Integrand size = 27, antiderivative size = 145 \[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (5 d+7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (15 d+19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \]
4/5*e*(e*x+d)/d/(-e^2*x^2+d^2)^(5/2)+1/5*e*(7*e*x+5*d)/d^3/(-e^2*x^2+d^2)^ (3/2)-3*e*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^5+1/5*e*(19*e*x+15*d)/d^5/(-e^ 2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^5/x
Time = 0.41 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81 \[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (5 d^3-39 d^2 e x+57 d e^2 x^2-24 e^3 x^3\right )}{x (-d+e x)^3}-15 \sqrt {d^2} e \log (x)+15 \sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{5 d^6} \]
((d*Sqrt[d^2 - e^2*x^2]*(5*d^3 - 39*d^2*e*x + 57*d*e^2*x^2 - 24*e^3*x^3))/ (x*(-d + e*x)^3) - 15*Sqrt[d^2]*e*Log[x] + 15*Sqrt[d^2]*e*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(5*d^6)
Time = 0.51 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {532, 25, 2336, 27, 2336, 27, 534, 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.
| \(\displaystyle \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int -\frac {5 d^3+15 e x d^2+16 e^2 x^2 d}{x^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {5 d^3+15 e x d^2+16 e^2 x^2 d}{x^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int -\frac {3 \left (5 d^3+15 e x d^2+14 e^2 x^2 d\right )}{x^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {5 d^3+15 e x d^2+14 e^2 x^2 d}{x^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {\frac {e (15 d+19 e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {5 d^2 (d+3 e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{d^2}}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5 \int \frac {d+3 e x}{x^2 \sqrt {d^2-e^2 x^2}}dx+\frac {e (15 d+19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle \frac {\frac {5 \left (3 e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )+\frac {e (15 d+19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )+\frac {e (15 d+19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {3 \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )+\frac {e (15 d+19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {5 \left (-\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )+\frac {e (15 d+19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {4 e (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\) |
(4*e*(d + e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + ((e*(5*d + 7*e*x))/(d*(d^2 - e^2*x^2)^(3/2)) + ((e*(15*d + 19*e*x))/(d*Sqrt[d^2 - e^2*x^2]) + 5*(-(Sqr t[d^2 - e^2*x^2]/(d*x)) - (3*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/d^2)/(5 *d^2)
3.1.90.3.1 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]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Time = 0.43 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.43
| method | result | size |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{5} x}-\frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{4} \sqrt {d^{2}}}+\frac {4 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{5 d^{4} e \left (x -\frac {d}{e}\right )^{2}}-\frac {19 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{5 d^{5} \left (x -\frac {d}{e}\right )}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{5 d^{3} e^{2} \left (x -\frac {d}{e}\right )^{3}}\) | \(208\) |
| default | \(\frac {e}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+d^{3} \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )+3 d \,e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+3 d^{2} e \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )\) | \(309\) |
-(-e^2*x^2+d^2)^(1/2)/d^5/x-3/d^4*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(- e^2*x^2+d^2)^(1/2))/x)+4/5/d^4/e/(x-d/e)^2*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^ (1/2)-19/5/d^5/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)-1/5/d^3/e^2/(x -d/e)^3*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)
Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {24 \, e^{4} x^{4} - 72 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} - 24 \, d^{3} e x + 15 \, {\left (e^{4} x^{4} - 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} - d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (24 \, e^{3} x^{3} - 57 \, d e^{2} x^{2} + 39 \, d^{2} e x - 5 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{5} e^{3} x^{4} - 3 \, d^{6} e^{2} x^{3} + 3 \, d^{7} e x^{2} - d^{8} x\right )}} \]
1/5*(24*e^4*x^4 - 72*d*e^3*x^3 + 72*d^2*e^2*x^2 - 24*d^3*e*x + 15*(e^4*x^4 - 3*d*e^3*x^3 + 3*d^2*e^2*x^2 - d^3*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/ x) - (24*e^3*x^3 - 57*d*e^2*x^2 + 39*d^2*e*x - 5*d^3)*sqrt(-e^2*x^2 + d^2) )/(d^5*e^3*x^4 - 3*d^6*e^2*x^3 + 3*d^7*e*x^2 - d^8*x)
\[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {9 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {4 \, e}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {12 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {d}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x} + \frac {24 \, e^{2} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}} - \frac {3 \, e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{5}} + \frac {3 \, e}{\sqrt {-e^{2} x^{2} + d^{2}} d^{4}} \]
9/5*e^2*x/((-e^2*x^2 + d^2)^(5/2)*d) + 4/5*e/(-e^2*x^2 + d^2)^(5/2) + 12/5 *e^2*x/((-e^2*x^2 + d^2)^(3/2)*d^3) + e/((-e^2*x^2 + d^2)^(3/2)*d^2) - d/( (-e^2*x^2 + d^2)^(5/2)*x) + 24/5*e^2*x/(sqrt(-e^2*x^2 + d^2)*d^5) - 3*e*lo g(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^5 + 3*e/(sqrt(-e^2*x^2 + d^2)*d^4)
Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (129) = 258\).
Time = 0.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.10 \[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {3 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{5} {\left | e \right |}} - \frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{2 \, d^{5} x {\left | e \right |}} - \frac {{\left (5 \, e^{2} - \frac {121 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{x} + \frac {410 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{2} x^{2}} - \frac {610 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{4} x^{3}} + \frac {425 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{6} x^{4}} - \frac {125 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{8} x^{5}}\right )} e^{2} x}{10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{5} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]
-3*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/( d^5*abs(e)) - 1/2*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(d^5*x*abs(e)) - 1/1 0*(5*e^2 - 121*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/x + 410*(d*e + sqrt(-e^ 2*x^2 + d^2)*abs(e))^2/(e^2*x^2) - 610*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e)) ^3/(e^4*x^3) + 425*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^6*x^4) - 125*( d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e^8*x^5))*e^2*x/((d*e + sqrt(-e^2*x^ 2 + d^2)*abs(e))*d^5*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^5*a bs(e))
Timed out. \[ \int \frac {(d+e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^3}{x^2\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]