3.14 \(\int \frac {A+B x+C x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=103 \[ -\frac {\sqrt {d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{d e^3 (d+e x)}-\frac {(C d-B e) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3} \]

[Out]

-(-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)

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1639, 793, 217, 203} \[ -\frac {\sqrt {d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{d e^3 (d+e x)}-\frac {(C d-B e) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}-\frac {C \sqrt {d^2-e^2 x^2}}{e^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/((d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-((C*Sqrt[d^2 - e^2*x^2])/e^3) - ((C*d^2 - B*d*e + A*e^2)*Sqrt[d^2 - e^2*x^2])/(d*e^3*(d + e*x)) - ((C*d - B*e
)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^3

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

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]

Rule 793

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] + Dist[(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]

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(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 + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \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 {\int \frac {-A e^4+e^3 (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}-\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) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2}\\ &=-\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) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}\\ &=-\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) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 83, normalized size = 0.81 \[ \frac {(B e-C d) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {\sqrt {d^2-e^2 x^2} (e (A e-B d)+C d (2 d+e x))}{d (d+e x)}}{e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/((d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(-((Sqrt[d^2 - e^2*x^2]*(e*(-(B*d) + A*e) + C*d*(2*d + e*x)))/(d*(d + e*x))) + (-(C*d) + B*e)*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/e^3

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 155, normalized size = 1.50 \[ -\frac {2 \, C d^{3} - B d^{2} e + A d e^{2} + {\left (2 \, C d^{2} e - B d e^{2} + A e^{3}\right )} x - 2 \, {\left (C d^{3} - B d^{2} e + {\left (C d^{2} e - B d e^{2}\right )} x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (C d e x + 2 \, C d^{2} - B d e + A e^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{d e^{4} x + d^{2} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-(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)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(-4*A*exp(2)^2-4*C*d^2*exp(2)+4*B*d*
exp(1)*exp(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt
(-exp(1)^4+exp(2)^2)/d/exp(1)/exp(2)-1/4*(-4*B*exp(1)+4*C*d)*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)/exp(2)-4*e
xp(1)^2*C*1/4/exp(1)^5*sqrt(-exp(2)*x^2+d^2)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 149, normalized size = 1.45 \[ \frac {B \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e}-\frac {C d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C}{e^{3}}-\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{\left (x +\frac {d}{e}\right ) d \,e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-C*(-e^2*x^2+d^2)^(1/2)/e^3+1/e*B/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/e^2*C*d/(e^2)^(1/2)
*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-(A*e^2-B*d*e+C*d^2)/e^4/d/(x+d/e)*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1
/2)

________________________________________________________________________________________

maxima [A]  time = 0.99, size = 138, normalized size = 1.34 \[ -\frac {\sqrt {-e^{2} x^{2} + d^{2}} C d}{e^{4} x + d e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} A}{d e^{2} x + d^{2} e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{e^{3} x + d e^{2}} - \frac {C d \arcsin \left (\frac {e x}{d}\right )}{e^{3}} + \frac {B \arcsin \left (\frac {e x}{d}\right )}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} C}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {C\,x^2+B\,x+A}{\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x + C*x^2)/((d^2 - e^2*x^2)^(1/2)*(d + e*x)),x)

[Out]

int((A + B*x + C*x^2)/((d^2 - e^2*x^2)^(1/2)*(d + e*x)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x + C x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(e*x+d)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral((A + B*x + C*x**2)/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)), x)

________________________________________________________________________________________