4*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4-3/5*d^2*x^2*(-e^2*x^2+d^2)^(1 /2)/e^2-1/2*d*x^3*(-e^2*x^2+d^2)^(1/2)/e-1/5*x^4*(-e^2*x^2+d^2)^(1/2)-3/20 *d^3*(5*e*x+8*d)*(-e^2*x^2+d^2)^(1/2)/e^4
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (24 d^4+15 d^3 e x+12 d^2 e^2 x^2+10 d e^3 x^3+4 e^4 x^4\right )+30 d^5 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{20 e^4} \]
-1/20*(Sqrt[d^2 - e^2*x^2]*(24*d^4 + 15*d^3*e*x + 12*d^2*e^2*x^2 + 10*d*e^ 3*x^3 + 4*e^4*x^4) + 30*d^5*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]) ])/e^4
Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {541, 25, 27, 533, 27, 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.
| \(\displaystyle \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 541 |
| \(\displaystyle -\frac {\int -\frac {d e^2 x^3 (9 d+10 e x)}{\sqrt {d^2-e^2 x^2}}dx}{5 e^2}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {d e^2 x^3 (9 d+10 e x)}{\sqrt {d^2-e^2 x^2}}dx}{5 e^2}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} d \int \frac {x^3 (9 d+10 e x)}{\sqrt {d^2-e^2 x^2}}dx-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{5} d \left (\frac {\int \frac {6 d e x^2 (5 d+6 e x)}{\sqrt {d^2-e^2 x^2}}dx}{4 e^2}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \int \frac {x^2 (5 d+6 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {\int \frac {3 d e x (4 d+5 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {d \int \frac {x (4 d+5 e x)}{\sqrt {d^2-e^2 x^2}}dx}{e}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {d \left (\frac {\int \frac {d e (5 d+8 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {5 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{e}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {d \left (\frac {d \int \frac {5 d+8 e x}{\sqrt {d^2-e^2 x^2}}dx}{2 e}-\frac {5 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{e}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {d \left (\frac {d \left (5 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{e}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {d \left (\frac {d \left (5 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{e}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{5} d \left (\frac {3 d \left (\frac {d \left (\frac {d \left (\frac {5 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{e}-\frac {2 x^2 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}-\frac {5 x^3 \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}\) |
-1/5*(x^4*Sqrt[d^2 - e^2*x^2]) + (d*((-5*x^3*Sqrt[d^2 - e^2*x^2])/(2*e) + (3*d*((-2*x^2*Sqrt[d^2 - e^2*x^2])/e + (d*((-5*x*Sqrt[d^2 - e^2*x^2])/(2*e ) + (d*((-8*Sqrt[d^2 - e^2*x^2])/e + (5*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2] ])/e))/(2*e)))/e))/(2*e)))/5
3.1.34.3.1 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 = 0.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.67
| method | result | size |
| risch | \(-\frac {\left (4 e^{4} x^{4}+10 d \,e^{3} x^{3}+12 d^{2} e^{2} x^{2}+15 d^{3} e x +24 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{20 e^{4}}+\frac {3 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{4 e^{3} \sqrt {e^{2}}}\) | \(97\) |
| default | \(e^{2} \left (-\frac {x^{4} \sqrt {-e^{2} x^{2}+d^{2}}}{5 e^{2}}+\frac {4 d^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )}{5 e^{2}}\right )+d^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+2 d e \left (-\frac {x^{3} \sqrt {-e^{2} x^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )\) | \(222\) |
-1/20*(4*e^4*x^4+10*d*e^3*x^3+12*d^2*e^2*x^2+15*d^3*e*x+24*d^4)/e^4*(-e^2* x^2+d^2)^(1/2)+3/4*d^5/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2) ^(1/2))
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.65 \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (4 \, e^{4} x^{4} + 10 \, d e^{3} x^{3} + 12 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x + 24 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{20 \, e^{4}} \]
-1/20*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (4*e^4*x^4 + 10* d*e^3*x^3 + 12*d^2*e^2*x^2 + 15*d^3*e*x + 24*d^4)*sqrt(-e^2*x^2 + d^2))/e^ 4
Time = 0.46 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.17 \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {3 d^{5} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{4 e^{3}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {6 d^{4}}{5 e^{4}} - \frac {3 d^{3} x}{4 e^{3}} - \frac {3 d^{2} x^{2}}{5 e^{2}} - \frac {d x^{3}}{2 e} - \frac {x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d^{2} x^{4}}{4} + \frac {2 d e x^{5}}{5} + \frac {e^{2} x^{6}}{6}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]
Piecewise((3*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))/(4* e**3) + sqrt(d**2 - e**2*x**2)*(-6*d**4/(5*e**4) - 3*d**3*x/(4*e**3) - 3*d **2*x**2/(5*e**2) - d*x**3/(2*e) - x**4/5), Ne(e**2, 0)), ((d**2*x**4/4 + 2*d*e*x**5/5 + e**2*x**6/6)/sqrt(d**2), True))
Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{5} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{4} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x^{3}}{2 \, e} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x^{2}}{5 \, e^{2}} + \frac {3 \, d^{5} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{4 \, \sqrt {e^{2}} e^{3}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{4 \, e^{3}} - \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{5 \, e^{4}} \]
-1/5*sqrt(-e^2*x^2 + d^2)*x^4 - 1/2*sqrt(-e^2*x^2 + d^2)*d*x^3/e - 3/5*sqr t(-e^2*x^2 + d^2)*d^2*x^2/e^2 + 3/4*d^5*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt( e^2)*e^3) - 3/4*sqrt(-e^2*x^2 + d^2)*d^3*x/e^3 - 6/5*sqrt(-e^2*x^2 + d^2)* d^4/e^4
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.58 \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{4 \, e^{3} {\left | e \right |}} - \frac {1}{20} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left ({\left (2 \, x + \frac {5 \, d}{e}\right )} x + \frac {6 \, d^{2}}{e^{2}}\right )} x + \frac {15 \, d^{3}}{e^{3}}\right )} x + \frac {24 \, d^{4}}{e^{4}}\right )} \]
3/4*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) - 1/20*sqrt(-e^2*x^2 + d^ 2)*((2*((2*x + 5*d/e)*x + 6*d^2/e^2)*x + 15*d^3/e^3)*x + 24*d^4/e^4)
Timed out. \[ \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^3\,{\left (d+e\,x\right )}^2}{\sqrt {d^2-e^2\,x^2}} \,d x \]