Integrand size = 34, antiderivative size = 196 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {2 C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{5 d e^3 (d+e x)^4}+\frac {(2 C d-B e) \left (d^2-e^2 x^2\right )^{3/2}}{3 d e^3 (d+e x)^3}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 e^3 (d+e x)^3}-\frac {C \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]
-1/5*(A*e^2-B*d*e+C*d^2)*(-e^2*x^2+d^2)^(3/2)/d/e^3/(e*x+d)^4+1/3*(-B*e+2* C*d)*(-e^2*x^2+d^2)^(3/2)/d/e^3/(e*x+d)^3-1/15*(A*e^2-B*d*e+C*d^2)*(-e^2*x ^2+d^2)^(3/2)/d^2/e^3/(e*x+d)^3-C*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3-2*C *(-e^2*x^2+d^2)^(1/2)/e^3/(e*x+d)
Time = 0.90 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.64 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {-\frac {\sqrt {d^2-e^2 x^2} \left (3 C d^2 \left (8 d^2+19 d e x+13 e^2 x^2\right )+e (d-e x) (A e (4 d+e x)+B d (d+4 e x))\right )}{d^2 (d+e x)^3}+30 C \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^3} \]
(-((Sqrt[d^2 - e^2*x^2]*(3*C*d^2*(8*d^2 + 19*d*e*x + 13*e^2*x^2) + e*(d - e*x)*(A*e*(4*d + e*x) + B*d*(d + 4*e*x))))/(d^2*(d + e*x)^3)) + 30*C*ArcTa n[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(15*e^3)
Time = 0.40 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2168, 2009}
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.
| \(\displaystyle \int \frac {\sqrt {d^2-e^2 x^2} \left (A+B x+C x^2\right )}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2168 |
| \(\displaystyle \int \left (\frac {\sqrt {d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{e^2 (d+e x)^4}+\frac {\sqrt {d^2-e^2 x^2} (B e-2 C d)}{e^2 (d+e x)^3}+\frac {C \sqrt {d^2-e^2 x^2}}{e^2 (d+e x)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{15 d^2 e^3 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{5 d e^3 (d+e x)^4}-\frac {C \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2} (2 C d-B e)}{3 d e^3 (d+e x)^3}-\frac {2 C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}\) |
(-2*C*Sqrt[d^2 - e^2*x^2])/(e^3*(d + e*x)) - ((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3/2))/(5*d*e^3*(d + e*x)^4) + ((2*C*d - B*e)*(d^2 - e^2*x^2)^ (3/2))/(3*d*e^3*(d + e*x)^3) - ((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3 /2))/(15*d^2*e^3*(d + e*x)^3) - (C*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^3
3.1.7.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2)^p, (d + e*x)^m*Pq, x], x] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq, x] + 2*p + 1, 0] && ILtQ[m, 0]
Time = 0.58 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.49
| method | result | size |
| default | \(\frac {C \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{e^{4}}-\frac {\left (B e -2 C d \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{6} d \left (x +\frac {d}{e}\right )^{3}}+\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{6}}\) | \(293\) |
C/e^4*(-1/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)-e/d*((-(x+d/e )^2*e^2+2*d*e*(x+d/e))^(1/2)+d*e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e )^2*e^2+2*d*e*(x+d/e))^(1/2))))-1/3*(B*e-2*C*d)/e^6/d/(x+d/e)^3*(-(x+d/e)^ 2*e^2+2*d*e*(x+d/e))^(3/2)+(A*e^2-B*d*e+C*d^2)/e^6*(-1/5/d/e/(x+d/e)^4*(-( x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)-1/15/d^2/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e *(x+d/e))^(3/2))
Time = 0.29 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.55 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {24 \, C d^{5} + B d^{4} e + 4 \, A d^{3} e^{2} + {\left (24 \, C d^{2} e^{3} + B d e^{4} + 4 \, A e^{5}\right )} x^{3} + 3 \, {\left (24 \, C d^{3} e^{2} + B d^{2} e^{3} + 4 \, A d e^{4}\right )} x^{2} + 3 \, {\left (24 \, C d^{4} e + B d^{3} e^{2} + 4 \, A d^{2} e^{3}\right )} x - 30 \, {\left (C d^{2} e^{3} x^{3} + 3 \, C d^{3} e^{2} x^{2} + 3 \, C d^{4} e x + C d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (24 \, C d^{4} + B d^{3} e + 4 \, A d^{2} e^{2} + {\left (39 \, C d^{2} e^{2} - 4 \, B d e^{3} - A e^{4}\right )} x^{2} + 3 \, {\left (19 \, C d^{3} e + B d^{2} e^{2} - A d e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{6} x^{3} + 3 \, d^{3} e^{5} x^{2} + 3 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]
-1/15*(24*C*d^5 + B*d^4*e + 4*A*d^3*e^2 + (24*C*d^2*e^3 + B*d*e^4 + 4*A*e^ 5)*x^3 + 3*(24*C*d^3*e^2 + B*d^2*e^3 + 4*A*d*e^4)*x^2 + 3*(24*C*d^4*e + B* d^3*e^2 + 4*A*d^2*e^3)*x - 30*(C*d^2*e^3*x^3 + 3*C*d^3*e^2*x^2 + 3*C*d^4*e *x + C*d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (24*C*d^4 + B*d^3* e + 4*A*d^2*e^2 + (39*C*d^2*e^2 - 4*B*d*e^3 - A*e^4)*x^2 + 3*(19*C*d^3*e + B*d^2*e^2 - A*d*e^3)*x)*sqrt(-e^2*x^2 + d^2))/(d^2*e^6*x^3 + 3*d^3*e^5*x^ 2 + 3*d^4*e^4*x + d^5*e^3)
\[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (A + B x + C x^{2}\right )}{\left (d + e x\right )^{4}}\, dx \]
\[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int { \frac {\sqrt {-e^{2} x^{2} + d^{2}} {\left (C x^{2} + B x + A\right )}}{{\left (e x + d\right )}^{4}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (180) = 360\).
Time = 0.30 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.25 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {C \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{2} {\left | e \right |}} + \frac {2 \, {\left (24 \, C d^{2} + B d e + 4 \, A e^{2} + \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} A}{x} + \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} C d^{2}}{e^{2} x} + \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} B d}{e x} + \frac {165 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} C d^{2}}{e^{4} x^{2}} - \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} B d}{e^{3} x^{2}} + \frac {25 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} A}{e^{2} x^{2}} + \frac {75 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} C d^{2}}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} B d}{e^{5} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} A}{e^{4} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} C d^{2}}{e^{8} x^{4}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} A}{e^{6} x^{4}}\right )}}{15 \, d^{2} e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]
-C*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) + 2/15*(24*C*d^2 + B*d*e + 4*A *e^2 + 5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*A/x + 105*(d*e + sqrt(-e^2*x^ 2 + d^2)*abs(e))*C*d^2/(e^2*x) + 5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*B*d /(e*x) + 165*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*C*d^2/(e^4*x^2) - 5*(d* e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*B*d/(e^3*x^2) + 25*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*A/(e^2*x^2) + 75*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*C *d^2/(e^6*x^3) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*B*d/(e^5*x^3) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*A/(e^4*x^3) + 15*(d*e + sqrt(-e^2 *x^2 + d^2)*abs(e))^4*C*d^2/(e^8*x^4) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs (e))^4*A/(e^6*x^4))/(d^2*e^2*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))
Timed out. \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {\sqrt {d^2-e^2\,x^2}\,\left (C\,x^2+B\,x+A\right )}{{\left (d+e\,x\right )}^4} \,d x \]