Integrand size = 27, antiderivative size = 134 \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=-\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-e^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {13}{8} e^4 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]
-1/4*d*(-e^2*x^2+d^2)^(3/2)/x^4-e*(-e^2*x^2+d^2)^(3/2)/x^3-e^4*arctan(e*x/ (-e^2*x^2+d^2)^(1/2))+13/8*e^4*arctanh((-e^2*x^2+d^2)^(1/2)/d)-1/8*e^2*(8* e*x+13*d)*(-e^2*x^2+d^2)^(1/2)/x^2
Time = 0.43 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=-\frac {d \sqrt {d^2-e^2 x^2} \left (2 d^2+8 d e x+11 e^2 x^2\right )}{8 x^4}-\frac {13}{4} e^4 \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )+e \left (-e^2\right )^{3/2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right ) \]
-1/8*(d*Sqrt[d^2 - e^2*x^2]*(2*d^2 + 8*d*e*x + 11*e^2*x^2))/x^4 - (13*e^4* ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/4 + e*(-e^2)^(3/2)*Log[-( Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]
Time = 0.43 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {540, 25, 2338, 27, 537, 25, 538, 224, 216, 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.
| \(\displaystyle \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx\) |
| \(\Big \downarrow \) 540 |
| \(\displaystyle -\frac {\int -\frac {\sqrt {d^2-e^2 x^2} \left (12 e d^4+13 e^2 x d^3+4 e^3 x^2 d^2\right )}{x^4}dx}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sqrt {d^2-e^2 x^2} \left (12 e d^4+13 e^2 x d^3+4 e^3 x^2 d^2\right )}{x^4}dx}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 2338 |
| \(\displaystyle \frac {-\frac {\int -\frac {3 d^4 e^2 (13 d+4 e x) \sqrt {d^2-e^2 x^2}}{x^3}dx}{3 d^2}-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 e^2 \int \frac {(13 d+4 e x) \sqrt {d^2-e^2 x^2}}{x^3}dx-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 537 |
| \(\displaystyle \frac {d^2 e^2 \left (\frac {1}{2} e^2 \int -\frac {13 d+8 e x}{x \sqrt {d^2-e^2 x^2}}dx-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \int \frac {13 d+8 e x}{x \sqrt {d^2-e^2 x^2}}dx-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \left (8 e \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+13 d \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx\right )-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \left (13 d \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx+8 e \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \left (13 d \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx+8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \left (\frac {13}{2} d \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2+8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \left (8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {13 d \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e^2}\right )-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {d^2 e^2 \left (-\frac {1}{2} e^2 \left (8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-13 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\right )-\frac {(13 d+8 e x) \sqrt {d^2-e^2 x^2}}{2 x^2}\right )-\frac {4 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}}{4 d^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
-1/4*(d*(d^2 - e^2*x^2)^(3/2))/x^4 + ((-4*d^2*e*(d^2 - e^2*x^2)^(3/2))/x^3 + d^2*e^2*(-1/2*((13*d + 8*e*x)*Sqrt[d^2 - e^2*x^2])/x^2 - (e^2*(8*ArcTan [(e*x)/Sqrt[d^2 - e^2*x^2]] - 13*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]))/2))/(4*d ^2)
3.1.64.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) , x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 0.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87
| method | result | size |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (11 e^{2} x^{2}+8 d e x +2 d^{2}\right ) d}{8 x^{4}}-\frac {e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {13 e^{4} d \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}\) | \(116\) |
| default | \(d^{3} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 d^{2} x^{4}}+\frac {e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{2 d^{2}}\right )}{4 d^{2}}\right )+e^{3} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{2} x}-\frac {2 e^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{d^{2}}\right )+3 d \,e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{2 d^{2}}\right )-\frac {e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{x^{3}}\) | \(324\) |
-1/8*(-e^2*x^2+d^2)^(1/2)*(11*e^2*x^2+8*d*e*x+2*d^2)*d/x^4-e^5/(e^2)^(1/2) *arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+13/8*e^4*d/(d^2)^(1/2)*ln((2*d ^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)
Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83 \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=\frac {16 \, e^{4} x^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 13 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (11 \, d e^{2} x^{2} + 8 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{8 \, x^{4}} \]
1/8*(16*e^4*x^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 13*e^4*x^4*log (-(d - sqrt(-e^2*x^2 + d^2))/x) - (11*d*e^2*x^2 + 8*d^2*e*x + 2*d^3)*sqrt( -e^2*x^2 + d^2))/x^4
Result contains complex when optimal does not.
Time = 4.42 (sec) , antiderivative size = 544, normalized size of antiderivative = 4.06 \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=d^{3} \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) + 3 d^{2} e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 x} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]
d**3*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3* sqrt(d**2/(e**2*x**2) - 1)) - e**3/(8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*x* *5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1 )) + I*e**3/(8*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/ (8*d**3), True)) + 3*d**2*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x* *2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1) , (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x** 2) + 1)/(3*d**2), True)) + 3*d*e**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(2*x) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/ (2*e*x**3*sqrt(-d**2/(e**2*x**2) + 1)) - I*e/(2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**2*asin(d/(e*x))/(2*d), True)) + e**3*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1 + e**2*x**2/d **2)), Abs(e**2*x**2/d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin (e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**2/d**2)), True))
Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=-\frac {e^{5} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} + \frac {13}{8} \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {13 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{3}}{x} - \frac {13 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{4 \, x^{4}} \]
-e^5*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) + 13/8*e^4*log(2*d^2/abs(x) + 2 *sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 13/8*sqrt(-e^2*x^2 + d^2)*e^4/d - sqrt(- e^2*x^2 + d^2)*e^3/x - 13/8*(-e^2*x^2 + d^2)^(3/2)*e^2/(d*x^2) - (-e^2*x^2 + d^2)^(3/2)*e/x^3 - 1/4*(-e^2*x^2 + d^2)^(3/2)*d/x^4
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (118) = 236\).
Time = 0.30 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.43 \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=\frac {{\left (e^{5} + \frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e^{3}}{x} + \frac {24 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} e}{x^{2}} + \frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e x^{3}}\right )} e^{8} x^{4}}{64 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} {\left | e \right |}} - \frac {e^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} + \frac {13 \, e^{5} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{8 \, {\left | e \right |}} - \frac {\frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e^{5} {\left | e \right |}}{x} + \frac {24 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} e^{3} {\left | e \right |}}{x^{2}} + \frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} e {\left | e \right |}}{x^{3}} + \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} {\left | e \right |}}{e x^{4}}}{64 \, e^{4}} \]
1/64*(e^5 + 8*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e^3/x + 24*(d*e + sqrt(- e^2*x^2 + d^2)*abs(e))^2*e/x^2 + 8*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/( e*x^3))*e^8*x^4/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4*abs(e)) - e^5*arcsi n(e*x/d)*sgn(d)*sgn(e)/abs(e) + 13/8*e^5*log(1/2*abs(-2*d*e - 2*sqrt(-e^2* x^2 + d^2)*abs(e))/(e^2*abs(x)))/abs(e) - 1/64*(8*(d*e + sqrt(-e^2*x^2 + d ^2)*abs(e))*e^5*abs(e)/x + 24*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*e^3*ab s(e)/x^2 + 8*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*e*abs(e)/x^3 + (d*e + s qrt(-e^2*x^2 + d^2)*abs(e))^4*abs(e)/(e*x^4))/e^4
Timed out. \[ \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx=\int \frac {\sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^3}{x^5} \,d x \]