Integrand size = 24, antiderivative size = 106 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {6 x}{35 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {8 x}{35 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 x}{35 d^7 \sqrt {d^2-e^2 x^2}} \] Output:
6/35*x/d^3/(-e^2*x^2+d^2)^(5/2)-1/7/d/e/(e*x+d)/(-e^2*x^2+d^2)^(5/2)+8/35* x/d^5/(-e^2*x^2+d^2)^(3/2)+16/35*x/d^7/(-e^2*x^2+d^2)^(1/2)
Time = 0.37 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-5 d^6+30 d^5 e x+30 d^4 e^2 x^2-40 d^3 e^3 x^3-40 d^2 e^4 x^4+16 d e^5 x^5+16 e^6 x^6\right )}{35 d^7 e (d-e x)^3 (d+e x)^4} \] Input:
Integrate[1/((d + e*x)*(d^2 - e^2*x^2)^(7/2)),x]
Output:
(Sqrt[d^2 - e^2*x^2]*(-5*d^6 + 30*d^5*e*x + 30*d^4*e^2*x^2 - 40*d^3*e^3*x^ 3 - 40*d^2*e^4*x^4 + 16*d*e^5*x^5 + 16*e^6*x^6))/(35*d^7*e*(d - e*x)^3*(d + e*x)^4)
Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {470, 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.
\(\displaystyle \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {6 \int \frac {1}{\left (d^2-e^2 x^2\right )^{7/2}}dx}{7 d}-\frac {1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {6 \left (\frac {4 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}\right )}{7 d}-\frac {1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}+\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{5 d^2}+\frac {x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}\right )}{7 d}-\frac {1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {6 \left (\frac {x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 \left (\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{3 d^4 \sqrt {d^2-e^2 x^2}}\right )}{5 d^2}\right )}{7 d}-\frac {1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}\) |
Input:
Int[1/((d + e*x)*(d^2 - e^2*x^2)^(7/2)),x]
Output:
-1/7*1/(d*e*(d + e*x)*(d^2 - e^2*x^2)^(5/2)) + (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)
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[(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 = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87
method | result | size |
gosper | \(-\frac {\left (-e x +d \right ) \left (-16 e^{6} x^{6}-16 d \,e^{5} x^{5}+40 d^{2} e^{4} x^{4}+40 d^{3} e^{3} x^{3}-30 d^{4} e^{2} x^{2}-30 d^{5} e x +5 d^{6}\right )}{35 d^{7} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(92\) |
orering | \(-\frac {\left (-e x +d \right ) \left (-16 e^{6} x^{6}-16 d \,e^{5} x^{5}+40 d^{2} e^{4} x^{4}+40 d^{3} e^{3} x^{3}-30 d^{4} e^{2} x^{2}-30 d^{5} e x +5 d^{6}\right )}{35 d^{7} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(92\) |
trager | \(-\frac {\left (-16 e^{6} x^{6}-16 d \,e^{5} x^{5}+40 d^{2} e^{4} x^{4}+40 d^{3} e^{3} x^{3}-30 d^{4} e^{2} x^{2}-30 d^{5} e x +5 d^{6}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{35 d^{7} \left (e x +d \right )^{4} \left (-e x +d \right )^{3} e}\) | \(101\) |
default | \(\frac {-\frac {1}{7 d e \left (x +\frac {d}{e}\right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {6 e \left (-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{10 d^{2} e^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {-\frac {2 \left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right )}{15 d^{2} e^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {4 \left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right )}{15 e^{2} d^{4} \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}}{d^{2}}\right )}{7 d}}{e}\) | \(223\) |
Input:
int(1/(e*x+d)/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/35*(-e*x+d)*(-16*e^6*x^6-16*d*e^5*x^5+40*d^2*e^4*x^4+40*d^3*e^3*x^3-30* d^4*e^2*x^2-30*d^5*e*x+5*d^6)/d^7/e/(-e^2*x^2+d^2)^(7/2)
Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (90) = 180\).
Time = 0.16 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.23 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {5 \, e^{7} x^{7} + 5 \, d e^{6} x^{6} - 15 \, d^{2} e^{5} x^{5} - 15 \, d^{3} e^{4} x^{4} + 15 \, d^{4} e^{3} x^{3} + 15 \, d^{5} e^{2} x^{2} - 5 \, d^{6} e x - 5 \, d^{7} + {\left (16 \, e^{6} x^{6} + 16 \, d e^{5} x^{5} - 40 \, d^{2} e^{4} x^{4} - 40 \, d^{3} e^{3} x^{3} + 30 \, d^{4} e^{2} x^{2} + 30 \, d^{5} e x - 5 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{35 \, {\left (d^{7} e^{8} x^{7} + d^{8} e^{7} x^{6} - 3 \, d^{9} e^{6} x^{5} - 3 \, d^{10} e^{5} x^{4} + 3 \, d^{11} e^{4} x^{3} + 3 \, d^{12} e^{3} x^{2} - d^{13} e^{2} x - d^{14} e\right )}} \] Input:
integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")
Output:
-1/35*(5*e^7*x^7 + 5*d*e^6*x^6 - 15*d^2*e^5*x^5 - 15*d^3*e^4*x^4 + 15*d^4* e^3*x^3 + 15*d^5*e^2*x^2 - 5*d^6*e*x - 5*d^7 + (16*e^6*x^6 + 16*d*e^5*x^5 - 40*d^2*e^4*x^4 - 40*d^3*e^3*x^3 + 30*d^4*e^2*x^2 + 30*d^5*e*x - 5*d^6)*s qrt(-e^2*x^2 + d^2))/(d^7*e^8*x^7 + d^8*e^7*x^6 - 3*d^9*e^6*x^5 - 3*d^10*e ^5*x^4 + 3*d^11*e^4*x^3 + 3*d^12*e^3*x^2 - d^13*e^2*x - d^14*e)
\[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}} \left (d + e x\right )}\, dx \] Input:
integrate(1/(e*x+d)/(-e**2*x**2+d**2)**(7/2),x)
Output:
Integral(1/((-(-d + e*x)*(d + e*x))**(7/2)*(d + e*x)), x)
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {1}{7 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} e\right )}} + \frac {6 \, x}{35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {8 \, x}{35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {16 \, x}{35 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} \] Input:
integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")
Output:
-1/7/((-e^2*x^2 + d^2)^(5/2)*d*e^2*x + (-e^2*x^2 + d^2)^(5/2)*d^2*e) + 6/3 5*x/((-e^2*x^2 + d^2)^(5/2)*d^3) + 8/35*x/((-e^2*x^2 + d^2)^(3/2)*d^5) + 1 6/35*x/(sqrt(-e^2*x^2 + d^2)*d^7)
\[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} {\left (e x + d\right )}} \,d x } \] Input:
integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")
Output:
integrate(1/((-e^2*x^2 + d^2)^(7/2)*(e*x + d)), x)
Time = 6.45 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {17\,x}{70\,d^3}-\frac {1}{7\,d^2\,e}\right )}{{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^3}+\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {8\,x}{35\,d^5}+\frac {1}{56\,d^4\,e}\right )}{{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}-\frac {\sqrt {d^2-e^2\,x^2}}{56\,d^4\,e\,{\left (d+e\,x\right )}^4}+\frac {16\,x\,\sqrt {d^2-e^2\,x^2}}{35\,d^7\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \] Input:
int(1/((d^2 - e^2*x^2)^(7/2)*(d + e*x)),x)
Output:
((d^2 - e^2*x^2)^(1/2)*((17*x)/(70*d^3) - 1/(7*d^2*e)))/((d + e*x)^3*(d - e*x)^3) + ((d^2 - e^2*x^2)^(1/2)*((8*x)/(35*d^5) + 1/(56*d^4*e)))/((d + e* x)^2*(d - e*x)^2) - (d^2 - e^2*x^2)^(1/2)/(56*d^4*e*(d + e*x)^4) + (16*x*( d^2 - e^2*x^2)^(1/2))/(35*d^7*(d + e*x)*(d - e*x))
Time = 0.25 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {30 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}+30 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} e x -60 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} e^{2} x^{2}-60 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e^{3} x^{3}+30 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{4} x^{4}+30 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x^{5}-5 d^{6}+30 d^{5} e x +30 d^{4} e^{2} x^{2}-40 d^{3} e^{3} x^{3}-40 d^{2} e^{4} x^{4}+16 d \,e^{5} x^{5}+16 e^{6} x^{6}}{35 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{7} e \left (e^{5} x^{5}+d \,e^{4} x^{4}-2 d^{2} e^{3} x^{3}-2 d^{3} e^{2} x^{2}+d^{4} e x +d^{5}\right )} \] Input:
int(1/(e*x+d)/(-e^2*x^2+d^2)^(7/2),x)
Output:
(30*sqrt(d**2 - e**2*x**2)*d**5 + 30*sqrt(d**2 - e**2*x**2)*d**4*e*x - 60* sqrt(d**2 - e**2*x**2)*d**3*e**2*x**2 - 60*sqrt(d**2 - e**2*x**2)*d**2*e** 3*x**3 + 30*sqrt(d**2 - e**2*x**2)*d*e**4*x**4 + 30*sqrt(d**2 - e**2*x**2) *e**5*x**5 - 5*d**6 + 30*d**5*e*x + 30*d**4*e**2*x**2 - 40*d**3*e**3*x**3 - 40*d**2*e**4*x**4 + 16*d*e**5*x**5 + 16*e**6*x**6)/(35*sqrt(d**2 - e**2* x**2)*d**7*e*(d**5 + d**4*e*x - 2*d**3*e**2*x**2 - 2*d**2*e**3*x**3 + d*e* *4*x**4 + e**5*x**5))