Integrand size = 34, antiderivative size = 103 \[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}-\frac {\left (C d^2-B d e+A e^2\right ) \sqrt {d^2-e^2 x^2}}{d e^3 (d+e x)}-\frac {(C d-B e) \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]
-(-B*e+C*d)*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3-C*(-e^2*x^2+d^2)^(1/2)/e^ 3-(A*e^2-B*d*e+C*d^2)*(-e^2*x^2+d^2)^(1/2)/d/e^3/(e*x+d)
Time = 0.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {-\frac {\sqrt {d^2-e^2 x^2} (e (-B d+A e)+C d (2 d+e x))}{d (d+e x)}+2 (C d-B e) \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]
(-((Sqrt[d^2 - e^2*x^2]*(e*(-(B*d) + A*e) + C*d*(2*d + e*x)))/(d*(d + e*x) )) + 2*(C*d - B*e)*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^3
Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2170, 25, 27, 671, 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.
\(\displaystyle \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle -\frac {\int -\frac {e^3 (A e-(C d-B e) x)}{(d+e x) \sqrt {d^2-e^2 x^2}}dx}{e^4}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {e^3 (A e-(C d-B e) x)}{(d+e x) \sqrt {d^2-e^2 x^2}}dx}{e^4}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {A e-(C d-B e) x}{(d+e x) \sqrt {d^2-e^2 x^2}}dx}{e}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {-\frac {(C d-B e) \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx}{e}-\frac {\sqrt {d^2-e^2 x^2} \left (\frac {A}{d}+\frac {C d-B e}{e^2}\right )}{d+e x}}{e}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {-\frac {(C d-B e) \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}}{e}-\frac {\sqrt {d^2-e^2 x^2} \left (\frac {A}{d}+\frac {C d-B e}{e^2}\right )}{d+e x}}{e}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {-\frac {\sqrt {d^2-e^2 x^2} \left (\frac {A}{d}+\frac {C d-B e}{e^2}\right )}{d+e x}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) (C d-B e)}{e^2}}{e}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3}\) |
-((C*Sqrt[d^2 - e^2*x^2])/e^3) + (-(((A/d + (C*d - B*e)/e^2)*Sqrt[d^2 - e^ 2*x^2])/(d + e*x)) - ((C*d - B*e)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^2)/ e
3.1.14.3.1 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[((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 = 0.53 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.23
method | result | size |
risch | \(-\frac {C \sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}+\frac {\frac {\left (B e -C d \right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e \sqrt {e^{2}}}-\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{3} d \left (x +\frac {d}{e}\right )}}{e}\) | \(127\) |
default | \(\frac {\frac {B e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-\frac {C \sqrt {-e^{2} x^{2}+d^{2}}}{e}-\frac {C d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}}{e^{2}}-\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} d \left (x +\frac {d}{e}\right )}\) | \(149\) |
-C*(-e^2*x^2+d^2)^(1/2)/e^3+1/e*((B*e-C*d)/e/(e^2)^(1/2)*arctan((e^2)^(1/2 )*x/(-e^2*x^2+d^2)^(1/2))-(A*e^2-B*d*e+C*d^2)/e^3/d/(x+d/e)*(-(x+d/e)^2*e^ 2+2*d*e*(x+d/e))^(1/2))
Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.50 \[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 \, C d^{3} - B d^{2} e + A d e^{2} + {\left (2 \, C d^{2} e - B d e^{2} + A e^{3}\right )} x - 2 \, {\left (C d^{3} - B d^{2} e + {\left (C d^{2} e - B d e^{2}\right )} x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (C d e x + 2 \, C d^{2} - B d e + A e^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{d e^{4} x + d^{2} e^{3}} \]
-(2*C*d^3 - B*d^2*e + A*d*e^2 + (2*C*d^2*e - B*d*e^2 + A*e^3)*x - 2*(C*d^3 - B*d^2*e + (C*d^2*e - B*d*e^2)*x)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e* x)) + (C*d*e*x + 2*C*d^2 - B*d*e + A*e^2)*sqrt(-e^2*x^2 + d^2))/(d*e^4*x + d^2*e^3)
\[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {A + B x + C x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} C d}{e^{4} x + d e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} A}{d e^{2} x + d^{2} e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{e^{3} x + d e^{2}} - \frac {C d \arcsin \left (\frac {e x}{d}\right )}{e^{3}} + \frac {B \arcsin \left (\frac {e x}{d}\right )}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} C}{e^{3}} \]
-sqrt(-e^2*x^2 + d^2)*C*d/(e^4*x + d*e^3) - sqrt(-e^2*x^2 + d^2)*A/(d*e^2* x + d^2*e) + sqrt(-e^2*x^2 + d^2)*B/(e^3*x + d*e^2) - C*d*arcsin(e*x/d)/e^ 3 + B*arcsin(e*x/d)/e^2 - sqrt(-e^2*x^2 + d^2)*C/e^3
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {{\left (C d - B e\right )} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{2} {\left | e \right |}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} C}{e^{3}} + \frac {2 \, {\left (C d^{2} - B d e + A e^{2}\right )}}{d e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]
-(C*d - B*e)*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) - sqrt(-e^2*x^2 + d^ 2)*C/e^3 + 2*(C*d^2 - B*d*e + A*e^2)/(d*e^2*((d*e + sqrt(-e^2*x^2 + d^2)*a bs(e))/(e^2*x) + 1)*abs(e))
Timed out. \[ \int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {C\,x^2+B\,x+A}{\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \]