Integrand size = 25, antiderivative size = 136 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {d^3 x \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^2} \]
-1/6*d*x*(-e^2*x^2+d^2)^(3/2)/e-2/15*(-e^2*x^2+d^2)^(5/2)/e^2-1/3*(-e^2*x^ 2+d^2)^(7/2)/e^2/(e*x+d)^2-1/4*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^2-1/ 4*d^3*x*(-e^2*x^2+d^2)^(1/2)/e
Time = 0.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {e \sqrt {d^2-e^2 x^2} \left (-28 d^4+15 d^3 e x+16 d^2 e^2 x^2-30 d e^3 x^3+12 e^4 x^4\right )-15 d^5 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{60 e^3} \]
(e*Sqrt[d^2 - e^2*x^2]*(-28*d^4 + 15*d^3*e*x + 16*d^2*e^2*x^2 - 30*d*e^3*x ^3 + 12*e^4*x^4) - 15*d^5*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2* x^2]])/(60*e^3)
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {562, 541, 25, 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.
| \(\displaystyle \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 562 |
| \(\displaystyle \int x (d-e x)^2 \sqrt {d^2-e^2 x^2}dx\) |
| \(\Big \downarrow \) 541 |
| \(\displaystyle -\frac {\int -d e^2 x (7 d-10 e x) \sqrt {d^2-e^2 x^2}dx}{5 e^2}-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int d e^2 x (7 d-10 e x) \sqrt {d^2-e^2 x^2}dx}{5 e^2}-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} d \int x (7 d-10 e x) \sqrt {d^2-e^2 x^2}dx-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{5} d \left (\frac {\int -2 d e (5 d-14 e x) \sqrt {d^2-e^2 x^2}dx}{4 e^2}+\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{2 e}\right )-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} d \left (\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{2 e}-\frac {d \int (5 d-14 e x) \sqrt {d^2-e^2 x^2}dx}{2 e}\right )-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {1}{5} d \left (\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{2 e}-\frac {d \left (5 d \int \sqrt {d^2-e^2 x^2}dx+\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{2 e}\right )-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {1}{5} d \left (\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{2 e}-\frac {d \left (5 d \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{2 e}\right )-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{5} d \left (\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{2 e}-\frac {d \left (5 d \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{2 e}\right )-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{5} d \left (\frac {5 x \left (d^2-e^2 x^2\right )^{3/2}}{2 e}-\frac {d \left (5 d \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )}{2 e}\right )-\frac {1}{5} x^2 \left (d^2-e^2 x^2\right )^{3/2}\) |
-1/5*(x^2*(d^2 - e^2*x^2)^(3/2)) + (d*((5*x*(d^2 - e^2*x^2)^(3/2))/(2*e) - (d*((14*(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))))/(2*e)))/5
3.2.61.3.1 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]
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]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[c^(2*n)/a^n Int[x^m*((a + b*x^2)^(n + p)/(c - d*x)^n), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && IGtQ[n + p + 1/2, 0]
Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(-\frac {\left (-12 e^{4} x^{4}+30 d \,e^{3} x^{3}-16 d^{2} e^{2} x^{2}-15 d^{3} e x +28 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{60 e^{2}}-\frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{4 e \sqrt {e^{2}}}\) | \(97\) |
| default | \(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{e^{2}}-\frac {d \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{e^{3}}\) | \(438\) |
-1/60*(-12*e^4*x^4+30*d*e^3*x^3-16*d^2*e^2*x^2-15*d^3*e*x+28*d^4)/e^2*(-e^ 2*x^2+d^2)^(1/2)-1/4*d^5/e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2) ^(1/2))
Time = 0.32 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (12 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} + 16 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x - 28 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{60 \, e^{2}} \]
1/60*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (12*e^4*x^4 - 30* d*e^3*x^3 + 16*d^2*e^2*x^2 + 15*d^3*e*x - 28*d^4)*sqrt(-e^2*x^2 + d^2))/e^ 2
Time = 1.83 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.59 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=d^{2} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2}}{3 e^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{2} \sqrt {d^{2}}}{2} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \frac {d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2} x}{8 e^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{3} \sqrt {d^{2}}}{3} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{4}} - \frac {d^{2} x^{2}}{15 e^{2}} + \frac {x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]
d**2*Piecewise((sqrt(d**2 - e**2*x**2)*(-d**2/(3*e**2) + x**2/3), Ne(e**2, 0)), (x**2*sqrt(d**2)/2, True)) - 2*d*e*Piecewise((d**4*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)*(-d* *2*x/(8*e**2) + x**3/4), Ne(e**2, 0)), (x**3*sqrt(d**2)/3, True)) + e**2*P iecewise((sqrt(d**2 - e**2*x**2)*(-2*d**4/(15*e**4) - d**2*x**2/(15*e**2) + x**4/5), Ne(e**2, 0)), (x**4*sqrt(d**2)/4, True))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.23 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{4 \, e^{2}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{4 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{4 \, {\left (e^{3} x + d e^{2}\right )}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{2 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{12 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e^{2}} \]
1/4*I*d^5*arcsin(e*x/d + 2)/e^2 - 1/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^3* x/e - 1/4*(-e^2*x^2 + d^2)^(5/2)*d/(e^3*x + d*e^2) - 1/2*sqrt(e^2*x^2 + 4* d*e*x + 3*d^2)*d^4/e^2 + 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e - 5/12*(-e^2*x^2 + d^2)^(3/2)*d^2/e^2 + 1/5*(-e^2*x^2 + d^2)^(5/2)/e^2
Time = 0.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.59 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {{\left (480 \, d^{6} e^{6} \arctan \left (\sqrt {\frac {2 \, d}{e x + d} - 1}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + \frac {{\left (15 \, d^{6} e^{6} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 250 \, d^{6} e^{6} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 128 \, d^{6} e^{6} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 70 \, d^{6} e^{6} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 15 \, d^{6} e^{6} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} {\left (e x + d\right )}^{5}}{d^{5}}\right )} {\left | e \right |}}{960 \, d e^{9}} \]
1/960*(480*d^6*e^6*arctan(sqrt(2*d/(e*x + d) - 1))*sgn(1/(e*x + d))*sgn(e) + (15*d^6*e^6*(2*d/(e*x + d) - 1)^(9/2)*sgn(1/(e*x + d))*sgn(e) - 250*d^6 *e^6*(2*d/(e*x + d) - 1)^(7/2)*sgn(1/(e*x + d))*sgn(e) - 128*d^6*e^6*(2*d/ (e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) - 70*d^6*e^6*(2*d/(e*x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) - 15*d^6*e^6*sqrt(2*d/(e*x + d) - 1)*sgn (1/(e*x + d))*sgn(e))*(e*x + d)^5/d^5)*abs(e)/(d*e^9)
Timed out. \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\int \frac {x\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^2} \,d x \]