Integrand size = 24, antiderivative size = 133 \[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{7 d e (d+e x)^4}-\frac {3 \sqrt {d^2-e^2 x^2}}{35 d^2 e (d+e x)^3}-\frac {2 \sqrt {d^2-e^2 x^2}}{35 d^3 e (d+e x)^2}-\frac {2 \sqrt {d^2-e^2 x^2}}{35 d^4 e (d+e x)} \]
-1/7*(-e^2*x^2+d^2)^(1/2)/d/e/(e*x+d)^4-3/35*(-e^2*x^2+d^2)^(1/2)/d^2/e/(e *x+d)^3-2/35*(-e^2*x^2+d^2)^(1/2)/d^3/e/(e*x+d)^2-2/35*(-e^2*x^2+d^2)^(1/2 )/d^4/e/(e*x+d)
Time = 0.49 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-12 d^3-13 d^2 e x-8 d e^2 x^2-2 e^3 x^3\right )}{35 d^4 e (d+e x)^4} \]
(Sqrt[d^2 - e^2*x^2]*(-12*d^3 - 13*d^2*e*x - 8*d*e^2*x^2 - 2*e^3*x^3))/(35 *d^4*e*(d + e*x)^4)
Time = 0.24 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {461, 461, 461, 460}
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.
\(\displaystyle \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {3 \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}}dx}{7 d}-\frac {\sqrt {d^2-e^2 x^2}}{7 d e (d+e x)^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {3 \left (\frac {2 \int \frac {1}{(d+e x)^2 \sqrt {d^2-e^2 x^2}}dx}{5 d}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}\right )}{7 d}-\frac {\sqrt {d^2-e^2 x^2}}{7 d e (d+e x)^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {\int \frac {1}{(d+e x) \sqrt {d^2-e^2 x^2}}dx}{3 d}-\frac {\sqrt {d^2-e^2 x^2}}{3 d e (d+e x)^2}\right )}{5 d}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}\right )}{7 d}-\frac {\sqrt {d^2-e^2 x^2}}{7 d e (d+e x)^4}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {3 \left (\frac {2 \left (-\frac {\sqrt {d^2-e^2 x^2}}{3 d^2 e (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{3 d e (d+e x)^2}\right )}{5 d}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}\right )}{7 d}-\frac {\sqrt {d^2-e^2 x^2}}{7 d e (d+e x)^4}\) |
-1/7*Sqrt[d^2 - e^2*x^2]/(d*e*(d + e*x)^4) + (3*(-1/5*Sqrt[d^2 - e^2*x^2]/ (d*e*(d + e*x)^3) + (2*(-1/3*Sqrt[d^2 - e^2*x^2]/(d*e*(d + e*x)^2) - Sqrt[ d^2 - e^2*x^2]/(3*d^2*e*(d + e*x))))/(5*d)))/(7*d)
3.9.34.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 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])
Time = 2.35 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.45
method | result | size |
trager | \(-\frac {\left (2 e^{3} x^{3}+8 d \,e^{2} x^{2}+13 d^{2} e x +12 d^{3}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{35 d^{4} \left (e x +d \right )^{4} e}\) | \(60\) |
gosper | \(-\frac {\left (-e x +d \right ) \left (2 e^{3} x^{3}+8 d \,e^{2} x^{2}+13 d^{2} e x +12 d^{3}\right )}{35 \left (e x +d \right )^{3} d^{4} e \sqrt {-x^{2} e^{2}+d^{2}}}\) | \(66\) |
default | \(\frac {-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{7 d e \left (x +\frac {d}{e}\right )^{4}}+\frac {3 e \left (-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{7 d}}{e^{4}}\) | \(197\) |
Time = 0.35 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {12 \, e^{4} x^{4} + 48 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 48 \, d^{3} e x + 12 \, d^{4} + {\left (2 \, e^{3} x^{3} + 8 \, d e^{2} x^{2} + 13 \, d^{2} e x + 12 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{35 \, {\left (d^{4} e^{5} x^{4} + 4 \, d^{5} e^{4} x^{3} + 6 \, d^{6} e^{3} x^{2} + 4 \, d^{7} e^{2} x + d^{8} e\right )}} \]
-1/35*(12*e^4*x^4 + 48*d*e^3*x^3 + 72*d^2*e^2*x^2 + 48*d^3*e*x + 12*d^4 + (2*e^3*x^3 + 8*d*e^2*x^2 + 13*d^2*e*x + 12*d^3)*sqrt(-e^2*x^2 + d^2))/(d^4 *e^5*x^4 + 4*d^5*e^4*x^3 + 6*d^6*e^3*x^2 + 4*d^7*e^2*x + d^8*e)
\[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{4}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}}}{7 \, {\left (d e^{5} x^{4} + 4 \, d^{2} e^{4} x^{3} + 6 \, d^{3} e^{3} x^{2} + 4 \, d^{4} e^{2} x + d^{5} e\right )}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}}}{35 \, {\left (d^{2} e^{4} x^{3} + 3 \, d^{3} e^{3} x^{2} + 3 \, d^{4} e^{2} x + d^{5} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}}}{35 \, {\left (d^{3} e^{3} x^{2} + 2 \, d^{4} e^{2} x + d^{5} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}}}{35 \, {\left (d^{4} e^{2} x + d^{5} e\right )}} \]
-1/7*sqrt(-e^2*x^2 + d^2)/(d*e^5*x^4 + 4*d^2*e^4*x^3 + 6*d^3*e^3*x^2 + 4*d ^4*e^2*x + d^5*e) - 3/35*sqrt(-e^2*x^2 + d^2)/(d^2*e^4*x^3 + 3*d^3*e^3*x^2 + 3*d^4*e^2*x + d^5*e) - 2/35*sqrt(-e^2*x^2 + d^2)/(d^3*e^3*x^2 + 2*d^4*e ^2*x + d^5*e) - 2/35*sqrt(-e^2*x^2 + d^2)/(d^4*e^2*x + d^5*e)
Time = 0.29 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=\frac {2 \, {\left (\frac {49 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {147 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {210 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + \frac {210 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} + \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{10} x^{5}} + \frac {35 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{6}}{e^{12} x^{6}} + 12\right )}}{35 \, d^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{7} {\left | e \right |}} \]
2/35*(49*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 147*(d*e + sqrt(-e^ 2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 210*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e)) ^3/(e^6*x^3) + 210*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) + 105*( d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e^10*x^5) + 35*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^6/(e^12*x^6) + 12)/(d^4*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e) )/(e^2*x) + 1)^7*abs(e))
Time = 9.56 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(d+e x)^4 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}}{7\,d\,e\,{\left (d+e\,x\right )}^4}-\frac {3\,\sqrt {d^2-e^2\,x^2}}{35\,d^2\,e\,{\left (d+e\,x\right )}^3}-\frac {2\,\sqrt {d^2-e^2\,x^2}}{35\,d^3\,e\,{\left (d+e\,x\right )}^2}-\frac {2\,\sqrt {d^2-e^2\,x^2}}{35\,d^4\,e\,\left (d+e\,x\right )} \]