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

Optimal. Leaf size=148 \[ \frac {\sqrt {d^2-e^2 x^2} \left (C d^2-e (B d-2 A e)\right )}{2 e^3}+\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (C d^2-e (B d-2 A e)\right )}{2 e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2} (C d-B e)}{2 e^3 (d+e x)}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {1639, 795, 665, 217, 203} \[ \frac {\sqrt {d^2-e^2 x^2} \left (C d^2-e (B d-2 A e)\right )}{2 e^3}+\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (C d^2-e (B d-2 A e)\right )}{2 e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2} (C d-B e)}{2 e^3 (d+e x)}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C*d^2 - e*(B*d - 2*A*e))*Sqrt[d^2 - e^2*x^2])/(2*e^3) - (C*(d^2 - e^2*x^2)^(3/2))/(3*e^3) + ((C*d - B*e)*(d^
2 - e^2*x^2)^(3/2))/(2*e^3*(d + e*x)) + (d*(C*d^2 - e*(B*d - 2*A*e))*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(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] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

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

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Mathematica [A]  time = 0.22, size = 103, normalized size = 0.70 \[ \frac {\sqrt {d^2-e^2 x^2} \left (3 e (2 A e-2 B d+B e x)+C \left (4 d^2-3 d e x+2 e^2 x^2\right )\right )+3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (e (2 A e-B d)+C d^2\right )}{6 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(3*e*(-2*B*d + 2*A*e + B*e*x) + C*(4*d^2 - 3*d*e*x + 2*e^2*x^2)) + 3*d*(C*d^2 + e*(-(B*d)
+ 2*A*e))*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(6*e^3)

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fricas [A]  time = 0.89, size = 112, normalized size = 0.76 \[ -\frac {6 \, {\left (C d^{3} - B d^{2} e + 2 \, A d e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (2 \, C e^{2} x^{2} + 4 \, C d^{2} - 6 \, B d e + 6 \, A e^{2} - 3 \, {\left (C d e - B e^{2}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(C*d^3 - B*d^2*e + 2*A*d*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (2*C*e^2*x^2 + 4*C*d^2 - 6*B
*d*e + 6*A*e^2 - 3*(C*d*e - B*e^2)*x)*sqrt(-e^2*x^2 + d^2))/e^3

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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^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(4*A*d*exp(2)^3-4*A*d*exp(1)^4*exp(2
)+4*B*d^2*exp(1)^3*exp(2)-4*B*d^2*exp(1)*exp(2)^2)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+ex
p(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^4/exp(1)-1/4*(-2*C*d^3-4*A*d*exp(2)+2*B*d^2*ex
p(1))*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)/exp(2)+2*((16*exp(1)^4*C*1/96/exp(1)^5*x-(-24*exp(1)^4*B+24*exp(1
)^3*C*d)*1/96/exp(1)^5)*x-(-48*exp(1)^4*A+48*exp(1)^3*d*B-32*exp(1)^2*C*d^2)*1/96/exp(1)^5)*sqrt(d^2-x^2*exp(2
))

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maple [B]  time = 0.02, size = 384, normalized size = 2.59 \[ \frac {A d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}}-\frac {B \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e}+\frac {B \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e}+\frac {C \,d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e^{2}}-\frac {C \,d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, B x}{2 e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C d x}{2 e^{2}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, A}{e}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, B d}{e^{2}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, C \,d^{2}}{e^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} C}{3 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.09, size = 171, normalized size = 1.16 \[ \frac {C d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} - \frac {B d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{2}} + \frac {A d \arcsin \left (\frac {e x}{d}\right )}{e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} C d x}{2 \, e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B x}{2 \, e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B d}{e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} A}{e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C}{3 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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