Integrand size = 25, antiderivative size = 148 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {x^2 (d+e x)}{9 d e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d-3 e x)}{63 d e^3 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {2 x}{105 d^3 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 x}{315 d^5 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {16 x}{315 d^7 e^2 \sqrt {d^2-e^2 x^2}} \]
1/9*x^2*(e*x+d)/d/e/(-e^2*x^2+d^2)^(9/2)-2/63*(-3*e*x+d)/d/e^3/(-e^2*x^2+d ^2)^(7/2)-2/105*x/d^3/e^2/(-e^2*x^2+d^2)^(5/2)-8/315*x/d^5/e^2/(-e^2*x^2+d ^2)^(3/2)-16/315*x/d^7/e^2/(-e^2*x^2+d^2)^(1/2)
Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-10 d^8+10 d^7 e x+35 d^6 e^2 x^2+70 d^5 e^3 x^3-70 d^4 e^4 x^4-56 d^3 e^5 x^5+56 d^2 e^6 x^6+16 d e^7 x^7-16 e^8 x^8\right )}{315 d^7 e^3 (d-e x)^5 (d+e x)^4} \]
(Sqrt[d^2 - e^2*x^2]*(-10*d^8 + 10*d^7*e*x + 35*d^6*e^2*x^2 + 70*d^5*e^3*x ^3 - 70*d^4*e^4*x^4 - 56*d^3*e^5*x^5 + 56*d^2*e^6*x^6 + 16*d*e^7*x^7 - 16* e^8*x^8))/(315*d^7*e^3*(d - e*x)^5*(d + e*x)^4)
Time = 0.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {529, 27, 454, 209, 209, 208}
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.
\(\displaystyle \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {d (d+e x)}{9 e^3 \left (d^2-e^2 x^2\right )^{9/2}}-\frac {\int \frac {d (d+9 e x)}{e^2 \left (d^2-e^2 x^2\right )^{9/2}}dx}{9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d (d+e x)}{9 e^3 \left (d^2-e^2 x^2\right )^{9/2}}-\frac {\int \frac {d+9 e x}{\left (d^2-e^2 x^2\right )^{9/2}}dx}{9 e^2}\) |
\(\Big \downarrow \) 454 |
\(\displaystyle \frac {d (d+e x)}{9 e^3 \left (d^2-e^2 x^2\right )^{9/2}}-\frac {\frac {6 \int \frac {1}{\left (d^2-e^2 x^2\right )^{7/2}}dx}{7 d}+\frac {9 d+e x}{7 d e \left (d^2-e^2 x^2\right )^{7/2}}}{9 e^2}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {d (d+e x)}{9 e^3 \left (d^2-e^2 x^2\right )^{9/2}}-\frac {\frac {6 \left (\frac {4 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}\right )}{7 d}+\frac {9 d+e x}{7 d e \left (d^2-e^2 x^2\right )^{7/2}}}{9 e^2}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {d (d+e x)}{9 e^3 \left (d^2-e^2 x^2\right )^{9/2}}-\frac {\frac {6 \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}+\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{5 d^2}+\frac {x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}\right )}{7 d}+\frac {9 d+e x}{7 d e \left (d^2-e^2 x^2\right )^{7/2}}}{9 e^2}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {d (d+e x)}{9 e^3 \left (d^2-e^2 x^2\right )^{9/2}}-\frac {\frac {9 d+e x}{7 d e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {6 \left (\frac {x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 \left (\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{3 d^4 \sqrt {d^2-e^2 x^2}}\right )}{5 d^2}\right )}{7 d}}{9 e^2}\) |
(d*(d + e*x))/(9*e^3*(d^2 - e^2*x^2)^(9/2)) - ((9*d + e*x)/(7*d*e*(d^2 - e ^2*x^2)^(7/2)) + (6*(x/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (4*(x/(3*d^2*(d^2 - e^2*x^2)^(3/2)) + (2*x)/(3*d^4*Sqrt[d^2 - e^2*x^2])))/(5*d^2)))/(7*d))/(9 *e^2)
3.1.31.3.1 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)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a *(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L tQ[p, -1] && NeQ[p, -3/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82
method | result | size |
gosper | \(-\frac {\left (-e x +d \right ) \left (e x +d \right )^{2} \left (16 e^{8} x^{8}-16 d \,e^{7} x^{7}-56 d^{2} e^{6} x^{6}+56 d^{3} e^{5} x^{5}+70 d^{4} x^{4} e^{4}-70 d^{5} e^{3} x^{3}-35 d^{6} e^{2} x^{2}-10 d^{7} e x +10 d^{8}\right )}{315 d^{7} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {11}{2}}}\) | \(121\) |
trager | \(-\frac {\left (16 e^{8} x^{8}-16 d \,e^{7} x^{7}-56 d^{2} e^{6} x^{6}+56 d^{3} e^{5} x^{5}+70 d^{4} x^{4} e^{4}-70 d^{5} e^{3} x^{3}-35 d^{6} e^{2} x^{2}-10 d^{7} e x +10 d^{8}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{315 d^{7} \left (-e x +d \right )^{5} \left (e x +d \right )^{4} e^{3}}\) | \(123\) |
default | \(e \left (\frac {x^{2}}{7 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}-\frac {2 d^{2}}{63 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}\right )+d \left (\frac {x}{8 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}-\frac {d^{2} \left (\frac {x}{9 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}+\frac {\frac {8 x}{63 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}+\frac {8 \left (\frac {6 x}{35 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{7 d^{2}}\right )}{9 d^{2}}}{d^{2}}\right )}{8 e^{2}}\right )\) | \(199\) |
-1/315*(-e*x+d)*(e*x+d)^2*(16*e^8*x^8-16*d*e^7*x^7-56*d^2*e^6*x^6+56*d^3*e ^5*x^5+70*d^4*e^4*x^4-70*d^5*e^3*x^3-35*d^6*e^2*x^2-10*d^7*e*x+10*d^8)/d^7 /e^3/(-e^2*x^2+d^2)^(11/2)
Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (130) = 260\).
Time = 0.50 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.06 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=-\frac {10 \, e^{9} x^{9} - 10 \, d e^{8} x^{8} - 40 \, d^{2} e^{7} x^{7} + 40 \, d^{3} e^{6} x^{6} + 60 \, d^{4} e^{5} x^{5} - 60 \, d^{5} e^{4} x^{4} - 40 \, d^{6} e^{3} x^{3} + 40 \, d^{7} e^{2} x^{2} + 10 \, d^{8} e x - 10 \, d^{9} - {\left (16 \, e^{8} x^{8} - 16 \, d e^{7} x^{7} - 56 \, d^{2} e^{6} x^{6} + 56 \, d^{3} e^{5} x^{5} + 70 \, d^{4} e^{4} x^{4} - 70 \, d^{5} e^{3} x^{3} - 35 \, d^{6} e^{2} x^{2} - 10 \, d^{7} e x + 10 \, d^{8}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{315 \, {\left (d^{7} e^{12} x^{9} - d^{8} e^{11} x^{8} - 4 \, d^{9} e^{10} x^{7} + 4 \, d^{10} e^{9} x^{6} + 6 \, d^{11} e^{8} x^{5} - 6 \, d^{12} e^{7} x^{4} - 4 \, d^{13} e^{6} x^{3} + 4 \, d^{14} e^{5} x^{2} + d^{15} e^{4} x - d^{16} e^{3}\right )}} \]
-1/315*(10*e^9*x^9 - 10*d*e^8*x^8 - 40*d^2*e^7*x^7 + 40*d^3*e^6*x^6 + 60*d ^4*e^5*x^5 - 60*d^5*e^4*x^4 - 40*d^6*e^3*x^3 + 40*d^7*e^2*x^2 + 10*d^8*e*x - 10*d^9 - (16*e^8*x^8 - 16*d*e^7*x^7 - 56*d^2*e^6*x^6 + 56*d^3*e^5*x^5 + 70*d^4*e^4*x^4 - 70*d^5*e^3*x^3 - 35*d^6*e^2*x^2 - 10*d^7*e*x + 10*d^8)*s qrt(-e^2*x^2 + d^2))/(d^7*e^12*x^9 - d^8*e^11*x^8 - 4*d^9*e^10*x^7 + 4*d^1 0*e^9*x^6 + 6*d^11*e^8*x^5 - 6*d^12*e^7*x^4 - 4*d^13*e^6*x^3 + 4*d^14*e^5* x^2 + d^15*e^4*x - d^16*e^3)
Result contains complex when optimal does not.
Time = 16.45 (sec) , antiderivative size = 1401, normalized size of antiderivative = 9.47 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\text {Too large to display} \]
d*Piecewise((-105*I*d**6*x**3/(315*d**17*sqrt(-1 + e**2*x**2/d**2) - 1260* d**15*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 1890*d**13*e**4*x**4*sqrt(-1 + e**2*x**2/d**2) - 1260*d**11*e**6*x**6*sqrt(-1 + e**2*x**2/d**2) + 315*d* *9*e**8*x**8*sqrt(-1 + e**2*x**2/d**2)) + 126*I*d**4*e**2*x**5/(315*d**17* sqrt(-1 + e**2*x**2/d**2) - 1260*d**15*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 1890*d**13*e**4*x**4*sqrt(-1 + e**2*x**2/d**2) - 1260*d**11*e**6*x**6*s qrt(-1 + e**2*x**2/d**2) + 315*d**9*e**8*x**8*sqrt(-1 + e**2*x**2/d**2)) - 72*I*d**2*e**4*x**7/(315*d**17*sqrt(-1 + e**2*x**2/d**2) - 1260*d**15*e** 2*x**2*sqrt(-1 + e**2*x**2/d**2) + 1890*d**13*e**4*x**4*sqrt(-1 + e**2*x** 2/d**2) - 1260*d**11*e**6*x**6*sqrt(-1 + e**2*x**2/d**2) + 315*d**9*e**8*x **8*sqrt(-1 + e**2*x**2/d**2)) + 16*I*e**6*x**9/(315*d**17*sqrt(-1 + e**2* x**2/d**2) - 1260*d**15*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 1890*d**13*e **4*x**4*sqrt(-1 + e**2*x**2/d**2) - 1260*d**11*e**6*x**6*sqrt(-1 + e**2*x **2/d**2) + 315*d**9*e**8*x**8*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d **2) > 1), (105*d**6*x**3/(315*d**17*sqrt(1 - e**2*x**2/d**2) - 1260*d**15 *e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 1890*d**13*e**4*x**4*sqrt(1 - e**2*x **2/d**2) - 1260*d**11*e**6*x**6*sqrt(1 - e**2*x**2/d**2) + 315*d**9*e**8* x**8*sqrt(1 - e**2*x**2/d**2)) - 126*d**4*e**2*x**5/(315*d**17*sqrt(1 - e* *2*x**2/d**2) - 1260*d**15*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 1890*d**13 *e**4*x**4*sqrt(1 - e**2*x**2/d**2) - 1260*d**11*e**6*x**6*sqrt(1 - e**...
Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {x^{2}}{7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}} e} + \frac {d x}{9 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}} e^{2}} - \frac {2 \, d^{2}}{63 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}} e^{3}} - \frac {x}{63 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d e^{2}} - \frac {2 \, x}{105 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3} e^{2}} - \frac {8 \, x}{315 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5} e^{2}} - \frac {16 \, x}{315 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7} e^{2}} \]
1/7*x^2/((-e^2*x^2 + d^2)^(9/2)*e) + 1/9*d*x/((-e^2*x^2 + d^2)^(9/2)*e^2) - 2/63*d^2/((-e^2*x^2 + d^2)^(9/2)*e^3) - 1/63*x/((-e^2*x^2 + d^2)^(7/2)*d *e^2) - 2/105*x/((-e^2*x^2 + d^2)^(5/2)*d^3*e^2) - 8/315*x/((-e^2*x^2 + d^ 2)^(3/2)*d^5*e^2) - 16/315*x/(sqrt(-e^2*x^2 + d^2)*d^7*e^2)
\[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\int { \frac {{\left (e x + d\right )} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {11}{2}}} \,d x } \]
Time = 11.62 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.36 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}}{144\,d^3\,e^3\,{\left (d-e\,x\right )}^5}-\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {1}{252\,e^3}-\frac {17\,x}{252\,d\,e^2}\right )}{{\left (d+e\,x\right )}^4\,{\left (d-e\,x\right )}^4}-\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {5}{144\,d^2\,e^3}+\frac {131\,x}{5040\,d^3\,e^2}\right )}{{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^3}-\frac {8\,x\,\sqrt {d^2-e^2\,x^2}}{315\,d^5\,e^2\,{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}-\frac {16\,x\,\sqrt {d^2-e^2\,x^2}}{315\,d^7\,e^2\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \]
(d^2 - e^2*x^2)^(1/2)/(144*d^3*e^3*(d - e*x)^5) - ((d^2 - e^2*x^2)^(1/2)*( 1/(252*e^3) - (17*x)/(252*d*e^2)))/((d + e*x)^4*(d - e*x)^4) - ((d^2 - e^2 *x^2)^(1/2)*(5/(144*d^2*e^3) + (131*x)/(5040*d^3*e^2)))/((d + e*x)^3*(d - e*x)^3) - (8*x*(d^2 - e^2*x^2)^(1/2))/(315*d^5*e^2*(d + e*x)^2*(d - e*x)^2 ) - (16*x*(d^2 - e^2*x^2)^(1/2))/(315*d^7*e^2*(d + e*x)*(d - e*x))