Integrand size = 24, antiderivative size = 82 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {4 x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 x}{15 d^5 \sqrt {d^2-e^2 x^2}} \] Output:
4/15*x/d^3/(-e^2*x^2+d^2)^(3/2)-1/5/d/e/(e*x+d)/(-e^2*x^2+d^2)^(3/2)+8/15* x/d^5/(-e^2*x^2+d^2)^(1/2)
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-3 d^4+12 d^3 e x+12 d^2 e^2 x^2-8 d e^3 x^3-8 e^4 x^4\right )}{15 d^5 e (d-e x)^2 (d+e x)^3} \] Input:
Integrate[1/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]
Output:
(Sqrt[d^2 - e^2*x^2]*(-3*d^4 + 12*d^3*e*x + 12*d^2*e^2*x^2 - 8*d*e^3*x^3 - 8*e^4*x^4))/(15*d^5*e*(d - e*x)^2*(d + e*x)^3)
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {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.
\(\displaystyle \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {4 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}+\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}\right )}{5 d}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {4 \left (\frac {x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{3 d^4 \sqrt {d^2-e^2 x^2}}\right )}{5 d}-\frac {1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}\) |
Input:
Int[1/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]
Output:
-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)
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 = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85
method | result | size |
gosper | \(-\frac {\left (-e x +d \right ) \left (8 e^{4} x^{4}+8 d \,e^{3} x^{3}-12 d^{2} e^{2} x^{2}-12 d^{3} e x +3 d^{4}\right )}{15 d^{5} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\) | \(70\) |
orering | \(-\frac {\left (-e x +d \right ) \left (8 e^{4} x^{4}+8 d \,e^{3} x^{3}-12 d^{2} e^{2} x^{2}-12 d^{3} e x +3 d^{4}\right )}{15 d^{5} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\) | \(70\) |
trager | \(-\frac {\left (8 e^{4} x^{4}+8 d \,e^{3} x^{3}-12 d^{2} e^{2} x^{2}-12 d^{3} e x +3 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{5} \left (e x +d \right )^{3} \left (-e x +d \right )^{2} e}\) | \(79\) |
default | \(\frac {-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{6 d^{2} e^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e^{2} d^{4} \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}}{e}\) | \(164\) |
Input:
int(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(-e*x+d)*(8*e^4*x^4+8*d*e^3*x^3-12*d^2*e^2*x^2-12*d^3*e*x+3*d^4)/d^5 /e/(-e^2*x^2+d^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (70) = 140\).
Time = 0.12 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {3 \, e^{5} x^{5} + 3 \, d e^{4} x^{4} - 6 \, d^{2} e^{3} x^{3} - 6 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + 3 \, d^{5} + {\left (8 \, e^{4} x^{4} + 8 \, d e^{3} x^{3} - 12 \, d^{2} e^{2} x^{2} - 12 \, d^{3} e x + 3 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{5} e^{6} x^{5} + d^{6} e^{5} x^{4} - 2 \, d^{7} e^{4} x^{3} - 2 \, d^{8} e^{3} x^{2} + d^{9} e^{2} x + d^{10} e\right )}} \] Input:
integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")
Output:
-1/15*(3*e^5*x^5 + 3*d*e^4*x^4 - 6*d^2*e^3*x^3 - 6*d^3*e^2*x^2 + 3*d^4*e*x + 3*d^5 + (8*e^4*x^4 + 8*d*e^3*x^3 - 12*d^2*e^2*x^2 - 12*d^3*e*x + 3*d^4) *sqrt(-e^2*x^2 + d^2))/(d^5*e^6*x^5 + d^6*e^5*x^4 - 2*d^7*e^4*x^3 - 2*d^8* e^3*x^2 + d^9*e^2*x + d^10*e)
\[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \] Input:
integrate(1/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)
Output:
Integral(1/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e\right )}} + \frac {4 \, x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {8 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}} \] Input:
integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")
Output:
-1/5/((-e^2*x^2 + d^2)^(3/2)*d*e^2*x + (-e^2*x^2 + d^2)^(3/2)*d^2*e) + 4/1 5*x/((-e^2*x^2 + d^2)^(3/2)*d^3) + 8/15*x/(sqrt(-e^2*x^2 + d^2)*d^5)
\[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}} \,d x } \] Input:
integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")
Output:
integrate(1/((-e^2*x^2 + d^2)^(5/2)*(e*x + d)), x)
Time = 6.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (3\,d^4-12\,d^3\,e\,x-12\,d^2\,e^2\,x^2+8\,d\,e^3\,x^3+8\,e^4\,x^4\right )}{15\,d^5\,e\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \] Input:
int(1/((d^2 - e^2*x^2)^(5/2)*(d + e*x)),x)
Output:
-((d^2 - e^2*x^2)^(1/2)*(3*d^4 + 8*e^4*x^4 + 8*d*e^3*x^3 - 12*d^2*e^2*x^2 - 12*d^3*e*x))/(15*d^5*e*(d + e*x)^3*(d - e*x)^2)
Time = 0.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.12 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {12 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}+12 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e x -12 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2} x^{2}-12 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x^{3}-3 d^{4}+12 d^{3} e x +12 d^{2} e^{2} x^{2}-8 d \,e^{3} x^{3}-8 e^{4} x^{4}}{15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} e \left (-e^{3} x^{3}-d \,e^{2} x^{2}+d^{2} e x +d^{3}\right )} \] Input:
int(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)
Output:
(12*sqrt(d**2 - e**2*x**2)*d**3 + 12*sqrt(d**2 - e**2*x**2)*d**2*e*x - 12* sqrt(d**2 - e**2*x**2)*d*e**2*x**2 - 12*sqrt(d**2 - e**2*x**2)*e**3*x**3 - 3*d**4 + 12*d**3*e*x + 12*d**2*e**2*x**2 - 8*d*e**3*x**3 - 8*e**4*x**4)/( 15*sqrt(d**2 - e**2*x**2)*d**5*e*(d**3 + d**2*e*x - d*e**2*x**2 - e**3*x** 3))