Integrand size = 24, antiderivative size = 148 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {8 x}{63 d^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 x}{63 d^7 \sqrt {d^2-e^2 x^2}} \]
8/63*x/d^5/(-e^2*x^2+d^2)^(3/2)-1/9/d/e/(e*x+d)^3/(-e^2*x^2+d^2)^(3/2)-2/2 1/d^2/e/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2)-2/21/d^3/e/(e*x+d)/(-e^2*x^2+d^2)^( 3/2)+16/63*x/d^7/(-e^2*x^2+d^2)^(1/2)
Time = 0.60 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-19 d^6+6 d^5 e x+66 d^4 e^2 x^2+56 d^3 e^3 x^3-24 d^2 e^4 x^4-48 d e^5 x^5-16 e^6 x^6\right )}{63 d^7 e (d-e x)^2 (d+e x)^5} \]
(Sqrt[d^2 - e^2*x^2]*(-19*d^6 + 6*d^5*e*x + 66*d^4*e^2*x^2 + 56*d^3*e^3*x^ 3 - 24*d^2*e^4*x^4 - 48*d*e^5*x^5 - 16*e^6*x^6))/(63*d^7*e*(d - e*x)^2*(d + e*x)^5)
Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {461, 461, 470, 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.
| \(\displaystyle \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 461 |
| \(\displaystyle \frac {2 \int \frac {1}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{3 d}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 461 |
| \(\displaystyle \frac {2 \left (\frac {5 \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}}dx}{7 d}-\frac {1}{7 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{3 d}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 470 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (\frac {4 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}\right )}{7 d}-\frac {1}{7 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{3 d}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}+\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{5 d}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}\right )}{7 d}-\frac {1}{7 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{3 d}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (\frac {4 \left (\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{3 d^4 \sqrt {d^2-e^2 x^2}}\right )}{5 d}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}\right )}{7 d}-\frac {1}{7 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{3 d}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}\) |
-1/9*1/(d*e*(d + e*x)^3*(d^2 - e^2*x^2)^(3/2)) + (2*(-1/7*1/(d*e*(d + e*x) ^2*(d^2 - e^2*x^2)^(3/2)) + (5*(-1/5*1/(d*e*(d + e*x)*(d^2 - e^2*x^2)^(3/2 )) + (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)))/(7*d)))/(3*d)
3.9.44.3.1 Defintions of rubi rules used
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_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Time = 2.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.67
| method | result | size |
| gosper | \(-\frac {\left (-e x +d \right ) \left (16 e^{6} x^{6}+48 d \,e^{5} x^{5}+24 d^{2} e^{4} x^{4}-56 x^{3} d^{3} e^{3}-66 d^{4} e^{2} x^{2}-6 d^{5} e x +19 d^{6}\right )}{63 \left (e x +d \right )^{2} d^{7} e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}\) | \(99\) |
| trager | \(-\frac {\left (16 e^{6} x^{6}+48 d \,e^{5} x^{5}+24 d^{2} e^{4} x^{4}-56 x^{3} d^{3} e^{3}-66 d^{4} e^{2} x^{2}-6 d^{5} e x +19 d^{6}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{63 d^{7} \left (e x +d \right )^{5} \left (-e x +d \right )^{2} e}\) | \(101\) |
| default | \(\frac {-\frac {1}{9 d e \left (x +\frac {d}{e}\right )^{3} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {2 e \left (-\frac {1}{7 d e \left (x +\frac {d}{e}\right )^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {5 e \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{6 d^{2} e^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e^{2} d^{4} \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{7 d}\right )}{3 d}}{e^{3}}\) | \(268\) |
-1/63*(-e*x+d)*(16*e^6*x^6+48*d*e^5*x^5+24*d^2*e^4*x^4-56*d^3*e^3*x^3-66*d ^4*e^2*x^2-6*d^5*e*x+19*d^6)/(e*x+d)^2/d^7/e/(-e^2*x^2+d^2)^(5/2)
Time = 0.35 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {19 \, e^{7} x^{7} + 57 \, d e^{6} x^{6} + 19 \, d^{2} e^{5} x^{5} - 95 \, d^{3} e^{4} x^{4} - 95 \, d^{4} e^{3} x^{3} + 19 \, d^{5} e^{2} x^{2} + 57 \, d^{6} e x + 19 \, d^{7} + {\left (16 \, e^{6} x^{6} + 48 \, d e^{5} x^{5} + 24 \, d^{2} e^{4} x^{4} - 56 \, d^{3} e^{3} x^{3} - 66 \, d^{4} e^{2} x^{2} - 6 \, d^{5} e x + 19 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{63 \, {\left (d^{7} e^{8} x^{7} + 3 \, d^{8} e^{7} x^{6} + d^{9} e^{6} x^{5} - 5 \, d^{10} e^{5} x^{4} - 5 \, d^{11} e^{4} x^{3} + d^{12} e^{3} x^{2} + 3 \, d^{13} e^{2} x + d^{14} e\right )}} \]
-1/63*(19*e^7*x^7 + 57*d*e^6*x^6 + 19*d^2*e^5*x^5 - 95*d^3*e^4*x^4 - 95*d^ 4*e^3*x^3 + 19*d^5*e^2*x^2 + 57*d^6*e*x + 19*d^7 + (16*e^6*x^6 + 48*d*e^5* x^5 + 24*d^2*e^4*x^4 - 56*d^3*e^3*x^3 - 66*d^4*e^2*x^2 - 6*d^5*e*x + 19*d^ 6)*sqrt(-e^2*x^2 + d^2))/(d^7*e^8*x^7 + 3*d^8*e^7*x^6 + d^9*e^6*x^5 - 5*d^ 10*e^5*x^4 - 5*d^11*e^4*x^3 + d^12*e^3*x^2 + 3*d^13*e^2*x + d^14*e)
\[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )^{3}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{9 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{4} x^{3} + 3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3} x^{2} + 3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e\right )}} - \frac {2}{21 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3} x^{2} + 2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e\right )}} - \frac {2}{21 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e\right )}} + \frac {8 \, x}{63 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {16 \, x}{63 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} \]
-1/9/((-e^2*x^2 + d^2)^(3/2)*d*e^4*x^3 + 3*(-e^2*x^2 + d^2)^(3/2)*d^2*e^3* x^2 + 3*(-e^2*x^2 + d^2)^(3/2)*d^3*e^2*x + (-e^2*x^2 + d^2)^(3/2)*d^4*e) - 2/21/((-e^2*x^2 + d^2)^(3/2)*d^2*e^3*x^2 + 2*(-e^2*x^2 + d^2)^(3/2)*d^3*e ^2*x + (-e^2*x^2 + d^2)^(3/2)*d^4*e) - 2/21/((-e^2*x^2 + d^2)^(3/2)*d^3*e^ 2*x + (-e^2*x^2 + d^2)^(3/2)*d^4*e) + 8/63*x/((-e^2*x^2 + d^2)^(3/2)*d^5) + 16/63*x/(sqrt(-e^2*x^2 + d^2)*d^7)
\[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}^{3}} \,d x } \]
Time = 10.20 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {197\,x}{1008\,d^5}-\frac {155}{1008\,d^4\,e}\right )}{{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}-\frac {\sqrt {d^2-e^2\,x^2}}{36\,d^3\,e\,{\left (d+e\,x\right )}^5}-\frac {13\,\sqrt {d^2-e^2\,x^2}}{252\,d^4\,e\,{\left (d+e\,x\right )}^4}-\frac {23\,\sqrt {d^2-e^2\,x^2}}{336\,d^5\,e\,{\left (d+e\,x\right )}^3}+\frac {16\,x\,\sqrt {d^2-e^2\,x^2}}{63\,d^7\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \]
((d^2 - e^2*x^2)^(1/2)*((197*x)/(1008*d^5) - 155/(1008*d^4*e)))/((d + e*x) ^2*(d - e*x)^2) - (d^2 - e^2*x^2)^(1/2)/(36*d^3*e*(d + e*x)^5) - (13*(d^2 - e^2*x^2)^(1/2))/(252*d^4*e*(d + e*x)^4) - (23*(d^2 - e^2*x^2)^(1/2))/(33 6*d^5*e*(d + e*x)^3) + (16*x*(d^2 - e^2*x^2)^(1/2))/(63*d^7*(d + e*x)*(d - e*x))