Integrand size = 34, antiderivative size = 170 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=-\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \sqrt {d^2-e^2 x^2}}{2 d e^3}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{d e^3 (d+e x)^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}-\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]
-(A*e^2-B*d*e+C*d^2)*(-e^2*x^2+d^2)^(3/2)/d/e^3/(e*x+d)^2-1/2*C*(-e^2*x^2+ d^2)^(3/2)/e^3/(e*x+d)-1/2*(5*C*d^2-2*e*(-A*e+2*B*d))*arctan(e*x/(-e^2*x^2 +d^2)^(1/2))/e^3-1/2*(5*C*d^2-2*e*(-A*e+2*B*d))*(-e^2*x^2+d^2)^(1/2)/d/e^3
Time = 0.59 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.71 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (2 e (3 B d-2 A e+B e x)+C \left (-8 d^2-3 d e x+e^2 x^2\right )\right )}{d+e x}+2 \left (5 C d^2+2 e (-2 B d+A e)\right ) \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]
((Sqrt[d^2 - e^2*x^2]*(2*e*(3*B*d - 2*A*e + B*e*x) + C*(-8*d^2 - 3*d*e*x + e^2*x^2)))/(d + e*x) + 2*(5*C*d^2 + 2*e*(-2*B*d + A*e))*ArcTan[(e*x)/(Sqr t[d^2] - Sqrt[d^2 - e^2*x^2])])/(2*e^3)
Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2170, 27, 671, 466, 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.
| \(\displaystyle \int \frac {\sqrt {d^2-e^2 x^2} \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2170 |
| \(\displaystyle -\frac {\int \frac {e^2 \left (C d^2-2 A e^2+e (3 C d-2 B e) x\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2}dx}{2 e^4}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\left (C d^2-2 A e^2+e (3 C d-2 B e) x\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2}dx}{2 e^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}\) |
| \(\Big \downarrow \) 671 |
| \(\displaystyle -\frac {\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x}dx}{d}+\frac {2 \left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{d e (d+e x)^2}}{2 e^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}\) |
| \(\Big \downarrow \) 466 |
| \(\displaystyle -\frac {\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \left (d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {\sqrt {d^2-e^2 x^2}}{e}\right )}{d}+\frac {2 \left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{d e (d+e x)^2}}{2 e^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle -\frac {\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \left (d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e}\right )}{d}+\frac {2 \left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{d e (d+e x)^2}}{2 e^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\frac {\left (\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {\sqrt {d^2-e^2 x^2}}{e}\right ) \left (5 C d^2-2 e (2 B d-A e)\right )}{d}+\frac {2 \left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{d e (d+e x)^2}}{2 e^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}\) |
-1/2*(C*(d^2 - e^2*x^2)^(3/2))/(e^3*(d + e*x)) - ((2*(C*d^2 - B*d*e + A*e^ 2)*(d^2 - e^2*x^2)^(3/2))/(d*e*(d + e*x)^2) + ((5*C*d^2 - 2*e*(2*B*d - A*e ))*(Sqrt[d^2 - e^2*x^2]/e + (d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/d)/( 2*e^2)
3.1.5.3.1 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_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 2*(n + 2*p + 1))) Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 ] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Time = 0.54 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.15
| method | result | size |
| risch | \(\frac {\left (C x e +2 B e -4 C d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{3}}-\frac {\frac {2 A \,e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {5 C \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-\frac {4 B d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {4 \left (A \,e^{2}-B d e +C \,d^{2}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{2} \left (x +\frac {d}{e}\right )}}{2 e^{2}}\) | \(196\) |
| default | \(\frac {C \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{2}}+\frac {\left (B e -2 C d \right ) \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{3}}+\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{e^{4}}\) | \(287\) |
1/2*(C*e*x+2*B*e-4*C*d)/e^3*(-e^2*x^2+d^2)^(1/2)-1/2/e^2*(2*A*e^2/(e^2)^(1 /2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+5*C*d^2/(e^2)^(1/2)*arctan( (e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-4*B*d*e/(e^2)^(1/2)*arctan((e^2)^(1/2) *x/(-e^2*x^2+d^2)^(1/2))+4*(A*e^2-B*d*e+C*d^2)/e^2/(x+d/e)*(-(x+d/e)^2*e^2 +2*d*e*(x+d/e))^(1/2))
Time = 0.31 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.12 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=-\frac {8 \, C d^{3} - 6 \, B d^{2} e + 4 \, A d e^{2} + 2 \, {\left (4 \, C d^{2} e - 3 \, B d e^{2} + 2 \, A e^{3}\right )} x - 2 \, {\left (5 \, C d^{3} - 4 \, B d^{2} e + 2 \, A d e^{2} + {\left (5 \, C d^{2} e - 4 \, B d e^{2} + 2 \, A e^{3}\right )} x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (C e^{2} x^{2} - 8 \, C d^{2} + 6 \, B d e - 4 \, A e^{2} - {\left (3 \, C d e - 2 \, B e^{2}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{4} x + d e^{3}\right )}} \]
-1/2*(8*C*d^3 - 6*B*d^2*e + 4*A*d*e^2 + 2*(4*C*d^2*e - 3*B*d*e^2 + 2*A*e^3 )*x - 2*(5*C*d^3 - 4*B*d^2*e + 2*A*d*e^2 + (5*C*d^2*e - 4*B*d*e^2 + 2*A*e^ 3)*x)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (C*e^2*x^2 - 8*C*d^2 + 6 *B*d*e - 4*A*e^2 - (3*C*d*e - 2*B*e^2)*x)*sqrt(-e^2*x^2 + d^2))/(e^4*x + d *e^3)
\[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=\int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (A + B x + C x^{2}\right )}{\left (d + e x\right )^{2}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.16 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=-\frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{e^{4} x + d e^{3}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{e^{3} x + d e^{2}} - \frac {5 \, C d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {2 \, B d \arcsin \left (\frac {e x}{d}\right )}{e^{2}} - \frac {A \arcsin \left (\frac {e x}{d}\right )}{e} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{e^{2} x + d e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C x}{2 \, e^{2}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{e^{2}} \]
-2*sqrt(-e^2*x^2 + d^2)*C*d^2/(e^4*x + d*e^3) + 2*sqrt(-e^2*x^2 + d^2)*B*d /(e^3*x + d*e^2) - 5/2*C*d^2*arcsin(e*x/d)/e^3 + 2*B*d*arcsin(e*x/d)/e^2 - A*arcsin(e*x/d)/e - 2*sqrt(-e^2*x^2 + d^2)*A/(e^2*x + d*e) + 1/2*sqrt(-e^ 2*x^2 + d^2)*C*x/e^2 - 2*sqrt(-e^2*x^2 + d^2)*C*d/e^3 + sqrt(-e^2*x^2 + d^ 2)*B/e^2
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (156) = 312\).
Time = 0.31 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.91 \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=-\frac {{\left (8 \, C d^{3} e^{3} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 8 \, B d^{2} e^{4} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 8 \, A d e^{5} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 4 \, {\left (5 \, C d^{3} e^{3} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 4 \, B d^{2} e^{4} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 2 \, A d e^{5} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} \arctan \left (\sqrt {\frac {2 \, d}{e x + d} - 1}\right ) + \frac {{\left (5 \, C d^{3} e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 2 \, B d^{2} e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 3 \, C d^{3} e^{3} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 2 \, B d^{2} e^{4} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} {\left (e x + d\right )}^{2}}{d^{2}}\right )} {\left | e \right |}}{4 \, d e^{7}} \]
-1/4*(8*C*d^3*e^3*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e) - 8*B*d^ 2*e^4*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e) + 8*A*d*e^5*sqrt(2*d /(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e) - 4*(5*C*d^3*e^3*sgn(1/(e*x + d))* sgn(e) - 4*B*d^2*e^4*sgn(1/(e*x + d))*sgn(e) + 2*A*d*e^5*sgn(1/(e*x + d))* sgn(e))*arctan(sqrt(2*d/(e*x + d) - 1)) + (5*C*d^3*e^3*(2*d/(e*x + d) - 1) ^(3/2)*sgn(1/(e*x + d))*sgn(e) - 2*B*d^2*e^4*(2*d/(e*x + d) - 1)^(3/2)*sgn (1/(e*x + d))*sgn(e) + 3*C*d^3*e^3*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d) )*sgn(e) - 2*B*d^2*e^4*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e))*(e *x + d)^2/d^2)*abs(e)/(d*e^7)
Timed out. \[ \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx=\int \frac {\sqrt {d^2-e^2\,x^2}\,\left (C\,x^2+B\,x+A\right )}{{\left (d+e\,x\right )}^2} \,d x \]