Integrand size = 24, antiderivative size = 132 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=-\frac {35}{2} d x \sqrt {d^2-e^2 x^2}-\frac {35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac {35 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]
-35/3*(-e^2*x^2+d^2)^(3/2)/e-14*(-e^2*x^2+d^2)^(5/2)/e/(e*x+d)^2-2*(-e^2*x ^2+d^2)^(7/2)/e/(e*x+d)^4-35/2*d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-35/2 *d*x*(-e^2*x^2+d^2)^(1/2)
Time = 0.47 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.81 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-166 d^3-55 d^2 e x+13 d e^2 x^2-2 e^3 x^3\right )}{6 e (d+e x)}+\frac {35 d^3 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 \sqrt {-e^2}} \]
(Sqrt[d^2 - e^2*x^2]*(-166*d^3 - 55*d^2*e*x + 13*d*e^2*x^2 - 2*e^3*x^3))/( 6*e*(d + e*x)) + (35*d^3*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*Sq rt[-e^2])
Time = 0.39 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {463, 2346, 27, 2346, 27, 455, 224, 216}
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.
| \(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx\) |
| \(\Big \downarrow \) 463 |
| \(\displaystyle -\int \frac {15 d^3-11 e x d^2+5 e^2 x^2 d-e^3 x^3}{\sqrt {d^2-e^2 x^2}}dx-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\int -\frac {5 \left (3 d x^2 e^4-7 d^2 x e^3+9 d^3 e^2\right )}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {5 \int \frac {3 d x^2 e^4-7 d^2 x e^3+9 d^3 e^2}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle -\frac {5 \left (-\frac {\int -\frac {7 d^2 e^4 (3 d-2 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \int \frac {3 d-2 e x}{\sqrt {d^2-e^2 x^2}}dx-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \left (3 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {2 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \left (3 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {2 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \left (\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {2 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}-\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}-\frac {16 d^3 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
-1/3*(e*x^2*Sqrt[d^2 - e^2*x^2]) - (16*d^3*Sqrt[d^2 - e^2*x^2])/(e*(d + e* x)) - (5*((-3*d*e^2*x*Sqrt[d^2 - e^2*x^2])/2 + (7*d^2*e^2*((2*Sqrt[d^2 - e ^2*x^2])/e + (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/2))/(3*e^2)
3.9.7.3.1 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_)^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[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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x ))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ (2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 2.37 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89
| method | result | size |
| risch | \(-\frac {\left (2 x^{2} e^{2}-15 d e x +70 d^{2}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{6 e}-\frac {35 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {16 d^{3} \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{2} \left (x +\frac {d}{e}\right )}\) | \(117\) |
| default | \(\frac {-\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{d e \left (x +\frac {d}{e}\right )^{5}}-\frac {4 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}+\frac {5 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {7 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{7}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{12 e^{2}}+\frac {5 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}\right )\right )}{5 d}\right )}{d}\right )}{d}\right )}{d}}{e^{5}}\) | \(455\) |
-1/6*(2*e^2*x^2-15*d*e*x+70*d^2)/e*(-e^2*x^2+d^2)^(1/2)-35/2*d^3/(e^2)^(1/ 2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-16*d^3/e^2/(x+d/e)*(-e^2*(x+ d/e)^2+2*d*e*(x+d/e))^(1/2)
Time = 0.66 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.83 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=-\frac {166 \, d^{3} e x + 166 \, d^{4} - 210 \, {\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{3} x^{3} - 13 \, d e^{2} x^{2} + 55 \, d^{2} e x + 166 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{2} x + d e\right )}} \]
-1/6*(166*d^3*e*x + 166*d^4 - 210*(d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x ^2 + d^2))/(e*x)) + (2*e^3*x^3 - 13*d*e^2*x^2 + 55*d^2*e*x + 166*d^3)*sqrt (-e^2*x^2 + d^2))/(e^2*x + d*e)
\[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}{\left (d + e x\right )^{5}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.49 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=-\frac {35 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{3 \, {\left (e^{5} x^{4} + 4 \, d e^{4} x^{3} + 6 \, d^{2} e^{3} x^{2} + 4 \, d^{3} e^{2} x + d^{4} e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{6 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{6 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {35 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{2} x + d e} \]
-35/2*d^3*arcsin(e*x/d)/e + 1/3*(-e^2*x^2 + d^2)^(7/2)/(e^5*x^4 + 4*d*e^4* x^3 + 6*d^2*e^3*x^2 + 4*d^3*e^2*x + d^4*e) + 7/6*(-e^2*x^2 + d^2)^(5/2)*d/ (e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e) + 35/6*(-e^2*x^2 + d^2)^(3/2 )*d^2/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 35*sqrt(-e^2*x^2 + d^2)*d^3/(e^2*x + d*e)
Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.40 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=\frac {{\left (840 \, d^{4} e^{4} \arctan \left (\sqrt {\frac {2 \, d}{e x + d} - 1}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 384 \, d^{4} e^{4} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - \frac {{\left (87 \, d^{4} e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 136 \, d^{4} e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 57 \, d^{4} e^{4} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} {\left (e x + d\right )}^{3}}{d^{3}}\right )} {\left | e \right |}}{24 \, d e^{6}} \]
1/24*(840*d^4*e^4*arctan(sqrt(2*d/(e*x + d) - 1))*sgn(1/(e*x + d))*sgn(e) - 384*d^4*e^4*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e) - (87*d^4*e^ 4*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) + 136*d^4*e^4*(2*d/(e* x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) + 57*d^4*e^4*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e))*(e*x + d)^3/d^3)*abs(e)/(d*e^6)
Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^5} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{7/2}}{{\left (d+e\,x\right )}^5} \,d x \]