Integrand size = 27, antiderivative size = 117 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \]
2/5*(e*x+d)/d/(-e^2*x^2+d^2)^(5/2)+1/15*(8*e*x+5*d)/d^3/(-e^2*x^2+d^2)^(3/ 2)-arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^5+1/15*(16*e*x+15*d)/d^5/(-e^2*x^2+d^ 2)^(1/2)
Time = 0.48 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (26 d^3-22 d^2 e x-17 d e^2 x^2+16 e^3 x^3\right )}{(d-e x)^3 (d+e x)}+30 \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^5} \]
((Sqrt[d^2 - e^2*x^2]*(26*d^3 - 22*d^2*e*x - 17*d*e^2*x^2 + 16*e^3*x^3))/( (d - e*x)^3*(d + e*x)) + 30*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d ])/(15*d^5)
Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {532, 25, 27, 532, 25, 532, 27, 243, 73, 221}
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.
| \(\displaystyle \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int -\frac {d (5 d+8 e x)}{x \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {d (5 d+8 e x)}{x \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {5 d+8 e x}{x \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int -\frac {15 d+16 e x}{x \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {15 d+16 e x}{x \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}+\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {\frac {\frac {15 d+16 e x}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d}{x \sqrt {d^2-e^2 x^2}}dx}{d^2}}{3 d^2}+\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {15 \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx}{d}+\frac {15 d+16 e x}{d^2 \sqrt {d^2-e^2 x^2}}}{3 d^2}+\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {\frac {15 \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2}{2 d}+\frac {15 d+16 e x}{d^2 \sqrt {d^2-e^2 x^2}}}{3 d^2}+\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {15 d+16 e x}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {15 \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{d e^2}}{3 d^2}+\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {15 d+16 e x}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {15 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2}}{3 d^2}+\frac {5 d+8 e x}{3 d^2 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
(2*(d + e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + ((5*d + 8*e*x)/(3*d^2*(d^2 - e ^2*x^2)^(3/2)) + ((15*d + 16*e*x)/(d^2*Sqrt[d^2 - e^2*x^2]) - (15*ArcTanh[ Sqrt[d^2 - e^2*x^2]/d])/d^2)/(3*d^2))/(5*d)
3.1.50.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Time = 0.36 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.73
| method | result | size |
| default | \(\frac {1}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+d^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )+2 d e \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )\) | \(202\) |
1/5/(-e^2*x^2+d^2)^(5/2)+d^2*(1/5/d^2/(-e^2*x^2+d^2)^(5/2)+1/d^2*(1/3/d^2/ (-e^2*x^2+d^2)^(3/2)+1/d^2*(1/d^2/(-e^2*x^2+d^2)^(1/2)-1/d^2/(d^2)^(1/2)*l n((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x))))+2*d*e*(1/5*x/d^2/(-e^2* x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2 +d^2)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {26 \, e^{4} x^{4} - 52 \, d e^{3} x^{3} + 52 \, d^{3} e x - 26 \, d^{4} + 15 \, {\left (e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (16 \, e^{3} x^{3} - 17 \, d e^{2} x^{2} - 22 \, d^{2} e x + 26 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{5} e^{4} x^{4} - 2 \, d^{6} e^{3} x^{3} + 2 \, d^{8} e x - d^{9}\right )}} \]
1/15*(26*e^4*x^4 - 52*d*e^3*x^3 + 52*d^3*e*x - 26*d^4 + 15*(e^4*x^4 - 2*d* e^3*x^3 + 2*d^3*e*x - d^4)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (16*e^3*x^ 3 - 17*d*e^2*x^2 - 22*d^2*e*x + 26*d^3)*sqrt(-e^2*x^2 + d^2))/(d^5*e^4*x^4 - 2*d^6*e^3*x^3 + 2*d^8*e*x - d^9)
\[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, e x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {2}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {8 \, e x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {16 \, e x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{5}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} d^{4}} \]
2/5*e*x/((-e^2*x^2 + d^2)^(5/2)*d) + 2/5/(-e^2*x^2 + d^2)^(5/2) + 8/15*e*x /((-e^2*x^2 + d^2)^(3/2)*d^3) + 1/3/((-e^2*x^2 + d^2)^(3/2)*d^2) + 16/15*e *x/(sqrt(-e^2*x^2 + d^2)*d^5) - log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)* d/abs(x))/d^5 + 1/(sqrt(-e^2*x^2 + d^2)*d^4)
\[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{x\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]